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Contemporary
Abstract Algebra
SEVENTH EDITION

Joseph A
...
Gallian
VP/Editor-in-Chief: Michelle Julet
Publisher: Richard Stratton
Senior Sponsoring Editor: Molly Taylor
Associate Editor: Daniel Seibert
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© 2010, 2006 Brooks/Cole, Cengage Learning
ALL RIGHTS RESERVED
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Student Edition:
ISBN-13: 978-0-547-16509-7

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Printed in the United States of America
1 2 3 4 5 6 7 12 11 10 09 08

Contents

Preface xi
PART 1

Integers and Equivalence Relations 1
0 Preliminaries 3
Properties of Integers 3 | Modular Arithmetic 7 |
Mathematical Induction 12 | Equivalence Relations 15 |
Functions (Mappings) 18
Exercises 21
Computer Exercises 25

PART 2

Groups 27
1 Introduction to Groups 29
Symmetries of a Square 29 | The Dihedral Groups 32
Exercises 35
Biography of Niels Abel 39

2 Groups 40
Definition and Examples of Groups 40 | Elementary
Properties of Groups 48 | Historical Note 51
Exercises 52
Computer Exercises 55

3 Finite Groups; Subgroups 57
Terminology and Notation 57 | Subgroup Tests 58 |
Examples of Subgroups 61
Exercises 64
Computer Exercises 70

iii

iv

Contents

4 Cyclic Groups 72
Properties of Cyclic Groups 72 | Classification of Subgroups
of Cyclic Groups 77
Exercises 81
Computer Exercises 86
Biography of J
...
Sylvester 89
Supplementary Exercises for Chapters 1–4 91

5 Permutation Groups 95
Definition and Notation 95 | Cycle Notation 98 | Properties of
Permutations 100 | A Check Digit Scheme Based on D5 110
Exercises 113
Computer Exercises 118
Biography of Augustin Cauchy 121

6 Isomorphisms 122
Motivation 122 | Definition and Examples 122 | Cayley’s
Theorem 126 | Properties of Isomorphisms 128 |
Automorphisms 129
Exercises 133
Computer Exercise 136
Biography of Arthur Cayley 137

7 Cosets and Lagrange’s Theorem 138
Properties of Cosets 138 | Lagrange’s Theorem and
Consequences 141 | An Application of Cosets to Permutation
Groups 145 | The Rotation Group of a Cube and a Soccer Ball 146
Exercises 149
Computer Exercise 153
Biography of Joseph Lagrange 154

8 External Direct Products 155
Definition and Examples 155 | Properties of External Direct
Products 156 | The Group of Units Modulo n as an External Direct
Product 159 | Applications 161
Exercises 167
Computer Exercises 170
Biography of Leonard Adleman 173
Supplementary Exercises for Chapters 5–8 174

Contents

9 Normal Subgroups and Factor Groups 178
Normal Subgroups 178 | Factor Groups 180 | Applications of
Factor Groups 185 | Internal Direct Products 188
Exercises 193
Biography of Évariste Galois 199

10 Group Homomorphisms 200
Definition and Examples 200 | Properties of Homomorphisms
202 | The First Isomorphism Theorem 206
Exercises 211
Computer Exercise 216
Biography of Camille Jordan 217

11 Fundamental Theorem of Finite
Abelian Groups 218
The Fundamental Theorem 218 | The Isomorphism Classes of
Abelian Groups 218 | Proof of the Fundamental Theorem 223
Exercises 226
Computer Exercises 228
Supplementary Exercises for Chapters 9–11 230
PART 3

Rings 235
12 Introduction to Rings 237
Motivation and Definition 237 | Examples of Rings 238 |
Properties of Rings 239 | Subrings 240
Exercises 242
Computer Exercises 245
Biography of I
...
Herstein 248

13 Integral Domains 249
Definition and Examples 249 | Fields 250 | Characteristic of a
Ring 252
Exercises 255
Computer Exercises 259
Biography of Nathan Jacobson 261

14 Ideals and Factor Rings 262
Ideals 262 | Factor Rings 263 | Prime Ideals and Maximal
Ideals 267
Exercises 269

v

vi

Contents

Computer Exercises 273
Biography of Richard Dedekind 274
Biography of Emmy Noether 275
Supplementary Exercises for Chapters 12–14 276

15 Ring Homomorphisms 280
Definition and Examples 280 | Properties of Ring Homomorphisms
283 | The Field of Quotients 285
Exercises 287

16 Polynomial Rings 293
Notation and Terminology 293 | The Division Algorithm and
Consequences 296
Exercises 300
Biography of Saunders Mac Lane 304

17 Factorization of Polynomials 305
Reducibility Tests 305 | Irreducibility Tests 308 | Unique
Factorization in Z[x] 313 | Weird Dice: An Application of Unique
Factorization 314
Exercises 316
Computer Exercises 319
Biography of Serge Lang 321

18 Divisibility in Integral Domains 322
Irreducibles, Primes 322 | Historical Discussion of Fermat’s Last
Theorem 325 | Unique Factorization Domains 328 | Euclidean
Domains 331
Exercises 335
Computer Exercise 337
Biography of Sophie Germain 339
Biography of Andrew Wiles 340
Supplementary Exercises for Chapters 15–18 341
PART 4

Fields 343
19 Vector Spaces 345
Definition and Examples 345 | Subspaces 346 | Linear
Independence 347

Contents

vii

Exercises 349
Biography of Emil Artin 352
Biography of Olga Taussky-Todd 353

20 Extension Fields 354
The Fundamental Theorem of Field Theory 354 | Splitting
Fields 356 | Zeros of an Irreducible Polynomial 362
Exercises 366
Biography of Leopold Kronecker 369

21 Algebraic Extensions 370
Characterization of Extensions 370 | Finite Extensions 372 |
Properties of Algebraic Extensions 376 |
Exercises 378
Biography of Irving Kaplansky 381

22 Finite Fields 382
Classification of Finite Fields 382 | Structure of Finite Fields 383 |
Subfields of a Finite Field 387
Exercises 389
Computer Exercises 391
Biography of L
...
Dickson 392

23 Geometric Constructions 393
Historical Discussion of Geometric Constructions 393 |
Constructible Numbers 394 | Angle-Trisectors and
Circle-Squarers 396
Exercises 396
Supplementary Exercises for Chapters 19–23 399
PART 5

Special Topics 401
24 Sylow Theorems 403
Conjugacy Classes 403 | The Class Equation 404 | The
Probability That Two Elements Commute 405 | The Sylow
Theorems 406 | Applications of Sylow Theorems 411
Exercises 414
Computer Exercise 418
Biography of Ludwig Sylow 419

viii

Contents

25 Finite Simple Groups 420
Historical Background 420 | Nonsimplicity Tests 425 |
The Simplicity of A5 429 | The Fields Medal 430 |
The Cole Prize 430 |
Exercises 431
Computer Exercises 432
Biography of Michael Aschbacher 434
Biography of Daniel Gorenstein 435
Biography of John Thompson 436

26 Generators and Relations 437
Motivation 437 | Definitions and Notation 438 | Free
Group 439 | Generators and Relations 440 | Classification of
Groups of Order Up to 15 444 | Characterization of Dihedral
Groups 446 | Realizing the Dihedral Groups with Mirrors 447
Exercises 449
Biography of Marshall Hall, Jr
...
C
...
Conway 486

29 Symmetry and Counting 487
Motivation 487 | Burnside’s Theorem 488 | Applications 490 |
Group Action 493
Exercises 494
Biography of William Burnside 497

30 Cayley Digraphs of Groups 498
Motivation 498 | The Cayley Digraph of a Group 498 |
Hamiltonian Circuits and Paths 502 | Some Applications 508

Contents

Exercises 511
Biography of William Rowan Hamilton 516
Biography of Paul Erdös 517

31 Introduction to Algebraic Coding Theory 518
Motivation 518 | Linear Codes 523 | Parity-Check Matrix
Decoding 528 | Coset Decoding 531 | Historical Note: The
Ubiquitous Reed-Solomon Codes 535
Exercises 537
Biography of Richard W
...
I want students to receive a solid
introduction to the traditional topics
...
I want students to enjoy
reading the book
...
I want students to be
able to do computations and to write proofs
...

Changes for the seventh edition include 120 new exercises, new
theorems and examples, and a freshening of the quotations and biographies
...

These changes accentuate and enhance the hallmark features that
have made previous editions of the book a comprehensive, lively, and
engaging introduction to the subject:
• Extensive coverage of groups, rings, and fields, plus a variety of
non-traditional special topics
• A good mixture of now more than 1750 computational and theoretical exercises appearing in each chapter and in Supplementary
Exercise sets that synthesize concepts from multiple chapters
• Worked-out examples—now totaling 275—providing thorough
practice for key concepts
• Computer exercises performed using interactive software available
on my website
xi

xii

Preface

• A large number of applications from scientific and computing fields,
as well as from everyday life
• Numerous historical notes and biographies that illuminate the people and events behind the mathematics
• Annotated suggested readings and media for interesting further
exploration of topics
...
d
...
edu/~jgallian or through
Cengage’s book companion site at www
...
com/math/gallian—
offers a wealth of additional online resources supporting the book,
including:
• True/false questions
• Flash cards
• Essays on learning abstract algebra, doing proofs, and reasons why
abstract algebra is a valuable subject to learn
• Links to abstract algebra-related websites and software packages
•
...

Additionally, Cengage offers the following student and instructor
ancillaries to accompany the book:
• A Student Solutions Manual, available for purchase separately, with
worked-out solutions to the odd-numbered exercises in the book
(ISBN-13: 978-0-547-16539-4; ISBN-10: 0-547-16539-0)
• An online laboratory manual, written by Julianne Rainbolt and me,
with exercises designed to be done with the free computer algebra
system software GAP
• An online Instructor’s Solutions Manual with solutions to the evennumbered exercises in the book and additional test questions and
solutions
• Online instructor answer keys to the book’s computer exercises and
the exercises in the GAP lab manual
...
I am grateful to each of them for their careful reading of the
manuscript
...

I greatly valued the thoughtful input of the following people, who
kindly served as reviewers for the seventh edition:
Rebecca Berg, Bowie State University; Monte Boisen, University of
Idaho; Tara Brendle, Louisiana State University; Jeff Clark, Elon
University; Carl Eckberg, San Diego State University; Tom Farmer,
Miami University; Yuval Flicker, Ohio State University; Ed Hinson,

Preface

xiii

University of New Hampshire; Gizem Karaali, Pomona College; Mohan
Shrikhande, Central Michigan University; Ernie Stitzinger, North
Carolina State University
...
They have helped to make each edition
better
...
Please send any comments and suggestions you have to me
at jgallian@d
...
edu
...
Gallian

This page intentionally left blank

PA R T

1

Integers and
Equivalence Relations

For online student resources, visit this textbook’s website at
http://college
...
com/PIC/gallian7e

1

This page intentionally left blank

0 Preliminaries

The whole of science is nothing more than a refinement
of everyday thinking
...
In this
chapter we collect the properties we need for future reference
...
Since this property cannot be proved
from the usual properties of arithmetic, we will take it as an axiom
...


The concept of divisibility plays a fundamental role in the theory of
numbers
...
In this case, we write t | s (read “t
divides s”)
...
A prime is a
positive integer greater than 1 whose only positive divisors are 1 and
itself
...

As our first application of the Well Ordering Principle, we establish
a fundamental property of integers that we will use often
...
1 Division Algorithm
Let a and b be integers with b
...
Then there exist unique integers q
and r with the property that a 5 bq 1 r, where 0 # r , b
...
Consider
the set S 5 {a 2 bk | k is an integer and a 2 bk $ 0}
...

Now assume 0 n S
...
0, a 2 b ? 0 [ S; if a ,
0, a 2 b(2a) 5 a(1 2 2b) [ S; a 0 since 0 n S], we may apply the
Well Ordering Principle to conclude that S has a smallest member, say
r 5 a 2 bq
...

If r $ b, then a 2 b(q 1 1) 5 a 2 bq 2 b 5 r 2 b $ 0, so that
a 2 b(q 1 1) [ S
...
So, r , b
...


For convenience, we may also suppose that r9 $ r
...
So, b divides r9 2 r and 0 # r9 2 r #
r9 , b
...

The integer q in the division algorithm is called the quotient upon dividing a by b; the integer r is called the remainder upon dividing a by b
...

Several states use linear functions to encode the month and date of
birth into a three-digit number that is incorporated into driver’s license numbers
...
For instance, the last three digits of a Florida male
driver’s license number are those given by the formula 40(m 2 1) 1 b,
where m is the number of the month of birth and b is the day of birth
...
For New York licenses issued prior to
September of 1992, the last two digits indicate the year of birth, and
the three preceding digits code the month and date of birth
...
So, since 701 5
63 ? 11 1 2 ? 4, a license that ends with 70174 indicates that the
holder is a male born on November 4, 1974
...
) Incidentally, Wisconsin uses the same method
as Florida to encode birth information, but the numbers immediately
precede the last pair of digits
...
We denote this integer by
gcd(a, b)
...


The following property of the greatest common divisor of two integers plays a critical role in abstract algebra
...

Theorem 0
...
Moreover, gcd(a, b) is the smallest positive integer
of the form as 1 bt
...
0}
...
We claim that d 5 gcd(a, b)
...
If r
...

So, r 5 0 and d divides a
...
This proves that d is a common divisor of a and b
...
Then d 5 as 1 bt 5 (d9h)s 1 (d9k)t 5 d9(hs 1 kt),
so that d9 is a divisor of d
...

The special case of Theorem 0
...

Corollary
If a and b are relatively prime, than there exist integers s and t such
that as 1 bt 5 1
...
Note that 4 and 15 are relatively prime, whereas 4 and 10 are
not
...

72)

The next lemma is frequently used
...

Euclid’s Lemma p | ab Implies p | a or p | b
If p is a prime that divides ab, then p divides a or p divides b
...
We
must show that p divides b
...
Then b 5 abs 1 ptb, and since
p divides the right-hand side of this equation, p also divides b
...

Our next property shows that the primes are the building blocks for
all integers
...

Theorem 0
...
This
product is unique, except for the order in which the factors appear
...
pr and n 5 q1q2
...


We will prove the existence portion of Theorem 0
...
The uniqueness portion is a consequence of Euclid’s Lemma
(Exercise 27)
...

Definition Least Common Multiple

The least common multiple of two nonzero integers a and b is the
smallest positive integer that is a multiple of both a and b
...


We leave it as an exercise (Exercise 12) to prove that every common
multiple of a and b is a multiple of lcm(a, b)
...


Modular Arithmetic
Another application of the division algorithm that will be important to
us is modular arithmetic
...
For example, if it is now
September, what month will it be 25 months from now? Of course, the
answer is October, but the interesting fact is that you didn’t arrive at the
answer by starting with September and counting off 25 months
...
Similarly, if it
is now Wednesday, you know that in 23 days it will be Friday
...
If your
electricity is off for 26 hours, you must advance your clock 2 hours,
since 26 5 2 ? 12 1 2
...
You will see
a few of them in this section
...

When a 5 qn 1 r, where q is the quotient and r is the remainder
upon dividing a by n, we write a mod n 5 r
...

In general, if a and b are integers and n is a positive integer, then
a mod n 5 b mod n if and only if n divides a 2 b (Exercise 9)
...

When you wish to compute ab mod n or (a 1 b) mod n, and a or b is
greater than n, it is easier to “mod first
...
(See Exercise 11
...
We present two such applications
...
1 has an identification number consisting of 10 digits together
with an extra digit called a check
...
Thus, the number 3953988164 has the check digit 2, since

8

Integers and Equivalence Relations

Figure 0
...
† If the number 39539881642 were incorrectly
entered into a computer (programmed to calculate the check digit) as,
say, 39559881642 (an error in the fourth position), the machine would
calculate the check digit as 4, whereas the entered check digit would be
2
...

EXAMPLE 5 Airline companies, United Parcel Service, and the
rental car companies Avis and National use the modulo 7 values of
identification numbers to assign check digits
...
2) has the check digit 3 appended

Figure 0
...
If N 5 9q 1 r, where r is
the remainder upon dividing N by 9, then on a calculator screen N 4 9 appears as
q
...
, so the first decimal digit is the check digit
...
222, so 2 is the check digit
...
Thus, 3953988164 mod 9 5
56 mod 9 5 2
...


0 | Preliminaries

9

Figure 0
...
Similarly, the UPS pickup
record number 768113999, shown in Figure 0
...

The methods used by the Postal Service and the airline companies do
not detect all single-digit errors (see Exercises 35 and 39)
...
One method that does
this is the one used to assign the so-called Universal Product Code (UPC)
to most retail items (see Figure 0
...
A UPC identification number has 12
digits
...
(For many items, the 12th digit is not
printed, but it is always bar-coded
...
4, the check digit is 8
...
4

To explain how the check digit is calculated, it is convenient to introduce the dot product notation for two k-tuples:
(a1, a2,
...
, wk) 5 a1w1 1 a2w2 1 ? ? ? 1 akwk
...
, a12) ? (3, 1, 3, 1,
...

To verify that the number in Figure 0
...

The fixed k-tuple used in the calculation of check digits is called the
weighting vector
...
4 into a computer
...
Then the computer
calculates
0?312?111?310?110?310?119?3
1 5 ? 1 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1 5 99
...

In general, any single error will result in a sum that is not 0 modulo 10
...
For doubters, let us say that the identification number given in Figure 0
...
Notice
that the last two digits preceding the check digit have been transposed
...
In fact, the only undetected transposition errors of adjacent digits a and b are those where |a 2 b| 5 5
...
, ai11, ai,
...
, 3, 1) mod 10 5 0
...
, ai11, ai,
...
, 3, 1) mod 10
5 (a1, a2,
...
, a12) ? (3, 1, 3, 1,
...

This equality simplifies to either
(3ai11 1 ai) mod 10 5 (3ai 1 ai11) mod 10
or
(ai11 1 3ai) mod 10 5 (ai 1 3ai11) mod 10

0 | Preliminaries

11

depending on whether i is even or odd
...
It follows that |ai11 2 ai| 5 5, if ai11 ai
...
The weighing vector for 13-digit numbers is
(1, 3, 1, 3,
...

Identification numbers printed on bank checks (on the bottom left
between the two colons) consist of an eight-digit number a1a2 ? ? ? a8
and a check digit a9, so that
(a1, a2,
...

As is the case for the UPC scheme, this method detects all singledigit errors and all errors involving the transposition of adjacent digits a
and b except when |a 2 b| 5 5
...

In Chapter 5, we will examine more sophisticated means of assigning check digits to numbers
...
(Think of
it
...
One
of the check digits determines the magnitude of any single-digit error,
while the other check digit locates the position of the error
...
To illustrate the idea, let
us say that we have the eight-digit identification number a1a2 ? ? ? a8
...
, a9, a10) ? (1, 2, 3,
...

Let’s do an example
...
Then a9 and a10 are chosen to satisfy
(7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 1 a9 1 a10) mod 11 5 0

(1)

12

Integers and Equivalence Relations

and
(7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1 0 ? 6
1 1 ? 7 1 8 ? 8 1 a9 ? 9 1 a10 ? 10) mod 11 5 0
...


(19)

Likewise, since (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1
0 ? 6 1 1 ? 7 1 8 ? 8) mod 11 5 10, Equation (2) reduces to
(10 1 9a9 1 10a10) mod 11 5 0
...
Thus a9 5 7
...
So, the number is encoded as 7324501877
...
Since the sum of the digits of the received number
mod 11 is 5, we know that some digit is 5 too large or 6 too small
(assuming only one error has been made)
...
Then the second dot product has the form a1 ? 1 1
a2 ? 2 1 ? ? ? 1 (ai 1 5)i 1 ai11 ? (i 1 1) 1 ? ? ? 1 a10 ? 10 5
(a1, a2, ? ? ? , a10) ? (1, 2, ? ? ? , 10) 1 5i
...
Since the left-hand
side mod 11 is 10, we see that i 5 2
...
We have successfully corrected the error
...
Both are equivalent to the Well Ordering Principle
...
Francisco Maurolycus (1494–1575), a teacher of Galileo, used
it in 1575 to prove that 1 1 3 1 5 1 ? ? ? 1 (2n 2 1) 5 n2, and Blaise
Pascal (1623–1662) used it when he presented what we now call
Pascal’s triangle for the coefficients of the binomial expansion
...


0 | Preliminaries

13

Theorem 0
...
Suppose S has the property that
whenever some integer n $ a belongs to S, then the integer n 1 1 also
belongs to S
...


PROOF The proof is left as an exercise (Exercise 29)
...
We then assume the statement is true for
the integer n and use this assumption to prove that the statement is true
for the integer n 1 1
...
Recall that
given a straightedge and compass, we can construct a right angle
...
The case when n 5 1 is given
...
Then use
the straightedge and compass to construct a right triangle with height 1
and base "n
...
So,
by induction, we can construct a line segment of length "n for every
positive integer n
...

Obviously, the statement is true for n 5 1
...

We must prove that (cos u 1 i sin u)n11 5 cos(n 1 1)u 1 i sin(n 1 1)u
...

Now, using trigonometric identities for cos(a 1 b) and sin(a 1 b), we
see that this last term is cos(n 1 1)u 1 i sin(n 1 1)u
...

In many instances, the assumption that a statement is true for an integer n does not readily lend itself to a proof that the statement is true

14

Integers and Equivalence Relations

for the integer n 1 1
...
Some authors call this formulation
the strong form of induction
...
5 Second Principle of Mathematical Induction
Let S be a set of integers containing a
...
Then, S contains every integer greater than or
equal to a
...

To use this form of induction, we first show that the statement is true
for the integer a
...

EXAMPLE 8 We will use the Second Principle of Mathematical
Induction with a 5 2 to prove the existence portion of the Fundamental
Theorem of Arithmetic
...
Clearly, 2 [ S
...
We must
show that n [ S
...
If n is not a
prime, then n can be written in the form ab, where 1 , a , n and 1 ,
b , n
...
Thus, n is also a
product of primes
...

Notice that it is more natural to prove the Fundamental Theorem of
Arithmetic with the Second Principle of Mathematical Induction than
with the First Principle
...
(Does knowing that 5280 is a product of primes help you
to factor 5281 as a product of primes?)
The following problem appeared in the “Brain Boggler” section of
the January 1988 issue of the science magazine Discover
...
00 and red chips worth $8
...
What is the largest bet that
cannot be made?

0 | Preliminaries

15

To gain insight into this problem, we try various combinations of
blue and red chips and obtain 5, 8, 10, 13, 15, 16, 18, 20, 21, 23, 24, 25,
26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40
...
But how can we be sure? Well, we need only prove that
every integer greater than 27 can be written in the form a ? 5 1
b ? 8, where a and b are nonnegative integers
...
For the purpose of contrast,
we will give two proofs—one using the First Principle of Mathematical
Induction and one using the Second Principle
...
Obviously, 28 [ S
...
We must show
that n 1 1 [ S
...
If a $ 3, then
n 1 1 5 (a ? 5 1 b ? 8) 1 (23 ? 5 1 2 ? 8)
5 (a 2 3) ? 5 1 (b 1 2) ? 8
...
) If b $ 3, then
n 1 1 5 (a ? 5 1 b ? 8) 1 (5 ? 5 2 3 ? 8)
5 (a 1 5) ? 5 1 (b 2 3) ? 8
...
) This completes the proof
...
Now assume that for
some integer n
...
We
must show that n [ S
...
But then n 5
(a 1 1) ? 5 1 b ? 8
...


Equivalence Relations
In mathematics, things that are considered different in one context may
be viewed as equivalent in another context
...
Indeed, the sums 2 1 1 and 4 1 4 are certainly different
in ordinary arithmetic, but are the same under modulo 5 arithmetic
...

In physics, vectors of the same magnitude and direction can produce

16

Integers and Equivalence Relations

different effects—a 10-pound weight placed 2 feet from a fulcrum produces a different effect than a 10-pound weight placed 1 foot from a
fulcrum
...
What is needed to make these
distinctions precise is an appropriate generalization of the notion of
equality; that is, we need a formal mechanism for specifying whether or
not two quantities are the same in a given setting
...

Definition Equivalence Relation

An equivalence relation on a set S is a set R of ordered pairs of
elements of S such that
1
...

2
...

3
...


When R is an equivalence relation on a set S, it is customary to write
aRb instead of (a, b) [ R
...
Using this notation, the three conditions for an equivalence relation become a , a; a , b implies
b , a; and a , b and b , c imply a , c
...

EXAMPLE 10 Let S be the set of all triangles in a plane
...
Then, , is an equivalence relation on S
...
If f, g [ S, define f , g if f 9 5 g9, where f 9 is the derivative of f
...
Since two polynomials with
equal derivatives differ by a constant, we see that for any f in S, [ f ] 5
{ f 1 c | c is real}
...
If a, b [ S, define a ; b if a mod n 5 b mod n (that is, if a 2 b is
divisible by n)
...
Since this particular relation is important in abstract algebra, we will take the trouble to verify that it is indeed an equivalence
relation
...

Next, assume that a ; b, say, a 2 b 5 rn
...
Finally, assume that a ; b and b ; c, say, a 2 b 5 rn
and b 2 c 5 sn
...

EXAMPLE 13 Let ; be as in Example 12 and let n 5 7
...
Also, [1] 5 {
...
} and [4] 5 {
...

EXAMPLE 14 Let S 5 {(a, b) | a, b are integers, b 2 0}
...
Then < is an equivalence relation on S
...
In fact, the pairs (a, b) and (c, d) are equivalent if the fractions a/b
and c/d are equal
...
Next, we assume that (a, b) < (c, d),
so that ad 5 bc
...
Finally, we assume that (a, b) <
(c, d ) and (c, d) < (e, f ) and prove that (a, b) < (e, f )
...
Multiplying both sides
of ad 5 bc by f and replacing cf by de, we obtain adf 5 bcf 5 bde
...

Definition Partition

A partition of a set S is a collection of nonempty disjoint subsets of S
whose union is S
...
5 illustrates a partition of a set into four
subsets
...
5 Partition of S into four subsets
...
}, and {
...

EXAMPLE 16 The set of nonnegative integers and the set of nonpositive integers do not partition the integers, since both contain 0
...


18

Integers and Equivalence Relations

Theorem 0
...
Conversely, for any partition P of S, there
is an equivalence relation on S whose equivalence classes are the
elements of P
...
For any a [ S, the
reflexive property shows that a [ [a]
...
Now, suppose that [a] and [b] are distinct
equivalence classes
...

assume c [ [a] > [b]
...
To this end, let x [ [a]
...
By the symmetric property, we
also have a , c
...
This proves [a] # [b]
...
Thus, [a] 5 [b],
in contradiction to our assumption that [a] and [b] are distinct equivalence classes
...
Define a , b if a and b belong to the
same subset in the collection
...


Functions (Mappings)
Although the concept of a function plays a central role in nearly every
branch of mathematics, the terminology and notation associated with
functions vary quite a bit
...

Definition Function (Mapping)

A function (or mapping) f from a set A to a set B is a rule that assigns
to each element a of A exactly one element b of B
...
If f assigns b to a, then
b is called the image of a under f
...


We use the shorthand f: A → B to mean that f is a mapping from
A to B
...

There are often different ways to denote the same element of a set
...
For example, the

0 | Preliminaries

19

correspondence f from the rational numbers to the integers given by
f(a/b) 5 a 1 b does not define a function since 1/2 5 2/4 but f (1/2) ?
f (2/4)
...

Definition Composition of Functions

Let f: A → B and c: B → C
...
The composition
function cf can be visualized as in Figure 0
...


φ

ψ
φ (a)
ψ (φ (a))

a

ψφ

Figure 0
...


In calculus courses, the composition of f with g is written ( f 8 g)(x) and
is defined by ( f 8 g)(x) 5 f (g(x))
...
”
There are several kinds of functions that occur often enough to be
given names
...


The term one-to-one is suggestive, since the definition ensures that
one element of B can be the image of only one element of A
...
That is, different elements of A map to different elements of B
...
7
...
7

20

Integers and Equivalence Relations

Definition Function from A onto B

A function f from a set A to a set B is said to be onto B if each element
of B is the image of at least one element of A
...

See Figure 0
...

φ

ψ

φ is onto

ψ is not onto

Figure 0
...

Theorem 0
...
g(ba) 5 (gb)a (associativity)
...
If a and b are one-to-one, then ba is one-to-one
...
If a and b are onto, then ba is onto
...
If a is one-to-one and onto, then there is a function a21 from B
onto A such that (a21a)(a) 5 a for all a in A and (aa21)(b) 5 b
for all b in B
...
The remaining parts are left as exercises
(Exercise 51)
...
Then (g(ba))(a) 5 g((ba)(a)) 5 g(b(a(a)))
...
So, g(ba) 5
(gb)a
...
7 has the property that if a(s) 5 t,
then a21(t) 5 s
...
In effect, a21 “undoes” what a does
...
The following table illustrates the properties of one-to-one and onto
...
Clearly, the mapping from Z to Z given by x → x3 is not
onto, since 2 is the cube of no integer
...
The remaining verifications are left to
the reader
...

Immediately the host interrupted me and asked: “Are there still infinitely
many primes?”
NOGA ALON

1
...

2
...

3
...

4
...
Show that s and t
are not unique
...
In Florida, the fourth and fifth digits from the end of a driver’s license
number give the year of birth
...
For
females the digits are 40(m 2 1) 1 b 1 500
...

6
...
For females the digits are 63m 1 2b 1 1
...


22

Integers and Equivalence Relations

7
...

8
...
If a and b are
relatively prime, show that ab divides c
...

9
...

10
...
If a 5 da9 and b 5 db9,
show that gcd(a9, b9) 5 1
...
Let n be a fixed positive integer greater than 1
...
(This exercise is referred to in Chapters
6, 8, and 15
...
Let a and b be positive integers and let d 5 gcd(a, b) and m 5
lcm(a, b)
...
If s is a
multiple of both a and b, prove that s is a multiple of m
...
Let n and a be positive integers and let d 5 gcd(a, n)
...
(This
exercise is referred to in Chapter 2
...
Show that 5n 1 3 and 7n 1 4 are relatively prime for all n
...
Prove that every prime greater than 3 can be written in the form
6n 1 1 or 6n 1 5
...
Determine 71000 mod 6 and 61001 mod 7
...
Let a, b, s, and t be integers
...
What condition on s
and t is needed to make the converse true? (This exercise is referred
to in Chapter 8
...
Determine 8402 mod 5
...
Show that gcd(a, bc) 5 1 if and only if gcd(a, b) 5 1 and
gcd(a, c) 5 1
...
)
20
...
, pn be primes
...

21
...
(Hint: Use Exercise 20
...
For every positive integer n, prove that 1 1 2 1 ? ? ? 1 n 5
n(n 1 1)/2
...
For every positive integer n, prove that a set with exactly n elements
has exactly 2n subsets (counting the empty set and the entire set)
...
For any positive integer n, prove that 2n 32n 2 1 is always divisible
by 17
...
Prove that there is some positive integer n such that n, n 1 1,
n 1 2, ? ? ? , n 1 200 are all composite
...
(Generalized Euclid’s Lemma) If p is a prime and p divides
a1a2 ? ? ? an, prove that p divides ai for some i
...
Use the Generalized Euclid’s Lemma (see Exercise 26) to establish
the uniqueness portion of the Fundamental Theorem of Arithmetic
...
What is the largest bet that cannot be made with chips worth $7
...
00? Verify that your answer is correct with both forms of
induction
...
Prove that the First Principle of Mathematical Induction is a consequence of the Well Ordering Principle
...
The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34,
...
Prove that the nth Fibonacci number fn satisfies fn , 2n
...
In the cut “As” from Songs in the Key of Life, Stevie Wonder mentions the equation 8 3 8 3 8 3 8 5 4
...

32
...

33
...
M
...
Determine the check digit for a money order with identification
number 7234541780
...
Suppose that in one of the noncheck positions of a money order
number, the digit 0 is substituted for the digit 9 or vice versa
...
Prove that all
other errors involving a single position are detected
...
Suppose that a money order identification number and check digit
of 21720421168 is erroneously copied as 27750421168
...
A transposition error involving distinct adjacent digits is one of the
form
...
ba
...
Prove that the money
order check digit scheme will not detect such errors unless the
check digit itself is transposed
...
Determine the check digit for the Avis rental car with identification
number 540047
...
)
39
...

40
...

41
...


24

Integers and Equivalence Relations

42
...

43
...
, a10) ? (10, 9, 8, 7,
6, 5, 4, 3, 2, 1) mod 11 5 0
...
When
a10 is required to be 10 to make the dot product 0, the character X is
used as the check digit
...

44
...
Determine the missing
digit
...
Suppose three consecutive digits abc of an ISBN-10 are scrambled as
bca
...
The ISBN-10 0-669-03925-4 is the result of a transposition of two
adjacent digits not involving the first or last digit
...

47
...
Explain how this would affect the check digit
...
Use the two-check-digit error-correction method described in this
chapter to append two check digits to the number 73445860
...
Suppose that an eight-digit number has two check digits appended
using the error-correction method described in this chapter and it is
incorrectly transcribed as 4302511568
...

50
...
a8 so that (9a1 1 8a2 1 7a3 1 6a4 1 5a5 1
4a6 1 3a7 1 2a8 1 a9) mod 10 5 0
...

51
...
7
...
Let S be the set of real numbers
...
Show that , is an equivalence relation on S
...

53
...
If a, b [ S, define aRb if ab $ 0
...
Let S be the set of integers
...

Prove that R is an equivalence relation and determine the equivalence
classes of S
...
Complete the proof of Theorem 0
...


0 | Preliminaries

25

56
...
is a
square of an integer
...
(Cancellation Property) Suppose a, b and g are functions
...


Computer Exercises
There is nothing more practical than a good theory
...
d
...
edu/~jgallian
1
...
Use it to verify that 39539881642 is valid
...
Was the error detected? Enter
the number with the 9 in position 2 replaced with a 0
...
Enter the number with two digits transposed
...

2
...
Use it to verify
that 090146003386 is valid
...
Was the error detected? Enter the number with two
consecutive digits transposed
...
Was the error detected? Explain why or why not
...
Was the error detected? Explain why or why not
...
This software checks the validity of a UPS number
...
Now enter the same number with one digit
incorrect
...
Was the error detected? Enter the number with
the 8 replaced by 1
...

4
...
Use it to verify that 091902049 is valid
...
Was the error detected?
Enter the number with two consecutive digits transposed
...

Was the error detected? Explain why or why not
...
This software checks the validity of an ISBN-10
...
Now enter the same number with one digit incorrect
...
Was the error detected?

26

Integers and Equivalence Relations

6
...
Run the
program with the input 21355432, 20965744, 10033456
...
Was the error corrected?

Suggested Readings
Linda Deneen, “Secret Encryption with Public Keys,” The UMAP Journal
8 (1987): 9–29
...
They range from a
simple scheme used by Julius Caesar to a highly sophisticated scheme
invented in 1978 and based on modular n arithmetic, where n has more
than 200 digits
...
A
...

This article describes various methods used by the states to assign driver’s license numbers
...

This article can be downloaded at http://www
...
umn
...
pdf
J
...
Gallian, “The Mathematics of Identification Numbers,” The College
Mathematics Journal 22 (1991): 194–202
...
This article can be downloaded
at http://www
...
umn
...
pdf
J
...
Gallian and S
...

This article provides a more detailed analysis of the check digit
schemes presented in this chapter
...
This article can be downloaded
at http://www
...
umn
...
pdf

PA R T

2

Groups

For online student resources, visit this textbook’s website at
http://college
...
com/PIC/gallian7e

27

This page intentionally left blank

Introduction
1 to Groups
Symmetry is a vast subject, significant in art and nature
...

HERMANN WEYL, Symmetry

Symmetries of a Square
Suppose we remove a square region from a plane, move it in some way,
then put the square back into the space it originally occupied
...
More specifically, we want to describe
the possible relationships between the starting position of the square
and its final position in terms of motions
...
Thus, for
example, we consider a 908 rotation and a 4508 rotation as equal, since
they have the same net effect on every point
...

To begin, we can think of the square region as being transparent
(glass, say), with the corners marked on one side with the colors blue,
white, pink, and green
...
With this marking scheme, we are now
in a position to describe, in simple fashion, all possible ways in which
a square object can be repositioned
...
1
...
To verify this claim, observe that the final position of the square
is completely determined by the location and orientation (that is, face
up or face down) of any particular corner
...


29

30

Groups

P

W

P

G

G

P

G

B

G

W

P

G

P

B

W

G

B

P

W

W

P

B

G

P

G

W

B

B

B

P

W

G

P

R90

W

R180 = Rotation of 180°

R180
G

P

R270 = Rotation of 270°

B

W

G
P

= Flip about a horizontal axis

R270

B
W

H
G
P

= Flip about a vertical axis

B
W

G

D

B

P

W

G

V

B

W

B

R90 = Rotation of 90° (counterclockwise)

H

W

R0

R0 = Rotation of 0° (no change in position)

B

P

= Flip about the main diagonal

W

G

D

B

P

DЈ = Flip about the other diagonal

V

W

G

DЈ

B

Figure 1
...
1
...
In pictures,
P
G

W
B

R90

W

B

P
H

P

G

G

W

B

Thus, we see that this pair of motions—taken together—is equal to
the single motion D
...
And indeed we can, since the

31

1 | Introduction to Groups

eight motions may be viewed as functions from the square region to
itself, and as such we can combine them using function composition
...
The eight motions R0,
R90, R180, R270, H, V, D, and D9, together with the operation composition,
form a mathematical system called the dihedral group of order 8 (the
order of a group is the number of elements it contains)
...
Rather than introduce the formal definition of a group here, let’s
look at some properties of groups by way of the example D4
...
The
circled entry represents the fact that D 5 HR90
...
)
R0
R0
R90
R180
R270
H
V
D
D9

R90

R180

R270

H

V

D

R0
R90
R180
R270
H
V
D
D9

R90
R180
R270
R0
D
⅜
D9
V
H

R180
R270
R0
R90
V
H
D9
D

R270
R0
R90
R180
D9
D
H
V

H
D9
V
D
R0
R180
R270
R90

V
D
H
D9
R180
R0
R90
R270

D
H
D9
V
R90
R270
R0
R180

D9
D9
V
D
H
R270
R90
R180
R0

Notice how orderly this table looks! This is no accident
...
Of course, this is because, as
we have already pointed out, any sequence of motions turns out to be
the same as one of these eight
...
This property is called closure, and it is one of
the requirements for a mathematical system to be a group
...
Thus, combining
any element A on either side with R0 yields A back again
...
Moreover, we see that for each element A in D4, there is exactly
one element B in D4 such that AB 5 BA 5 R0
...
For example, R90 and R270 are
inverses of each other, and H is its own inverse
...
Another striking feature

32

Groups

of the table is that every element of D4 appears exactly once in each
row and column
...

Another property of D4 deserves special comment
...
Thus, in a group, ab may or may not
be the same as ba
...

Otherwise, we say the group is non-Abelian
...
The remaining condition required for a group
is associativity; that is, (ab)c 5 a(bc) for all a, b, c in the set
...
In practice, however,
this is rarely done! Here, for example, we simply observe that the eight
motions are functions and the operation is function composition
...


The Dihedral Groups
The analysis carried out above for a square can similarly be done for
an equilateral triangle or regular pentagon or, indeed, any regular n-gon
(n $ 3)
...

The dihedral groups arise frequently in art and nature
...
Corporation logos
are rich sources of dihedral symmetry [1]
...
The ubiquitous
five-pointed star has symmetry group D5
...

Chemists classify molecules according to their symmetry
...
The symmetry group of a pyramidal molecule such as ammonia (NH3), depicted in Figure 1
...


1 | Introduction to Groups

33

N

H
H
H

Figure 1
...


Mineralogists determine the internal structures of crystals (that is,
rigid bodies in which the particles are arranged in three-dimensional
repeating patterns—table salt and table sugar are two examples) by
studying two-dimensional x-ray projections of the atomic makeup
of the crystals
...
Commonly occurring
symmetry patterns are D4 and D6 (see Figure 1
...
Interestingly, it is
mathematically impossible for a crystal to possess a Dn symmetry pattern with n 5 5 or n
...


Figure 1
...


The dihedral group of order 2n is often called the group of symmetries of a regular n-gon
...
(The term symmetry is from the Greek word
symmetros, meaning “of like measure
...
Symmetries in
three dimensions are defined analogously
...
Similarly, any translation of a plane or of three-dimensional
space is a symmetry
...
4)
...
Notice that the restriction of a 1808 rotation about a line L in three dimensions to a plane containing L is a
reflection across L in the plane
...
Just as a reflection across a line is a plane symmetry that
cannot be achieved by a physical motion of the plane in two dimensions, a reflection across a plane is a three-dimensional symmetry that
cannot be achieved by a physical motion of three-dimensional space
...

q
q9
L

Figure 1
...
A symmetry group consisting of the rotational symmetries of
08, 3608/n, 2(3608)/n,
...
Cyclic
rotation groups, along with dihedral groups, are favorites of artists, designers, and nature
...
5 illustrates with corporate logos the cyclic
rotation groups of orders 2, 3, 4, 5, 6, 8, 16, and 20
...
A study of symmetry in greater depth is given in Chapters 27 and 28
...
5 Logos with cyclic rotation symmetry groups
...

PAUL HALMOS, Hilbert Space Problem Book

1
...

2
...

3
...
Describe in pictures or words the elements of D5 (symmetries of a
regular pentagon)
...
For n $ 3, describe the elements of Dn
...
) How many elements
does Dn have?
6
...

7
...

8
...

9
...
Describe an analogy between multiplying these two
numbers and multiplying elements of Dn
...
If r1, r2, and r3 represent rotations from Dn and f1, f2, and f3 represent
reflections from Dn, determine whether r1r2 f1r3 f2 f3r3 is a rotation
or a reflection
...
Find elements A, B, and C in D4 such that AB 5 BC but A Z C
...
)
12
...

1

1

n

F
2

n

R360 / n
2
2

1

2

n–1
2

R360/ n
n

1

n

F
1

3

3

1

13
...
Construct the
corresponding Cayley table
...
Describe the symmetries of a parallelogram that is neither a rectangle nor a rhombus
...

15
...
Do the same for
a hyperbola
...
Consider an infinitely long strip of equally spaced H’s:
???HHHH???
Describe the symmetries of this strip
...
For each of the snowflakes in the figure, find the symmetry group
and locate the axes of reflective symmetry (disregard imperfections)
...


1 | Introduction to Groups

37

18
...


19
...
Bottle caps that are pried off typically have 22 ridges around the
rim
...

21
...
For each design below, determine the symmetry group (ignore
imperfections)
...
What would the effect be if a six-bladed ceiling fan were designed
so that the centerlines of two of the blades were at a 708 angle and
all the other blades were set at a 588 angle?

Reference
1
...
B
...


Suggested Reading
Michael Field and Martin Golubitsky, Symmetry in Chaos, Oxford University Press, 1992
...


Niels Abel
He [Abel] has left mathematicians
something to keep them busy for five
hundred years
...


NIELS HENRIK ABEL, one of the foremost
mathematicians of the 19th century, was
born in Norway on August 5, 1802
...
When Abel was 18 years old, his
father died, and the burden of supporting the
family fell upon him
...
At the age of 19,
Abel solved a problem that had vexed leading mathematicians for hundreds of years
...

Although Abel died long before the advent of the subjects that now make up abstract algebra, his solution to the quintic
problem laid the groundwork for many of
these subjects
...
He died on
April 6, 1829, at the age of 26
...


In recognition of the fact that there is no
Nobel Prize for mathematics, in 2002 Norway
established the Abel Prize as the “Nobel Prize
in mathematics” in honor of its native son
...

To find more information about Abel, visit:
http://www-groups
...
st-and

...
uk/~history/

39

2 Groups

A good stock of examples, as large as possible, is indispensable
for a thorough understanding of any concept, and when I want
to learn something new, I make it my first job to build one
...
HALMOS

Definition and Examples of Groups
The term group was used by Galois around 1830 to describe sets of
one-to-one functions on finite sets that could be grouped together to
form a set closed under composition
...
Although this definition was given by both Heinrich Weber and Walter von Dyck in 1882,
it did not gain universal acceptance until the 20th century
...
A binary operation on G is a function that assigns each
ordered pair of elements of G an element of G
...
This condition is called closure
...
Division of integers is not a binary operation on the integers because an integer divided by an integer need not
be an integer
...
, n 2 1}, which we denote by Zn, play an
extremely important role in abstract algebra
...
It will be clear

40

2 | Groups

41

from the context whether we are using addition only or addition and
multiplication
...

Definition Group

Let G be a set together with a binary operation (usually called multiplication) that assigns to each ordered pair (a, b) of elements of G an element in G denoted by ab
...

1
...
The operation is associative; that is, (ab)c 5 a(bc) for
all a, b, c in G
...
Identity
...

3
...
For each element a in G, there is an element b in G
(called an inverse of a) such that ab 5 ba 5 e
...
Be
sure to verify closure when testing for a group (see Example 5)
...

If a group has the property that ab 5 ba for every pair of elements
a and b, we say the group is Abelian
...
When encountering a particular group for the first time, one should determine whether
or not it is Abelian
...
These examples will be used
throughout the text to illustrate the theorems
...
) As we
progress, the reader is bound to have hunches and conjectures that
can be tested against the stock of examples
...

EXAMPLE 1 The set of integers Z (so denoted because the German
word for numbers is Zahlen), the set of rational numbers Q (for quotient), and the set of real numbers R are all groups under ordinary addition
...


42

Groups

EXAMPLE 2 The set of integers under ordinary multiplication is not
a group
...
For example, there is no integer b such that 5b 5 1
...
Note that 21 is its own inverse,
whereas the inverse of i is 2i, and vice versa
...
The inverse of any a is 1/a 5 a21
...
Indeed, "2 ? "2 5 2, so S is not closed
under multiplication
...
The set of all 2 3 2 matrices with real entries is a group
under componentwise addition
...

0 0
c d
2c 2d

EXAMPLE 7 The set Zn 5 {0, 1,
...
For any j
...

This group is usually referred to as the group of integers modulo n
...
But what about multiplication? In each case, the existence of
some elements that do not have inverses prevents the set from being a
group under the usual multiplication
...
Examples 8, 9, and 11
illustrate this
...
The identity is 1
...


2 | Groups

43

a b
d is the
c d
number ad 2 bc
...
The set
EXAMPLE 9† The determinant of the 2 3 2 matrix c

GL(2, R) 5 e c

a b
d ` a, b, c, d
c d

P R, ad

2 bc ? 0 f

of 2 3 2 matrices with real entries and nonzero determinant is a nonAbelian group under the operation
c

a a 1 b1c2 a1b2 1 b1d2
a1 b1 a2 b2
dc
d 5 c 1 2
d
c1 d1 c2 d2
c1a2 1 d1c2 c1b2 1 d1d2
...
This follows from the fact that for any pair of 2 3 2
matrices A and B, det (AB) 5 (det A)(det B)
...
The identity is c
d ; the inverse of c
d is
0 1
c d
d
2b
ad 2 bc ad 2 bc
≥
¥
2c
a
ad 2 bc ad 2 bc
(explaining the requirement that ad 2 bc 2 0)
...

EXAMPLE 10 The set of all 2 3 2 matrices with real number entries
is not a group under the operation defined in Example 9
...

Now that we have shown how to make subsets of the real numbers
and subsets of the set of 2 3 2 matrices into multiplicative groups, we
next consider the integers under multiplication modulo n
...
However,
readers who have had linear algebra can readily generalize to n 3 n matrices
...
Euler, 1761) By Exercise 13 in Chapter 0, an
integer a has a multiplicative inverse modulo n if and only if a and n are
relatively prime
...
1, we define U(n) to be the set of all
positive integers less than n and relatively prime to n
...
(We leave it to the reader to
check that this set is closed under this operation
...
The Cayley table for
U(10) is
mod 10

1

3

7

9

1
3
7
9

1
3
7
9

3
9
1
7

7
1
9
3

9
7
3
1

(Recall that ab mod n is the unique integer r with the property a ? b 5
nq 1 r, where 0 # r , n and a ? b is ordinary multiplication
...
, n 2 1}
...

EXAMPLE 12 The set {0, 1, 2, 3} is not a group under multiplication modulo 4
...

EXAMPLE 13 The set of integers under subtraction is not a group,
since the operation is not associative
...
Study actively!
Don’t just read along and be spoon-fed by the book
...
, n 2 1 f
n
n

(i
...
, complex zeros of xn 2 1) is a group under multiplication
...
) Compare this group
with the one in Example 3
...
The distance from the point a 1 bi to the origin is "a 2 1b 2 and is often denoted by Ua 1 bi|
...
Thus, the six complex
zeros of x6 5 1 are located at points around the circle of radius 1, 60°
apart, as shown in Figure 2
...

Imaginary
– 1 + √3 i
2
2

1 + √3
2
2 i

√3
2
60°

–1

1
1
2

– 1 – √3 i
2
2

Real

1 – √3
2
2 i

Figure 2
...
, an) U a1, a2,
...
e
...
, an) 1
(b1, b2,
...
, an 1 bn)]
...
Then G 5 {Ta,b U a, b [ R} is a group
under function composition
...
From this formula we may observe that G is
closed, T0,0 is the identity, the inverse of Ta,b is T2a,2b, and G is Abelian
...
The elements of G are
called translations
...
This group is called
the special linear group of 2 3 2 matrices over Q, R, C, or Zp, respectively
...
For the group SL(2, F), the formula given in Example 9 for
a b
d 2b
the inverse of c
d simplifies to c
d
...
To illustrate the case SL(2, Z5),
3 4
consider the element A 5 c
d
...
Note
24
3
1 3
3 4 4 1
1 0
that c
dc
d 5 c
d when the arithmetic is done modulo 5
...

EXAMPLE 18 Let F be any of Q, R, C, or Zp ( p a prime)
...
As in
Example 17, when F is Zp, modulo p arithmetic is used to calculate
determinants, the matrix products, and inverses
...
For example, in GL(2, Z7),
4 5
d
...

mod 7 5 1]
...

[The reader should check that c
6 3 5 6
0 1
EXAMPLE 19 The set {1, 2,
...

EXAMPLE 20 The set of all symmetries of the infinite ornamental
pattern in which arrowheads are spaced uniformly a unit apart along

47

2 | Groups

a line is an Abelian group under composition
...
Then, every member of the
group is of the form x1x2 ? ? ? xn, where each xi [
{T, T 21, H}
...

Table 2
...

As the examples above demonstrate, the notion of a group is a very
broad one indeed
...

The goal of abstract algebra is to discover truths about algebraic
systems (that is, sets with one or more binary operations) that are independent of the specific nature of the operations
...
1 Summary of Group Examples (F can be any of Q, R, C, or Zp; L is a reflection)
Group

Operation

Z
Q1

Addition
Multiplication

0
1

Zn
R*
GL(2, F)

Addition mod n
Multiplication
Matrix
multiplication

0
1

U(n)
Rn
SL(2, F)

Dn

Multiplication
mod n
Componentwise
addition
Matrix
multiplication
Composition

Identity

c

1
0

0
d
1

Form of
Element
k
m/n,
m, n
...
We then seek to deduce consequences of these
properties
...
It must be remembered, however, that when a specific group
is being discussed, a specific operation must be given (at least
implicitly)
...
The definition itself raises
some fundamental questions
...
Could a
group have more than one? Every group element has an inverse
...
But examples can only suggest
...
We are forced to
restrict ourselves to the properties that all groups have; that is, we must
view groups as abstract entities rather than argue by example
...

Theorem 2
...


PROOF Suppose both e and e9 are identities of G
...
ae 5 a for all a in G, and
2
...

The choices of a 5 e9 in (1) and a 5 e in (2) yield e9e 5 e9 and
e9e 5 e
...

Because of this theorem, we may unambiguously speak of “the identity” of a group and denote it by “e” (because the German word for
identity is Einheit)
...
2 Cancellation
In a group G, the right and left cancellation laws hold; that is,
ba 5 ca implies b 5 c, and ab 5 ac implies b 5 c
...
Let a9 be an inverse of a
...
Associativity yields
b(aa9) 5 c(aa9)
...
Similarly, one can prove that ab 5 ac implies b 5 c by multiplying by a9 on
the left
...
Another consequence of the
cancellation property is the uniqueness of inverses
...
3 Uniqueness of Inverses
For each element a in a group G, there is a unique element b in G
such that ab 5 ba 5 e
...
Then ab 5 e and
ac 5 e, so that ab 5 ac
...

As was the case with the identity element, it is reasonable, in view
of Theorem 2
...
This notation is suggested by that used for ordinary real numbers under multiplication
...

n factors

g0

We define 5 e
...
Unlike for real numbers, in an abstract group we
do not permit noninteger exponents such as g1/2
...

Although the way one manipulates the group expressions gmgn and
(gm)n coincides with the laws of exponents for real numbers, the laws
of exponents fail to hold for expressions involving two group elements
...

Also, one must be careful with this notation when dealing with a
specific group whose binary operation is addition and is denoted by

50

Groups

“1
...
For
example, the inverse of g is written as 2g
...
2
Multiplicative Group

Additive Group

a ? b or ab
e or 1
a21
an

Multiplication
Identity or one
Multiplicative inverse of a
Power of a

a1b
0
2a
na

Addition
Zero
Additive inverse of a
Multiple of a

ab21

Quotient

a2b

Difference

means g 1 g 1 g and is usually written as 3g, whereas g23 means
(2g) 1 (2g) 1 (2g) and is written as 23g
...
2 shows
the common notation and corresponding terminology for groups under multiplication and groups under addition
...

Because of the associative property, we may unambiguously write
the expression abc, for this can be reasonably interpreted as only (ab)c
or a(bc), which are equal
...
Thus,
a2(bcdb2) 5 a2b(cd )b2 5 (a2b)(cd )b2 5 a(abcdb)b,
and so on
...

Theorem 2
...


PROOF Since (ab)(ab)21 5 e and (ab)(b21a21) 5 a(bb21)a21 5
aea21 5 aa21 5 e, we have by Theorem 2
...


2 | Groups

51

Historical Note
We conclude this chapter with a bit of history concerning the noncommutativity of matrix multiplication
...
It was Werner
Heisenberg who recognized the cause
...
For all his boldness, this shook Heisenberg
...
94]:
In my paper the fact that XY was not equal to YX was very disagreeable to me
...
But this difficulty had worried me and I was not able to solve it
...
Born was fascinated and deeply impressed by Heisenberg’s new
approach
...
217]:
After having sent off Heisenberg’s paper to the Zeitschrift für Physik for publication, I began to ponder over his symbolic multiplication, and was soon so involved
in it that I thought about it for the whole day and could hardly sleep at night
...
And one morning, about the 10 July 1925, I suddenly saw light:
Heisenberg’s symbolic multiplication was nothing but the matrix calculus, wellknown to me since my student days from Rosanes’ lectures in Breslau
...
In his autobiography, Born laments [1, p
...

In fact, Heisenberg knew at that time very little of matrices and had to study
them
...
220]:
If I have not written to you for such a long time, and have not thanked you for your
congratulations, it was partly because of my rather bad conscience with respect to
you
...
I am, of course, glad that our common efforts are now appreciated, and I enjoy the recollection of the beautiful time of collaboration
...
Yet I myself can do nothing but thank you again for all the fine
collaboration, and feel a little ashamed
...


52

Groups

Exercises
“For example,” is not proof
...
Give two reasons why the set of odd integers under addition is not
a group
...
Referring to Example 13, verify the assertion that subtraction is not
associative
...
Show that {1, 2, 3} under multiplication modulo 4 is not a group
but that {1, 2, 3, 4} under multiplication modulo 5 is a group
...
Show that the group GL(2, R) of Example 9 is non-Abelian by exhibiting a pair of matrices A and B in GL(2, R) such that AB 2 BA
...
Find the inverse of the element c
d in GL(2, Z11)
...
Give an example of group elements a and b with the property that
a21ba 2 b
...
Translate each of the following multiplicative expressions into its
additive counterpart
...

a
...
a22(b21c)2
c
...
Show that the set {5, 15, 25, 35} is a group under multiplication
modulo 40
...
List the members of H 5 {x 2 | x [ D4} and K 5 {x [ D4 | x 2 5 e}
...
Prove that the set of all 2 3 2 matrices with entries from R and determinant 11 is a group under matrix multiplication
...
For any integer n
...


2 | Groups

53

13
...
Instead,
one of the nine integers was inadvertently left out, so that the list appeared as 1, 9, 16, 22, 53, 74, 79, 81
...
Let G be a group with the following property: Whenever a, b, and
c belong to G and ab 5 ca, then b 5 c
...

(“Cross cancellation” implies commutativity
...
(Law of Exponents for Abelian Groups) Let a and b be elements of
an Abelian group and let n be any integer
...

Is this also true for non-Abelian groups?
16
...
Find an example that shows that in a group, it is possible
to have (ab)22 2 b22a22
...

17
...

18
...

19
...

20
...
, an belong to a group, what is the inverse of a1a2
...
The integers 5 and 15 are among a collection of 12 integers that
form a group under multiplication modulo 56
...

22
...
Give two examples
of groups with 44 elements
...
Prove that every group table is a Latin square†; that is, each element of the group appears exactly once in each row and each column
...
)
24
...

25
...
Fill in the blank entries
...
There is also a close connection between Latin squares and finite geometries
...
Prove that if (ab)2 5 a2b2 in a group G, then ab 5 ba
...
Let a, b, and c be elements of a group
...
Solve a21xa 5 c for x
...
Prove that the set of all rational numbers of the form 3m6n, where
m and n are integers, is a group under multiplication
...
Let G be a finite group
...
Show that the number of elements x of G
such that x2 2 e is even
...
Give an example of a group with elements a, b, c, d, and x such
that axb 5 cxd but ab 2 cd
...
)
31
...
Prove that RFR 5 F
...
Let R be any rotation in some dihedral group and F, any reflection
in the same group
...

33
...
Prove that G is
Abelian
...
)
34
...

Write each of the following products in the form Ri or RiF, where
0 # i , n
...
In D4, FR22FR5
b
...
In D6, FR5FR22F
35
...
(This exercise is referred
to in Chapter 26
...
Prove that the set of all 3 3 3 matrices with real entries of the form
1
£0
0

a
1
0

b
c§
1

is a group
...

1

This group, sometimes called the Heisenberg group after the
Nobel Prize–winning physicist Werner Heisenberg, is intimately related to the Heisenberg Uncertainty Principle of quantum physics
...
Prove the assertion made in Example 19 that the set {1, 2,
...

38
...
If the stipulation
that the group be finite is omitted, what can you say about the
number of nonidentity elements that satisfy the equation x 5 5 e?
a a
39
...
Show that G is a group under
a a
matrix multiplication
...
(Compare with Example 10
...
By producing examples and by
observing the properties of special mathematical objects, one could hope to
obtain clues as to the behavior of general statements which have been
tested on examples
...
M
...
d
...
edu/~jgallian
1
...

2
...
Run the program for
k 5 9, 27, 81, 243, 25, 125, 49, 121
...
Run the program for k 5 18, 54, 162, 486, 50,
250, 98, 242
...

3
...

4
...
(The technical term for the number of elements in a group
is the order of the group
...


56

Groups

Do you see a relationship between the orders of GL (2, Zp) and
SL(2, Zp) and p 2 1? Does this relationship hold for p 5 2? Based
on these examples, does it appear that p always divides the order
of SL (2, Z p)? What about p 2 1? What about p 1 1? Guess a
formula for the order of SL (2, Zp)
...


References
1
...

2
...
Mehra and H
...
3, New York: Springer-Verlag, 1982
...

Chapter 3 of this book describes how the dihedral group of order 8 can
be used to encode the social structure of the kin system of family relationships among a tribe of native people of Australia
...

The author uses properties of groups to analyze the peg board game
central solitaire (which also goes by the name peg solitaire)
...
E
...

Students interested in the physical sciences may find this article worthwhile
...


Finite Groups;
3
Subgroups
In our own time, in the period 1960–1980, we have seen particle physics
emerge as the playground of group theory
...
To facilitate
the study of finite groups, it is convenient to introduce some terminology and notation
...
We will use |G| to denote the order of G
...

Definition Order of an Element

The order of an element g in a group G is the smallest positive integer
n such that gn 5 e
...
) If no
such integer exists, we say that g has infinite order
...


So, to find the order of a group element g, you need only compute the
sequence of products g, g2, g3,
...
The exponent of this product (or coefficient if the operation is addition) is the order of g
...

EXAMPLE 1 Consider U(15) 5 {1, 2, 4, 7, 8, 11, 13, 14} under
multiplication modulo 15
...
To find the order of
57

58

Groups

the element 7, say, we compute the sequence 71 5 7, 72 5 4, 73 5 13,
74 5 1, so |7| 5 4
...
Similar computations show that |1| 5 1, |2| 5 4,
|4| 5 2, |8| 5 4, |13| 5 4, |14| 5 2
...
Rather than compute the sequence 131, 132, 133,
134, we may observe that 13 5 22 mod 15, so that 132 5 (22)2 5 4,
133 5 22 ? 4 5 28, 134 5 (22)(28) 5 1
...
Since 1 ? 2 5 2,
2 ? 2 5 4, 3 ? 2 5 6, 4 ? 2 5 8, 5 ? 2 5 0, we know that |2| 5 5
...
(Here 2 ? 2 is
an abbreviation for 2 1 2, 3 ? 2 is an abbreviation for 2 1 2 1 2, etc
...
Here every nonzero
element has infinite order, since the sequence a, 2a, 3a,
...

The perceptive reader may have noticed among our examples of
groups in Chapter 2 that some are subsets of others with the same
binary operation
...
Similarly, the group of
complex numbers {1, 21, i, 2i} under multiplication is a subset of
the group described in Example 14 for n equal to any multiple of 4
...

Definition Subgroup

If a subset H of a group G is itself a group under the operation of G, we
say that H is a subgroup of G
...
If we
want to indicate that H is a subgroup of G but is not equal to G itself,
we write H , G
...
The
subgroup {e} is called the trivial subgroup of G; a subgroup that is not
{e} is called a nontrivial subgroup of G
...


Subgroup Tests
When determining whether or not a subset H of a group G is a subThe website www
...
com provides a convenient way to do modular arithmetic
...


†

3 | Finite Groups; Subgroups

59

group of G, one need not directly verify the group axioms
...

Theorem 3
...
If ab21 is in H
whenever a and b are in H, then H is a subgroup of G
...
)

PROOF Since the operation of H is the same as that of G, it is clear
that this operation is associative
...
Since H
is nonempty, we may pick some x in H
...
To verify that x21 is
in H whenever x is in H, all we need to do is to choose a 5 e and b 5
x in the statement of the theorem
...
Well, we have already shown that y21 is in H
whenever y is; so, letting a 5 x and b 5 y21, we have xy 5 x(y21)21 5
ab21 is in H
...
1 the “One-Step Subgroup
Test,’’ there are actually four steps involved in applying the theorem
...
)
Notice the similarity between the last three steps listed below and the
three steps involved in the Principle of Mathematical Induction
...
Identify the property P that distinguishes the elements of H; that is,
identify a defining condition
...
Prove that the identity has property P
...
)
3
...

4
...

The procedure is illustrated in Examples 4 and 5
...
Then H 5
{x [ G | x2 5 e} is a subgroup of G
...
So, we first note that e2 5 e, so that H is nonempty
...
This means that a2 5 e
and b2 5 e
...
Since G is
Abelian, (ab21)2 5 ab21ab21 5 a2(b21)2 5 a2(b2)21 5 ee21 5 e
...


60

Groups

In many instances, a subgroup will consist of all elements that have
a particular form
...
This is illustrated in the following example
...
Then H 5 {x2 | x [ G} is a subgroup of G
...
”) Since e2 5 e, the identity has the correct form
...
We
must show that a2(b2)21 also has the correct form; that is, a2(b2)21 is the
square of some element
...
Thus, H is a subgroup of G
...
1
...
2 Two-Step Subgroup Test
Let G be a group and let H be a nonempty subset of G
...


PROOF By Theorem 3
...
So, we suppose that a, b [ H
...
Thus, ab21 [ H by closure under multiplication
...

How do you prove that a subset of a group is not a subgroup? Here
are three possible ways, any one of which guarantees that the subset is
not a subgroup:
1
...

2
...

3
...

EXAMPLE 6 Let G be the group of nonzero real numbers under
multiplication, H 5 {x [ G | x 5 1 or x is irrational} and K 5
{x [ G | x $ 1}
...
Also, K is not a subgroup, since 2 [ K but
221 o K
...

Theorem 3
...
If H is closed under
the operation of G, then H is a subgroup of G
...
2, we need only prove that a21 [ H
whenever a [ H
...
If a e,
consider the sequence a, a2,
...
Since H is finite, not all of these elements are distinct
...
j
...
1
...
But, i 2 j 2 1 $ 1
implies ai2j21 [ H and we are done
...
We first introduce an important notation
...
In particular, observe
that the exponents of a include all negative integers as well as 0 and the
positive integers (a0 is defined to be the identity)
...
4 ͗a͘ Is a Subgroup
Let G be a group, and let a be any element of G
...


PROOF Since a [ ͗a͘, ͗a͘ is not empty
...
Then,
an(am)21 5 an2m [ ͗a͘; so, by Theorem 3
...

The subgroup ͗a͘ is called the cyclic subgroup of G generated by a
...

(A cyclic group may have many generators
...
, a22, a21, a0, a1, a2,
...
Also note that,
since aia j 5 ai1j 5 a j1i 5 a j a i , every cyclic group is Abelian
...
; 321 5 7

62

Groups

(since 3 ? 7 5 1), 322 5 9, 323 5 3, 324 5 1, 325 5 324 ? 321 5
1 ? 7, 326 5 324 ? 322 5 1 ? 9 5 9,
...
Remember, an means na
when the operation is addition
...
Here each entry in the list
...
represents a distinct group
element
...
Then,
Rn 5 R360° 5 e,

Rn11 5 R,

Rn12 5 R2,
...
, so that ͗R͘ 5 {e, R,
...
We see, then, that the powers of R “cycle back” periodically
with period n
...

Rn 5 e
Rn11 5 R

Rn12 5 R2

R21 5 Rn21

R22 5 Rn22

In Chapter 4 we will show that |͗a͘| 5 |a|; that is, the order of the
subgroup generated by a is the order of a itself
...
)
We next consider one of the most important subgroups
...
In symbols,

Z(G) 5 {a [ G | ax 5 xa for all x in G}
...
The term was coined by J
...
de Seguier in 1904
...
5 Center Is a Subgroup
The center of a group G is a subgroup of G
...
2 to prove this result
...
Now, suppose a, b [ Z(G)
...

Next, assume that a [ Z(G)
...

What we want is a21x 5 xa21 for all x in G
...
(Be careful here; groups need not be commutative
...
) Formally, the desired equation
can be obtained from the original one by multiplying it on the left and
right by a21, like so:
a21(ax)a21 5 a21(xa)a21,
(a21a)xa21 5 a21x(aa21),
exa21 5 a21xe,
xa21 5 a21x
...

For practice, let’s determine the centers of the dihedral groups
...


To verify this, first observe that since every rotation in Dn is a power
of R360/n, rotations commute with rotations
...
Let R be any rotation in Dn and let
F be any reflection in Dn
...
Thus it follows that R and F commute
if and only if FR 5 RF 5 FR21
...
But R 5 R21 only when R 5 R0 or R 5 R180, and R180 is in Dn
only when n is even
...

Although an element from a non-Abelian group does not necessarily
commute with every element of the group, there are always some
elements with which it will commute
...
This observation prompts the next definition and theorem
...
The centralizer of a in G, C(a), is
the set of all elements in G that commute with a
...


EXAMPLE 12 In D4, we have the following centralizers:
C(R0) 5 D4 5 C(R180),
C(R90) 5 {R0, R90, R180, R270} 5 C(R270),
C(H) 5 {R0, H, R180, V} 5 C(V),
C(D) 5 {R0, D, R180, D9} 5 C(D9)
...
The next theorem shows that this was not a coincidence
...
6 C(a) Is a Subgroup
For each a in a group G, the centralizer of a is a subgroup of G
...
5 is left to the reader to
supply (Exercise 25)
...
Also,
observe that G is Abelian if and only if C(a) 5 G for all a in G
...

ARNOLD ROSS

1
...
What relation do you
see between the orders of the elements of a group and the order of
the group?
Z12,

U(10),

U(12),

U(20),

D4

2
...

1
In Q, list the elements in ͗2 ͘
...

2

3 | Finite Groups; Subgroups

65

3
...
Find the order of each element
in Q and in Q*
...
Prove that in any group, an element and its inverse have the
same order
...
Without actually computing the orders, explain why the two elements in each of the following pairs of elements from Z30 must
have the same order: {2, 28}, {8, 22}
...

6
...
What are the possibilities for |a|? Provide reasons for your answer
...
If a is a group element and a has infinite order, prove that am ? an
when m ? n
...
Let x belong to a group
...
What can we say about the order of x?
9
...

10
...
[Hence, U(14) is cyclic
...
Show that U(20) 2 ͗k͘ for any k in U(20)
...
]
12
...

13
...
|ab| 5 3,
b
...
|ab| 5 5
...
Suppose that H is a proper subgroup of Z under addition and H
contains 18, 30, and 40
...

15
...
What are the possibilities for H?
16
...

17
...
1 of n, let Uk(n) 5 {x [ U(n) | x mod k 5 1}
...
]
List the elements of U4(20), U5(20), U5(30), and U10(30)
...
Let H 5 {x [ U(10) | x mod 3 5 1}
...
)
18
...
(Can you see that the same proof shows that the intersection
of any number of subgroups of G, finite or infinite, is again a
subgroup of G?)

66

Groups

19
...
Show that Z(G) 5 >a[GC(a)
...
]
20
...
Prove that C(a) 5 C(a21)
...
For any group element a and any integer k, show that C(a) # C(a k)
...
” Is the converse true?
22
...

1
1
2
3
4
5
6
7
8

2

3

4

5

6

7

8

1
2
3
4
5
6
7
8

2
1
4
3
6
5
8
7

3
4
2
1
8
7
5
6

4
3
1
2
7
8
6
5

5
6
7
8
1

6
5
8
7

7
8
6
5

8
7
5
6

1
1
1

23
...

1
1
2
3
4
5
6
7
8

24
...

26
...

28
...


2

3

4

5

6

7

8

1
2
3
4
5
6
7
8

2
1
4
3
6
5
8
7

3
8
5
2
7
4
1
6

4
7
6
1
8
3
2
5

5
6
7
8
1
2
3
4

6
5
8
7
2
1
4
3

7
4
1
6
3
8
5
2

8
3
2
5
4
7
6
1

a
...

b
...

c
...
How are these orders arithmetically related to the order of the group?
If a and b are distinct group elements, prove that either a2 2 b2 or
a3 2 b3
...
6
...
Prove that C(H) is a subgroup of G
...
Prove that the set of all elements of G that satisfy the equation

3 | Finite Groups; Subgroups

30
...

32
...

34
...


67

xn 5 e is a subgroup of G
...
(This exercise is referred to in Chapter 11
...
Prove that C(a) 5 C(a3)
...

Determine all finite subgroups of R*, the group of nonzero real
numbers under multiplication
...

Prove that either every member of H is even or exactly half of the
members of H are even
...
Find |ab|
...
Determine |b|
...
Prove that |ad | 5 n/d
...
Find |A|, |B|, and |AB|
...
Consider the elements A 5 c
37
...
What is the order of
0 1

1 1
d as a member of SL(2, Zp) (p is a prime),
0 1
what is the order of A?
38
...

cos nu

2 sin 60°
cos "2°
d and c
cos 60°
sin "2°

2 sin "2°
d
...
)
39
...
Show that G has elements
of every finite order as well as elements of infinite order
...
Let x belong to a group and |x| 5 6
...
Let
y belong to a group and |y| 5 9
...
, 8
...
D4 has seven cyclic subgroups
...
Find a subgroup of D4 of
order 4 that is not cyclic
...
U(15) has six cyclic subgroups
...

43
...

44
...

How many subgroups of order 3 does G have?
45
...
Suppose that g belongs to
G and n is the smallest positive integer such that gn [ H
...

46
...

a
...
U(5), U(7), U(35)
c
...
U(3), U(5), U(15)
On the basis of your answers, make a conjecture about the relationship among |U(r)|, |U(s)|, and |U(rs)|
...
Let R* be the group of nonzero real numbers under multiplication
and let H 5 {x [ R* | x2 is rational}
...
Can the exponent 2 be replaced by any positive integer and still
have H be a subgroup?
48
...
Do these groups provide a
counterexample to your answer to Exercise 46? If so, revise your
conjecture
...
Find a cyclic subgroup of order 4 in U(40)
...
Find a noncyclic subgroup of order 4 in U(40)
...
Let G 5 e c

a b
d 0 a, b, c, d
c d

[Z f

under addition
...
Prove that H is a subgroup of G
...
Let H 5 {A [ GL(2, R)| det A is an integer power of 2}
...

53
...
Let K 5 {2a | a [ H}
...

ec

3 | Finite Groups; Subgroups

69

54
...
Let H 5 { f [ G | f(2) 5 1}
...
Can 2 be replaced by any real
number?
a 0
55
...
Prove or

56
...


58
...


60
...


62
...

Let H 5 {a 1 bi | a, b [ R, ab $ 0}
...

Let H 5 {a 1 bi | a, b [ R, a2 1 b2 5 1}
...
Describe the elements
of H geometrically
...
(So, the smallest subgroup containing S is
contained in every subgroup that contains S
...
In the group Z, find
a
...
͗8, 13͘
c
...
͗m, n͘
e
...

In each part, find an integer k such that the subgroup is ͗k͘
...

1 1
a
...

1 0
0 1
b
...

1 0
c
...

Let G be a finite group with more than one element
...

Let a belong to a group and |a| 5 m
...

Let G be a finite Abelian group and let a and b belong to G
...
What can
you say about |Ka, bL| in terms of |a| and |b|?

70

Groups

Computer Exercises
A Programmer’s Lament
I really hate this damned machine;
I wish that they would sell it
It never does quite what I want
but only what I tell it
...
VAN TASSEL, The Compleat Computer

Software for the computer exercises in this chapter is available at the
website:
http://www
...
umn
...
This software determines the cyclic subgroups of U(n) generated
by each k in U(n) (n , 100)
...
Compare the order of the subgroups with the order of the group
itself
...
The program lists the elements of Zn that generate all of Zn—that is,
those elements k, 0 # k # n 2 1, for which Zn 5 ͗k͘
...

3
...
Run
the program for several values of n
...
This exercise repeats Exercise 3 for Zn using a 1 b in place of ab
...
This software computes the order of elements in GL(2, Zp)
...
The software returns |A|, |B|,
|AB|, |BA|, |A21BA|, and |B21AB|
...
Test your conjecture for several other choices for A and B
...
Test your conjecture for several other choices for A and B
...


3 | Finite Groups; Subgroups

71

In this note, the author investigates groups obtained from U(n) by multiplying each element by some k in U(n)
...

J
...
Reid, “Abelian Forcing Sets,” American Mathematical
Monthly 100 (1993): 580–582
...

This paper characterizes the Abelian forcing sets
...

This is a delightful nontechnical article that discusses how group theory and computers were used to solve a difficult problem about shuffling a deck of cards
...


Suggested Software
Allen Hibbard and Kenneth Levasseur, Exploring Abstract Algebra with
Mathematica, New York: Springer-Verlag, 1999
...
The software uses
the Mathematica language, and only a basic familiarity with the program
is required
...
central
...
This article can be downloaded at http://www
...
umn
...
pdf

4 Cyclic Groups
The notion of a “group,” viewed only 30 years ago as the epitome of
sophistication, is today one of the mathematical concepts most widely
used in physics, chemistry, biochemistry, and mathematics itself
...
Such an element a is called a
generator of G
...

In this chapter, we examine cyclic groups in detail and determine
their important characteristics
...

EXAMPLE 1 The set of integers Z under ordinary addition is cyclic
...
(Recall that, when the operation is addition, 1n is interpreted as
1 1 1 1 ??? 1 1
n terms

when n is positive and as
(21) 1 (21) 1 ? ? ? 1 (21)
|n| terms

when n is negative
...
, n 2 1} for n $ 1 is a
cyclic group under addition modulo n
...


72

4 | Cyclic Groups

73

Unlike Z, which has only two generators, Zn may have many generators (depending on which n we are given)
...
To verify, for instance, that
Z8 5 k3l, we note that k3l 5 {3, 3 1 3, 3 1 3 1 3,
...
Thus, 3 is a generator of Z8
...

EXAMPLE 4 (See Example 11 in Chapter 2
...
Also, {1, 3, 7, 9} 5
{70, 73, 71, 72} 5 ͗7͘
...

Quite often in mathematics, a “nonexample” is as helpful in understanding a concept as an example
...
How can we verify this? Well, note that U(8) 5 {1, 3, 5, 7}
...

With these examples under our belts, we are now ready to tackle
cyclic groups in an abstract way and state their key properties
...
1 Criterion for ai 5 a j
Let G be a group, and let a belong to G
...
If a has finite order, say, n, then ͗a͘ 5
{e, a, a2,
...


PROOF If a has infinite order, there is no nonzero n such that an is the
identity
...

Now assume that |a| 5 n
...
, an21}
...
, an21 are in ͗a͘
...
By the division
algorithm, there exist integers q and r such that
k 5 qn 1 r with 0 # r , n
...
, an21}
...
, an21}
...
We
begin by observing that ai 5 a j implies ai2j 5 e
...


Then ai2j 5 aqn1r, and therefore e 5 ai2j 5 aqn1r 5 (an)qar 5 eqar 5
ear 5 ar
...

Conversely, if i 2 j 5 nq, then ai2j 5 anq 5 eq 5 e, so that
i 5 a j
...
1 reveals the reason for the dual use of the notation and
terminology for the order of an element and the order of a group
...


One special case of Theorem 4
...

Corollary 2 a k 5 e Implies That |a| Divides k
Let G be a group and let a be an element of order n in G
...


PROOF Since ak 5 e 5 a0, we know by Theorem 4
...

Theorem 4
...
1
...
1 in the finite case is that it says
that multiplication in ͗a͘ is essentially done by addition modulo n
...
Thus, no matter what group G
is, or how the element a is chosen, multiplication in ͗a͘ works the same
as addition in Zn whenever |a| 5 n
...
a –6 = a 0 = a 6
...
a – 5 = a = a 7
...
a –1 = a 5 = a 11
...
a – 4 = a 2 = a 8
...
a –2 = a 4 = a 10
...
a –3 = a 3 = a 9
...
1

then multiplication in ͗a͘ works the same as addition in Z, since aia j 5
ai1j and no modular arithmetic is done
...
What is meant by this is that, although
there may be many different sets of the form {an | n [ Z}, there is
essentially only one way to operate on these sets
...
We will return to this theme in the chapter
on isomorphisms (Chapter 6)
...

Theorem 4
...
Then ͗ak͘ 5 ͗agcd(n,k)͘ and |ak| 5 n/gcd(n,k)
...

Since ak 5 (ad)r, we have by closure that ͗ak͘ # ͗ad͘
...
2
(the gcd theorem), there are integers s and t such that d 5 ns 1 kt
...
This proves
͗ad͘ # ͗ak͘
...

We prove the second part of the theorem by showing first that |ad| 5
n/d for any divisor d of n
...
On
the other hand, if i is a positive integer less than n/d, then (ad)i 2 e by definition of |a|
...

The advantage of Theorem 4
...
For example,

76

Groups

if |a| 5 30, we have ͗a26͘ 5 ͗a2͘, ͗a23͘ 5 ͗a͘, ͗a22͘ 5 ͗a2͘, ͗a21͘ 5 ͗a3͘
...
Moreover, if
one wants to list the elements of, say, ͗a21͘, it is easier to list the elements
of ͗a3͘ instead
...

Theorem 4
...

Corollary 1 Orders of Elements in Finite Cyclic Groups
In a finite cyclic group, the order of an element divides the order
of the group
...
Then ͗ai͘ 5 ͗a j͘ if and only if gcd(n, i) 5 gcd(n, j)
and |ai| 5 |a j| if and only if gcd(n, i) 5 gcd(n, j)
...
2 shows that ͗ai͘ 5 ͗agcd(n,i)͘ and ͗a j͘ 5 ͗agcd(n,j)͘,
so that the proof reduces to proving that ͗agcd(n,i)͘ 5 ͗agcd(n,j)͘ if and
only if gcd(n, i) 5 gcd(n, j)
...
On the other hand, ͗agcd(n,i)͘ 5 ͗agcd(n,j)͘
implies that |agcd(n,i)|5 |agcd(n,j)|, so that by the second conclusion of
Theorem 4
...

The second part of the corollary follows from the first part and
Corollary 1 of Theorem 4
...

The next two corollaries are important special cases of the preceding
corollary
...
Then ͗a͘ 5 ͗a j͘ if and only if gcd(n, j) 5 1 and
|a| 5 |͗a j͘| if and only if gcd(n, j) 5 1
...


The value of Corollary 3 is that once one generator of a cyclic group has
been found, all generators of the cyclic group can easily be determined
...
Clearly, one
generator is R60
...
Of course, we could have readily deduced
this information without the aid of Corollary 3 by direct calculations
...
First, note that direct computations show
that |U(50)| 5 20 and that 3 is one of its generators
...


Admittedly, we had to do some arithmetic here, but it certainly entailed
much less work than finding all the generators by simply determining
the order of each element of U(50) one by one
...
2 and its corollaries
apply only to elements of finite order
...

Theorem 4
...
Moreover, if |͗a͘| 5 n,
then the order of any subgroup of ͗a͘ is a divisor of n; and, for each
positive divisor k of n, the group ͗a͘ has exactly one subgroup of
order k—namely, ͗an/ k ͘
...
Understanding what a theorem means is a prerequisite to understanding its proof
...
The first and second parts of the
theorem say that if H is any subgroup of G, then H has the form ͗a30/k͘ for
some k that is a divisor of 30
...
The proof will also show how to find these subgroups
...
We must
show that H is cyclic
...
So we may assume that H 2 {e}
...
Since G 5 ͗a͘, every
element of H has the form at; and when at belongs to H with t , 0, then
a2t belongs to H also and 2t is positive
...
Now
let m be the least positive integer such that am [ H
...

We next claim that H 5 ͗am͘
...
Since b [ G 5 ͗a͘, we
have b 5 ak for some k
...
Then ak 5
amq1r 5 amqar, so that ar 5 a2mqak
...
But, m is the least positive integer such that
am [ H, and 0 # r , m, so r must be 0
...
This proves the assertion of the theorem that every subgroup of a cyclic group is cyclic
...
We have already shown that H 5 ͗am͘, where
m is the least positive integer such that am [ H
...

Finally, let k be any positive divisor of n
...
From Theorem 4
...
Now let H be any
subgroup of ͗a͘ of order k
...
Then m 5 gcd(n, m) and k 5 |am| 5 |agcd(n,m)| 5
n/gcd (n, m) 5 n/m
...

Returning for a moment to our discussion of the cyclic group ͗a͘,
where a has order 30, we may conclude from Theorem 4
...
Moreover, if k is a divisor of 30, the subgroup of order k is
͗a30/k͘
...
, a29}
͗a2͘ 5 {e, a2, a4,
...
, a27}
͗a5͘ 5 {e, a5, a10, a15, a20, a25}
͗a6͘ 5 {e, a6, a12, a18, a24}
͗a10͘ 5 {e, a10, a20}
͗a15͘ 5 {e, a15}
͗a30͘ 5 {e}

order 30,
order 15,
order 10,
order 6,
order 5,
order 3,
order 2,
order 1
...

Taking the group in Theorem 4
...


4 | Cyclic Groups

79

Corollary Subgroups of Zn
For each positive divisor k of n, the set ͗n/k͘ is the unique subgroup
of Zn of order k; moreover, these are the only subgroups of Zn
...
, 29}
͗2͘ 5 {0, 2, 4,
...
, 27}
͗5͘ 5 {0, 5, 10, 15, 20, 25}
͗6͘ 5 {0, 6, 12, 18, 24}
͗10͘ 5 {0, 10, 20}
͗15͘ 5 {0, 15}
͗30͘ 5 {0}

order 30,
order 15,
order 10,
order 6,
order 5,
order 3,
order 2,
order 1
...
2 and 4
...
For convenience,
we introduce an important number-theoretic function called the Euler
phi function
...
1, let f(n) denote
the number of positive integers less than n and relatively prime to n
...
The first 12
values of f(n) are given in Table 4
...

Table 4
...
4 Number of Elements of Each Order in a Cyclic Group
If d is a positive divisor of n, the number of elements of order d in
a cyclic group of order n is f(d)
...
3, the group has exactly one subgroup of
order d—call it ͗a͘
...
2, an element a k generates
͗a͘ if and only if gcd(k, d) 5 1
...


80

Groups

Notice that for a finite cyclic group of order n, the number of elements
of order d for any divisor d of n depends only on d
...

Although there is no formula for the number of elements of each
order for arbitrary finite groups, we still can say something important
in this regard
...


PROOF If a finite group has no elements of order d, the statement is
true, since f(d) divides 0
...
By
Theorem 4
...
If all
elements of order d in G are in ͗a͘, we are done
...
Then, ͗b͘ also has f(d)
elements of order d
...
If there is an element c of order d that belongs to both ͗a͘ and
͗b͘, then we have ͗a͘ 5 ͗c͘ 5 ͗b͘, so that b [ ͗a͘, which is a contradiction
...

On its face, the value of Theorem 4
...
However, the following properties of the
f function make computing f(n) simple: For any prime p, f(pn ) 5
pn 2 pn21 (see Exercise 71) and for relatively prime m and n, f(mn)
5 f(m)f(n)
...

The relationships among the various subgroups of a group can be
illustrated with a subgroup lattice of the group
...
Although there are many
ways to draw such a diagram, the connections between the subgroups
must be the same
...
The lattice diagram for Z30 is shown in Figure 4
...

Notice that ͗10͘ is a subgroup of both ͗2͘ and ͗5͘, but ͗6͘ is not a subgroup of ͗10͘
...
2 Subgroup lattice of Z30
...
3 can be appreciated by comparing the
ease with which we are able to identify the subgroups of Z30 with that of
doing the same for, say, U(30) or D30
...

We will prove in Chapter 7 that a certain portion of Theorem 4
...
We will also see, however, that a finite
group need not have exactly one subgroup corresponding to each divisor
of the order of the group
...
Indeed, D4, the dihedral
group of order 8, has five subgroups of order 2 and three of order 4
...
Although cyclic groups constitute a very narrow class of finite
groups, we will see in Chapter 11 that they play the role of building
blocks for all finite Abelian groups in much the same way that primes
are the building blocks for the integers and that chemical elements are
the building blocks for the chemical compounds
...

ARNOLD ROSS

1
...

2
...
Find all generators of ͗a͘, ͗b͘, and ͗c͘
...
List the elements of the subgroups ͗20͘ and ͗10͘ in Z30
...
List the elements of the subgroups ͗a20 ͘
and ͗a10 ͘
...
List the elements of the subgroups ͗3͘ and ͗15͘ in Z18
...
List the elements of the subgroups ͗a3͘
and ͗a15͘
...
List the elements of the subgroups ͗3͘ and ͗7͘ in U(20)
...
What do Exercises 3, 4, and 5 have in common? Try to make a generalization that includes these three cases
...
Find an example of a noncyclic group, all of whose proper subgroups are cyclic
...
Let a be an element of a group and let |a| 5 15
...

a
...
a5, a10
c
...
How many subgroups does Z20 have? List a generator for each of
these subgroups
...
How many
subgroups does G have? List a generator for each of these subgroups
...
In Z24 list all generators for the subgroup of order 8
...
List all generators for the subgroup of order 8
...
Let G be a group and let a [ G
...

12
...
If a has infinite order,
find all generators of the subgroup ͗a3 ͘
...
In Z24 find a generator for ͗21͘ > ͗10͘
...
Find
a generator for ͗a21͘ > ͗a10͘
...
Suppose that a cyclic group G has exactly three subgroups: G
itself, {e}, and a subgroup of order 7
...
Let G be an Abelian group and let H 5 {g [ G| |g| divides 12}
...
Is there anything special about 12
here? Would your proof be valid if 12 were replaced by some other
positive integer? State the general result
...
Find a collection of distinct subgroups ͗a1͘, ͗a2͘,
...

17
...

18
...
List the cyclic subgroups of U(30)
...
Suppose that G is an Abelian group of order 35 and every element
of G satisfies the equation x35 5 e
...
Does
your argument work if 35 is replaced with 33?
21
...

a
...
If am 5 e, what can we say about the order of a?
c
...
If a8 2 e and a12 2 e,
show that ͗a͘ 5 G
...
Prove that a group of order 3 must be cyclic
...
Let Z denote the group of integers under addition
...
Let a be
a group element with infinite order
...

24
...

25
...
How many elements of
order 2 does Dn have?
26
...
Let a be a group element that has infinite
order
...

27
...

28
...
Prove that ͗ai͘ 5
͗a j ͘ if and only if i 5 Ϯj
...
List all the elements of order 8 in Z8000000
...

List all elements of order 8 in ͗a͘
...
Suppose a and b belong to a group, a has odd order, and aba21 5
b21
...

31
...
Show that there exists a fixed positive integer
n such that an 5 e for all a in G
...
)
32
...

33
...

34
...

35
...

36
...

37
...


84

Groups

38
...
Show that this set is a group under
multiplication modulo 20 by constructing its Cayley table
...

39
...
Generalize to exactly n
subgroups for any positive integer n
...
Let m and n be elements of the group Z
...

41
...
If ͗a͘ > ͗b͘ 5 {e}, prove that the group contains an
element whose order is the least common multiple of m and n
...

42
...

43
...
If a group has more than p 2 1 elements of order p,
why can’t the group be cyclic?
44
...
How many
elements of order 6 does G have? If 8 divides |G|, how many elements of order 8 does G have? If a is one element of order 8, list
the other elements of order 8
...
List all the elements of Z40 that have order 10
...
List all
the elements of ͗x͘ that have order 10
...
Reformulate the corollary of Theorem 4
...

47
...

48
...
If 20 divides
the order of G, determine the number of solutions of x20 5 e
...

49
...

50
...

51
...

52
...


4 | Cyclic Groups

85

53
...
If |a| 5 10 and |b| 5 21, show
that ͗a͘ > ͗b͘ 5 {e}
...
Let a and b belong to a group
...

55
...
If |a| 5 24 and |b| 5 10, what are
the possibilities for |͗a͘ > ͗b͘|?
56
...

57
...
Can you conclude that G is not cyclic? What if G has at least five
elements x such that x4 5 e? Generalize
...
Prove that Zn has an even number of generators if n
...
What
does this tell you about f(n)?
59
...
Suppose that |x| 5 n
...

61
...

Determine |a|
...
Let a be group element such that |a| 5 48
...
͗a21 ͘ 5 ͗ak ͘
b
...
͗a18 ͘ 5 ͗ak ͘
...
Let p be a prime
...

1 n
d 0 n [ Z f is a cyclic subgroup of
64
...

65
...
If |a| 5 12, |b| 5 22, and ͗a͘ > ͗b͘ 2
{e}, prove that a6 5 b11
...
Suppose that G is a finite group with the property that every nonidentity element has prime order (for example, D3 and D5)
...

67
...
We can make G a group under
addition by adding the polynomials in the usual way, except that
we use modulo 3 to combine the coefficients
...


86

Groups

68
...
Prove that the group G 5
5n1r1 1 n2r2 |n1 and n2 are integers6 under addition is cyclic
...
, rk rationals
...
Let a and b belong to some group
...
If ak 5 bk for some integer k, prove that mn divides k
...
For every integer n greater than 2, prove that the group U(n2 2 1)
is not cyclic
...
Prove that for any prime p and positive integer n, f(pn ) 5
pn 2 pn21
...
Give an example of an infinite group that has exactly two elements
of order 4
...

JOE PISCOPO

Software for the computer exercises in this chapter is available at the
website:
http://www
...
umn
...
This software determines if U(n) is cyclic
...
Make a conjecture
...
Make a conjecture
...
Make a conjecture
...
For any pair of positive integers m and n, let Zm % Zn 5 {(a, b) |
a [ Zm, b [ Zn}
...
[For
example, in Z3 % Z4, we have (1, 2) 1 (2, 3) 5 (0, 1)
...
Run the program for
the following choices of m and n: (2, 2), (2, 3), (2, 4), (2, 5), (3, 4),
(3, 5), (3, 6), (3, 7), (3, 8), (3, 9), and (4, 6)
...

3
...
Define ͗a, b͘ 5 {aib j | 0 # i , |a|,
0 # j , |b|}
...
Run the program for the following choices of a, b,
and n: (21, 101, 550), (21, 49, 550), (7, 11, 100), (21, 31, 100), and

4 | Cyclic Groups

87

(63, 77, 100)
...

4
...
Do you see any relationship
between the order of U(n) and the order of its elements? Run the
program for n 5 8, 16, 32, 64, and 128
...

Make a conjecture about the number of elements of order 4 in
U(2k) when k is at least 4
...
Make a conjecture
about the maximum order of any element in U(2k) when k is at least
3
...

5
...
For each order d of some element of
U(n), this software lists f(d) and the number of elements of order d
...
Do you see any relationship
between the number of elements of order d and f(d)? Run the program for n 5 3, 9, 27, 81, 5, 25, 125, 7, 49, and 343
...
Run the program for n 5 6, 18, 54, 162,
10, 50, 250, 14, 98, and 686
...
Make a conjecture about the number of elements of various
orders in U( pk) and U(2pk) where p is an odd prime
...
For each positive integer n, this software gives the order of U(n)
...
Try to guess a formula
for the order of U(3k) when k is at least 2
...
How does the order of U(2 ? 3k) appear to be related to the order of U(3k)? Run the program for n 5 25, 125, and
625
...

Run the program for n 5 50, 250, and 1250
...
Try to guess a formula for the order of U(7k)
when k is at least 2
...
How does
the order of U(2 ? 7k) appear to be related to the order of U(7k)?
Based on your guesses for U(3k), U(5k), and U(7k), guess a formula
for the order of U( pk) when p is an odd prime and k is at least 2
...
Does your formula also work when k is 1?

88

Groups

Suggested Reading
Deborah L
...

In this easy-to-read paper, it is shown that the probability of a randomly chosen element from a cyclic group being a generator of the
group depends only on the set of prime divisors of the order of the
group, and not on the order itself
...


J
...
Sylvester
I really love my subject
...
J
...
Sylvester was born on September 3,
1814, in London and showed his mathematical genius early
...

After receiving B
...
and M
...
degrees
from Trinity College in Dublin in 1841,
Sylvester began a professional life that was
to include academics, law, and actuarial careers
...
During
his seven years at Johns Hopkins, Sylvester
pursued research in pure mathematics
with tremendous vigor and enthusiasm
...
Sylvester
returned to England in 1884 to a professorship at Oxford, a position he held until his
death on March 15, 1897
...
His writings and lectures—flowery
and eloquent, pervaded with poetic flights,
emotional expressions, bizarre utterances,
and paradoxes—reflected the personality of
this sensitive, excitable, and enthusiastic

†F
...
We quote three of his students
...
W
...

Sylvester’s methods! He had none
...
” It did last all the
rest of that year
...
We
all got the text
...
Then he began to think
about matrices again
...
He could not confine himself to the hour, nor to the one lecture
a week
...
Statements like the following were not
infrequent in his lectures: “I haven’t proved
this, but I am as sure as I can be of anything
that it must be so
...
” At the next lecture it turned out that what
he was so sure of was false
...
Afterward he would
go back and work it all over again, and surprise us with all sorts of side lights
...


Cajori, Teaching and History of Mathematics in the U
...
, Washington, 1890, 265–266
...

Sylvester’s capacious head was ever lost in
the highest cloud-lands of pure mathematics
...

Everything he saw suggested to him something new in the higher algebra
...
They began to do it themselves
...
He
seldom remembered theorems, propositions,
etc
...
In this he was the very
antithesis of Cayley, who was thoroughly
conversant with everything that had been
done in every branch of mathematics
...
To his astonishment, I showed him a paper of his own in
which he had proved the proposition; in fact, I
believe the object of his paper had been the
very proof which was so strange to him
...
P
...


For more information about Sylvester,
visit:
http://www-groups
...
st-and

...
uk/~history/

Sylvester’s enthusiasm for teaching and his
influence on his students are captured in the
following passage written by Sylvester’s first
student at Johns Hopkins, G
...
Halsted
...
d
...
edu/~jgallian/TF
1
...
For any fixed x in
G, define xHx21 5 {xhx21 | h [ H}
...

a
...

b
...

c
...

The group xHx21 is called a conjugate of H
...
)
2
...
Define N(H) 5
{x [ G | xHx21 5 H}
...
†
3
...
For each a [ G, define cl(a) 5 {xax21 | x [ G}
...
[cl(a) is called the
conjugacy class of a
...
The group defined by the following table is called the group of
quaternions
...
The center
b
...
cl(b)
d
...
Sylow in 1872 to prove the existence of certain kinds of subgroups in a group
...


92

Groups

5
...
) Prove that, in any group, |xax21| 5
|a|
...
)
6
...

7
...

8
...
Prove that there is an element x in
G such that xax 5 b if and only if ab 5 c2 for some element c in G
...
Prove that if a is the only element of order 2 in a group, then a lies
in the center of the group
...
Let G be the plane symmetry group of the infinite strip of equally
spaced H’s shown below
...

12
...

14
...


16
...


18
...

20
...
Calculate |x|, |y|, and |xy|
...

Prove that a group of order 5 must be cyclic
...

Let G be an Abelian group and let n be a fixed positive integer
...
Prove that Gn is a subgroup of G
...
(This exercise is referred to in Chapter 11
...
Prove that G is a group under ordinary multiplication
...
Does the statement remain true if “two” is
replaced by “three”?
Prove that the subset of elements of finite order in an Abelian
group forms a subgroup
...
) Is the same thing true for non-Abelian groups?
Let p be a prime and let G be an Abelian group
...

Suppose that a and b are group elements
...


4 | Supplementary Exercises for Chapters 1–4

93

21
...
Given that a35 b2 5 e and ba2 5 ab, construct
the Cayley table for the group
...
Name it
...
If a is an element of a group and |a| 5 n, prove that C(a) 5 C(ak )
when k is relatively prime to n
...
Let x and y belong to a group G
...

24
...
Show that H > K is a nontrivial subgroup of Q
...
Let H be a subgroup of G and let g be an element of G
...
See Exercise 2 for the notation
...
Let H be a subgroup of a group G and let |g| 5 n
...

27
...

28
...

How many cyclic subgroups of order 10 does G have?
29
...
Assume that for every a in S the
set {an | n 5 1, 2, 3,
...
Must S be a group?
30
...
be a sequence of subgroups of a group with the
property that H1 # H2 # H3
...

31
...
Prove that H is a subgroup of R*
...
Suppose that a and b belong to a group, a and b commute, and |a|
and |b| are relatively prime
...
Give an example showing that |ab| need not be |a||b| when a and b commute but
|a| and |b| are not relatively prime
...
)
33
...
Prove or disprove that
H is a subgroup of GL(2, R)
...
Suppose that G is a group that has exactly one nontrivial proper
subgroup
...

35
...
Prove that G is cyclic and |G| 5 pq, where p and q are
distinct primes, or that G is cyclic and |G| 5 p3, where p is prime
...
If |a2| 5 |b2|, prove or disprove that |a| 5 |b|
...
(1995 Putnam Competition) Let S be a set of real numbers that is
closed under multiplication
...
Given that the product of any three (not necessarily distinct) elements of T is in T and that the product of any
three elements of U is in U, show that at least one of the two subsets T and U is closed under multiplication
...
If p is an odd prime, prove that there is no group that has exactly p
elements of order p
...
Give an example of a group G with infinitely many distinct subgroups H1, H2, H3,
...

40
...
If a21ba 5 b2 and
|a| 5 3, find |b|
...
Let a and b belong to a group G
...

42
...
Let
H 5 {x [ G| |x| is relatively prime to 3}
...
Is your
argument valid if 3 is replaced by an arbitrary positive integer n?
Explain why or why not
...
Let G be a finite group and let S be a subset of G that contains
more than half of the elements of G
...

44
...
Show
that H 5 {g [ G| f (xg) 5 f (x) for all x [ G} is a subgroup of G
...

45
...
Show that H 5 {x [ G| |x| divides d}
...
Let a be an element of maximum order from a finite Abelian group
G
...
Show by
example that this need not be true for finite non-Abelian groups
...
Define an operation * on the set of integers by a * b 5 a 1 b 2 1
...

48
...
Find a noncyclic subgroup of
U(4n) of order 4 that contains the element 2n 2 1
...
He and others found many similar applications and nowadays
group theoretical methods—especially those involving characters and
representations—pervade all branches of quantum mechanics
...
In the early and mid-19th century, groups
of permutations were the only groups investigated by mathematicians
...

Definitions Permutation of A, Permutation Group of A

A permutation of a set A is a function from A to A that is both oneto-one and onto
...


Although groups of permutations of any nonempty set A of objects
exist, we will focus on the case where A is finite
...
, n} for some positive integer n
...
For example, we define a permutation a of the set {1, 2, 3, 4} by
specifying
a(1) 5 2,

a(2) 5 3,

a(3) 5 1,

a(4) 5 4
...

2 3 1 4

Here a( j) is placed directly below j for each j
...

5 3 1 6 2 4

Composition of permutations expressed in array notation is carried
out from right to left by going from top to bottom, then again from top
to bottom
...
The remainder of the bottom row gs is obtained in
a similar fashion
...

EXAMPLE 1 Symmetric Group S3 Let S3 denote the set of all oneto-one functions from {1, 2, 3} to itself
...
The six elements are
1 2 3
e5c
d,
1 2 3

a5c

1 2 3
d,
2 3 1

a2 5 c

1 2 3
d,
3 1 2

5 | Permutation Groups

b5 c

1 2 3
d,
1 3 2

Note that ba 5 c

ab 5 c

1 2 3
d,
2 1 3

a2b 5 c

97

1 2 3
d
...

3 2 1

The relation ba 5 a 2b can be used to compute other products in S3
without resorting to the arrays
...

Example 1 can be generalized as follows
...
, n}
...
Elements of Sn have the form
a5 c

1
2 c n
d
...
There are n choices of a(1)
...
After choosing a(2), there
are exactly n 2 2 possibilities for a(3)
...
We leave it to the
reader to prove that Sn is non-Abelian when n $ 3 (Exercise 41)
...
The group S4 has 30
subgroups, and S5 has well over 100 subgroups
...
For example, if we label the four corner
positions as in the figure below and keep these labels fixed for reference,
we may describe a 90° counterclockwise rotation by the permutation
3

2

4

1

r 5 c

1 2 3 4
d,
2 3 4 1

98

Groups

whereas a reflection across a horizontal axis yields
f 5 c

1 2 3 4
d
...

When D4 is represented in this way, we see that it is a subgroup
of S4
...
It is
called cycle notation and was first introduced by the great French mathematician Cauchy in 1815
...

As an illustration of cycle notation, let us consider the permutation
a5c

1 2 3 4 5 6
d
...
Instead, we leave out the arrows and simply write a 5 (1, 2)
(3, 4, 6)(5)
...

5 3 1 6 2 4
In cycle notation, b can be written (2, 3, 1, 5)(6, 4) or (4, 6)(3, 1, 5, 2),
since both of these unambiguously specify the function b
...
, am) is called a cycle of length m or an
m-cycle
...

Thus, the cycle (4, 6) can be thought of as representing the
1 2 3 4 5 6
d
...
Consider the
following example from S8
...
(When the domain consists of single-digit integers, it is
common practice to omit the commas between the digits
...
Well, keeping in mind that function composition is done from right to left and that each cycle that does not contain a symbol fixes the symbol, we observe that: (5) fixes 1; (648) fixes 1;
(1237) sends 1 to 2; (8) fixes 2; (456) fixes 2; (27) sends 2 to 7; and (13)
fixes 7
...
Thus we begin
ab 5 (17 ? ? ?) ? ? ?
...
Ultimately, we have ab 5 (1732)(48)(56)
...
(Warning: Some authors compose cycles from left to right
...
)
To be sure you understand how to switch from one notation to the
other and how to multiply permutations, we will do one more example
of each
...

To put ab in disjoint cycle form, observe that (24) fixes 1; (153)
sends 1 to 5; (45) sends 5 to 4; and (3) and (12) both fix 4
...
Continuing in this way we obtain ab 5 (14)(253)
...

One final remark about cycle notation: Mathematicians prefer not to
write cycles that have only one entry
...
With this convention, the permutation
a above can be written as (12)(45)
...
Of course, the identity permutation consists
only of cycles with one entry, so we cannot omit all of these! In this
case, one usually writes just one cycle
...
Just remember that missing
elements are mapped to themselves
...
The proof of the first theorem is implicit in our discussion of
writing permutations in cycle form
...
1 Products of Disjoint Cycles
Every permutation of a finite set can be written as a cycle or as a
product of disjoint cycles
...
, n}
...
We know that such
an m exists because the sequence a1, a(a1), a 2(a1), ? ? ? must be finite;
so there must eventually be a repetition, say a i(a1) 5 a j(a1) for some
i and j with i , j
...
We express this
relationship among a1, a2,
...
, am) ? ? ?
...
In such a case, we merely choose
any element b1 of A not appearing in the first cycle and proceed to create a new cycle as before
...
This new cycle will have no

5 | Permutation Groups

101

elements in common with the previously constructed cycle
...
But then a i2j(a1) 5 b1, and therefore b1 5 at for some t
...

Continuing this process until we run out of elements of A, our permutation will appear as
a 5 (a1, a2,
...
, bk) ? ? ? (c1, c2,
...

In this way, we see that every permutation can be written as a product
of disjoint cycles
...
2 Disjoint Cycles Commute
If the pair of cycles a 5 (a1, a2,
...
, bn )
have no entries in common, then ab 5 ba
...
, am, b1, b2,
...
, ck}
where the c’s are the members of S left fixed by both a and b (there
may not be any c’s)
...
If x is one of the a elements, say ai, then
(ab)(ai) 5 a(b(ai)) 5 a(ai) 5 ai11,
since b fixes all a elements
...
) For the
same reason,
(ba)(ai) 5 b(a(ai)) 5 b(ai11) 5 ai11
...
A similar
argument shows that ab and ba agree on the b elements as well
...
Then, since both a and b
fix c elements, we have
(ab)(ci) 5 a(b(ci)) 5 a(ci) 5 ci
and
(ba)(ci) 5 b(a(ci)) 5 b(ci) 5 ci
...

In demonstrating how to multiply cycles, we showed that the
product (13)(27)(456)(8)(1237)(648)(5) can be written in disjoint cycle

102

Groups

form as (1732)(48)(56)
...
The next theorem
shows that the disjoint cycle form has the enormous advantage of
allowing us to “eyeball” the order of the permutation
...
3 Order of a Permutation (Ruffini—1799)
The order of a permutation of a finite set written in disjoint cycle
form is the least common multiple of the lengths of the cycles
...
(Verify this
yourself
...
It follows from
Theorem 4
...
Thus, we know by
Corollary 2 to Theorem 4
...
But then (ab)t 5 a tb t 5 e,
so that a t 5 b 2t
...
But, if a t and b 2t are equal and have
no common symbol, they must both be the identity, because every symbol in a t is fixed by b 2t and vice versa (remember that a symbol not appearing in a permutation is fixed by the permutation)
...
This means that k, the least common
multiple of m and n, divides t also
...

Thus far, we have proved that the theorem is true in the cases
where the permutation is a single cycle or a product of two disjoint
cycles
...

Theorem 5
...
We demonstrate this in the next example
...
For convenience, we denote an n-cycle by (n)
...

Now, from Theorem 5
...
To do the same for the 10! 5 3,628,800
elements of S10 would be nearly as simple
...
Many authors call these permutations transpositions, since the
effect of (ab) is to interchange or transpose a and b
...
4 Product of 2-Cycles
Every permutation in Sn, n
...


PROOF First, note that the identity can be expressed as (12)(12), and
so it is a product of 2-cycles
...
1, we know that every permutation can be written in the form
(a1a2 ? ? ? ak)(b1b2 ? ? ? bt) ? ? ? (c1c2 ? ? ? cs)
...

This completes the proof
...


104

Groups

EXAMPLE 5
(12345) 5 (15)(14)(13)(12)
(1632)(457) 5 (12)(13)(16)(47)(45)
The decomposition of a permutation into a product of 2-cycles given
in the proof of Theorem 5
...
Although the next example shows that
even the number of 2-cycles may vary from one decomposition to another, we will prove in Theorem 5
...

EXAMPLE 6
(12345) 5 (54)(53)(52)(51)
(12345) 5 (54)(52)(21)(25)(23)(13)
We isolate a special case of Theorem 5
...

Lemma
If e 5 b1b2 ? ? ? br , where the b’s are 2-cycles, then r is even
...
If r 5 2, we
are done
...
2, and we proceed by induction
...

If the first case occurs, we may delete br21 br from the original product
to obtain e 5 b1 b2 ? ? ? br22
...
We now repeat the procedure just described with br22 br21, and, as before, we obtain a product of
(r 2 2) 2-cycles equal to the identity or a new product of r 2-cycles,
where the rightmost occurrence of a is in the third 2-cycle from the right
...
Hence,
by the Second Principle of Mathematical Induction, r 2 2 is even, and r
is even as well
...
5 Always Even or Always Odd
If a permutation a can be expressed as a product of an even (odd)
number of 2-cycles, then every decomposition of a into a product of
2-cycles must have an even (odd) number of 2-cycles
...


PROOF Observe that b1b2 ? ? ? br 5 g1g2 ? ? ? gs implies
e 5 g1g2 ? ? ? gsbr 21 ? ? ? b221b121
5 g1g2 ? ? ? gsbr ? ? ? b2b1,
since a 2-cycle is its own inverse
...
It follows that r and s are both even or both
odd
...
A permutation that can
be expressed as a product of an odd number of 2-cycles is called an
odd permutation
...
4 and 5
...
The significance of this
observation is given in Theorem 5
...

Theorem 5
...


PROOF This proof is left to the reader (Exercise 13)
...


106

Groups

Definition Alternating Group of Degree n

The group of even permutations of n symbols is denoted by An and is
called the alternating group of degree n
...
1)
are even permutations
...
7
For n
...


PROOF For each odd permutation a, the permutation (12)a is even
and (12)a 2 (12)b when a 2 b
...
On the other hand, for each
even permutation a, the permutation (12)a is odd and (12)a 2 (12)b
when a 2 b
...
It follows that there are equal numbers of even and odd
permutations
...

The names for the symmetric group and the alternating group of degree
n come from the study of polynomials over n variables
...
, xn is one that is unchanged under
any transposition of two of the variables
...
For
example, the polynomial x1 x2 x3 is unchanged by any transposition of two
of the three variables, whereas the polynomial (x12x2)(x12x3)(x22x3)
changes signs when any two of the variables are transposed
...
Likewise, since any member of the alternating group is the
product of an even number of transpositions, the alternating polynomials
are those that are unchanged by members of the alternating group and
change sign by the other permutations of Sn
...
The groups A4 and A5 will arise on several occasions in later
chapters
...

A geometric interpretation of A4 is given in Example 7, and a multiplication table for A4 is given as Table 5
...

EXAMPLE 7 ROTATIONS OF A TETRAHEDRON The 12 rotations of a regular tetrahedron can be conveniently described with the
elements of A4
...
1 illustrates the identity and
three 180° “edge” rotations about axes joining midpoints of two edges
...
1 The Alternating Group A4 of Even Permutations of {1, 2, 3, 4}
(In this table, the permutations of A4 are designated as a1, a2,
...
For example, a3 a8 5 a6
...
1 Rotations of a regular tetrahedron
...
The third row consists of 2120° (or
240°) “face” rotations
...

Many molecules with chemical formulas of the form AB4, such as
methane (CH4) and carbon tetrachloride (CCl4), have A4 as their symmetry group
...
2 shows the form of one such molecule
...


Figure 5
...


EXAMPLE 8 (Loren Larson) A Sliding Disk Puzzle
Consider the puzzle shown below (the space in the middle is empty)
...
Let r denote the
following operation: Move the disk in position 1 to the center position,
then move the disk in position 6 to position 1, the disk in position 5 to
position 6, the disk in position 4 to position 5, the disk in position 3 to
position 4, then the disk in the middle position to position 3
...

In permutation notation, we have r 5 (13456) and s 5 (132)
...
Clearly, if we can
express (16523) as a string of r’s and s’s, we can achieve the desired
arangement
...
One such software program is GAP (see
Suggested Software at the end of this chapter)
...
G :5 SymmetricGroup(6);
gap
...
K :5 Subgroup(G,[r,s]);
gap
...
The fourth
line requests that (16523) be expressed in terms of r and s
...
Size (K);
tells us that the order of the subgroup generated by r and s is 360
...


110

Groups

GAP can even compute the 43,252,003,274,489,856,000 (431 quintillion) permutations of the Rubik’s Cube! Labeling the faces of the
cube as shown here,
1
4
6
9
12
14

10
left
15

11
13
16

2
top
7

3
5
8

17
20
22

18
front
23

19
21
24

42
bottom
47

43
45
48

41
44
46

25
28
30

26
right
31

27
29
32

33
36
38

34
rear
39

35
37
40

the group of permutations of the cube is generated by the following rotations of the six layers:
top 5 (1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19)
left 5 (9,11,16,14)(10,13,15,12)(1,17,41,40)(4,20,44,37)(6,22,46,35)
front 5 (17,19,24,22)(18,21,23,20)(6,25,43,16)(7,28,42,13)(8,30,41,11)
right 5 (25,27,32,30)(26,29,31,28)(3,38,43,19)(5,36,45,21)(8,33,48,24)
rear 5 (33,35,40,38)(34,37,39,36)(3,9,46,32)(2,12,47,29)(1,14,48,27)
bottom 5 (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)
(16,24,32,40)

A Check Digit Scheme Based on D5
In Chapter 0, we presented several schemes for appending a check digit
to an identification number
...

However, recall that this success was achieved by introducing the alphabetical character X to handle the case where 10 was required to
make the dot product 0 modulo 11
...
Verhoeff [2] devised a method utilizing the
dihedral group of order 10 that detects all single-digit errors and all
transposition errors involving adjacent digits without the necessity of
avoiding certain numbers or introducing a new character
...
2
...
)

5 | Permutation Groups

111

Table 5
...

In particular, to any string of digits a1a2
...
[Here s 2 (x) 5 s(s(x)), s 3 (x) 5 s(s 2 (x)), and so on
...
Also, because
a p s(b) 2 b p s(a)

if a 2 b,

(1)

as can be checked on a case-by-case basis (see Exercise 49), it follows
that all transposition errors involving adjacent digits are detected [since
Equation (1) implies that s i(a) p s i11(b) 2 si(b) p s i11(a) if a 2 b]
...
Table 5
...
, s10 needed for the computations
...
]
Since the serial numbers on the banknotes use 10 letters of the alphabet in
addition to the 10 decimal digits, it is necessary to assign numerical values to the letters to compute the check digit
...
4
...
a10 corresponding to a banknote serial
number, the check digit a11 is chosen such that s (a1) p s 2(a2) p ? ? ? p
s9(a9) p s10(a10) p a11 5 0 [instead of s(a1) p s 2 (a2) p ? ? ? p s10(a10) p
s11(a11) 5 0 as in the Verhoeff scheme]
...
3 Powers of s
0

2

3

4

5

6

7

8

9

1
5
8
9
4
2
7
0
1
5

s
s2
s3
s4
s5
s6
s7
s8
s9
s10

1
5
8
9
4
2
7
0
1
5
8

7
0
1
5
8
9
4
2
7
0

6
3
6
3
6
3
6
3
6
3

2
7
0
1
5
8
9
4
2
7

8
9
4
2
7
0
1
5
8
9

3
6
3
6
3
6
3
6
3
6

0
1
5
8
9
4
2
7
0
1

9
4
2
7
0
1
5
8
9
4

4
2
7
0
1
5
8
9
4
2

Table 5
...
3 with the number
AG8536827U7
...
[To illustrate how to use the multiplication table
for D5, we compute 1 p 0 p 2 p 2 5 (1 p 0) p 2 p 2 5 1 p 2 p 2 5
(1 p 2) p 2 5 3 p 2 5 0
...
3 German banknote with serial number AG8536827U and check digit 7
...
Thus, a

5 | Permutation Groups

113

substitution of 7 for U (or vice versa) and the transposition of 7 and U
are not detected by the check digit
...
For example, the transposition of D and 8 in positions 10 and 11 is not detected
...
Using this method to append a check
character, all single-position errors and all transposition errors involving adjacent digits will be detected
...
Find the order of each of the following permutations
...
(14)
b
...
(14762)
d
...
ak )
2
...

a
...
(13256)(23)(46512)
c
...
What is the order of each of the following permutations?
a
...
(124)(3567)
c
...
(124)(357869)
e
...
(345)(245)

114

Groups

4
...
c
2 1 5 4 6 3
1 2 3 4 5 6 7
d
7 6 1 2 3 4 5
What is the order of the product of a pair of disjoint cycles of
lengths 4 and 6?
Show that A8 contains an element of order 15
...
)
What is the maximum order of any element in A10?
Determine whether the following permutations are even or odd
...
(135)
b
...
(13567)
d
...
(1243)(3521)
Show that a function from a finite set S to itself is one-to-one if and
only if it is onto
...
)
Let n be a positive integer
...
If a is odd, prove that a21 is odd
...
6
...
Complete
the following statements: ab is even if and only if r 1 s is ______;
abg is even if and only if r 1 s 1 t is ______
...
Prove that ab is even if and only if a
and b are both even or both odd
...
Draw an analogy between the
result of multiplying two permutations and the result of multiplying their corresponding numbers 11 or 21
...

2 1 3 5 4 6
6 1 2 4 3 5

b
...

6
...

8
...


10
...


12
...

14
...

16
...


5 | Permutation Groups

115

Compute each of the following
...
a21
b
...
ab
18
...


20
...

22
...


24
...

26
...

28
...

30
...


32
...

2 3 4 5 1 7 8 6
1 3 8 7 6 5 2 4

Write a, b, and ab as
a
...
products of 2-cycles
...

(This exercise is referred to in Chapter 25
...
What arithmetic relationship do these orders have with the order of A4?
Give two reasons why the set of odd permutations in Sn is not a
subgroup
...
Prove that a21b21ab is an even
permutation
...
1 to compute the following
...
The centralizer of a3 5 (13)(24)
...
The centralizer of a12 5 (124)
...

Let b [ S7 and suppose b 4 5 (2143567)
...

Let b 5 (123)(145)
...

Find three elements s in S9 with the property that s 3 5
(157)(283)(469)
...
Let a [ X and define
stab(a) 5 {a [ G|a(a) 5 a}
...
Prove
that stab(a) is a subgroup of G
...
) This exercise is referred to in Chapter 7
...
What is the smallest positive
integer n for which bn 5 b25?

116

Groups

33
...
If am is a 5-cycle, what can
you say about m?
34
...
Prove that H is a subgroup of S5
...
How many elements of order 5 are there in A6?
36
...

37
...
For which integers i between 2 and
10 is b i also a 10-cycle?
38
...

39
...

40
...

41
...

42
...
Prove that bab21 and a are both even or
both odd
...
Show that A5 has 24 elements of order 5, 20 elements of order 3, and
15 elements of order 2
...
)
44
...

45
...

46
...
Prove that H is a
subgroup of An
...
Show that every element in An for n $ 3 can be expressed as a
3-cycle or a product of three cycles
...
Show that for n $ 3, Z(Sn) 5 {e}
...
Verify the statement made in the discussion of the Verhoeff check
digit scheme based on D5 that a * s(b) 2 b * s(a) for distinct a and
b
...

Prove that this implies that all transposition errors involving adjacent
digits are detected
...
Use the Verhoeff check-digit scheme based on D5 to append a
check digit to 45723
...
Prove that every element of Sn (n
...

52
...
All of the hearts arranged in order

5 | Permutation Groups

53
...

55
...

57
...


59
...
If the cards
emerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in what
order were the cards after the first shuffle?
Show that a permutation with odd order is an even permutation
...
Prove or disprove that H 5 {g2 | g [ G} is a subgroup of G
...
)
Determine integers n for which H 5 5a [ An |a2 5 e6 is a subgroup of An
...

Why does the fact that the orders of the elements of A4 are 1, 2, and
3 imply that |Z(A4)| 5 1?
Label the four locations of tires on an automobile with the labels
1, 2, 3, and 4, clockwise
...
Let b represent the operation of rotating the tires in positions 2, 3, and 4 clockwise and leaving the tire in position 1 as is
...
How
many elements are in G?
Shown below are four tire rotation patterns recommended by the
Dunlop Tire Company
...
Is the subgroup Abelian?

Rear Wheel Drive
Vehicles

X Tires to
the Driven Axle
Front Wheel Drive
Vehicles

FRONT

FRONT
Modified
X

Modified X

4 Wheel Drive
Vehicles

Alternate Pattern

FRONT

FRONT

X

Normal

118

Groups

Computer Exercises
Science is what we understand well enough to explain to a computer
...

DONALD KNUTH, The Art of Computer Programming, 1969

Software for Computer Exercise 1 in this chapter is available at the
website:
http://www
...
umn
...
This software determines whether the two permutations (1x) and
(123 ? ? ? n) generate Sn for various choices of x and n (that is,
whether every element of Sn can be expressed as some product of
these permutations)
...
For n 5 5, run the program for x 5 2, 3, 4, and 5
...
For n 5 8, run the program
for x 5 2, 3, 4, 5, 6, 7, and 8
...

2
...
(For GAP, the prompt brk
...
In this situation, use Control-D to
return to the main prompt
...
)
a
...
(23)
c
...
(12)(34)(56)
3
...

1
6

2

5

3
4

5 | Permutation Groups

119

References
1
...
A
...

2
...
Verhoeff, Error Detecting Decimal Codes, Amsterdam: Math-ematisch
Centrum, 1969
...
Ensely, “Invariants Under Actions to Amaze Your Friends,” Mathematics Magazine, Dec
...

This article explains some card tricks that are based on permutation
groups
...
/Dec
...

In this article, permutation groups are used to analyze various sorts of
checkerboard tiling problems
...
A
...

This article gives a comprehensive survey of error-detection methods that
use check digits
...
d
...
edu/~jgallian/detection
...
N
...
Kaplansky, Matters Mathematical, New York: Chelsea,
1978
...

Douglas Hofstadter, “The Magic Cube’s Cubies Are Twiddled by Cubists and
Solved by Cubemeisters,” Scientific American 244 (1981): 20–39
...
In particular,
permutation groups, subgroups, conjugates (elements of the form xyx21),
commutators (elements of the form xyx21y21), and the “always even or
always odd” theorem (Theorem 5
...
At one
point, Hofstadter says, “It is this kind of marvelously concrete illustration
of an abstract notion of group theory that makes the Magic Cube one of
the most amazing things ever invented for teaching mathematical ideas
...
Kiltinen, Oval Track & Other Permutation Puzzles & Just Enough
Group Theory to Solve Them, Mathematical Association of America,
Washington, D
...
, 2003
...
The
book provides the background needed to use the puzzle software to its fullest
potential, and also gives the reader a gentle, not-too-technical introduction to
the theory of permutation groups that is a prerequisite to a full understanding
of how to solve puzzles of this type
...
nmu

...
It also has news about puzzle software—modules that add
functionality and fun to puzzles
...
/Dec
...

The author uses permutation groups to analyze solutions to the 15 puzzle,
Rubik’s Cube, and Rubik’s Clock
...
White and R
...

This article explains how permutation groups are used in bell ringing
...
Winters, “Error-Detecting Schemes Using Dihedral Groups,” UMAP
Journal 11, no
...

This article discusses error-detection schemes based on Dn for n odd
...


Suggested Software
GAP is free for downloading
...
gap-system
...

Spoken by Lagrange
to Laplace About the
11-year-old Cauchy

This stamp was issued by France
in Cauchy’s honor
...
By the time
he was 11, both Laplace and Lagrange had
recognized Cauchy’s extraordinary talent
for mathematics
...
At the age
of 21, he was given a commission in
Napoleon’s army as a civil engineer
...

In 1815, at the age of 26, Cauchy was
made Professor of Mathematics at the École
Polytechnique and was recognized as the
leading mathematician in France
...

Cauchy introduced a new level of rigor
into mathematical analysis
...
He gave the first proof of the
Fundamental Theorem of Calculus
...
His total written
output of mathematics fills 24 large volumes
...
Cauchy died at the age of
67 on May 23, 1857
...
dcs

...
ac
...

JACOB BRONOWSKI

Motivation
Suppose an American and a German are asked to count a handful of objects
...
,” whereas the
German says “Eins, zwei, drei, vier, fünf,
...
They are both counting the objects, but they are using different terminology to do so
...
An analogous situation often occurs
with groups; the same group is described with different terminology
...
In Chapter 1, we described the symmetries of a square in geometric terms (e
...
, R90), whereas in Chapter 5 we
described the same group by way of permutations of the corners
...
In Chapter 4,
we observed that when we have a cyclic group of order n generated by a,
the operation turns out to be essentially that of addition modulo n, since
aras 5 ak, where k 5 (r 1 s) mod n
...
So, each has the form ͗a͘, where aras 5 a (r 1 s) mod 42
...
When this is the
case, we say that there is an isomorphism between the two groups
...
The term
isomorphism is derived from the Greek words isos, meaning “same” or
“equal,” and morphe, meaning “form
...
Allenby has colorfully
122

6 | Isomorphisms

123

defined an algebraist as “a person who can’t tell the difference between
isomorphic systems
...

That is,
f(ab) 5 f(a)f(b)

for all a, b in G
...


This definition can be visualized as shown in Figure 6
...
The pairs
of dashed arrows represent the group operations
...
1

It is implicit in the definition of isomorphism that isomorphic
groups have the same order
...
The four
cases involving ? and 1 are shown in Table 6
...

Table 6
...

Step 1 “Mapping
...

Step 2 “1–1
...


124

Groups

Step 3 “Onto
...

Step 4 “O
...
” Prove that f is operation-preserving; that is, show that
f(ab) 5 f(a)f(b) for all a and b in G
...
The only one that may appear
novel is the fourth one
...
Roughly speaking, this says
that the two processes—operating and mapping—can be done in either
order without affecting the result
...

a

a

a

Before going any further, let’s consider some examples
...
Then G and G are isomorphic under the mapping f(x) 5 2x
...
To prove that it is one-to-one, suppose that 2x 5 2y
...
For “onto,” we must find for any positive
real number y some real number x such that f(x) 5 y; that is, 2x 5 y
...
Finally,
f(x 1 y) 5 2x1y 5 2x ? 2y 5 f(x)f(y)
for all x and y in G, so that f is operation-preserving as well
...
Indeed, if
a is a generator of the cyclic group, the mapping ak → k is an
isomorphism
...
That these correspondences are
functions and are one-to-one is the essence of Theorem 4
...
Obviously,
the mappings are onto
...


6 | Isomorphisms

125

EXAMPLE 3 The mapping from R under addition to itself given by
f(x) 5 x3 is not an isomorphism
...

EXAMPLE 4 U(10) < Z4 and U(5) < Z4
...
Then appeal to Example 2
...
This is a bit trickier to prove
...
Now, suppose that f is an isomorphism from U(10) onto U(12)
...

Thus, f(9) 5 f(1), but 9 2 1, which contradicts the assumption that
f is one-to-one
...
If f were such a mapping, there would be a rational number a such that f(a) 5 21
...


However, no rational number squared is 21
...
Let M be any 2 3 2 real matrix with determinant 1
...
To verify that fM is an isomorphism, we carry out the
four steps
...
Here, we must show that fM(A)
is indeed an element of G whenever A is
...

Thus, MAM21 is in G
...
Suppose that fM(A) 5 fM(B)
...


126

Groups

Step 3 fM is onto
...
We must find a matrix A in G
such that fM(A) 5 B
...
But this tells us exactly
what A must be! For we can solve for A to obtain A 5 M21BM and
verify that fM(A) 5 MAM21 5 M(M21BM)M21 5 B
...
Let A and B belong to G
...

The mapping fM is called conjugation by M
...
An important generalization of it will be given in Chapter 25
...
1 Cayley’s Theorem (1854)
Every group is isomorphic to a group of permutations
...
We must find a group G of
permutations that we believe is isomorphic to G
...
For any g in G,
define a function Tg from G to G by
Tg(x) 5 gx

for all x in G
...
) We leave it as an
exercise (Exercise 23) to prove that Tg is a permutation on the set of
elements of G
...
Then, G is a group under
the operation of function composition
...
From this it follows that Te is the
identity and (Tg)21 5 Tg21 (see Exercise 9)
...

The isomorphism f between G and G is now ready-made
...
If Tg 5 Th, then Tg(e) 5 Th(e) or ge 5 he
...
By the way G was constructed, we
see that f is onto
...
To this end, let a and b belong to G
...


127

6 | Isomorphisms

The group G constructed above is called the left regular representation of G
...
Writing the permutations of
U(12) in array form, we have (remember, Tx is just multiplication by x)
1 5 7 11
d,
1 5 7 11

T5 5 c

1 5 7 11
d,
7 11 1 5

T11 5 c

T1 5 c
T7 5 c

1 5 7 11
d,
5 1 11 7
1 5 7 11
d
...

U(12)

1

5

7

11

1
5
7
11

1
5
7
11

5
1
11
7

7
11
1
5

11
7
5
1

U(12)

T1

T5

T7

T11

T1
T5
T7
T11

T1
T5
T7
T11

T5
T1
T11
T7

T7
T11
T1
T5

T11
T7
T5
T1

It should be abundantly clear from these tables that U(12) and U(12)
are only notationally different
...
One is
that it allows us to represent an abstract group in a concrete way
...
Indeed, Cayley’s Theorem tells us that abstract
groups are not different from permutation groups
...
It is this difference of viewpoint that has
stimulated the tremendous progress in group theory and many other
branches of mathematics in the 20th century
...
For example, it
requires somewhat sophisticated techniques to prove the surprising fact
that the group of real numbers under addition is isomorphic to the
group of complex numbers under addition
...
In geometric terms, this says
that, as groups, the punctured plane and the unit circle are isomorphic
...
)

128

Groups

Properties of Isomorphisms
Our next two theorems give a catalog of properties of isomorphisms
and isomorphic groups
...
2 Properties of Isomorphisms Acting on Elements
Suppose that f is an isomorphism from a group G onto a group G
...
f carries the identity of G to the identity of G
...
For every integer n and for every group element a in G, f(an) 5
[f(a)]n
...
For any elements a and b in G, a and b commute if and only if
f(a) and f(b) commute
...
G 5 ͗a͘ if and only if G 5 ͗f(a)͘
...
|a| 5 |f(a)| for all a in G (isomorphisms preserve orders)
...
For a fixed integer k and a fixed group element b in G, the
equation xk 5 b has the same number of solutions in G as does
the equation xk 5 f(b) in G
...
If G is finite, then G and G have exactly the same number of
elements of every order
...
For
convenience, let us denote the identity in G by e and the identity in G
by e
...

Also, because f(e) [ G, we have f(e) 5 ef(e), as well
...
This proves property 1
...
If n is negative, then 2n is
positive, and we have from property 1 and the observation about the
positive integer case that e 5 f(e) 5 f(gng2n) 5 f(gn)f(g2n) 5
f(gn)(f(g))2n
...
Property 1 takes care of the case n 5 0
...
Because f is onto, for any element b in G, there is an element ak in
G such that f(ak) 5 b
...
This
proves that G 5 ͗f(a)͘
...
Clearly, ͗a͘ # G
...
So, for some integer k we have

6 | Isomorphisms

129

f(b) 5 (f(a))k 5 f(ak)
...
This proves
that ͗a͘ 5 G
...
2 is
f(na) 5 nf(a); property 4 says that an isomorphism between two
cyclic groups takes a generator to a generator
...
Often b is picked to be the identity
...
Because the equation x4 5 1 has four solutions in C* but only
two in R*, no matter how one attempts to define an isomorphism from
C* to R*, property 6 cannot hold
...
3 Properties of Isomorphisms Acting on Groups
Suppose that f is an isomorphism from a group G onto a group G
...

2
...

4
...

G is Abelian if and only if G is Abelian
...

If K is a subgroup of G, then f(K) 5 {f(k) | k [ K} is a
subgroup of G
...

Property 2 is a direct consequence of property 3 of Theorem 6
...

Property 3 follows from property 4 of Theorem 6
...
3
...
2 and 6
...
Actually, the definition is precisely formulated so
that isomorphic groups have all group-theoretic properties in common
...
This is why algebraists speak of isomorphic groups as “equal” or “the same
...


Automorphisms
Certain kinds of isomorphisms are referred to so often that they have
been given special names
...


The isomorphism in Example 7 is an automorphism of SL(2, R)
...

EXAMPLE 9 The function f from C to C given by f(a 1 bi) 5
a 2 bi is an automorphism of the group of complex numbers under
addition
...
(See
Exercise 25
...
Then f(a, b) 5 (b, a)
is an automorphism of the group R2 under componentwise addition
...

More generally, any reflection across a line passing through the
origin or any rotation of the plane about the origin is an automorphism of R2
...

Definition Inner Automorphism Induced by a

Let G be a group, and let a [ G
...


We leave it for the reader to show that fa is actually an automorphism of G
...
)
EXAMPLE 11 The action of the inner automorphism of D4 induced
by R90 is given in the following table
...
The reason these sets are noteworthy is demonstrated by
the next theorem
...
4 Aut(G) and Inn(G) Are Groups†
The set of automorphisms of a group and the set of inner
automorphisms of a group are both groups under the operation
of function composition
...
4 is left as an exercise (Exercise 15)
...
If G 5 {e, a, b, c
...
This latter list may have duplications,
however, since fa may be equal to fb even though a 2 b (see Exercise
33)
...
On the other
hand, the determination of Aut(G) is, in general, quite involved
...
Our job is
to determine the repetitions in this list
...
Also, fR270(x) 5
R270 xR27021 5 R90R180 xR18021R9021 5 R90 xR9021 5 fR90(x)
...

This proves that the previous list can be pared down to fR0, fR90, fH,
and fD
...

EXAMPLE 13 Aut(Z10)
To compute Aut(Z10), we try to discover enough information about an
element a of Aut(Z10) to determine how a must be defined
...
To begin with, observe that once
we know a(1), we know a(k) for any k, because

†The

group Aut(G) was first studied by O
...
H
...


132

Groups

a(k) 5 a(1 1 1 1 ? ? ? 1 1)
k terms

5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 ka(1)
...
Since property 5 of Theorem 6
...


To distinguish among the four possibilities, we refine our notation by
denoting the mapping that sends 1 to 1 by a 1, 1 to 3 by a 3, 1 to 7 by a 7,
and 1 to 9 by a 9
...
But are all these automorphisms? Clearly, a 1 is the identity
...
Since x mod 10 5 y mod 10 implies 3x mod 10 5 3y mod 10,
a3 is well defined
...
Finally, since a3(a 1 b) 5 3(a 1 b) 5 3a 1 3b 5 a3(a) 1 a3(b),
we see that a 3 is operation-preserving as well
...
The
same argument shows that a7 and a9 are also automorphisms
...
For instance, what is a 3a 3? Well, (a3a3)(1) 5 a 3(3) 5 3 ? 3 5 9 5 a9(1), so
a 3 a 3 5 a 9
...
Thus, Aut(Z10 ) is cyclic
...

U(10)

1

3

7

9

Aut(Z10)

a1

a3

a7

a9

1
3
7
9

1
3
7
9

3
9
1
7

7
1
9
3

9
7
3
1

a1
a3
a7
a9

a1
a3
a7
a9

a3
a9
a1
a7

a7
a1
a9
a3

a9
a7
a3
a1

With Example 13 as a guide, we are now ready to tackle the group
Aut(Zn)
...

Theorem 6
...


6 | Isomorphisms

133

PROOF As in Example 13, any automorphism a is determined by the
value of a(1), and a(1) [ U(n)
...
The fact that a(k) 5 ka(1)
(see Example 13) implies that T is a one-to-one mapping
...

To prove that T is onto, let r [ U(n) and consider the mapping a
from Zn to Zn defined by a(s) 5 sr (mod n) for all s in Zn
...

Then, since T(a) 5 a(1) 5 r, T is onto U(n)
...
Let a,
b [ Aut(Zn)
...

This completes the proof
...

STEVEN G
...
Find an isomorphism from the group of integers under addition to
the group of even integers under addition
...
Find Aut(Z)
...
Let R1 be the group of positive real numbers under multiplication
...

4
...

5
...

6
...
That is, if
G, H, and K are groups and G < H and H < K, then G < K
...
Prove that S4 is not isomorphic to D12
...
Show that the mapping a S log10 a is an isomorphism from R+
under multiplication to R under addition
...
In the notation of Theorem 6
...

21

134

Groups

10
...
Prove that the mapping a(g) 5 g21 for all g in G
is an automorphism if and only if G is Abelian
...
For inner automorphisms fg , fh , and fgh, prove that fgfh 5 fgh
...
Find two groups G and H such that G ] H, but Aut(G) < Aut(H)
...
Prove the assertion in Example 12 that the inner automorphisms
fR , fR , fH, and fD of D4 are distinct
...
Find Aut(Z6)
...
If G is a group, prove that Aut(G) and Inn(G) are groups
...
Prove that the mapping from U(16) to itself given by x → x3 is an
automorphism
...

17
...
Prove that the mapping a: Zn → Zn defined by a(s) 5
sr mod n for all s in Zn is an automorphism of Zn
...
)
1 a
18
...
If f and g are isomorphisms from the cyclic group ͗a͘ to some
group and f(a) 5 g(a) , prove that f 5 g
...
Suppose that f: Z 50 S Z 50 is an automorphism with f(11) 5 13
...

21
...
3
...
Prove Property 4 of Theorem 6
...

23
...
1, prove that Tg is indeed a permutation on
the set G
...
Prove or disprove that U(20) and U(24) are isomorphic
...
Show that the mapping f(a 1 bi) 5 a 2 bi is an automorphism of
the group of complex numbers under addition
...
(This exercise is referred to in Chapter 15
...
Let
G 5 {a 1 b"2 | a, b rational}
and
H5e c

a
b

2b
d ` a, b rational f
...
Prove that G
and H are closed under multiplication
...
)

6 | Isomorphisms

135

27
...

28
...

29
...


31
...


33
...

35
...

37
...


39
...


a
b

2b
d ` a, b [ R f
...

Let Rn 5 {(a1, a2,
...
Show that the mapping f:
(a1, a2,
...
, 2an) is an automorphism of
the group Rn under componentwise addition
...
Describe the action of f geometrically
...
Suppose you could prove this for finite
permutation groups
...

Suppose that G is a finite Abelian group and G has no element of
order 2
...

Show, by example, that if G is infinite the mapping need not be an
automorphism
...
If z [ Z(G), show that the inner
automorphism induced by g is the same as the inner automorphism
induced by zg (that is, that the mappings fg and fzg are equal)
...

Suppose that g and h induce the same inner automorphism of a
group G
...

Combine the results of Exercises 33 and 35 into a single “if and
only if” theorem
...
Let fa be the automorphism of G given by fa(x) 5 axa21
...

Exhibit an element a from a group for which 1 , |fa| , |a|
...
} and H 5 {0, 63, 66, 69,
...
Does
your isomorphism preserve multiplication? Generalize to the case
when G 5 ͗m͘ and H 5 ͗n͘, where m and n are integers
...
Determine f(D) and f(H)
...
Write a5 and a 8 as permutations of {0, 1,
...
[For example, a2 5 (0)(124875)(36)
...
Write the permutation corresponding to R90 in the left regular representation of D4 in cycle form
...
Show that every automorphism f of the rational numbers Q under
addition to itself has the form f(x) 5 xf(1)
...
Prove that Q1, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup
...
Prove that Q, the group of rational numbers under addition, is not
isomorphic to a proper subgroup of itself
...
Prove that every automorphism of R*, the group of nonzero real
numbers under multiplication, maps positive numbers to positive
numbers and negative numbers to negative numbers
...
Let G be a finite group
...

47
...


Reference
1
...
R
...


Computer Exercise
There is only one satisfying way to boot a computer
...
H
...
d
...
edu/~jgallian
1
...
Run the program for
n 5 3, 5, 7, and 11
...
Run the program for n 5 4, 8, 16, and 32
...
Run the program when n 5
6, 10, 14, and 22
...
Run the program for n 5 9, 15, 21, and 33
...
Try to deduce a general formula for the order of Aut(Dn)
...

PETER TAIT

ARTHUR CAYLEY was born on August 16,
1821, in England
...
He published his first research
paper while an undergraduate of 20, and in
the next year he published eight papers
...

After graduating from Trinity College,
Cambridge, Cayley stayed on for three years
as a tutor
...
During this period,
he published approximately 200 mathematical papers, many of which are now classics
...
He spent
the rest of his life in that position
...


Among Cayley’s many innovations in
mathematics were the notions of an abstract
group and a group algebra, and the matrix
concept
...
Cayley and his
lifelong friend and collaborator J
...
Sylvester
were the founders of the theory of invariants,
which was later to play an important role in
the theory of relativity
...

He died on January 26, 1895
...
dcs

...
ac
...
As we penetrate the subject
more deeply they will become more and more aware of its basic character
...
N
...
But first, we introduce a new and
powerful tool for analyzing a group—the notion of a coset
...
A
...

Definition Coset of H in G

Let G be a group and let H be a subset of G
...
Analogously, Ha 5 {ha | h [ H} and
aHa21 5 {aha21 | h [ H}
...
In this case, the element a is called the coset
representative of aH (or Ha)
...


EXAMPLE 1 Let G 5 S3 and H 5 {(1), (13)}
...


138

7 | Cosets and Lagrange’s Theorem

139

EXAMPLE 2 Let _ 5 {R0, R180} in D4, the dihedral group of order 8
...

EXAMPLE 3 Let H 5 {0, 3, 6} in Z9 under addition
...
Then the cosets of H in Z9 are
0 1 H 5 {0, 3, 6} 5 3 1 H 5 6 1 H,
1 1 H 5 {1, 4, 7} 5 4 1 H 5 7 1 H,
2 1 H 5 {2, 5, 8} 5 5 1 H 5 8 1 H
...
First, cosets are usually not subgroups
...

Third, since in Example 1 (12)H 5 {(12), (132)} whereas H(12) 5
{(12), (123)}, aH need not be the same as Ha
...
When does
aH 5 bH? Do aH and bH have any elements in common? When does
aH 5 Ha? Which cosets are subgroups? Why are cosets important? The
next lemma and theorem answer these questions
...
)
Lemma Properties of Cosets
Let H be a subgroup of G, and let a and b belong to G
...

2
...

4
...

6
...

8
...


PROOF
1
...

2
...
Then a 5
ae [ aH 5 H
...

4
...

6
...

8
...
The first inclusion follows directly from the closure of
H
...
Then, since a [ H and h [ H, we
know that a21h [ H
...

If aH 5 bH, then a 5 ae [ aH 5 bH
...

Property 4 follows directly from property 3, for if there is an element c in aH y bH, then cH 5 aH and cH 5 bH
...
The result now
follows from property 2
...
Obviously, the correspondence ah → bh
maps aH onto bH
...

Note that aH 5 Ha if and only if (aH)a21 5 (Ha)a21 5 H(aa–1) 5
H—that is, if and only if aHa21 5 H
...
Thus, aH >
eH 2 0; and, by property 4, we have aH 5 eH 5 H
...
Conversely, if a [ H, then, again by
property 2, aH 5 H
...
In the preceding lemma,
property 1 says simply that the left coset of H containing a does contain a
...
Property 3 shows that a left coset of H is uniquely
determined by any one of its elements
...
Property 4 says—and this is
very important—that two left cosets of H are either identical or disjoint
...
Property 6 says
that all left cosets of H have the same size
...
The last property of the lemma says that
H itself is the only coset of H that is a subgroup of G
...

Indeed, we may view the cosets of H as a partitioning of G into equivalence classes under the equivalence relation defined by a , b if
aH 5 bH (see Theorem 0
...

In practice, the subgroup H is often chosen so that the cosets partition the group in some highly desirable fashion
...
Thus, the cosets of H constitute a partition of 3-space into planes parallel to H
...
Thus,
c

2 0
d H is the set of all 2 3 2 matrices of determinant 2
0 1

and
c

1 2
d H is the set of all 2 3 2 matrices of determinant 23
...
We illustrate this in the next example
...
We can find a second coset by choosing any element not
in H, say 3, as a coset representative
...

We find our next coset by choosing a representative not already appearing in the two previously chosen cosets, say 5
...
We continue to form cosets by picking elements from U(32)
that have not yet appeared in the previous cosets as representatives of
the cosets until we have accounted for every element of U(32)
...


Lagrange’s Theorem and Consequences
We are now ready to prove a theorem that has been around for more
than 200 years—longer than group theory itself! (This theorem was not
originally stated in group theoretic terms
...

Theorem 7
...

Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|
...


142

Groups

PROOF Let a1H, a2H,
...
Then, for each a in G, we have aH 5 aiH for some i
...
Thus, each member of G belongs to one
of the cosets aiH
...

Now, property 4 of the lemma shows that this union is disjoint, so that
|G| 5 |a1H| 1 |a2H| 1 ? ? ? 1 |ar H|
...

We pause to emphasize that Lagrange’s Theorem is a subgroup candidate criterion; that is, it provides a list of candidates for the orders of
the subgroups of a group
...
Warning! The converse
of Lagrange’s Theorem is false
...
We prove this in Example 5
...
The index of a subgroup
H in G is the number of distinct left cosets of H in G
...
As an immediate consequence of the proof of
Lagrange’s Theorem, we have the following useful formula for the
number of distinct left (or right) cosets of H in G
...


Corollary 2 |a| Divides |G|
In a finite group, the order of each element of the group divides the
order of the group
...

Corollary 3 Groups of Prime Order Are Cyclic
A group of prime order is cyclic
...
Let a [ G and a 2 e
...
Thus, |͗a͘| 5 |G| and the corollary
follows
...
Then, a|G| 5 e
...
Thus,
a|G| 5 a|a|k 5 ek 5 e
...


PROOF By the division algorithm, a 5 pm 1 r, where 0 # r , p
...
If r 5 0,
the result is trivial, so we may assume that r [ U( p)
...
, p 2 1} under multiplication modulo p
...

Fermat’s Little Theorem has been used in conjunction with computers to test for primality of certain numbers
...
If p is prime, then we know from Fermat’s Little
Theorem that 10 p mod p 5 10 mod p and, therefore, 10 p11 mod p 5
100 mod p
...

The result was not 100, and so p is not prime
...
To verify this,
recall that A4 has eight elements of order 3 (a5 through a12 in the notation of Table 5
...
Let a be
any element of order 3 in A4
...
But equality of any pair of these
three implies that aH 5 H, so that a [ H
...
) Thus, a subgroup of A4 of order 6 would have
to contain all eight elements of order 3, which is absurd
...


144

Groups

For any prime p
...
This naturally raises the question of whether there
could be other possible groups of these orders
...

Theorem 7
...
Then
G is isomorphic to Z2p or Dp
...
We begin by first showing that G must have an
element of order p
...
Thus, to verify our assertion, we may assume that every nonidentity element of G has order 2
...
Then, for any nonidentity elements a, b [ G
with a 2 b, the set {e, a, b, ab} is closed and therefore is a subgroup of
G of order 4
...

Now let b be any element not in kal
...
We next claim that |b| 5 2
...
We may rule out b2kal 5 bkal, for then bkal 5
kal
...
But |b2| 5 p and |b| 2 2p imply that |b| 5 p
...
Thus, any
element of G not in kal has order 2
...
Since ab o kal, our argument above shows that
|ab| 5 2
...
Moreover, this relation
completely determines the multiplication table for G
...
] Since the multiplication table for all noncyclic groups
of order 2p is uniquely determined by the relation ab 5 ba21, all
noncyclic groups of order 2p must be isomorphic to each other
...

As an immediate corollary, we have that S3, the symmetric group of
degree 3, is isomorphic to D3
...
We next consider an application of
cosets to permutation groups
...
For each i in S, let stabG(i) 5
{f [ G | f(i) 5 i}
...


The student should verify that stabG(i) is a subgroup of G
...
)
Definition Orbit of a Point

Let G be a group of permutations of a set S
...
The set orbG(s) is a subset of S called the orbit of s
under G
...


Example 6 should clarify these two definitions
...

Then,
orbG(1) 5 {1, 3, 2},
orbG(2) 5 {2, 1, 3},
orbG(4) 5 {4, 6, 5},
orbG(7) 5 {7, 8},

stabG(1) 5 {(1), (78)},
stabG(2) 5 {(1), (78)},
stabG(4) 5 {(1), (78)},
stabG(7) 5 {(1), (132)(465), (123)(456)}
...
Figure 7
...
1(b) illustrates the orbit of the point q under D4
...

4

4

p

q
(a)

(b)

Figure 7
...


146

Groups

Theorem 7
...
Then, for
any i from S, |G| 5 |orbG (i)| |stabG(i)|
...
Thus, it suffices to establish a oneto-one correspondence between the left cosets of stabG(i) and the
elements in the orbit of i
...
To show that T is a welldefined function, we must show that astabG(i) 5 bstabG(i) implies a(i) 5
b(i)
...
Reversing the argument from
the last step to the first step shows that T is also one-to-one
...
Let j [ orbG(i)
...

We leave as an exercise the proof of the important fact that the orbits
of the elements of a set S under a group partition S (Exercise 33)
...
3 and Lagrange’s Theorem
(Theorem 7
...
† They enable us to determine the
numbers of elements in various sets
...
3 works, we
will determine the order of the rotation group of a cube and a soccer ball
...

EXAMPLE 8 Let G be the rotation group of a cube
...
Since any rotation of the cube must carry
each face of the cube to exactly one other face of the cube and different
rotations induce different permutations of the faces, G can be viewed as
a group of permutations on the set {1, 2, 3, 4, 5, 6}
...
Next, we consider
stabG(1)
...
Surely, there are only four such motions—
rotations of 0°, 90°, 180°, and 270°—about the line perpendicular to
†People

who don’t count won’t count (Anatole France)
...
2)
...
3, |G| 5 |orbG(1)| |stabG(1)| 5 6 ? 4 5 24
...
2 Axis of rotation of a cube
...
Recall that
S4 is the symmetric group of degree 4
...
4 The Rotation Group of a Cube
The group of rotations of a cube is isomorphic to S4
...
To this end, observe that a cube has four diagonals and
that the rotation group induces a group of permutations on the four diagonals
...
To see that this is so, all we need
do is show that all 24 permutations of the diagonals arise from rotations
...
See Figure 7
...
So, the group of permutations induced by the rotations contains the eight-element subgroup
{e, a, a2, a3, b2, b2a, b2a2, b2a3} (see Exercise 37) and ab, which has
order 3
...

EXAMPLE 9 A traditional soccer ball has 20 faces that are regular
hexagons and 12 faces that are regular pentagons
...
) To determine the number of rotational symmetries of a soccer ball using Theorem 7
...
3

our set S to be the 20 hexagons or the 12 pentagons
...
Since any pentagon can be carried to any other
pentagon by some rotation, the orbit of any pentagon is S
...
Thus, by
the Orbit-Stabilizer Theorem, there are 12 ? 5 5 60 rotational symmetries
...
)

In 1985, chemists Robert Curl, Richard Smalley, and Harold Kroto
caused tremendous excitement in the scientific community when they
created a new form of carbon by using a laser beam to vaporize graphite
...
Buckminster Fuller, Curl, Smalley, and Kroto named their discovery
“buckyballs
...
Group theory has been particularly useful in illuminating
the properties of buckyballs, since the absorption spectrum of a molecule
depends on its symmetries and chemists classify various molecular states

7 | Cosets and Lagrange’s Theorem

149

according to their symmetry properties
...
In 1996, Curl, Smalley, and Kroto
received the Nobel Prize in chemistry for their discovery
...
Trying is the first step towards failure
...
Let H 5 {(1), (12)(34), (13)(24), (14)(23)}
...
1 on page 107)
...
Let H be as in Exercise 1
...
)
3
...
Find all the left cosets of H in Z
...
Rewrite the condition a21b [ H given in property 5 of the lemma
on page 139 in additive notation
...

5
...
Use Exercise 4 to decide whether or not
the following cosets of H are the same
...
11 1 H and 17 1 H
b
...
7 1 H and 23 1 H
6
...
Let H 5 {0, 6n, 62n, 63n,
...
How many are there?
7
...

8
...
Find all of the left cosets of ͗a5͘ in ͗a͘
...
Let |a| 5 30
...

10
...
Prove that the only subgroup of G that contains
a and b is G itself
...
Let H be a subgroup of R*, the group of nonzero real numbers under multiplication
...

12
...
Give a geometric description of the coset (3 + 4i)H
...

13
...
What are the possible orders for the
subgroups of G?
14
...
If |K| 5 42 and |G| 5 420, what are the possible
orders of H?

150

Groups

15
...
Prove
that every proper subgroup of G is cyclic
...
Recall that, for any integer n greater than 1, f(n) denotes the number of positive integers less than n and relatively prime to n
...

17
...

18
...
1) to prove
that the order of U(n) is even when n
...

19
...

If g [ G and gm 5 e, prove that g 5 e
...
Suppose H and K are subgroups of a group G
...
Generalize
...
Suppose that H is a subgroup of S4 and that H contains (12) and
(234
...

22
...
Prove that H 8 K
...
Suppose that G is an Abelian group with an odd number of elements
...

24
...
Prove that |G| is prime
...
)
25
...
If G has only one subgroup of order 3 and only one
of order 5, prove that G is cyclic
...

26
...
Prove that G is cyclic or g5 5 e for
all g in G
...
Let |G| 5 33
...

28
...
Show that G must have an element of order 2
...
Can a group of order 55 have exactly 20 elements of order 11?
Give a reason for your answer
...
Determine all finite subgroups of C*, the group of nonzero complex numbers under multiplication
...
Let H and K be subgroups of a finite group G with H # K # G
...

32
...


7 | Cosets and Lagrange’s Theorem

151

33
...
Prove that the orbits of
the members of S constitute a partition of S
...
)
34
...

35
...

a
...

b
...

c
...

36
...
Prove that the center
of G cannot have order pn21
...
Prove that the eight-element set in the proof of Theorem 7
...

38
...

39
...

What is the minimum possible order of the group?
40
...
Show that the mapping a → an is an automorphism of G
...
Show that in a group G of odd order, the equation x2 5 a has a
unique solution for all a in G
...
Let G be a group of order pqr, where p, q, and r are distinct primes
...

43
...
Let A [ G and suppose that
det A 5 2
...

44
...
Thinking of G as a group of permutations of the plane, describe the orbit of a point Q in the plane
...
”)
45
...
Label the faces of the cube
1 through 6, and let H be the subgroup of elements of G that carry
face 1 to itself
...

46
...
(The axes of symmetry are drawn for
reference purposes
...
In each case, determine
the stabilizer of the indicated point
...
Let G 5 GL(2, R), the group of 2 3 2 matrices over R with nonzero
determinant
...

If a, b [ G and aH 5 bH, what can be said about det (a) and
det (b)? Prove or disprove the converse
...
Calculate the orders of the following (refer to Figure 27
...
The group of rotations of a regular tetrahedron (a solid with
four congruent equilateral triangles as faces)
b
...
The group of rotations of a regular dodecahedron (a solid with
12 congruent regular pentagons as faces)
d
...
If G is a finite group with fewer than 100 elements and G has subgroups of orders 10 and 25, what is the order of G?
50
...
Use Theorem 7
...


7 | Cosets and Lagrange’s Theorem

153

Computer Exercise
In the fields of observation chance favors only the prepared mind
...
d
...
edu/~jgallian
1
...
Run the
software for n 5 3 ? 5, 3 ? 7, 3 ? 11, 3 ? 13, 3 ? 17, 3 ? 31, 5 ? 7, 5 ? 11,
5 ? 13, 5 ? 17, 5 ? 31, 7 ? 11, 7 ? 13, 7 ? 17, 7 ? 19, and 7 ? 43
...


Joseph Lagrange
Lagrange is the Lofty Pyramid of the
Mathematical Sciences
...
He became captivated by mathematics at an early
age when he read an essay by Halley on
Newton’s calculus
...
Lagrange made significant contributions to many branches of
mathematics and physics, among them the
theory of numbers, the theory of equations,
ordinary and partial differential equations, the
calculus of variations, analytic geometry,
fluid dynamics, and celestial mechanics
...

Lagrange was a very careful writer with a
clear and elegant style
...
In offering this appointment, Frederick
the Great proclaimed that the “greatest king
in Europe” ought to have the “greatest mathematician in Europe” at his court
...
In 1793, Lagrange
headed a commission, which included
Laplace and Lavoisier, to devise a new system

154

This stamp was issued by
France in Lagrange’s honor
in 1958
...
Out of this came
the metric system
...
Lagrange died on
April 10, 1813
...
dcs

...
ac
...

STEVEN WEINBERG†

Definition and Examples
In this chapter, we show how to piece together groups to make larger
groups
...
These methods will later be used to give us a simple way to construct all finite Abelian groups
...
, Gn be a finite collection of groups
...
, Gn, written as G1 % G2 % ? ? ? % Gn, is the set of
all n-tuples for which the ith component is an element of Gi and the
operation is componentwise
...
, gn) | gi [ Gi},
where (g1, g2,
...
, g9) is defined to be (g1g9,
1
2
n
1
g2g9,
...
It is understood that each product gig9 is performed
2
n
i
with the operation of Gi
...

This construction is not new to students who have had linear algebra or
physics
...
Of course, there is also scalar multiplication, but
†Weinberg

received the 1979 Nobel Prize in physics with Sheldon Glashow and Abdus
Salam for their construction of a single theory incorporating weak and electromagnetic
interactions
...

EXAMPLE 1
U(8) % U(10) 5 {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3),
(3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9),
(7, 1),(7, 3), (7, 7), (7, 9)}
...

EXAMPLE 2
Z2 % Z3 5 {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}
...
Is this group related to another Abelian group of order 6 that we know, namely, Z6? Consider the
subgroup of Z2 % Z3 generated by (1, 1)
...
Hence
Z2 % Z3 is cyclic
...

In Theorem 7
...
Now that we have defined Z2 % Z2, it is easy to classify the
groups of order 4
...
To verify this, let G 5
{e, a, b, ab}
...
Then the mapping e S (0, 0), a S (1, 0),
b S (0, 1), and ab S (1, 1) is an isomorphism from G onto Z2 % Z2
...
Theorem 8
...


Properties of External Direct Products
Our first theorem gives a simple method for computing the order of an
element in a direct product in terms of the orders of the component
pieces
...
1 Order of an Element in a Direct Product
The order of an element in a direct product of a finite number of
finite groups is the least common multiple of the orders of the
components of the element
...
, gn)| 5 lcm(|g1|, |g2|,
...

PROOF Denote the identity of Gi by ei
...
, |gn|)
and t 5|(g1, g2,
...
Because s is a multiple of each |gi| implies that
s
s
(g1, g2,
...
, gn) 5 (e1, e2,
...
On
2
t , g t ,
...
, g )t 5(e , e ,
...
, |gn|
...

The next two examples are applications of Theorem 8
...

EXAMPLE 4 We determine the number of elements of order 5 in
Z25 % Z5
...
1, we may count the number of elements
(a, b) in Z25 % Z5 with the property that 5 5 |(a, b)| 5 lcm(|a|, |b|)
...
We consider two mutually exclusive cases
...
Here there are four choices for a
(namely, 5, 10, 15, and 20) and five choices for b
...

Case 2 |a| 5 1 and |b| 5 5
...

Thus, Z25 % Z5 has 24 elements of order 5
...
We begin by counting the number of elements (a, b) of
order 10
...
Since Z100 has a unique cyclic subgroup of order 10 and any cyclic group of order 10 has four generators
(Theorem 4
...
Similarly, there are five
choices for b
...

Case 2 |a| 5 2 and |b| 5 5
...
4), there is only one
choice for a
...
So, this case
yields four more possibilities for (a, b)
...
Because each cyclic
subgroup of order 10 has four elements of order 10 and no two of the
cyclic subgroups can have an element of order 10 in common, there
must be 24/4 5 6 cyclic subgroups of order 10
...
)
The direct product notation is convenient for specifying certain subgroups of a direct product
...
To find a subgroup of say Z30 % Z12 isomorphic to Z6 % Z4 we observe that ͗5͘ is a
subgroup of Z30 of order 6 and ͗3͘ is a subgroup of Z12 of order 4, so
͗5͘ % ͗3͘ is the desired subgroup
...

Theorem 8
...
Then G % H is cyclic if and only
if |G| and |H| are relatively prime
...
To prove the
first half of the theorem, we assume G % H is cyclic and show that
m and n are relatively prime
...
Since (g, h)mn/d 5 ((gm)n/d, (hn)m/d ) 5 (e, e), we
have mn 5 |(g, h)| # mn/d
...

To prove the other half of the theorem, let G 5 ͗g͘ and H 5 ͗h͘ and
suppose gcd(m, n) 5 1
...

As a consequence of Theorem 8
...
2
...


8 | External Direct Products

159

Corollary 2 Criterion for Zn1n2
...
% Znk
Let m 5 n1 n2 ? ? ? nk
...


By using the results above in an iterative fashion, one can express
the same group (up to isomorphism) in many different forms
...

Similarly,
Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5
< Z2 % Z3 % Z2 % Z5 < Z6 % Z10
...
Note, however, that Z2 % Z30 ] Z60
...
We first introduce some notation
...

For example, U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92}
...
(See Exercise 17 in
Chapter 3
...
3 U(n) as an External Direct Product
Suppose s and t are relatively prime
...
In short,
U(st) < U(s) % U(t)
...


PROOF An isomorphism from U(st) to U(s) % U(t) is x → (x mod s,
x mod t); an isomorphism from Us(st) to U(t) is x → x mod t; an isomorphism from Ut (st) to U(s) is x → x mod s
...
(See Exercises 11, 17, and 19 in Chapter 0; see also [1]
...
3, we have the following result
...
Then,
U(m) < U(n1) % U(n2) % ? ? ? % U(nk)
...
We obtain
U(105) < U(7) % U(15)
U(105) < U(21) % U(5)
U(105) < U(3) % U(5) % U(7)
...

Among all groups, surely the cyclic groups Zn have the simplest
structures and, at the same time, are the easiest groups with which to
compute
...
Because of this, algebraists endeavor to describe a finite Abelian group as such a direct
product
...
With this goal in mind, let us reexamine the
U-groups
...
3 and the facts (see
[2, p
...
For example,
U(105) 5 U(3 ? 5 ? 7) < U(3) % U(5) % U(7)
and

< Z2 % Z4 % Z6
U(720) 5 U(16 ? 9 ? 5) < U(16) % U(9) % U(5)
< Z2 % Z4 % Z6 % Z4
...
This follows from the observations that an
element from Z2 % Z4 % Z6 % Z4 has the form (a, b, c, d), where
|a| 5 1 or 2, |b| 5 1, 2, or 4, |c| 5 1, 2, 3, or 6, and |d| 5 1, 2, or 4, and
that |(a, b, c, d)| 5 lcm(|a|, |b|, |c|, |d|)
...

Because U(720) is isomorphic to Z2 % Z4 % Z6 % Z4, it suffices to calculate the number of elements of order 12 in Z2 % Z4 % Z6 % Z4
...
By Theorem 8
...
Since |a| 5 1 or 2, it does not matter
how a is chosen
...
By Theorem 4
...
So, in
this case, we have 2 ? 4 ? 4 5 32 choices
...
This gives 2 ? 4 ?
2 5 16 new choices
...

These calculations tell us more
...
Imagine trying to deduce this information directly from
U(720) or, worse yet, from Aut(Z720)! These results beautifully illustrate the advantage of being able to represent a finite Abelian group as
a direct product of cyclic groups
...
After all, theorems are laborsaving devices
...


Applications
We conclude this chapter with five applications of the material presented here—three to cryptography, the science of sending and deciphering secret messages, one to genetics, and one to electric circuits
...
A binary string of length n can naturally be thought of
as an element of Z2 % Z2 % ? ? ? % Z2 (n copies) where the parentheses and the commas have been deleted
...
Similarly, two binary strings a1a2 ? ? ? an
and b1b2 ? ? ? bn are added componentwise modulo 2 just as their
corresponding elements in Z2 % Z2 % ? ? ? % Z2 are
...

The fact that the sum of two binary sequences a1a2 ? ? ? an 1 b1b2 ? ? ?
bn 5 00 ? ? ? 0 if and only if the sequences are identical is the basis for
a data security system used to protect internet transactions
...
Amazon
...
Need you be concerned that a hacker will intercept
your credit-card number during the transaction? As you might expect,
your credit-card number is sent to Amazon in a way that protects the
data
...
When you place an order with Amazon the company sends
your computer a randomly generated string of 0’s and 1’s called a key
...
The resulting sum is then transmitted
to Amazon
...

To illustrate the idea, say you want to send an eight-digit binary string
such as s 5 10101100 to Amazon (actual credit-card numbers have very
long strings) and Amazon sends your computer the key
k 5 00111101
...
If someone intercepts the number
s 1 k 5 10010001 during transmission it is no value without knowing k
...
You can tell when you are using an encryption scheme on a web transaction by looking to see if the web address begins with “https” rather than the customary “http
...

Application to Public Key Cryptography
In the mid-1970s, Ronald Rivest, Adi Shamir, and Leonard Adleman
devised an ingenious method that permits each person who is to
receive a secret message to tell publicly how to scramble messages

8 | External Direct Products

163

sent to him or her
...
The idea is based on the fact
that there exist efficient methods for finding very large prime numbers
(say about 100 digits long) and for multiplying large numbers, but no
one knows an efficient algorithm for factoring large integers (say
about 200 digits long)
...
This person calculates n 5 pq and announces
that a message M is to be sent to him or her publicly as Mr mod n
...

To present a simple example that nevertheless illustrates the principal features of the method, say we wish to send the message “YES
...
, Z by 26, and a blank by 00
...
To keep the numbers involved from becoming too unwieldy,
we send the message in blocks of four digits and fill in with blanks
when needed
...
The person to whom the message is to be sent has
picked two primes p and q, say p 5 37 and q 5 73 (in actual practice,
p and q would have 100 or so digits), and a number r that has no prime
divisors in common with lcm( p 2 1, q 2 1) 5 72, say r 5 5, and has
published n 5 37 ? 73 5 2701 and r 5 5 in a public directory
...

We show the work involved for us and the receiver only for the block
2505
...

So, (2505)5 mod 2701 5 (2505)(470) mod 2701 5 2415
...
2228
...
Provided that the numbers are not too large, the Google
search engine at http://www
...
com will do modular arithmetic
...
Be careful, however, because entering 2505^5 mod 2701 computes the wrong value since 2505 5 is too large
...


164

Groups

Thus, the number 2415 is sent to the receiver
...
To do so, the receiver
takes the two factors of 2701, p 5 37 and q 5 73, and calculates the
least common multiple of p 2 1 5 36 and q 2 1 5 72, which is 72
...
) Next, the receiver must find s 5 r21 in U(72)—that is, solve the equation 5 ? s 5 1
mod 72
...
(There is a simple algorithm for finding
this number
...
This calculation can be simplified as follows:
2415 mod 2701 5 2415
(2415)2 mod 2701 5 766
(2415)4 mod 2701 5 (766)2 mod 2701 5 639
(2415)8 mod 2701 5 (639)2 mod 2701 5 470
(2415)16 mod 2701 5 (470)2 mod 2701 5 2119
So, (2415)29 mod 2701 5 (2415)16(2415)8(2415)4(2415) mod 2701 5
(2119)(470)(639)(2415) mod 2701 5 ((2119)(470) mod 2701 3
(639)(2415) mod 2701) mod 2701 5 (1962)(914) mod 2701 5 2505
...
]
Thus the receiver correctly determines the code for “YE
...

The procedure just described is called the RSA public key encryption
scheme in honor of the three people (Rivest, Shamir, and Adleman) who
discovered the method
...
The algorithm is summarized below
...
Pick very large primes p and q and compute n 5 pq
...
Compute the least common multiple of p 2 1 and q 2 1; let us call
it m
...
Pick r relatively prime to m
...
Find s such that rs mod m 5 1
...
Publicly announce n and r
...
Convert the message to a string of digits
...
)
2
...
, Mk
...
Check to see that the greatest common divisor of each Mi and n is
1
...
(In practice, the
primes p and q are so large that they exceed all Mi, so this step may
be omitted
...
Calculate and send Ri 5 Mi r mod n
...
For each received message Ri, calculate Ri s mod n
...
Convert the string of digits back to a string of characters
...
Thus an element of the form xm in U(n) corresponds under an isomorphism to one of the form (mx1, mx2) in Zp21 %
Zq21
...
Then
(mx1, mx2) 5 (u( p 2 1)x1, v(q 2 1)x2) 5 (0, 0) in Zp21 % Zq21, and it
follows that xm 5 1 for all x in U(n)
...

In 2002, Ronald Rivest, Adi Shamir, and Leonard Adleman received
the Association for Computing Machinery A
...
Turing Award which
is considered the “Nobel Prize of Computing” for their contribution to
public key cryptography
...

Digital Signatures
With so many financial transactions now taking place electronically, the
problem of authenticity is paramount
...
Let us say that person
A wants to send a secret message to person B in such a way that only B
can decode the message and B will know that only A could have sent it
...
Here
we assume that EA and EB are available to the public, whereas DA is
known only to A and DB is known only to B and that DBEB and EADA
applied to any message leaves the message unchanged
...

Notice that only A can execute the first step [i
...
, create DA(M)] and
only B can implement the last step (i
...
, apply EADB to the received
message)
...

Application to Genetics†
The genetic code can be conveniently modeled using elements of Z4 %
Z4 % ? ? ? % Z4 where we omit the parentheses and the commas and
just use strings of 0s, 1s, 2s, and 3s and add componentwise modulo 4
...
Each strand is made up of strings of the four nitrogen
bases adenine (A), thymine (T), guanine (G), and cytosine (C)
...

Adenine always is bound to thymine, and guanine always is bound to
cytosine
...
Thus, the DNA segment ACGTAACAGGA and its complement segment TGCATTGTCCT are denoted by 03120030110 and
21302212332
...
So, for any DNA segment a1a2 ? ? ? an represented by elements of Z4 % Z4 % ? ? ? % Z4, we see that its complementary segment
is represented by a1a2 ? ? ? an 1 22 ? ? ? 2
...

They are wired so that when either switch is thrown the light changes
its status (from on to off or vice versa)
...
We can
conveniently think of the states of the two switches as being matched
with the elements of Z2 % Z2 with the two switches in the up position
corresponding to (0, 0) and the two switches in the down position corresponding to (1, 1)
...
We then see that the
lights are on when the switches correspond to the elements of the subgroup ͗(1, 1)͘ and are off when the switches correspond to the elements
†This

discussion is adapted from [3]
...
A similar analysis applies in the case of
three switches with the subgroup {(0, 0, 0), (1, 1, 0), (0, 1, 1), (1, 0, 1)}
corresponding to the lights-on situation
...
“Trying to prove
theorems,” she said
...
”

1
...
(This exercise is referred to in this chapter
...
Show that Z2 % Z2 % Z2 has seven subgroups of order 2
...
Let G be a group with identity eG and let H be a group with identity eH
...

4
...

State the general case
...
Prove or disprove that Z % Z is a cyclic group
...
Prove, by comparing orders of elements, that Z8 % Z2 is not isomorphic to Z4 % Z4
...
Prove that G1 % G2 is isomorphic to G2 % G1
...

8
...
Is Z3 % Z5 isomorphic to Z15? Why?
10
...
)
11
...
) Explain why Z4 % Z4 has the same
number of elements of order 4 as does Z8000000 % Z400000
...

12
...
Explain why Dn cannot be isomorphic to the external direct product of two such groups
...
Prove that the group of complex numbers under addition is isomorphic to R % R
...
Suppose that G1 < G2 and H1 < H2
...
State the general case
...
If G % H is cyclic, prove that G and H are cyclic
...


168

Groups

16
...

17
...

18
...

19
...
State a necessary and sufficient condition for ͗(g, h)͘ 5 ͗g͘ % ͗h͘
...
Determine the number of elements of order 15 and the number of
cyclic subgroups of order 15 in Z30 % Z20
...
What is the order of any nonidentity element of Z3 % Z3 % Z3?
Generalize
...
Let m
...
2 be an odd integer
...

23
...
Let
N 5 R % R % R % R under componentwise addition
...
What is the corresponding theorem for
the group of m 3 n matrices under addition?
24
...
Determine which one by elimination
...
Let G be a group, and let H 5 {(g, g) | g [ G}
...
(This subgroup is called the diagonal of
G % G
...

26
...

27
...

28
...

29
...
How many elements of order 2 are in Z2000000 % Z4000000? Generalize
...
Find a subgroup of Z800 % Z200 that is isomorphic to Z2 % Z4
...
Find a subgroup of Z12 % Z4 % Z15 that has order 9
...
Prove that R* % R* is not isomorphic to C*
...
)
34
...


(See Exercise 36 in Chapter 2 for the definition of multiplication
...
Is H isomorphic to Z9
or to Z3 % Z3?

8 | External Direct Products

169

35
...
Prove that G is isomorphic to Z % Z
...
Let (a1, a2,
...
Give a necessary and
sufficient condition for |(a1, a2,
...

37
...

38
...

39
...
For any Abelian group G and any positive integer n, let Gn 5 {gn |
g [ G} (see Exercise 15, Supplementary Exercises for Chapters
1– 4)
...

41
...

42
...

43
...
, nk are positive even integers
...
% Zn have ? How many are
1
2
k
there if we drop the requirement that n1, n2,
...
Is Z10 % Z12 % Z6 ^ Z60 % Z6 % Z2?
45
...
Find an isomorphism from Z12 to Z4 % Z3
...
How many isomorphisms are there from Z12 to Z4 % Z3?
48
...
Find the element in Z3 % Z5 that maps to 1
...
Let (a, b) belong to Zm % Zn
...

50
...
Add elements of G as you
would polynomials with integer coefficients, except use modulo 3
addition
...
Generalize
...
Use properties of U-groups to determine all cyclic groups that have
exactly two generators
...
Explain a way that a string of length n of the four nitrogen bases A,
T, G, and C could be modeled with the external direct product of n
copies of Z2 % Z2
...
Let p be a prime
...

54
...

55
...

56
...

57
...


170

Groups

58
...
How many have order 2?
59
...

60
...

61
...
Let p and q be odd primes and let m and n be positive integers
...

63
...

64
...

65
...
2, prove that U(n)2 5 {x2 | x [ U(n)} is a proper
subgroup of U(n)
...
Show that U(55)3 5 {x3 | x [ U(55)} is U(55)
...
Find an integer n such that U(n) contains a subgroup isomorphic to
Z5 % Z5
...
Find a subgroup of order 6 in U(700)
...
Show that there is a U-group containing a subgroup isomorphic
to Z3 % Z3
...
Show that no U-group has order 14
...
Show that there is a U-group containing a subgroup isomorphic
to Z14
...
Show that no U-group is isomorphic to Z4 % Z4
...
Show that there is a U-group containing a subgroup isomorphic to
Z4 % Z4
...
Using the RSA scheme with p 5 37, q 5 73, and r 5 5, what number would be sent for the message “RM”?
75
...
”

Computer Exercises
The geek shall inherit the earth
...
d
...
edu/~jgallian

8 | External Direct Products

171

1
...
Run the program for (s, t) 5 (5, 16), (16, 5), (8, 25),
(5, 9), (9, 5), (9, 10), (10, 9), and (10, 25)
...
This software computes the elements of the subgroup U(n)k 5
{xk | x [ U(n)} of U(n) and its order
...
Do you see a relationship
connecting |U(n)| and |U(n)k|, f(n), and k? Make a conjecture
...

Do you see a relationship connecting |U(n)| and |U(n)k|, f(n), and
k? Make a conjecture
...
Do you see a relationship connecting |U(n)| and
|U(n)k|, f(n), and k? Make a conjecture
...
Run the program for
(n, k) 5 (77, 2), (77, 3), (77, 5), (77, 6), (77, 10), and (77, 15)? Do
you see a relationship among U(77, 6), U(77, 2), and U(77, 3)?
What about U(77, 10) U(77, 2), and U(77, 5)? What about U(77,
15), U(77, 3), and U(77, 5)? Make a conjecture
...

3
...

Run the program for n 5 3 ? 5 ? 7, 16 ? 9 ? 5, 8 ? 3 ? 25, 9 ? 5 ? 11,
and 2 ? 27 ? 125
...
This software allows you to input positive integers n1, n2, n3,
...
Use this software to verify the values obtained in Examples 4 and 5 and in Exercise 20
...

5
...
The user enters two primes p and q, an r that is relatively
prime to m 5 lcm (p 2 1, q 2 1), and the message M to be sent
...
Also, the user can input those numbers and have the computer raise the numbers to the s power to obtain the original input
...
This software determines the order of Aut(Zp % Zp), where p is a
prime
...
Is the result always
divisible by p? Is the result always divisible by p 2 1? Is the result
always divisible by p 1 1? Make a conjecture about the order of
Aut(Zp % Zp) for all primes p
...
This software determines the order of Aut(Zp % Zp % Zp), where p
is a prime
...
What is the highest
power of p that divides the order? What is the highest power of p 2 1
that divides the order? What is the highest power of p 1 1 that divides the order? Make a conjecture about the order of
Aut(Zp % Zp % Zp) for all primes p
...
J
...
Gallian and D
...

2
...
Shanks, Solved and Unsolved Problems in Number Theory, 2nd ed
...

3
...
Washburn, T
...
Ryan, Discrete Mathematics, Reading,
MA: Addison-Wesley, 1999
...
Cheng, “Decomposition of U-groups,” Mathematics Magazine 62
(1989): 271–273
...

David J
...

This article provides a simple proof that U(n) is not cyclic when n is
not of the form 1, 2, 4, pk, or 2pk, where p is an odd prime
...
Guichard, “When Is U(n) Cyclic? An Algebra Approach,”
Mathematics Magazine 72 (1999): 139–142
...


Leonard Adleman
“
...
Adleman [has played] a central role
in some of the most surprising, and
provocative, discoveries in theoretical
computer science
...


LEONARD ADLEMAN grew up in San Francisco
...
He enrolled at the
University of California at Berkeley intending
to be a chemist, then changed his mind and
said he would be a doctor
...
“I had gone through a
zillion things and finally the only thing that
was left where I could get out in a reasonable
time was mathematics,” he said
...
” He took a job as a computer programmer at the Bank of America
...

Once again, Adleman lost interest
...
Later, he returned to
Berkeley with the aim of getting a Ph
...
in
computer science
...
D
...

But, while in graduate school, something
else happened to Adleman
...
He discovered, he said, that
mathematics “is less related to accounting
than it is to philosophy
...
But, he
added, “the point when you become a mathematician is where you somehow see through
this and see the beauty and power of mathematics
...
D
...
There he met Ronald
Rivest and Adi Shamir, who were trying to
invent an unbreakable public key system
...
Nevertheless, Adleman agreed to
try to break the codes Rivest and Shamir proposed
...
Finally, on their 43rd attempt, they hit
upon what is now called the RSA scheme
...

He does not read mathematics journals, he
says, because he does not want to be influenced by other people’s ideas
...

Mathematicians are trained and inclined to
sit and think
...
” The only prop he needs,
he said, is a blackboard to stare at
...


For more information about Adleman,
visit:
http://www
...
com

Supplementary Exercises for Chapters 5–8
My mind rebels at stagnation
...

SHERLOCK HOLMES, The Sign of Four

True/False questions for Chapters 5–8 are available on the Web at:
www
...
umn
...
A subgroup N of a group G is called a characteristic subgroup if
f(N) 5 N for all automorphisms f of G
...
Frobenius in 1895
...

2
...

3
...
(That is, every element of G9
has the form a1i1a2i2 ? ? ? a kik, where each aj has the form x21y21xy,
each ij 5 61, and k is any positive integer
...
(This subgroup was first introduced by
G
...
Miller in 1898
...
Prove that the property of being a characteristic subgroup is transitive
...

5
...

1

8 | Supplementary Exercises for Chapters 5–8

175

(See Exercise 36 in Chapter 2 for the definition of multiplication
...
Are G
and H isomorphic? (This exercise shows that two groups with the
same number of elements of each order need not be isomorphic
...
Let H and K be subgroups of a group G and let HK 5 {hk | h [ H,
k [ K} and KH 5 {kh | k [ K, h [ H}
...

7
...
Prove that
0 HK 0 5
8
...

10
...


12
...

14
...

16
...

18
...

20
...

0H d K0

(This exercise is referred to in Chapters 10, 11, and 24
...
Prove that every finite group has an exponent that divides the order of the group
...

Suppose that H and K are subgroups of a group and that |H| and |K|
are relatively prime
...

Let R1 denote the multiplicative group of positive real numbers and
let T 5 {a 1 bi [ C*| a2 1 b2 5 1} be the multiplicative group of
complex numbers of norm 1
...

Use a group-theoretic proof to show that Q* under multiplication is
not isomorphic to R* under multiplication
...

Prove that R under addition is not isomorphic to R* under
multiplication
...

Suppose that G 5 {e, x, x2, y, yx, yx2} is a non-Abelian group with
|x| 5 3 and |y| 5 2
...

Let p be an odd prime
...

Let G be an Abelian group under addition
...
Show that H is a subgroup of
G % G
...

Find five subgroups of Z12 % Z20 1 Z10 isomorphic to Z4 % Z5
...
Prove that Z(G) 5
Z(G1) % Z(G2) % ? ? ? % Z(Gn)
...
Exhibit four nonisomorphic groups of order 18
...
What is the order of the largest cyclic subgroup in Aut(Z720)? (Hint:
It is not necessary to consider automorphisms of Z720
...
Let G be the group of all permutations of the positive integers
...
Prove that H is a subgroup of G
...
Let H be a subgroup of G
...

25
...
(This exercise is referred to in
Chapter 24
...
Show that D33 ] D11 % Z3
...
)
27
...
(This exercise is referred to in Chapter 24
...
Exhibit four nonisomorphic groups of order 66
...
)
29
...

30
...

31
...

32
...

33
...
Find an element in S10 that commutes with b but is not a power of b
...
Prove or disprove that Z4 % Z15 < Z6 % Z10
...
Prove or disprove that D12 < Z3 % D4
...
Describe a three-dimensional solid whose symmetry group is isomorphic to D5
...
Let G 5 U(15) % Z10 % S5
...
Find
the inverse of (2, 3, (123)(15))
...
Let G 5 Z % Z10 and let H 5 {g [ G| |g| 5 ` or |g| 5 1}
...

39
...
% Gn where
each Gi is a nontrivial group and n
...
Prove that G is not cyclic
...
For any s in Sn and any k-cycle (i1i2
...

ik)s21 5 s(i1)s(i2)
...

41
...

42
...

43
...
Prove that S3 % S4 is not isomorphic to a subgroup of S6
...
Find a permutation b such that b2 5 (13579)(268)
...
In R % R under componentwise addition, let H 5 {(x, 3x) | x [ R}
...
) Show
that (2, 5) 1 H is a straight line passing through the point (2, 5) and
parallel to the line y 5 3x
...
In R % R, suppose that H is the subgroup of all points lying on a
line through the origin
...

48
...
, n}
...
If g sends 1 to k, prove that
g stabG(1) 5 {b [ G | b(1) 5 k}
...
Let H be a subgroup of G and let a, b [ G
...

50
...
Prove that G is
cyclic
...
Let p be a prime
...

52
...

53
...
Determine the number of subgroups of Zp2 % Zp2
isomorphic to Zp2
...
Find a group of order 32 ? 52 ? 72 ? 28 that contains a subgroup isomorphic to A8
...
Let p and q be distinct odd primes
...

Prove that xn 5 1 for all x [ U( pq)
...
Prove that D6 is not isomorphic to a subgroup of S4
...
Prove that the permutations (12) and (123
...
(That
is, every member of Sn can be expressed as some combination of
these elements
...
Suppose that n is even and s is an (n 2 1)-cycle in Sn
...

59
...
Prove that s does
not commute with any element of order 2
...
Very
often in mathematics the crucial problem is to recognize and to discover
what are the relevant concepts; once this is accomplished the job may be
more than half done
...
N
...
There are certain situations where
this does hold, however, and these cases turn out to be of critical importance in the theory of groups
...

Definition Normal Subgroup

A subgroup H of a group G is called a normal subgroup of G if aH 5
Ha for all a in G
...


Many students make the mistake of thinking that “H is normal in G”
means ah 5 ha for a [ G and h [ H
...
Think of it this
way: You can switch the order of a product of an element from the group
and an element from the normal subgroup, but you must “fudge” a bit on
the element from the normal subgroup by using h9 or h0 rather than h
...
)
There are several equivalent formulations of the definition of normality
...

However, to verify that a subgroup is normal, it is usually better to use
Theorem 9
...
It allows us to substitute a condition about two subgroups of
G for a condition about two cosets of G
...
1 Normal Subgroup Test
A subgroup H of G is normal in G if and only if xHx21 # H
for all x in G
...
Thus, xhx21 5 h9, and therefore xHx21 # H
...
On the other hand, letting x 5 a21, we have
a21H(a21)21 5 a21Ha # H or Ha # aH
...
(In this
case, ah 5 ha for a in the group and h in the subgroup
...
[Again,
ah 5 ha for any a [ G and any h [ Z(G)
...
[Note, for example, that for (12) [ Sn and (123) [
An, we have (12)(123) 2 (123)(12) but (12)(123) 5 (132)(12) and
(132) [ An
...
(For
any rotation r and any reflection f, we have fr 5 r21f, whereas for any
rotations r and r9, we have rr9 5 r9r
...
To verify this, we use the normal subgroup test
given in Theorem 9
...
Let x [ GL(2, R) 5 G, h [ SL(2, R) 5 H and
note that det xhx21 5 (det x)(det h)(det x)21 5 (det x)(det x)21 5 1
...

EXAMPLE 6 Referring to the group table for A4 given in Table 5
...
To see that H is normal, simply note that for any b in A4, bHb21 is
a subgroup of order 4 and H is the only subgroup of A4 of order 4
since all other elements of A4 have order 3
...
In contrast, a2a5a21 5 a7, so that a2Ka 221 s K
...

The reason is simple
...
Quite often, one can obtain
information about a group by studying one of its factor groups
...

Theorem 9
...
Hölder, 1889)
Let G be a group and let H be a normal subgroup of G
...
†

PROOF Our first task is to show that the operation is well defined; that
is, we must show that the correspondence defined above from G/H 3
G/H into G/H is actually a function
...
That is, verify that abh 5 a9b9H
...
) From aH 5 a9H and bH 5
b9H , we have a9 5 ah1 and b9 5 bh2 for some h1, h2 in H, and therefore
a9b9H 5 ah1bh2H 5 ah1bH 5 ah1Hb 5 aHb 5 abH
...
The rest is easy: eH 5 H is the identity; a21H is the
inverse of aH; and (aHbH)cH 5 (ab)HcH 5 (ab)cH 5 a(bc)H 5
aH(bc)H 5 aH(bHcH)
...

Although it is merely a curiosity, we point out that the converse of
Theorem 9
...

The next few examples illustrate the factor group concept
...
To construct Z/4Z, we first
must determine the left cosets of 4Z in Z
...
},
1 1 4Z 5 {1, 5, 9,
...
},

†The

notation G/H was first used by C
...


181

9 | Normal Subgroups and Factor Groups

2 1 4Z 5 {2, 6, 10,
...
},
3 1 4Z 5 {3, 7, 11,
...

We claim that there are no others
...
Now that
we know the elements of the factor group, our next job is to determine
the structure of Z/4Z
...
More generally, if for any n
...
}, then Z/nZ is isomorphic to Zn
...
Then G/H 5
{0 1 H, 1 1 H, 2 1 H, 3 1 H, 4 1 H, 5 1 H}
...
This
should be one of the six elements listed in the set G/H
...

A few words of caution about notation are warranted here
...
One could be thinking of aH as a set of elements and |aH|
as the size of the set; or, as is more often the case, one could be thinking of aH as a group element of the factor group G/H and |aH| as the
order of the element aH in G/H
...
But the group element
3 1 H has order 2, since (3 1 H) 1 (3 1 H) 5 6 1 H 5 0 1 H
...

EXAMPLE 9 Let _ 5 {R0, R180}, and consider the factor group of
the dihedral group D4 (see page 31 for the multiplication table for D4)
D4/_ 5 {_, R90_, H_, D_}
...
1
...
1 for H_R90_
because D9_ 5 D_
...
1
_

_
H_

_
D_

_
R90_
H_
D_

_
R90_
_
H_
_
D_

R90_
R90_
_
D_
H_

H_
D_
_
R90_

D_
H_
R90_
_

D4/_ provides a good opportunity to demonstrate how a factor
group of G is related to G itself
...
2)
...

For example, when we pass from D4 to D4/_, the box
H
V

V
H

in Table 9
...
1
...

Table 9
...
In particular, all the
elements in the coset of H containing a collapse to the single group element aH in G/H
...
1
on page 107
...
) Let H 5 {1, 2, 3, 4}
...
(In this case, rearrangement of the headings is unnecessary
...
3) with
the cosets 1H, 5H, and 9H, we obtain the Cayley table for G/H given
in Table 9
...

Table 9
...
4
1H
1H
5H
9H

5H

9H

1H
5H
9H

5H
9H
1H

9H
1H
5H

This procedure can be illustrated more vividly with colors
...
Then, in Table 9
...
We could then think of

184

Groups

the factor group as consisting of the three colors that define a group
table isomorphic to G/H
...
Say, for instance, we were to take G to be A4
and H 5 {1, 5, 9}
...
Then the first three rows of the
rearranged Cayley table for A4 would be
1
1
5
9

5

9

2

6

10

3

7

11

4

8

12

1
5
9

5
9
1

9
1
5

2
8
11

6
12
3

10
4
7

3
6
12

7
10
4

11
2
8

4
7
10

8
11
2

12
3
6

But already we are in trouble, for blocking these off into 3 3 3 boxes
yields boxes that contain elements of different cosets
...
Had we printed the rearranged table in four colors with all
members of the same coset having the same color, we would see multicolored boxes rather than the uniformly colored boxes produced by a
normal subgroup
...
In particular, an
Abelian group of order 8 is isomorphic to one of Z8, Z4 % Z2, or Z2 %
Z2 % Z2
...

EXAMPLE 11 Let G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,
23, 25, 27, 29, 31} and H 5 U16(32) 5 {1, 17}
...
Which of the three Abelian groups of order 8
is it—Z8, Z4 % Z2, or Z2 % Z2 % Z2? To answer this question, we need

9 | Normal Subgroups and Factor Groups

185

only determine the elements of G/H and their orders
...
Clearly,
(3H)2 5 9H 2 H, and so 3H has order at least 4
...
On the other hand, direct computations show that both
7H and 9H have order 2, so that G/H cannot be Z8 either, since a cyclic
group of even order has exactly one element of order 2 (Theorem 4
...

This proves that U(32)/U16(32) L Z4 % Z2, which (not so incidentally!)
is isomorphic to U(16)
...
Then |G/K| 5 8, and
we ask, which of the three Abelian groups of order 8 is G/K? Since
(3K)4 5 81K 5 17K 2 K, |3K| 5 8
...

It is crucial to understand that when we factor out by a normal subgroup H, what we are essentially doing is defining every element in H
to be the identity
...
Likewise, R270_ 5 R90R180_ 5 R90_
...
This
is why 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z, and so on
...
Algebraists often refer to the process of
creating the factor group G/H as “killing” H
...
At the same time, G/H simulates G in many ways
...
14 for the irrational
number p)
...
We illustrate this by giving another proof that A4 has no subgroup of order 6
...
To see this, suppose that A4 does have a subgroup H

186

Groups

of order 6
...
Thus,
the factor group A4/H exists and has order 2
...
Thus, a2 [ H for all a in A4
...
1 on page 107, however, we observe that A4 has nine different elements of the form a2, all
of which must belong to H, a subgroup of order 6
...
†
The next three theorems illustrate how knowledge of a factor group
of G reveals information about G itself
...
3 The G/Z Theorem
Let G be a group and let Z(G) be the center of G
...


PROOF Let gZ(G) be a generator of the factor group G/Z(G), and let
a, b [ G
...

Thus, a 5 gix for some x in Z(G) and b 5 gjy for some y in Z(G)
...

A few remarks about Theorem 9
...
First, our proof shows
that a better result is possible: If G/H is cyclic, where H is a subgroup of
Z(G), then G is Abelian
...
For example, it follows immediately from this
statement and Lagrange’s Theorem that a non-Abelian group of order
pq, where p and q are primes, must have a trivial center
...


†How

often have I said to you that when you have eliminated the impossible, whatever
remains, however improbable, must be the truth
...
4 G/Z(G) < Inn(G)
For any group G, G/Z(G) is isomorphic to Inn(G)
...
First, we
show that T is well defined
...
(This shows that the image
of a coset of Z(G) depends only on the coset itself and not on the element representing the coset
...
Then, for all x in G, h21gx 5 xh21g
...
Reversing this
argument shows that T is one-to-one, as well
...

That T is operation-preserving follows directly from the fact that
fgfh 5 fgh for all g and h in G
...
3 and 9
...
Thus, |D6 /Z (D6)| 5 6
...
2), we know that Inn(D6) is isomorphic to D3
or Z6
...
4, D6 /Z(D6)
would be also
...
3 would tell us that D6 is Abelian
...

The next theorem demonstrates one of the most powerful proof techniques available in the theory of finite groups—the combined use of
factor groups and induction
...
5 Cauchy’s Theorem for Abelian Groups
Let G be a finite Abelian group and let p be a prime that divides the
order of G
...


PROOF Clearly, this statement is true for the case in which G has
order 2
...
That is, we assume that the statement is true
for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well
...
So let x be an element of G of some prime order q, say
...
Since every subgroup of
an Abelian group is normal, we may construct the factor group G 5
G/͗x͘
...
By
induction, then, G has an element—call it y͗x͘—of order p
...


Internal Direct Products
As we have seen, the external direct product provides a way of putting
groups together into a larger group
...
It is occasionally
possible to do this
...
We define the set HK 5 {hk | h [ H, k [ K}
...
Then, HK 5 {1, 13, 17, 5}, since 5 5 17 ? 13
mod 24
...
Then,
HK 5 {(1), (13), (12), (12)(13)} 5 {(1), (13), (12), (132)}
...

Definition Internal Direct Product of H and K

We say that G is the internal direct product of H and K and write
G 5 H 3 K if H and K are normal subgroups of G and
G 5 HK

and H > K 5 {e}
...

We want to call G the internal direct product of H and K if H and K are
subgroups of G, and if G is naturally isomorphic to the external direct
product of H and K
...
(The definition ensures that this is the case—see Theorem 9
...
)
On the other hand, one forms an external direct product by starting with
any two groups H and K, related or not, and proceeding to produce the
larger group H % K
...
(See Figures 9
...
2
...
1 For the internal direct product,
H and K must be subgroups of the same group
...
2 For the external
direct product, H and K can
be any groups
...
Just as we may take any (finite) collection of
integers and form their product, we may also take any collection of
groups and form their external direct product
...

EXAMPLE 17 In D6, the dihedral group of order 12, let F denote
some reflection and let Rk denote a rotation of k degrees
...

Students should be cautioned about the necessity of having all conditions of the definition of internal direct product satisfied to ensure
that HK L H % K
...
But G is not isomorphic to H % K,
since, by Theorem 8
...
Note that K
is not normal
...


190

Groups

Definition Internal Direct Product H1 3 H2 3 ? ? ? 3 Hn

Let H1, H2,
...
We
say that G is the internal direct product of H1, H2,
...
G 5 H1H2 ? ? ? Hn 5 {h1h2 ? ? ? hn | hi [ Hi}
2
...
, n 2 1
...
The student may wonder about the motivation for it—
that is, why should we want the subgroups to be normal and why is it
desirable for each subgroup to be disjoint from the product of all previous ones? The reason is quite simple
...
As the next theorem
shows, the conditions in the definition of internal direct product were
chosen to ensure that the two products are isomorphic
...
6 H1 3 H2 3 ? ? ? 3 Hn L H1 % H2 % ? ? ? % Hn
If a group G is the internal direct product of a finite number of
subgroups H1, H2,
...
, Hn
...
For if hi [ Hi and hj [ Hj with i 2 j, then
(hihj hi21)hj21 [ Hj hj 21 5 Hj
and
hi (hj hi21hj21) [ hiHi 5 Hi
...
We next claim that each member of G can be expressed
uniquely in the form h1h2 ? ? ? hn, where hi [ Hi
...
To
prove uniqueness, suppose that g 5 h1h2 ? ? ? hn and g 5 h1 h2 ? ? ? hn
9 9
9,
where hi and h9 belong to Hi for i 5 1,
...
Then, using the fact that
i
the h’s from different Hi’s commute, we can solve the equation
h1h2 ? ? ? hn 5 h9 h9 ? ? ? h9
1 2
n
for h9 h n21 to obtain
n
h9n hn21 5 (h9)21h1(h2 21h2 ? ? ? (h9 )21hn21
...
At this point, we can cancel
n
hn and h9 from opposite sides of the equal sign in Equation (1) and repeat
n
the preceding argument to obtain hn21 5 hn21
...
, n
...
, hn)
...

The next theorem provides an important application of Theorem 9
...

Theorem 9
...


PROOF Let G be a group of order p2 , where p is a prime
...
So, by Corollary 2 of
Lagrange’s Theorem, we may assume that every nonidentity element of
G has order p
...
If this is not the case then there is an element b in G such
that bab21 is not in ͗a͘
...
Since ͗a͘ x ͗bab21 ͘ is a subgroup of both ͗a͘ and ͗bab21 ͘ ,
we have that ͗a͘ x ͗bab21 ͘ 5 5e6
...
,
ap21 ͗bab21 ͘
...
Canceling the b21 terms, we obtain e 5 aibaj and therefore b 5 a2i2j [ ͗a͘
...
To complete the proof, let x be any nonidentity element in G and y be any element of G not in ͗x͘
...
6, we see that G 5 ͗x͘ 3 ͗y͘ < Z p % Z p
...
7, we have the following
important fact
...


192

Groups

We mention in passing that if G 5 H1 % H2 % ? ? ? % Hn, then G can
be expressed as the internal direct product of subgroups isomorphic to
H1, H2,
...
For example, if G 5 H1 % H2, then G 5 H1 3 H2,
where H1 5 H1 % {e} and H2 5 {e} % H2
...
Many authors use H 3 K to denote both the internal
direct product and the external direct product of H and K, making no
notational distinction between the two products
...
Many people reserve the notation
H % K for the situation where H and K are Abelian groups under addition and call it the direct sum of H and K
...

The U-groups provide a convenient way to illustrate the preceding
ideas and to clarify the distinction between internal and external direct
products
...
3 and its corollary and
Theorem 9
...

Let us return to the examples given following Theorem 8
...

U(105) 5 U(15 ? 7) 5 U15(105) 3 U7(105)
5 {1, 16, 31, 46, 61, 76} 3 {1, 8, 22, 29, 43, 64, 71, 92}
L U(7) % U(15),
U(105) 5 U(5 ? 21) 5 U5(105) 3 U21(105)
5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101}
3 {1, 22, 43, 64} L U(21) % U(5),
U(105) 5 U(3 ? 5 ? 7) 5 U35(105) 3 U21(105) 3 U15(105)
5 {1, 71} 3 {1, 22, 43, 64} 3 {1, 16, 31, 46, 61, 76}
L U(3) % U(5) % U(7)
...

CARL SAGAN

1
...
Is H normal in S3?
2
...

3
...
, Hn and
i 2 j with 1 # i # n, 1 # j # n, then Hi > Hj 5 {e}
...
)
a b
d ` a, b, d P R, ad 2 0 f
...
Let H 5 e c
0 d
group of GL(2, R)?
5
...
Prove that H 5
{A [ G| det A [ K} is a normal subgroup of G
...
Viewing ͗3͘ and ͗12͘ as subgroups of Z, prove that ͗3͘/͗12͘ is isomorphic to Z4
...

Generalize to arbitrary integers k and n
...
Prove that if H has index 2 in G, then H is normal in G
...
)
8
...

a
...

b
...
1 on page
107, show that, although a6H 5 a7H and a9H 5 a11H, it is not
true that a6a9H 5 a7a11H
...

9
...
Show that G/H
is not isomorphic to G/K
...
)
10
...

11
...
If H and G/H are Abelian, must
G be Abelian?
12
...

13
...

14
...
What is the order of the element 4U5(105) in the factor group
U(105)/U5(105)?
16
...
What is the order of the element
R60Z(D6) in the factor group D6/Z(D6)?

194

Groups

17
...

19
...

21
...

23
...

25
...


27
...


29
...

31
...

33
...

35
...


Let G 5 Z/͗20͘ and H 5 ͗4͘/͗20͘
...

What is the order of the factor group Z60/͗15͘?
What is the order of the factor group (Z10 % U(10))/͗(2, 9)͘?
Construct the Cayley table for U(20)/U5(20)
...

Determine the order of (Z % Z)/͗(2, 2)͘
...
Is the group cyclic?
The group (Z4 % Z12)/͗(2, 2)͘ is isomorphic to one of Z8, Z4 % Z2, or
Z2 % Z2 % Z2
...

Let G 5 U(32) and H 5 {1, 31}
...
Determine which one by
elimination
...
Is G/H isomorphic to Z4 or Z2 % Z2?
Let G 5 U(16), H 5 {1, 15}, and K 5 {1, 9}
...

Is G/H isomorphic to Z4 or Z2 % Z2? Is G/K isomorphic to Z4 or
Z2 % Z2?
Prove that A 4 % Z 3 has no subgroup of order 18
...

Let R* denote the group of all nonzero real numbers under multiplication
...
Prove that R* is the internal direct product of R1
and the subgroup {1, 21}
...

Let H and K be subgroups of a group G
...
Prove that Z 5 HK
...
Prove that G 5 ͗3͘ 3
͗6͘ 3 ͗10͘, whereas H 2 ͗3͘ 3 ͗6͘ 3 ͗12͘
...


9 | Normal Subgroups and Factor Groups

195

37
...
Prove
that the order of the element gH in G/H must divide the order
of g in G
...
Let H be a normal subgroup of G and let a belong to G
...
If H is a normal subgroup of a group G, prove that C(H), the centralizer of H in G, is a normal subgroup of G
...
An element is called a square if it can be expressed in the form b2
for some b
...
If every element of H is a square and every element of
G/H is a square, prove that every element of G is a square
...
Show, by example, that in a factor group G/H it can happen that
aH 5 bH but |a| 2 |b|
...
)
42
...
1 on page 107 that
the subgroup given in Example 6 of this chapter is the only subgroup of A4 of order 4
...

43
...
Show that if H is a subgroup of a group of order
2p that is not normal, then H has order 2
...
Show that D 13 is isomorphic to Inn(D 13 )
...
Suppose that N is a normal subgroup of a finite group G and H is a
subgroup of G
...

46
...

47
...
Prove that |Z(G)| 5 p
...
If |G| 5 pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| 5 1 or pq
...
Let N be a normal subgroup of G and let H be a subgroup of G
...

50
...
Prove that every nonidentity element in G/H has infinite order
...
Determine all subgroups of R* that have finite index
...
Let G 5 {61, 6i, 6j, 6k}, where i2 5 j2 5 k2 5 21, 2i 5 (21)i,
12 5 (21)2 5 1, ij 5 2ji 5 k, jk 5 2kj 5 i, and ki 5 2ik 5 j
...
Construct the Cayley table for G
...
Show that H 5 {1, 21} v G
...
Construct the Cayley table for G>H
...

i

k

53
...

55
...

57
...


j

Going clockwise, the product of two consecutive elements is the
third one
...
) This is the group of
quaternions that was given in another form in Exercise 4 in the
Supplementary Exercises for Chapters 1–4
...
The quaternions are used to describe
rotations in three-dimensional space, and they are used in physics
...

In D4, let K 5 {R0, D} and let L 5 {R0, D, D9, R180}
...
(Normality is not transitive
...
)
Show that the intersection of two normal subgroups of G is a normal subgroup of G
...

Let N be a normal subgroup of G and let H be any subgroup of G
...
Give an example to show that
NH need not be a subgroup of G if neither N nor H is normal
...
)
If N and M are normal subgroups of G, prove that NM is also a normal subgroup of G
...
If N is cyclic, prove that
every subgroup of N is also normal in G
...
)
Without looking at inner automorphisms of Dn, determine the number of such automorphisms
...
Let H be a normal subgroup of a finite group G and let x [ G
...
(This exercise is referred to
in Chapter 25
...
Let G be a group and let G9 be the subgroup of G generated by the
set S 5 {x21y21xy | x, y [ G}
...
)
a
...

b
...

c
...

d
...

61
...

62
...
Prove that the
intersection of all subgroups of G of order n is a normal subgroup
of G
...
If G is non-Abelian, show that Aut(G) is not cyclic
...
Let |G| 5 pnm, where p is prime and gcd( p, m) 5 1
...
If K is a subgroup of G of
order pk, show that K # H
...
Suppose that H is a normal subgroup of a finite group G
...

Show, by example, that the assumption that G is finite is necessary
...
)
66
...
If N is a characteristic
subgroup of G, show that N is a normal subgroup of G
...
In D4, let _ 5 {R0, H}
...
Is the result a group table? Does your answer
contradict Theorem 9
...
Show that S4 has a unique subgroup of order 12
...
If |G| 5 30 and |Z(G)| 5 5, what is the structure of G/Z(G)?
70
...

71
...

72
...
Show that the product of all the elements of G (taken in
any order) cannot belong to H
...
Let G be a group and p a prime
...
Show that H is normal and that every nonidentity
element of G>H has order p
...
Suppose that H is a normal subgroup of G
...

75
...
Show that H
contains every element of G of odd order
...

The authors offer 11 proofs that A4 has no subgroup of order 6
...

J
...
Gallian, R
...
Johnson, and S
...
“On the Quotient Structure of
Zn,” Pi Mu Epsilon Journal, 9 (1993): 524–526
...
This article can be downloaded at http://www
...
umn

...
pdf
Tony Rothman, “Genius and Biographers: The Fictionalization of Évariste
Galois,” The American Mathematical Monthly 89 (1982): 84–106
...

Paul F
...

The author discusses how group-theoretic notions such as subgroups,
cosets, factor groups, and isomorphisms of Z12 and Z20 relate to musical
scales, tuning, temperament, and structure
...

E
...
BELL, Men of Mathematics

This French stamp was issued as part of
the 1984 “Celebrity Series” in support of
the Red Cross Fund
...

Although he had mastered the works of
Legendre and Lagrange at age 15, Galois
twice failed his entrance examination to
l’Ecole Polytechnique
...

At 18, Galois wrote his important research
on the theory of equations and submitted it to
the French Academy of Sciences for publication
...
Cauchy, impressed by the paper,
agreed to present it to the academy, but he
never did
...

The paper was given to Fourier, who died
shortly thereafter
...

Galois spent most of the last year and a
half of his life in prison for revolutionary political offenses
...
On May 30, 1832, Galois was shot in a
duel and died the next day at the age of 20
...

His work provided a method for disposing
of several famous constructability problems,
such as trisecting an arbitrary angle and doubling a cube
...

To find more information about Galois,
visit:
http://www-groups
...
st-and
...
uk/~history/

199

Group
10 Homomorphisms
All modern theories of nuclear and electromagnetic interactions are based
on group theory
...
The term homomorphism comes from the
Greek words homo, “like,” and morphe, “form
...
The concept of group homomorphisms
was introduced by Camille Jordan in 1870, in his influential book
Traité des Substitutions
...


Before giving examples and stating numerous properties of
homomorphisms, it is convenient to introduce an important subgroup
that is intimately related to the image of a homomorphism
...
1
...
The kernel of f is denoted by
Ker f
...
The kernel of an isomorphism is the trivial subgroup
...
Then the determinant mapping A → det A is a
homomorphism from GL(2, R) to R*
...

EXAMPLE 3 The mapping f from R* to R*, defined by f(x) 5 |x|,
is a homomorphism with Ker f 5 {1, 21}
...
For any f in R[x], let f9 denote the derivative of f
...
The kernel of the derivative mapping is the set of all constant polynomials
...
The kernel
of this mapping is ͗n͘
...
(See Exercise 5
...

EXAMPLE 7 The mapping f(x) 5 x2 from R, the real numbers
under addition, to itself is not a homomorphism, since f(a 1 b) 5
(a 1 b)2 5 a2 1 2ab 1 b2, whereas f(a) 1 f(b) 5 a2 1 b2
...
(The term well-defined is
often used in this context
...
But it is not a function, since 0 1 ͗3͘ 5 3 1
͗3͘ in Z/͗3͘ but 3 ? 0 2 3 ? 3 in Z6
...
An invertible linear transformation is a group
isomorphism
...
1 Properties of Elements Under Homomorphisms
Let f be a homomorphism from a group G to a group G and let g be
an element of G
...
f carries the identity of G to the identity of G
...
f(gn) 5 (f(g))n for all n in Z
...
If |g| is finite, then |f(g)| divides |g|
...
Ker f is a subgroup of G
...
f(a) 5 f(b) if and only if aKer f 5 bKer f
...
If f(g) 5 g9, then f21(g9) 5 {x [ G | f(x) 5 g9} 5 gKer f
...
2
...
So, by Corollary 2 to Theorem 4
...

By property 1 we know that Ker f is not empty
...

Since f(a) 5 e and f(b) 5 e, we have f(ab21) 5 f(a)f(b21) 5
f(a)(f(b))21 5 ee21 5 e
...

To prove property 5, first assume that f(a) 5 f(b)
...

It now follows from property 5 of the lemma in Chapter 7 that
bKer f 5 aKer f
...

To prove property 6, we must show that f21(g9) # gKer f and that
gKer f # f21(g9)
...
Then f(g) 5 f(x) and by property 5 we have gKer f 5
xKer f and therefore x [ gKer f
...
To prove that gKer f # f21(g9), suppose that k [
Ker f
...
Thus, by definition, gk [
f21(g9)
...


10 | Group Homomorphisms

203

Theorem 10
...
Then
1
...

2
...

3
...

4
...

5
...

6
...

7
...

8
...

9
...


PROOF First note that the proofs of properties 1, 2, and 3 are identical to the proofs of properties 4, 3, and 2, respectively, of Theorem
6
...

To prove property 4, let f(h) [ f(H) and f(g) [ f(G)
...

Property 5 follows directly from property 6 of Theorem 10
...

To prove property 6, let fH denote the restriction of f to the
elements of H
...

Suppose |Ker fH| 5 t
...
So,
|f(H)|t 5 |H|
...
Clearly,
e [ f21(K ), so that f21(K ) is not empty
...
Then,
by the definition of f21(K ), we know that f(k1), f(k2) [ K
...
So, by definition
21
of f21(K ), we have k1k2 [ f21(K )
...
1
...
Thus, since K is normal in G, f(xkx21) 5 f(x)f(k)(f(x))21 [ K ,
and, therefore, xkx21 [ f21(K )
...


204

Groups

A few remarks about Theorems 10
...
2 are in order
...
For
example, properties 2 and 3 of Theorem 10
...
Property 4 of Theorem 10
...
Property 5 of Theorem 10
...
The set f21(g9) defined in property 6 of Theorem 10
...
Note that the inverse image of an element is a coset of the kernel
and that every element in that coset has the same image
...
2 is called the inverse
image of K (or the pullback of K )
...
1 is reminiscent of something from linear
algebra and differential equations
...
In reality, this
statement is just a special case of property 6
...
1 and property 5 of Theorem 10
...
1
...
2, where K 5 {e}, is
of such importance that we single it out
...
Then Ker f is a normal subgroup of G
...
1
and 10
...

EXAMPLE 8 Consider the mapping f from C* to C* given by
f(x) 5 x4
...
Clearly,
Ker f 5 {x | x4 5 1} 5 {1, 21, i, 2i}
...
2, we know that f is a 4-to-1 mapping
...
Certainly, f("2 ) 5 2
...
1, the set of all elements that map to 2 is "2 Ker f 5
4
4
4
4
{"2 , 2"2 , "2 i, 2"2 i}
...
, gn

G

gKer φ = φ 21(g9)
g, gg2,
...
1

Finally, we verify a specific instance of property 3 of Theorem 10
...
2
...
It follows from DeMoivre’s Theorem (Example 7
in Chapter 0) that |H| 5 12, f(H) 5 ͗cos 120° 1 i sin 120°͘, and
|f(H)| 5 3
...
To verify that f is a
homomorphism, we observe that in Z12, 3(a 1 b) 5 3a 1 3b (since the
group operation is addition modulo 12)
...
Thus, we know from property 5 of Theorem 10
...
Since f(2) 5 6, we have by property 6 of
Theorem 10
...
Notice also that ͗2͘
is cyclic and f(͗2͘) 5 {0, 6} is cyclic
...
1
...
This verifies property 7 of Theorem 10
...

The next example illustrates how one can easily determine all homomorphisms from a cyclic group to a cyclic group
...

By property 2 of Theorem 10
...
That is, if 1 maps to a, then x maps to xa
...
1 require that |a| divide both 12 and 30
...
Thus, a 5 0, 15, 10, 20,
5, or 25
...
That
each of these six possibilities yields an operation-preserving, welldefined function can now be verified by direct calculations
...
This is not a coincidence!]

206

Groups

)
3

3

)

(1

)
2
3

)

(1
)

(1
)

(1

(2

(1

3

(2

)

3

)

3

2

)

(1

(1

3

)

2

(1

3

2

)

3

)

(1

)
3
2
(1

)
(1
)

)
2

)

)
(1
)

1

1

(2

3

)

(1

(2

3

3

)

(1

(1

2

)

3

3

)

(1

(1

2

(1
3
2
)

(1

2

)

(2

2
3
(1

(1

3

3
)

2

)

(1
)
(1

)
2
3
(1
)

3
2
(1

2

)
2
(1

)
3

)
2
3
(1

)
3
2

)
(1

3

)

(2

2
(1

)
2
3
(1

)
3
2
(1
(1

)
(1
)
(1
2
3
)

(1

(1

)

(1

2

(1

)

3

(1

)

2

)

(1

3

)

(2

3

)

(2

3

)

EXAMPLE 11 The mapping from Sn to Z2 that takes an even permutation to 0 and an odd permutation to 1 is a homomorphism
...
2
illustrates the telescoping nature of the mapping
...
2 Homomorphism from S3 to Z2
...
The next theorem shows that for any homomorphism f of G and
the normal subgroup Ker f, the same process produces a Cayley table
isomorphic to the homomorphic image of G
...
This can be likened to viewing a group
through the reverse end of a telescope—the general features of the
group are present, but the apparent size is diminished
...

Theorem 10
...
Then the mapping
from G/Ker f to f(G), given by gKer f → f(g), is an isomorphism
...


PROOF Let us use c to denote the correspondence gKer f S f(g)
...
1
...

The next corollary follows directly from Theorem 10
...
2, and Lagrange’s Theorem
...


EXAMPLE 10 To illustrate Theorem 10
...
3 is
R0Ker f S R0, R90Ker f S H, HKer f S R180, DKer f S V
...

Mathematicians often give a pictorial representation of Theorem
10
...
The mapping g
is called the natural mapping from G to G/Ker f
...
3 shows that cg 5 f
...

As a consequence of Theorem 10
...
We may simply consider the various
factor groups of G
...
We know that the number of homomorphic images of a cyclic
group G of order n is the number of divisors of n, since there is exactly one
subgroup of G (and therefore one factor group of G) for each divisor of n
...
)
An appreciation for Theorem 10
...

EXAMPLE 13 Z/8N9 L ZN
Consider the mapping from Z to Zn defined in Example 5
...
So, by Theorem 10
...

EXAMPLE 14 The Wrapping Function
Recall the wrapping function W from trigonometry
...
3)
...
This mapping is a homomorphism from the group R under addition onto the
circle group (the group of complex numbers of magnitude 1 under
multiplication)
...
Since W is
periodic of period 2p, Ker W 5 ͗2p͘
...

W(2)

W(3)

W(1)
(1, 0) W(0)

(0, 0)
W(21)

Figure 10
...

EXAMPLE 15 The N/C Theorem
Let H be a subgroup of a group G
...
Consider the mapping from
N(H) to Aut(H) given by g S fg, where fg is the inner automorphism of
H induced by g [that is, fg(h) 5 ghg21 for all h in H]
...
So, by Theorem 10
...

As an application of the N/C Theorem, we will show that every
group of order 35 is cyclic
...
By Lagrange’s
Theorem, every nonidentity element of G has order 5, 7, or 35
...
So we may assume that all
nonidentity elements have order 5 or 7
...

Similarly, since 6 does not divide 34, not all nonidentity elements can
have order 7
...
Since G has
an element of order 7, it has a subgroup of order 7
...
In
fact, H is the only subgroup of G of order 7, for if K is another subgroup of G of order 7, we have by Exercise 7 of the Supplementary
Exercises for Chapters 5–8 that |HK| 5 |H||K|/|H > K| 5 7 ? 7/1 5 49
...
Since for every
a in G, aHa21 is also a subgroup of G of order 7 (see Exercise 1 of the
Supplementary Exercises for Chapters 1–4), we must have aHa21 5 H
...
Since H has prime order, it is cyclic and therefore
Abelian
...
So, 7 divides |C(H)| and
|C(H)| divides 35
...
If
C(H) 5 G, then we may obtain an element x of order 35 by letting
x 5 hk, where h is a nonidentity element of H and k has order 5
...

However, 5 does not divide |Aut(H)| 5 |Aut(Z7)| 5 6
...

The corollary of Theorem 10
...
We conclude
this chapter by verifying that the converse of this statement is also true
...
4 Normal Subgroups Are Kernels
Every normal subgroup of a group G is the kernel of a homomorphism of G
...


PROOF Define g:G S G/N by g(g) 5 gN
...
) Then, g(xy) 5 (xy)N 5 xNyN 5
g(x)g(y)
...

Examples 13, 14, and 15 illustrate the utility of the First Isomorphism
Theorem
...
One measure
of the likeness of a group and its homomorphic image is the size of the
kernel
...
On the other hand, if the kernel of the homomorphism is G itself, then the image tells us nothing about G
...

The utility of a particular homomorphism lies in its ability to preserve
the group properties we want, while losing some inessential ones
...
For example, if G is
a group of order 60 and G has a homomorphic image of order 12 that is
cyclic, then we know from properties 5, 7, and 8 of Theorem 10
...
To illustrate
further, suppose we are asked to find an infinite group that is the union of
three proper subgroups
...
Observing that Z2 % Z2 is the union of H1 5
͗1, 0͘, H2 5 ͗0, 1͘, and H3 5 ͗1, 1͘, we have found our finite group
...
Clearly, the mapping from Z2 % Z2 % Z onto Z2 % Z2 given by
f(a, b, c) 5 (a, b) is such a mapping, and therefore Z2 % Z2 % Z is the
union of f21(H1) 5 {(a, 0, c,) | a [ Z2, c [ Z}, f21(H2) 5 {(0, b, c) | b
[ Z2, c [ Z}, and f21(H3) 5 {(a, a, c) | a [ Z2, c [ Z}
...
Whereas isomorphisms
allow us to look at a group in an alternative way, homomorphisms act as
investigative tools
...
† A photograph of a person cannot
tell us the person’s exact height, weight, or age
...
In the same way, a homomorphic image of a group gives us some information about the group
...
Here, we may carry our analogy with photography one
step further by saying that this is like wanting photographs of a person
from many different angles (front view, profile, head-to-toe view, closeup, etc
...


Exercises
The greater the difficulty, the more glory in surmounting it
...

EPICURUS

1
...

3
...

5
...

Prove that the mapping given in Example 3 is a homomorphism
...

Prove that the mapping given in Example 11 is a homomorphism
...
Show that the mapping that takes x
to xr is a homomorphism from R* to R* and determine the kernel
...
Let G be the group of all polynomials with real coefficients under addition
...
Show that the mapping f S ͐f from G to G is
a homomorphism
...
Henry David Thoreau, Journal
...
If f is a homomorphism from G to H and s is a homomorphism
from H to K, show that sf is a homomorphism from G to K
...

8
...
For each s in G, define
sgn(s) 5 e

11 if s is an even permutation,
21 if s is an odd permutation
...
What is the kernel? Why does this homomorphism allow you to conclude that An is a normal subgroup of S n of
index 2?
9
...
What is the kernel? This mapping is called the
projection of G % H onto G
...
Let G be a subgroup of some dihedral group
...

12
...

14
...

16
...

18
...

20
...


11
21

if x is a rotation,
if x is a reflection
...
What is the kernel?
Prove that (Z % Z)/(͗(a, 0)͘ 3 ͗(0, b)͘) is isomorphic to Za % Zb
...
Prove that Zn/͗k͘ L Zk
...

Explain why the correspondence x → 3x from Z12 to Z10 is not a homomorphism
...
If f(23) 5 9, determine all elements that map to 9
...

Prove that there is no homomorphism from Z16 % Z2 onto Z4 % Z4
...

Suppose that there is a homomorphism f from Z17 to some group
and that f is not one-to-one
...

How many homomorphisms are there from Z20 onto Z8? How many
are there to Z8?
If f is a homomorphism from Z30 onto a group of order 5, determine the kernel of f
...
Suppose that f is a homomorphism from a finite group G onto G
and that G has an element of order 8
...
Generalize
...
Suppose that f is a homomorphism from Z36 to a group of order 24
...
Determine the possible homomorphic images
...
For each image in part a, determine the corresponding kernel of f
...
Suppose that f: Z50 S Z15 is a group homomorphism with f(7) 5 6
...
Determine f(x)
...
Determine the image of f
...
Determine the kernel of f
...
Determine f21(3)
...

25
...
Determine all homomorphisms from Z4 to Z2 % Z2
...
Determine all homomorphisms from Zn to itself
...
Suppose that f is a homomorphism from S4 onto Z2
...
Determine all homomorphisms from S4 to Z2
...
Suppose that there is a homomorphism from a finite group G onto
Z10
...

30
...
Explain why G must have normal subgroups of orders 5, 10, 15, 20, 30, and 60
...
Suppose that f is a homomorphism from U(30) to U(30) and
that Ker f 5 {1, 11}
...

32
...

33
...
If f(11) 5 11, find all elements of U(40)
that map to 11
...
Find a homomorphism f from U(40) to U(40) with kernel {1, 9,
17, 33} and f(11) 5 11
...
Prove that the mapping f: Z % Z S Z given by (a, b) S a 2 b is a
homomorphism
...

36
...
Determine f((4, 4)) in
terms of a and b
...


214

Groups

37
...

What is the kernel?
38
...
What is
the kernel when (m, n) 5 (3, 4)? What is the kernel when (m, n) 5
(6, 4)? Generalize
...
(Second Isomorphism Theorem) If K is a subgroup of G and N is
a normal subgroup of G, prove that K/(K > N) is isomorphic
to KN/N
...
(Third Isomorphism Theorem) If M and N are normal subgroups of
G and N # M, prove that (G/N)/(M/N) L G/M
...
Let f(d) denote the Euler phi function of d (see page 79)
...
[It follows
from number theory that this sum is actually gcd(n, k)
...
Let k be a divisor of n
...
What is the relationship between this
homomorphism and the subgroup Uk(n) of U(n)?
43
...

44
...
Use the theorems
of this chapter to prove that the order of the group element gN in
G/N divides the order of g
...
Suppose that G is a finite group and that Z10 is a homomorphic
image of G
...

46
...
What can be said about |G|? Generalize
...
Suppose that for each prime p, Zp is the homomorphic image of a
group G
...

48
...
) Suppose that x is a
particular solution to a system of linear equations and that S is the
entire solution set of the corresponding homogeneous system of
linear equations
...
1 guarantees that x 1 S is the entire solution set of the nonhomogeneous
system
...

49
...
Use property 7 of
Theorem 10
...
(This exercise is referred to in
Chapter 24
...
Show that a homomorphism defined on a cyclic group is completely determined by its action on a generator of the group
...
Use the First Isomorphism Theorem to prove Theorem 9
...

52
...
Prove or disprove that H is a subgroup of G
...
Let Z[x] be the group of polynomials in x with integer coefficients
under addition
...
Give a geometric description of
the kernel of this homomorphism
...

54
...

56
...


58
...

60
...


62
...
What is the kernel of the homomorphism?
Suppose there is a homomorphism f from G onto Z2 % Z2
...

If H and K are normal subgroups of G and H > K 5 {e}, prove that
G is isomorphic to a subgroup of G/H % G/K
...
Prove
that H > K is a normal subgroup of G of index 4 and that G/(H > K)
is not cyclic
...

How many homomorphisms are there from G to H % H % ? ? ? % H
(s terms)? When H is Abelian, how many homomorphisms are there
from G % G % ? ? ? % G (s terms) to H?
Prove that every group of order 77 is cyclic
...
Determine all
homomorphisms from Z to S3
...

Show that every proper subgroup of G is properly contained in a
proper subgroup of G
...

Let p be a prime
...


216

Groups

Computer Exercise
A computer lets you make more mistakes faster than any invention in
human history—with the possible exceptions of handguns and tequila
...
d
...
edu/~jgallian
1
...

(Recall that a homomorphism from Zm is completely determined by
the image of 1
...
Run the program for m 5 15 with various choices for n
...
Try to see the relationship between this
image and the values of m and n
...


Camille Jordan
Although these contributions [to
analysis and topology] would have been
enough to rank Jordan very high among
his mathematical contemporaries, it is
chiefly as an algebraist that he reached
celebrity when he was barely thirty; and
during the next forty years he was
universally regarded as the undisputed
master of group theory
...
DIEUDONNÉ, Dictionary of

Scientific Biography

CAMILLE JORDAN was born into a well-to-do
family on January 5, 1838, in Lyons, France
...

Nearly all of his 120 research papers in
mathematics were written before his retirement from engineering in 1885
...

In the great French tradition, Jordan was
a universal mathematician who published in
nearly every branch of mathematics
...


His classic book Traité des Substitutions,
published in 1870, was the first to be devoted solely to group theory and its applications to other branches of mathematics
...
This book gave the first
clear definitions of the notions of volume
and multiple integral
...
L
...
” Jordan died in Paris on
January 22, 1922
...
dcs

...
ac
...

MIGUEL DE CERVANTES, Don Quixote

The Fundamental Theorem
In this chapter, we present a theorem that describes to an algebraist’s
eye (that is, up to isomorphism) all finite Abelian groups in a standardized way
...
The first
proof of the theorem was given by Leopold Kronecker in 1858
...
1 Fundamental Theorem of Finite Abelian Groups
Every finite Abelian group is a direct product of cyclic groups of
prime-power order
...


Since a cyclic group of order n is isomorphic to Zn, Theorem 11
...
, pknk are uniquely determined by G
...


The Isomorphism Classes
of Abelian Groups
The Fundamental Theorem is extremely powerful
...
Let’s look at groups whose orders have the form pk, where p is
218

11 | Fundamental Theorem of Finite Abelian Groups

219

prime and k # 4
...

Partitions of k
1
2

Possible direct
products for G
Zp
Zp2

p3

111
3

Zp % Zp
Zp3

p4

211
11111
4

Zp2 % Zp
Zp % Zp % Zp
Zp4

311
212

Zp3 % Zp
Zp2 % Zp2

21111
1111111

Zp2 % Zp % Zp
Zp % Zp % Zp % Zp

Order of G
p
p2

Furthermore, the uniqueness portion of the Fundamental Theorem
guarantees that distinct partitions of k yield distinct isomorphism
classes
...

A reliable mnemonic for comparing external direct products is the cancellation property: If A is finite, then
A%BLA%C

if and only if

BLC

(see [1])
...

To appreciate fully the potency of the Fundamental Theorem, contrast
the ease with which the Abelian groups of order pk, k # 4, were
determined with the corresponding problem for non-Abelian groups
...

Now that we know how to construct all the Abelian groups of primepower order, we move to the problem of constructing all Abelian

220

Groups

groups of a certain order n, where n has two or more distinct prime
divisors
...
Next, we individually form all Abelian groups of
order p1n 1, then p2n 2, and so on, as described earlier
...
For example, let n 5
1176 5 23 ? 3 ? 72
...

If we are given any particular Abelian group G of order 1176, the
question we want to answer about G is: Which of the preceding six isomorphism classes represents the structure of G? We can answer this
question by comparing the orders of the elements of G with the orders of
the elements in the six direct products, since it can be shown that two finite Abelian groups are isomorphic if and only if they have the same
number of elements of each order
...
If so, then G must be isomorphic
to the first or fourth group above, since these are the only ones with elements of order 8
...

What if we have some specific Abelian group G of order p1n 1p2 n 2
? ? ? pk n k, where the pi’s are distinct primes? How can G be expressed as
an internal direct product of cyclic groups of prime-power order? For
simplicity, let us say that the group has 2n elements
...
After this is done, pick an element
of maximum order 2r, call it a1
...
If G 2 ͗a1͘, choose an element a2 of
maximum order 2s such that s # n 2 r and none of a2, a2 2, a2 4,
...
Then ͗a2͘ is a second direct factor
...
, a32 is in ͗a1͘ 3 ͗a2͘ 5 {a1i a 2 j | 0 # i , 2r, 0 #
j , 2s}
...
We continue in this fashion
until our direct product has the same order as G
...


221

11 | Fundamental Theorem of Finite Abelian Groups

Greedy Algorithm for an Abelian Group of Order pn
1
...

2
...

Set i 5 1
...
If |G| 5 |Gi|, stop
...

4
...
, ai p is in Gi21, and define
Gi 5 Gi21 3 ͗ai͘
...
Return to step 3
...
The direct product of all of these pieces is the
desired factorization of G
...

EXAMPLE 1 Let G 5 {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51,
53, 57, 64} under multiplication modulo 65
...

To decide which one, we dirty our hands to calculate the orders of the
elements of G
...
Finally, we observe that since this latter
group has a subgroup isomorphic to Z2 % Z2 % Z2, it has more than
three elements of order 2, and therefore we must have G L Z4 % Z4
...
Pick an element of maximum order, say the element 8
...
Next, choose a second element, say a, so that a has order 4 and
a and a2 are not in ͗8͘ 5 {1, 8, 64, 57}
...


222

Groups

Example 1 illustrates how quickly and easily one can write an Abelian
group as a direct product given the orders of the elements of the group
...

EXAMPLE 2 Let G 5 {1, 8, 17, 19, 26, 28, 37, 44, 46, 53, 62,
64, 71, 73, 82, 89, 91, 98, 107, 109, 116, 118, 127, 134} under multiplication modulo 135
...

Consider the element 8
...
(Be sure to mod as you go
...
) But now we
know G
...
At
the same time, |109| 5 2 5 |134| (remember, 134 5 21 mod 135) implies that G is not Z24 (see Theorem 4
...
Thus, G L Z12 % Z2, and G 5
͗8͘ 3 ͗134͘
...
For example, Z4 % Z4 % Z2 % Z9 % Z3 % Z5 would be written
as Z180 % Z12 % Z2 (see Exercise 11)
...
,
ai m21 is in Gi21, and define Gi 5 Gi21 3 ͗ai͘
...

Corollary Existence of Subgroups of Abelian Groups
If m divides the order of a finite Abelian group G, then G has a
subgroup of order m
...
Let us say
that G is an Abelian group of order 72 and we wish to produce a subgroup

223

11 | Fundamental Theorem of Finite Abelian Groups

of order 12
...


Obviously, Z8 % Z9 L Z72 and Z4 % Z2 % Z3 % Z3 L Z12 % Z6 both
have a subgroup of order 12
...
A subgroup of order
12 in Z8 % Z3 % Z3 is given by {(a, b, 0) | a [ {0, 2, 4, 6}, b [ Z3}
...


Proof of the Fundamental Theorem
Because of the length and complexity of the proof of the Fundamental
Theorem of Finite Abelian Groups, we will break it up into a series of
lemmas
...
Then G 5 H 3 K, where H 5 {x [ G | x p 5 e}
and K 5 {x [ G | x m 5 e}
...


PROOF It is an easy exercise to prove that H and K are subgroups of G
(see Exercise 29 in Chapter 3)
...
Since we
have gcd(m, pn) 5 1, there are integers s and t such that 1 5 sm 1 tpn
...
1), x sm [ H and x tp [ K
...
Now suppose that some x [ H > K
...
1, |x| divides both pn and m
...

To prove the second assertion of the lemma, note that pnm 5
|HK| 5 |H||K|/|H > K| 5 |H||K| (see Exercise 7 in the Supplementary
Exercises for Chapters 5–8)
...
5 and
Corollary 2 to Theorem 4
...

Given an Abelian group G with |G| 5 p1 n 1 p2 n 2 ? ? ? pk n k, where the
n
p’s are distinct primes, we let G(pi) denote the set {x [ G | x pi 5 e}
...
Hence, we turn our
attention to groups of prime-power order
...
Then G can be written in the form
͗a͘ 3 K
...
If n 5 1, then G 5
͗a͘ 3 ͗e͘
...
Among all the elements of G, choose
a of maximal order pm
...
We may assume
that G 2 ͗a͘, for otherwise there is nothing to prove
...
We
claim that ͗a͘ > ͗b͘ 5 {e}
...
Say b p 5 ai
...
Thus, ai is not a generator of
͗a͘ and, therefore, by Corollary 3 to Theorem 4
...

This proves that p divides i, so that we can write i 5 pj
...
Consider the element c 5 a2jb
...
Also, c p 5 a2jpb p 5 a2ib p 5 b2pb p 5 e
...
Since
b was chosen to have smallest order such that b o ͗a͘, we conclude
that b also has order p
...

Now consider the factor group G 5 G/͗b͘
...
If |a| , |a| 5 pm, then ap 5 e
...
Thus, |a| 5 |a| 5 pm, and therefore
a is an element of maximal order in G
...
Let K be
the pullback of K under the natural homomorphism from G to G (that
is, K 5 {x [ G | x [ K })
...
For if x [ ͗a͘
> K, then x [ ͗a͘ > K 5 {e} 5 ͗b͘ and x [ ͗a͘ > ͗b͘ 5 {e}
...

Lemma 2 and induction on the order of the group now give the
following
...


Let us pause to determine where we are in our effort to prove the
Fundamental Theorem of Finite Abelian Groups
...
Thus, we have
proved that G is an internal direct product of cyclic groups of primepower order
...

Certainly the groups G( pi) are uniquely determined by G, since they
comprise the elements of G whose orders are powers of pi
...

Lemma 4
Suppose that G is a finite Abelian group of prime-power order
...


PROOF We proceed by induction on |G|
...
Now suppose that the statement is true for all Abelian groups
of order less than |G|
...
5, is a proper subgroup if p
divides |L|
...
p, and n9 is the largest integer j such that |Kj|
...
(This ensures
that our two direct products for G p do not have trivial factors
...
,
m9
...
, m9
...
This follows directly from the facts that |H1||H2| ? ? ?
|Hm9|pm2m9 5 |G| 5 |K1||K2| ? ? ? |Kn9|pn2n9, |Hi| 5 |Ki|, and m9 5 n9
...

JOHN LENNON AND PAUL MCCARTNEY,

“The Ballad of John and Yoko”

1
...

2
...

3
...

4
...
Do the same for the elements of order 4
...
Prove that any Abelian group of order 45 has an element of order 15
...
Show that there are two Abelian groups of order 108 that have exactly one subgroup of order 3
...
Show that there are two Abelian groups of order 108 that have exactly four subgroups of order 3
...
Show that there are two Abelian groups of order 108 that have exactly 13 subgroups of order 3
...
Suppose that G is an Abelian group of order 120 and that G has
exactly three elements of order 2
...

10
...

11
...
, nt,
where ni11 divides ni for i 5 1, 2,
...
(This exercise is referred to in this chapter and in Chapter 22
...
Suppose that the order of some finite Abelian group is divisible by
10
...

13
...

14
...


11 | Fundamental Theorem of Finite Abelian Groups

227

15
...
of order 6?
b
...
of order 42?
d
...
of order pqr, where p, q, and r are distinct primes?
f
...

16
...
n 5 32 and m 5 52?
b
...
n 5 pr and m 5 qr, where p and q are prime?
d
...
n 5 pr and m 5 prq2, where p and q are distinct primes?
17
...
Is it isomorphic to Z4 or Z2 % Z2?
18
...

19
...
Determine the isomorphism class of this group
...
Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|
...

21
...

22
...

23
...

24
...
Express G as an external and an internal direct product of cyclic
groups
...
Let G 5 {1, 7, 43, 49, 51, 57, 93, 99, 101, 107, 143, 149, 151, 157,
193, 199} under multiplication modulo 200
...

26
...
Write G as an external and an internal direct product of cyclic groups of prime-power order
...
Suppose that G is an Abelian group of order 9
...
Suppose that G is an Abelian group of order 16, and in computing
the orders of its elements, you come across an element of order 8
and two elements of order 2
...

29
...
Suppose that there are elements a and b in G such that |a| 5 |b| 5 4 and a2 2 b2
...

30
...

31
...

32
...
What is the isomorphism class of G?
33
...

34
...
Prove that G has order pn,
where p is prime, if and only if the order of every element of G is a
power of p
...
Dirichlet’s Theorem says that, for every pair of relatively prime integers a and b, there are infinitely many primes of the form at 1 b
...

36
...

37
...


Computer Exercises
The purpose of computation is insight, not numbers
...
d
...
edu/~jgallian
1
...
Run the program for n 5 16, 24,
512, 2048, 441000, and 999999
...
This software determines how many integers in a given interval are
the order of exactly one Abelian group, of exactly two Abelian
groups, and so on, up to exactly nine Abelian groups
...
Then from 10001 to 11000
...
Is there
much difference in the results?
3
...
Run the
program for the groups U(32), U(80), and U(65)
...
R
...


Suggested Readings
J
...
Gallian, “Computers in Group Theory,” Mathematics Magazine
49 (1976): 69–73
...

J
...

In this article, the author determines the percentages of integers k between
1 and n, for sufficiently large n, that have exactly one isomorphism class of
Abelian groups of order k, exactly two isomorphism classes of Abelian
groups of order k, and so on, up to 13 isomorphism classes
...
Mackiw, “Computing in Abstract Algebra,” The College Mathematics
Journal 27 (1996): 136–142
...


Suggested Website
To find more information about the development of group theory, visit:

http://www-groups
...
st-and
...
uk/~history/

230

Groups

Supplementary Exercises for Chapters 9–11
Every prospector drills many a dry hole, pulls out his rig, and moves on
...
HESS

True/false questions for Chapters 9–11 are available on the Web at:
http://www
...
umn
...
Suppose that H is a subgroup of G and that each left coset of H in
G is some right coset of H in G
...

2
...

3
...
Prove that diag(G) v G % G if
and only if G is Abelian
...
Let H be any group of rotations in Dn
...

5
...

6
...
Prove that H is a normal subgroup if and
only if, for all a and b in G, ab [ H implies ba [ H
...
The factor group GL(2, R)/SL(2, R) is isomorphic to some very
familiar group
...
Let k be a divisor of n
...
What is the group?
9
...

a
...

b
...

c
...

d
...
Prove that D4/Z(D4) is isomorphic to Z2 % Z2
...
Prove that Q/Z under addition is an infinite group in which every
element has finite order
...
Show that the intersection of any collection of normal subgroups of
a group is a normal subgroup
...
Let n
...
If the set H 5
{x [ G| |x| 5 n} together with the identity forms a subgroup of
G, prove that it is a normal subgroup of G
...
Give an example
of a group G and a prime n for which the set H together with the
identity is not a subgroup
...
Show that Q/Z has a unique subgroup of order n for each positive
integer n
...
If H and K are normal Abelian subgroups of a group and H > K 5
{e}, prove that HK is Abelian
...
Let G be a group of odd order
...

17
...
If |G| 5 60
and orbG(5) 5 {1, 5}, prove that stabG(5) is normal in G
...
Suppose that G 5 H 3 K and that N is a normal subgroup of H
...

19
...

20
...

21
...
Prove that S4/H is isomorphic to S3
...
Suppose that f is a homomorphism of U(36), Ker f 5 {1, 13, 25},
and f(5) 5 17
...

23
...
How many elements of order 2
does Dn/Z(Dn) have? How many elements are in the subgroup
͗R360/n͘/Z(Dn)? How do these numbers compare with the number
of elements of order 2 in Dm?
24
...
Show that H # Z(G)
...
Let G be an Abelian group and let n be a positive integer
...
Prove that G/Gn is isomorphic
to Gn
...
Let R1 denote the multiplicative group of positive reals and let T 5
{a 1 bi [ C| a2 1 b2 5 1} be the multiplicative group of complex
numbers of norm 1
...


232

Groups

27
...
If p2
...

28
...
Show
that H is a subgroup of G
...
To which
familiar group is G/H isomorphic?
29
...
Prove that every element of order n in
Q/Z is contained in ͗1/n 1 Z͘
...
(1997 Putnam Competition) Let G be a group and let f : G S G be
a function such that
f(g1)f(g2)f(g3) 5 f(h1)f(h2)f(h3)

31
...

33
...

35
...


37
...


whenever g1g2g3 5 e 5 h1h2h3
...

Prove that every homomorphism from Z % Z into Z has the form
(x, y) S ax 1 by, where a and b are integers
...

Prove that Q/Z is not isomorphic to a proper subgroup of itself
...

Show that any group with more than two elements has an automorphism other than the identity mapping
...
Prove that Q under addition has
no maximal subgroups
...
Determine
whether G>H is isomorphic to Z4 or Z2 % Z2
...

Prove that N is normal in D8
...


11 | Supplementary Exercises for Chapters 9–11

39
...
Show that H is a subgroup of G
...

40
...

41
...
Prove that if H is a characteristic subgroup of K, and K is a normal subgroup of G, then H is a normal
subgroup of G
...
hmco
...

EDMUND BURKE, On a Regicide Peace

Motivation and Definition
Many sets are naturally endowed with two binary operations: addition
and multiplication
...
When considering these sets as groups, we simply used addition
and ignored multiplication
...
One abstract concept that does this is the concept of a ring
...

Definition Ring

A ring R is a set with two binary operations, addition (denoted by
a 1 b) and multiplication (denoted by ab), such that for all a, b, c in R:
1
...

2
...

3
...
That is, there is an element 0 in R
such that a 1 0 5 a for all a in R
...
There is an element 2a in R such that a 1 (2a) 5 0
...
a(bc) 5 (ab)c
...
a(b 1 c) 5 ab 1 ac and (b 1 c) a 5 ba 1 ca
...

Note that multiplication need not be commutative
...
Also, a ring need not have an identity
†The

term ring was first applied in 1897 by the German mathematician David Hilbert
(1862–1943)
...
A unity (or identity) in a ring is a nonzero element
that is an identity under multiplication
...
When it
does, we say that it is a unit of the ring
...

The following terminology and notation are convenient
...
If a does not divide b, we write a B b
...
When dealing with rings, this notation can cause
confusion, since we also use juxtaposition for the ring multiplication
...

For an abstraction to be worthy of study, it must have many diverse
concrete realizations
...


Examples of Rings
EXAMPLE 1 The set Z of integers under ordinary addition and
multiplication is a commutative ring with unity 1
...

EXAMPLE 2 The set Zn 5 {0, 1,
...
The set of
units is U(n)
...

EXAMPLE 4 The set M2(Z) of 2 3 2 matrices with integer entries
is a noncommutative ring with unity c

1 0
d
...

EXAMPLE 6 The set of all continuous real-valued functions of a
real variable whose graphs pass through the point (1, 0) is a commutative ring without unity under the operations of pointwise addition and

12 | Introduction to Rings

239

multiplication [that is, the operations ( f 1 g)(a) 5 f(a) 1 g(a) and
( fg)(a) 5 f(a)g(a)]
...
, Rn be rings
...
Let
R1 % R2 % ? ? ? % Rn 5 {(a1, a2,
...
, an) 1 (b1, b2,
...
, an 1 bn)
and
(a1, a2,
...
, bn) 5 (a1b1, a2b2,
...

This ring is called the direct sum of R1, R2,
...


Properties of Rings
Our first theorem shows how the operations of addition and multiplication intertwine
...

Theorem 12
...
Then
1
...

3
...


a0 5 0a 5 0
...

(2a)(2b) 5 ab
...


Furthermore, if R has a unity element 1, then
5
...

6
...


PROOF We will prove rules 1 and 2 and leave the rest as easy exercises
(see Exercise 11)
...
1, we
need only “play off ” the distributive property against the fact that R is a
group under addition with additive identity 0
...
Clearly,
0 1 a0 5 a0 5 a(0 1 0) 5 a0 1 a0
...
Similarly, 0a 5 0
...

The reason for this we need not discuss
...
H
...
So, adding 2(ab) to both sides yields a(2b) 5 2(ab)
...

Recall that in the case of groups, the identity and inverses are unique
...
The proofs
are identical to the ones given for groups and therefore are omitted
...
2 Uniqueness of the Unity and Inverses
If a ring has a unity, it is unique
...


Many students have the mistaken tendency to treat a ring as if it were
a group under multiplication
...
The two most common errors are
the assumptions that ring elements have multiplicative inverses—they
need not—and that a ring has a multiplicative identity—it need not
...
Similarly, if a2 5 a, we cannot conclude that a 5 0
or 1 (as is the case with real numbers)
...
There is an important class of
rings wherein multiplicative identities exist and for which multiplicative cancellation holds
...


Subrings
In our study of groups, subgroups played a crucial role
...
Nevertheless, subrings are important
...


Just as was the case for subgroups, there is a simple test for subrings
...
3 Subring Test
A nonempty subset S of a ring R is a subring if S is closed under
subtraction and multiplication—that is, if a 2 b and ab are in S
whenever a and b are in S
...
1) that S
is an Abelian group under addition
...
Thus, the only condition remaining to be checked
is that multiplication is a binary operation on S
...

We leave it to the student to confirm that each of the following examples is a subring
...
{0} is called the
trivial subring of R
...
Note that although 1 is the unity in Z6, 4 is the unity in
{0, 2, 4}
...
}
is a subring of the integers Z
...

EXAMPLE 12 Let R be the ring of all real-valued functions of a single real variable under pointwise addition and multiplication
...

EXAMPLE 13 The set
ec

a 0
d 0 a, b [ Z f
0 b

of diagonal matrices is a subring of the ring of all 2 3 2 matrices
over Z
...
In such a diagram, any ring
is a subring of all the rings that it is connected to by one or more upward lines
...
1 shows the relationships among some of the
rings we have already discussed
...
1 Partial subring lattice diagram of C

In the next several chapters, we will see that many of the fundamental concepts of group theory can be naturally extended to rings
...


Exercises
There is no substitute for hard work
...
Give an example of a finite noncommutative ring
...

2
...
Find it
...
Give an example of a subset of a ring that is a subgroup under
addition but not a subring
...
Show, by example, that for fixed nonzero elements a and b in a
ring, the equation ax 5 b can have more than one solution
...
Prove Theorem 12
...

6
...

a
...

b
...

c
...

Is the n you found prime?

12 | Introduction to Rings

243

7
...

8
...

9
...

10
...

11
...
1
...
Let a, b, and c be elements of a commutative ring, and suppose that
a is a unit
...

13
...

14
...
Prove that
m ? (ab) 5 (m ? a)b 5 a(m ? b)
...
Show that if m and n are integers and a and b are elements from a
ring, then (m ? a)(n ? b) 5 (mn) ? (ab)
...
)
16
...

17
...

18
...
Let S 5 {x [ R | ax 5 0}
...

19
...
The center of R is the set {x [ R | ax 5 xa for all
a in R}
...

20
...

21
...
, Rn are rings that contain nonzero elements
...

22
...
Prove that U(R) is a group under the multiplication of
R
...
)
23
...

24
...
, Rn are commutative rings with unity, show that
U(R1 % R2 % ? ? ? % Rn) 5 U(R1) % U(R2) % ? ? ? % U(Rn)
...
Determine U(Z[x])
...
)
26
...

27
...

28
...


244

Rings

29
...
If
a is a unit of R and b2 5 0, show that a 1 b is a unit of R
...
Suppose that there is an integer n
...
If m is a positive integer and am 5 0 for some a, show
that a 5 0
...
Give an example of ring elements a and b with the properties that
ab 5 0 but ba 2 0
...
Let n be an integer greater than 1
...

33
...
Prove that
6x 5 0 for all x in R
...
Suppose that a belongs to a ring and a4 5 a2
...

35
...
1 such that an 5 a for all a in Z6
...
Show that no such n exists for Zm when m is divisible by the
square of some prime
...
Let m and n be positive integers and let k be the least common multiple of m and n
...

37
...

38
...
Suppose that R is a ring with unity 1 and a is an element of R such
that a2 5 1
...
Prove that S is a subring of R
...
Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5
a
a1b
ec
d ` a, b [ Z f
...


42
...

44
...

Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5
a
a2b
ec
d ` a, b [ Z f
...

a a
Let R 5 e c
d ` a, b [ Z f
...

Let R 5 Z % Z % Z and S 5 {(a, b, c) [ R | a 1 b 5 c}
...

Suppose that there is a positive even integer n such that an 5 a for
all elements a of some ring
...


12 | Introduction to Rings

245

45
...
Show that S 5 {n ? 1| n [ Z} is a subring of R
...
Show that 2Z < 3Z is not a subring of Z
...
Determine the smallest subring of Q that contains 1/2
...
)
48
...

49
...
Prove that a2 2 b2 5 (a 1 b)(a 2 b) for all a, b in
R if and only if R is commutative
...
Suppose that R is a ring and that a2 5 a for all a in R
...
[A ring in which a2 5 a for all a is called a
Boolean ring, in honor of the English mathematician George Boole
(1815–1864)
...
Give an example of a Boolean ring with four elements
...


Computer Exercises
Theory is the general; experiments are the soldiers
...
d
...
edu/~jgallian
1
...

Run the software for all odd primes up to 37
...

2
...
This software finds the group of units of this ring
and the order of each element of the group
...
Is the group of units cyclic for these cases? Try
to guess a formula for the order of the group of units of Zn[i] as a
function of n when n is a prime and n mod 4 5 3
...
Are the groups cyclic? Try to guess a formula for
the order when n 5 3k
...

Is the group cyclic for these cases? What is the largest order of any
element in the group? Try to guess a formula for the order of the
group of units of Zn[i] as a function of n when n is a prime and
n mod 4 5 1
...
On the basis of the orders of the elements
of the group of units, try to guess the isomorphism class of the
group
...
Is this group cyclic? Based on
the number of elements in this group and the orders of the elements,
try to guess the isomorphism class of the group
...
This software determines the isomorphism class of the group of
units of Zn[i]
...
Make
a conjecture
...

Make a conjecture
...
Make
a conjecture
...
Make a conjecture
...
Make a conjecture
...
Make a conjecture
...
Make a conjecture
...
Make
a conjecture about the case where n 5 pk where p is a prime and
p mod 4 5 3
...
This software determines the order of the group of units in the ring
of 2 3 2 matrices over Zn (that is, the group GL(2, Zn)) and the subgroup SL(2, Zn)
...

What relationship do you see between the order of GL(2, Zn) and the
order of SL(2, Zn) in these cases? Run the program for n 516, 27,
25, and 49
...
Run the program for n 5 32
...
) How do the orders of the two groups change each
time you increase the power of 2 by 1? Run the program for n 5 27
...
How do the orders
of the two groups change when you increase the power of 5 by 1?
Make a conjecture about the relationship between |SL(2, Zp i)| and
|SL(2, Zp i11)|
...
Run the program for n 5 12, 15, 20,
21, and 30
...
(Notice that when you run the program for st, the
table shows the values for st, s, and t
...
In the ring Zn, this software finds the number of solutions to the
equation x2 5 21
...


12 | Introduction to Rings

247

How does the answer depend on the prime? Make a conjecture about
the number of solutions when n is a prime greater than 2
...
Make a conjecture about the number of solutions when n is the square of a
prime greater than 2
...
Make a conjecture about the number of solutions
when n is any power of an odd prime
...
Make a conjecture about the number of solutions
when n is a power of 2
...
Make a conjecture about the number of solutions when n is a
multiple of 4
...
Make a conjecture about the
number of solutions when n 5 pq or n 5 2pq where p and q are odd
primes
...

Make a conjecture about the number of solutions when n 5 pqr or
n 5 2pqr where p, q, and r are odd primes
...
This software determines the number of solutions to the equation
X2 5 2I where X is a 2 3 2 matrix with entries from Zn and I is the
identity
...
Make a conjecture about the
number of solutions when n 5 2k where k
...
Run the program
for n 5 3, 11, 19, 23, and 31
...
Run the program for n 5 27 and 49
...
Run the program for n 5 5, 13, 17, 29, and 37
...
Run the program for n 5 6, 10, 14, 22, 15, 21, 33, 39,
30, 42
...
B
...

In this elementary paper, it is shown that there exists a noncommutative ring
of order m
...


I
...
Herstein
A whole generation of textbooks and an entire
generation of mathematicians, myself included,
have been profoundly influenced by that text
[Herstein’s Topics in Algebra]
...
N
...
His family moved to Canada
when he was seven
...
During his
school years he played football, hockey, golf,
tennis, and pool
...

Herstein received a B
...
degree from the
University of Manitoba, an M
...
from the
University of Toronto, and, in 1948, a Ph
...

degree from Indiana University under the supervision of Max Zorn
...

Herstein wrote more than 100 research
papers and a dozen books
...
His textbook Topics in Algebra,
first published in 1964, dominated the field
for 20 years and has become a classic
...
He had 30 Ph
...
students, and
traveled and lectured widely
...
He spoke Italian, Hebrew, Polish, and
Portuguese
...

To find more information about Herstein,
visit:
http://www-groups
...
stand
...
uk/~history/

13 Integral Domains

Don’t just read it! Ask your own questions, look for your own examples,
discover your own proofs
...
A
ring is not the appropriate abstraction of the integers, however, for too
much is lost in the process
...
In this chapter, we introduce integral domains—a particular
class of rings that have all three of these properties
...

Definition Zero-Divisors

A zero-divisor is a nonzero element a of a commutative ring R such
that there is a nonzero element b [ R with ab 5 0
...


Thus, in an integral domain, a product is 0 only when one of the
factors is 0; that is, ab 5 0 only when a 5 0 or b 5 0
...
For each example, the student should verify the
assertion made
...


249

250

Rings

EXAMPLE 2 The ring of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z}
is an integral domain
...

EXAMPLE 4 The ring Z["2 ] 5 {a 1 b"2 | a, b [ Z} is an integral
domain
...

EXAMPLE 6 The ring Zn of integers modulo n is not an integral domain when n is not prime
...

EXAMPLE 8 Z % Z is not an integral domain
...

This property is cancellation
...
1 Cancellation
Let a, b, and c belong to an integral domain
...


PROOF From ab 5 ac, we have a(b 2 c) 5 0
...

Many authors prefer to define integral domains by the cancellation
property—that is, as commutative rings with unity in which the cancellation property holds
...


Fields
In many applications, a particular kind of integral domain called a field
is necessary
...


13 | Integral Domains

251

To verify that every field is an integral domain, observe that if a and
b belong to a field with a 2 0 and ab 5 0, we can multiply both sides
of the last expression by a21 to obtain b 5 0
...
With this in
mind, a field can be thought of as simply an algebraic system that
is closed under addition, subtraction, multiplication, and division
(except by 0)
...
The abstract theory
of fields was initiated by Heinrich Weber in 1893
...
Theorem 13
...

Theorem 13
...


PROOF Let D be a finite integral domain with unity 1
...
We must show that a is a unit
...
Now consider the following
sequence of elements of D: a, a2, a3,
...
j and ai 5 a j
...
Since a 2 1, we know that i 2 j
...

Corollary Zp Is a Field
For every prime p, Zp, the ring of integers modulo p, is a field
...
2, we need only prove that Zp has
no zero-divisors
...
Then ab 5 pk
for some integer k
...
Thus, in Zp, a 5 0 or b 5 0
...
In Chapter 22, we will describe
how all finite fields can be constructed
...

EXAMPLE 9 Field with Nine Elements
Let
Z3[i] 5 {a 1 bi | a, b [ Z3}
5 {0, 1, 2, i, 1 1 i, 2 1 i, 2i, 1 1 2i, 2 1 2i},

252

Rings

where i2 5 21
...
Elements are added and multiplied as in the complex numbers, except that
the coefficients are reduced modulo 3
...
Table 13
...

Table 13
...
It is easy to see
that Q["2] is a ring
...
To verify that Q["2] is a field, we must show that 1/(a 1 b"2)
can be written in the form c 1 d"2
...
” Specifically,
1
1
a 2 b"2
a
b
5
5 2
"2
...
)

Characteristic of a Ring
Note that for any element x in Z3[i], we have 3x 5 x 1 x 1 x 5 0, since
addition is done modulo 3
...
This observation motivates
the following definition
...
If no such integer exists, we say that R has characteristic 0
...


Thus, the ring of integers has characteristic 0, and Zn has characteristic n
...
Indeed, the

13 | Preliminaries

253

ring Z2[x] of all polynomials with coefficients in Z2 has characteristic 2
...
)
When a ring has a unity, the task of determining the characteristic is
simplified by Theorem 13
...

Theorem 13
...
If 1 has infinite order under addition,
then the characteristic of R is 0
...


PROOF If 1 has infinite order, then there is no positive integer n such
that n ? 1 5 0, so R has characteristic 0
...
Then n ? 1 5 0, and n is the least positive integer with this
property
...

Thus, R has characteristic n
...

Theorem 13
...


PROOF By Theorem 13
...
Suppose that 1 has order n and that n 5 st,
where 1 # s, t # n
...

So, s ? 1 5 0 or t ? 1 5 0
...
Thus, n is
prime
...
The existence of zero-divisors in a ring causes unusual
results when one is finding zeros of polynomials with coefficients in
the ring
...
In the
integers, we could find all solutions by factoring
x2 2 4x 1 3 5 (x 2 3)(x 2 1) 5 0
and setting each factor equal to 0
...
In Z12, there are many pairs of
nonzero elements whose products are 0: 2 ? 6 5 0, 3 ? 4 5 0, 4 ? 6 5 0,
6 ? 8 5 0, and so on
...
Observe that we
can find all solutions of x2 2 4x 1 3 5 0 over Z11 or Z13, say, by setting
the two factors x 2 3 and x 2 1 equal to 0
...
Perhaps this will
convince you that integral domains are particularly advantageous rings
...
2 gives a summary of some of the rings we have introduced and
their properties
...
2 Summary of Rings and Their Properties
Ring
Z
Zn, n composite
Zp, p prime
Z[x]
nZ, n
...
But it so happens that you go on worrying
away at a problem in science and it seems to get tired, and lies down and
lets you catch it
...
Verify that Examples 1 through 8 are as claimed
...
Which of Examples 1 through 5 are fields?
3
...

4
...
Can you see a relationship between the
zero-divisors of Z20 and the units of Z20?
5
...

6
...

7
...
Prove that every
nonzero element of R is either a zero-divisor or a unit
...
Describe all zero-divisors and units of Z % Q % Z
...
Let d be an integer
...
(This exercise is referred to in Chapter 18
...
In Z7, give a reasonable interpretation for the expressions 1/2,
22/3, "23, and 21/6
...
Give an example of a commutative ring without zero-divisors that
is not an integral domain
...
Find two elements a and b in a ring such that both a and b are zerodivisors, a 1 b 2 0, and a 1 b is not a zero-divisor
...
Let a belong to a ring R with unity and suppose that an 5 0 for
some positive integer n
...
)
Prove that 1 2 a has a multiplicative inverse in R
...
]
14
...

15
...

16
...
Prove that the
only idempotents in an integral domain are 0 and 1
...
He
remains the youngest person ever to receive the Nobel Prize
...
Let a and b be idempotents in a commutative ring
...

18
...

19
...

20
...

21
...

22
...

23
...

24
...

a
...

b
...

c
...

25
...
Prove that K is a subfield of F if, for any
a, b (b 2 0) in K, a 2 b and ab21 belong to K
...
Let d be a positive integer
...

27
...
If the product of any pair of nonzero
elements of R is nonzero, prove that ab 5 1 implies ba 5 1
...
Let R 5 {0, 2, 4, 6, 8} under addition and multiplication modulo
10
...

29
...
Let D be an integral domain with unity 1
...
Show that P is contained in every subdomain
of D
...
Prove that there is no integral domain with exactly six elements
...

31
...
Prove that char F 5 2
...
Determine all elements of an integral domain that are their own inverses under multiplication
...
Characterize those integral domains for which 1 is the only element that is its own multiplicative inverse
...
Determine all integers n
...

35
...

a
...

b
...

36
...

37
...

38
...
Is this ring a field? Is it an integral domain?
39
...
Is it isomorphic to Z8, Z4 % Z2, or Z2 % Z2 % Z2?
40
...
For any
positive integer k and any prime p, determine a necessary and sufficient condition for Zp["k] 5 {a 1 b"k | a, b [ Zp} to be a field
...
Show that a finite commutative ring with no zero-divisors and at
least two elements has a unity
...
Suppose that a and b belong to a commutative ring and ab is a
zero-divisor
...

43
...
Show
that all the nonzero elements of R have the same additive order
...
Suppose that R is a commutative ring without zero-divisors
...

45
...

a
...

n
n
n
b
...

c
...
(This exercise is referred to in Chapter 20
...
Let R be a commutative ring with unity 1 and prime characteristic
...

47
...
Hint: Use
facts about finite Abelian groups
...
)
48
...

49
...
Explain why these two rings have the same characteristic
...
Let R be a ring with m elements
...

51
...

52
...
(See Example 9
...
Consider the equation x2 2 5x 1 6 5 0
...
How many solutions does this equation have in Z7?
b
...

c
...

d
...

54
...

55
...
(Recall that n ? 1 means the sum 1 1 1 1 ? ? ? 1 1 with
n terms
...
In a commutative ring of characteristic 2, prove that the idempotents form a subring
...
Describe the smallest subfield of the field of real numbers that contains "2
...
)
58
...
Prove that xn21 5 1 for all
nonzero x in F
...
Let F be a field of prime characteristic p
...

60
...
Prove that a 5 0 and b 5 0
...

61
...

Show that (x 1 y)3 2 x3 1 y3 for some x and y in F
...
Suppose that F is a field with characteristic not 2, and that the
nonzero elements of F form a cyclic group under multiplication
...

63
...
If x is a unit in D, show that f(x) 5 1
...
Let F be a field of order 32
...

65
...
Show that for every
element a [ F , 5a 5 2a
...

ARNOLD ROSS

Software for the computer exercises in this chapter is available at the
website:
http://www
...
umn
...
This software lists the idempotents (see Exercise 16 for the definition) in Zn
...
Use these data
to make conjectures about the number of idempotents in Zn as a
function of n
...
This software lists the nilpotent elements (see Exercise 13 for the
definition) in Zn
...
Use
these data to make conjectures about the number of nilpotent elements in Zn as a function of n
...
This software determines which rings of the form Zp[i] are fields
...
From these data, make a
conjecture about the form of the primes that yield a field
...
This software finds the idempotents in Zn[i] 5 {a 1 bi | a, b [ Zn}
(Gaussian integers modulo n)
...
Make a conjecture about the number of idempotents when n 5 2k
...
What do these values of
n have in common? Make a conjecture about the number of idempotents for these n
...

What do these values of n have in common? Make a conjecture about
the number of idempotents for these n
...
This software finds the nilpotent elements in Zn[i] 5 {a 1 bi |
a, b [ Zn}
...
Make a conjecture about the number of nilpotent elements when n 5 2k
...
What do these values of n
have in common? Make a conjecture about the number of nilpotent
elements for these n
...
Do you need to
revise the conjecture you made based on n 5 3, 5, 7, 11, 13, and 17?
Run the software for n 5 9, 25, and 49
...
Run the program for n 5 81
...
Run the program for n 5 27
...
On the basis of all of your data for this exercise, make a single conjecture in the case that n 5 pk where p is
any prime
...
Make a conjecture
...
Make a conjecture
...
On
the basis of all your data for this exercise, make a single conjecture
that covers all integers n
...

6
...
Use the software to formulate and test conjectures about the
number of zero-divisors in Zn[i] based on various conditions of n
...

In this article, the author uses properties of logarithms and exponents
to define recursively an infinite family of fields starting with the real
numbers
...
A
...

Here it is shown that a ring has nonzero characteristic n if and only
if n is the maximum of the orders of the elements of R
...
Robin McLean, “Groups in Modular Arithmetic,” The Mathematical
Gazette 62 (1978): 94–104
...
It shows how to construct
groups of integers in which the identity is not obvious; for example, 1977
is the identity of the group {1977, 5931} under multiplication modulo
7908
...

Citation for the Steele Prize
for Lifetime Achievement

NATHAN JACOBSON was born on September 8,
1910, in Warsaw, Poland
...
He
received a B
...
degree from the University of
Alabama in 1930 and a Ph
...
from Princeton
in 1934
...

Jacobson’s principal contributions to algebra were in the areas of rings, Lie algebras,
and Jordan algebras
...
He
was the author of nine books and numerous
articles, and he had 33 Ph
...
students
...
Among his many honors
were the presidency of the American
Mathematical Society, memberships in the
National Academy of Sciences and the
American Academy of Arts and Sciences, a
Guggenheim Fellowship, and an honorary
degree from the University of Chicago
...

To find more information about Jacobson,
visit:
http://www-groups
...
st-and
...
uk/~history/

261

14 Ideals and Factor Rings

The secret of science is to ask the right questions, and it is the choice of
problem more than anything else that marks the man of genius in the
scientific world
...
P
...
In this chapter, we introduce the analogous
concepts for rings—ideals and factor rings
...


So, a subring A of a ring R is an ideal of R if A “absorbs” elements
from R—that is, if rA 5 {ra| a [ A} # A and Ar 5 {ar| a [ A} # A
for all r [ R
...
In practice, one identifies ideals with the following test, which is
an immediate consequence of the definition of ideal and the subring
test given in Theorem 12
...

Theorem 14
...
a 2 b [ A whenever a, b [ A
...
ra and ar are in A whenever a [ A and r [ R
...
The ideal {0}
is called the trivial ideal
...
} is an ideal of Z
...

The set ͗a͘ 5 {ra | r [ R} is an ideal of R called the principal ideal
generated by a
...
However, the intended meaning
will always be clear from the context
...

EXAMPLE 4 Let R[x] denote the set of all polynomials with real coefficients and let A denote the subset of all polynomials with constant
term 0
...

EXAMPLE 5 Let R be a commutative ring with unity and let a1,
a2,
...
Then I 5 ͗a1, a2,
...
, an
...

EXAMPLE 6 Let Z[x] denote the ring of all polynomials with integer coefficients and let I be the subset of Z[x] of all polynomials with
even constant terms
...

EXAMPLE 7 Let R be the ring of all real-valued functions of a real
variable
...


Factor Rings
Let R be a ring and let A be an ideal of R
...
The natural question at this point is: How may
we form a ring of this group of cosets? The addition is already taken care
of, and, by analogy with groups of cosets, we define the product of two
cosets of s 1 A and t 1 A as st 1 A
...


264

Rings

Theorem 14
...
The set of cosets {r 1 A |
r [ R} is a ring under the operations (s 1 A) 1 (t 1 A) 5 s 1 t 1 A
and (s 1 A)(t 1 A) 5 st 1 A if and only if A is an ideal of R
...

Once we know that multiplication is indeed a binary operation on the
cosets, it is trivial to check that the multiplication is associative and
that multiplication is distributive over addition
...
To do this, let us suppose that A is an ideal and let s 1 A 5
s9 1 A and t 1 A 5 t9 1 A
...

Well, by definition, s 5 s9 1 a and t 5 t9 1 b, where a and b belong
to A
...
Thus, multiplication is well defined
when A is an ideal
...
Then there exist elements a [ A and r [ R such that ar o A
or ra o A
...
Consider the elements a 1 A 5
0 1 A and r 1 A
...
Since ar 1 A 2 A, the multiplication is not
well defined and the set of cosets is not a ring
...

EXAMPLE 8 Z/4Z 5 {0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z}
...

(2 1 4Z) 1 (3 1 4Z) 5 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z,
(2 1 4Z)(3 1 4Z) 5 6 1 4Z 5 2 1 4 1 4Z 5 2 1 4Z
...


14 | Ideals and Factor Rings

265

EXAMPLE 9 2Z/6Z 5 {0 1 6Z, 2 1 6Z, 4 1 6Z}
...
For example, (4 1 6Z) 1
(4 1 6Z) 5 2 1 6Z and (4 1 6Z)(4 1 6Z) 5 4 1 6Z
...

a1 a2
d ` a i P Z f and let I be the
a3 a4
subset of R consisting of matrices with even entries
...
Consider the factor
ring R/I
...

r3 r4
An example illustrates the typical situation
...

4 24
1 1
The general case is left to the reader (Exercise 23)
...
What does this ring look like? Of course, the elements
of R have the form a 1 bi 1 ͗2 2 i͘, where a and b are integers, but the
important question is: What do the distinct cosets look like? The fact
that 2 2 i 1 ͗2 2 i͘ 5 0 1 ͗2 2 i͘ means that when dealing with coset
representatives, we may treat 2 2 i as equivalent to 0, so that 2 5 i
...

Similarly, all the elements of R can be written in the form a 1 ͗2 2 i͘,
where a is an integer
...

Thus, the coset 3 1 4i 1 ͗2 2 i͘ 5 11 1 ͗2 2 i͘ 5 1 1 5 1 5 1 ͗2 2 i͘ 5
1 1 ͗2 2 i͘
...
Is any further reduction possible? To demonstrate
that there is not, we will show that these five cosets are distinct
...
Since 5(1 1 ͗2 2 i͘) 5
5 1 ͗2 2 i͘ 5 0 1 ͗2 2 i͘, 1 1 ͗2 2 i͘ has order 1 or 5
...
Thus, 1 5
(2 2 i) (a 1 bi) 5 2a 1 b 1 (2a 1 2b)i for some integers a and b
...
It should be
clear that the ring R is essentially the same as the field Z5
...

Then
R[x]/͗x2 1 1͘ 5 {g(x) 1 ͗x2 1 1͘ | g(x) [ R[x]}
5 {ax 1 b 1 ͗x2 1 1͘ | a, b [ R}
...
In
particular, r(x) 5 0 or the degree of r(x) is less than 2, so that r(x) 5
ax 1 b for some a and b in R
...

How is multiplication done? Since
x2 1 1 1 ͗x2 1 1͘ 5 0 1 ͗x2 1 1͘,
one should think of x2 1 1 as 0 or, equivalently, as x2 5 21
...

In view of the fact that the elements of this ring have the form ax 1
b 1 ͗x2 1 1͘, where x2 1 ͗x2 1 1͘ 5 21 1 ͗x2 1 1͘, it is perhaps not
surprising that this ring turns out to be algebraically the same ring as
the ring of complex numbers
...

Examples 11 and 12 illustrate one of the most important applications of factor rings—the construction of rings with highly desirable
properties
...


14 | Ideals and Factor Rings

267

Prime Ideals and Maximal Ideals
Definition Prime Ideal, Maximal Ideal

A prime ideal A of a commutative ring R is a proper ideal of R such
that a, b [ R and ab [ A imply a [ A or b [ A
...


So, the only ideal that properly contains a maximal ideal is the entire ring
...

EXAMPLE 13 Let n be an integer greater than 1
...

({0} is also a prime ideal of Z
...
1) shows that
only ͗2͘ and ͗3͘ are maximal ideals
...
To see this, assume that A is an ideal of R[x] that properly contains ͗x2 1 1͘
...
[This is the constant polynomial h(x) 5 c for all x
...
To this end, let
f(x) [ A, but f(x) o ͗x2 1 1͘
...
It follows that
r(x) 5 ax 1 b, where a and b are not both 0, and
ax 1 b 5 r(x) 5 f(x) 2 q(x)(x2 1 1) [ A
...
1

268

Rings

Thus,
a2x2 2 b2 5 (ax 1 b)(ax 2 b) [ A

and

a2(x2 1 1) [ A
...

EXAMPLE 16 The ideal ͗x2 1 1͘ is not prime in Z2[x], since it contains (x 1 1)2 5 x2 1 2x 1 1 5 x2 1 1 but does not contain x 1 1
...

Theorem 14
...

Then R/A is an integral domain if and only if A is prime
...
Then
(a 1 A)(b 1 A) 5 ab 1 A 5 A, the zero element of the ring R/A
...
Hence,
A is prime
...
Thus, our task is
simply to show that when A is prime, R/A has no zero-divisors
...
Then ab [ A
and, therefore, a [ A or b [ A
...

For maximal ideals, we can do even better
...
4 R/A Is a Field If and Only If A Is Maximal
Let R be a commutative ring with unity and let A be an ideal of R
...


PROOF Suppose that R/A is a field and B is an ideal of R that properly
contains A
...
Then b 1 A is a nonzero element
of R/A and, therefore, there exists an element c 1 A such that
(b 1 A) ? (c 1 A) 5 1 1 A, the multiplicative identity of R/A
...
Because
1 1 A 5 (b 1 A)(c 1 A) 5 bc 1 A,

14 | Ideals and Factor Rings

269

we have 1 2 bc [ A , B
...
By Exercise 15,
B 5 R
...

Now suppose that A is maximal and let b [ R but b o A
...
(All other properties
for a field follow trivially
...
This
is an ideal of R that properly contains A (Exercise 25)
...
Thus, 1 [ B, say, 1 5 bc 1 a9, where a9 [ A
...

When a commutative ring has a unity, it follows from Theorems
14
...
4 that a maximal ideal is a prime ideal
...

EXAMPLE 17 The ideal ͗x͘ is a prime ideal in Z[x] but not a maximal ideal in Z[x]
...
Thus, if g(x)h(x) [ ͗x͘,
then g(0)h(0) 5 0
...

To see that ͗x͘ is not maximal, we simply note that ͗x͘ , ͗x, 2͘ ,
Z[x] (see Exercise 37)
...
Verify that the set defined in Example 3 is an ideal
...
Verify that the set A in Example 4 is an ideal and that A 5 ͗x͘
...
Verify that the set I in Example 5 is an ideal and that if J is any
ideal of R that contains a1, a2,
...
(Hence, ͗a1,
a2,
...
, an
...
Find a subring of Z % Z that is not an ideal of Z % Z
...
Let S 5 {a 1 bi | a, b [ Z, b is even}
...

6
...
Z8,
b
...
Z12,
d
...

7
...
Show that aR 5 {ar | r [ R} is
an ideal of R
...

8
...


270

Rings

9
...
(This exercise is referred to in this
chapter
...
If A and B are ideals of a ring, show that the sum of A and B, A 1 B 5
{a 1 b | a [ A, b [ B}, is an ideal
...
In the ring of integers, find a positive integer a such that
a
...
͗a͘ 5 ͗6͘ 1 ͗8͘,
c
...

12
...

13
...
͗a͘ 5 ͗3͗͘4͘,
b
...
͗a͘ 5 ͗m͗͘n͘
...
Let A and B be ideals of a ring
...

15
...

(This exercise is referred to in this chapter
...
If A and B are ideals of a commutative ring R with unity and A 1 B 5 R,
show that A > B 5 AB
...
If an ideal I of a ring R contains a unit, show that I 5 R
...
Suppose that in the ring Z the ideal ͗35͘ is a proper ideal of J and J
is a proper ideal of I
...
Give an example of a ring that has exactly two maximal ideals
...
Suppose that R is a commutative ring and |R| 5 30
...

21
...
Prove that I is an ideal
of R
...
Let I 5 ͗2͘
...

23
...

24
...

25
...
4
is an ideal of R
...
)
26
...

27
...


14 | Ideals and Factor Rings

271

28
...

29
...
Prove that I 5 ͗x͘
...
)
30
...

Generalize
...

31
...
Show that
A 5 {f [ R | f(0) 5 0} is a maximal ideal of R
...
Let R 5 Z8 % Z30
...

33
...

34
...
Prove that I is not a maximal ideal
...
In Z % Z, let I 5 {(a, 0) | a [ Z}
...

36
...
Prove that the factor ring
R/I is commutative if and only if rs 2 sr [ I for all r and s in R
...
In Z[x], let I 5 { f(x) [ Z[x] | f (0) is an even integer}
...
Is I a prime ideal of Z[x]? Is I a maximal ideal? How
many elements does Z[x]/I have? (This exercise is referred to in
this chapter
...
Prove that I 5 ͗2 1 2i͘ is not a prime ideal of Z[i]
...
In Z5[x], let I 5 ͗x2 1 x 1 2͘
...

40
...
Show that Ip 5 {r [ R |
additive order of r is a power of p} is an ideal of R
...
An integral domain D is called a principal ideal domain if every
ideal of D has the form ͗a͘ 5 {ad | d [ D} for some a in D
...
(This exercise is referred to in
Chapter 18
...
Let R 5 e c

a
0

b
r
d 0 a, b, d [ Z6 and S 5 e c
d
0

s
d 0 r, s, t [ Z, s
t

is even f
...
If R and S are principal ideal domains, prove that R % S is a principal ideal ring
...
Let a and b belong to a commutative ring R
...

45
...
Show
that the annihilator of A, Ann(A) 5 {r [ R | ra 5 0 for all a in A},
is an ideal
...
Let R be a commutative ring and let A be any ideal of R
...
[N(͗0͘) is called the nil
radical of R
...
Let R 5 Z27
...
N(͗0͘),
b
...
N(͗9͘)
...
Let R 5 Z36
...
N(͗0͘),
b
...
N(͗6͘)
...
Let R be a commutative ring
...

50
...
Prove that N(N(A)) 5 N(A)
...
Let Z2[x] be the ring of all polynomials with coefficients in Z2 (that
is, coefficients are 0 or 1, and addition and multiplication of coefficients are done modulo 2)
...

52
...

53
...

54
...
Describe
the smallest ideal I of R that contains a (that is, if J is any ideal that
contains a, then I # J)
...
Let R be the ring of continuous functions from R to R
...
Show that A is a subring of R,
but not an ideal of R
...
Show that Z[i]/͗1 2 i͘ is a field
...
If R is a principal ideal domain and I is an ideal of R, prove that
every ideal of R/I is principal (see Exercise 41)
...
How many elements are in Z5[i]/͗1 1 i͘?
59
...
Let I be a prime ideal in R
...

60
...

Prove that I is the unique maximal ideal of R
...
Let I0 5 { f(x) [ Z[x] | f(0) 5 0}
...

62
...
)}, where each ai [ Z
...
)}, where only a finite number of terms are nonzero
...

63
...
Show that
͗a, b͘, the smallest ideal of R containing a and b, is I 5 {ra 1 sb|
r, s [ R}
...


Computer Exercises
What is the common denominator of intellectual accomplishment? In math,
science, economics, history, or any other subject, the answer is the same:
great thinkers notice patterns
...
d
...
edu/~jgallian
1
...
Run the program for several cases and
formulate a conjecture based on your data
...
This software determines the characteristic of the ring Z[i]/͗a 1 bi͘
(where i2 5 21)
...
Run the program for several
cases with d 5 1 and formulate a conjecture based on your data
...
1 and formulate a conjecture in terms of a, b, and d based on your data
...
1 also work in the case that d 5 1?
3
...
Run the program for several cases
and formulate a conjecture based on your data
...

EDMUND LANDAU,
Commemorative Address
to the Royal Society of Göttingen

This stamp was issued by East Germany
in 1981 to commemorate the 150th
anniversary of Dedekind’s birth
...


RICHARD DEDEKIND was born on October 6,
1831, in Brunswick, Germany, the birthplace of Gauss
...
His early
interests were in chemistry and physics, but
he obtained a doctor’s degree in mathematics at the age of 21 under Gauss at the
University of Göttingen
...


274

Dedekind spent the years 1858–1862 as a
professor in Zürich
...
Although this
school was less than university level,
Dedekind remained there for the next 50
years
...

During his career, Dedekind made numerous fundamental contributions to mathematics
...
His work on unique
factorization led to the modern theory of
algebraic numbers
...
The notion of
ideals as well as the term itself are attributed
to Dedekind
...
”
To find more information about
Dedekind, visit:
http://www-groups
...
st-and
...
uk/~history/

Emmy Noether
In the judgment of the most competent
living mathematicians, Fräulein Noether
was the most significant creative mathematical genius thus far produced since the
higher education of women began
...

ALBERT EINSTEIN,

The New York Times

EMMY NOETHER was born on March 23,
1882, in Germany
...

Noether completed her doctorate in 1907
...
While there, she made a major
contribution to physics with her theorem
that whenever there is a symmetry in nature,
there is also a conservation law, and vice
versa
...
After all, we are a
university and not a bathing establishment
...
Over the next
13 years, she used an axiomatic method to
develop a general theory of ideals and noncommutative algebras
...
Her approach was
even more important than the individual results
...
”
With the rise of Hitler in 1933, Noether, a
Jew, fled to the United States and took a position at Bryn Mawr College
...

To find more information about Noether,
visit:
http://www-groups
...
st-and
...
uk/~history/

275

276

Rings

Supplementary Exercises for Chapters 12–14
If at first you do succeed—try to hide your astonishment
...
BANKS

True/false questions for Chapters 12–14 are available on the Web at:
http://www
...
umn
...
Find all idempotent elements in Z10, Z20, and Z30
...
)
2
...

3
...
Show that
R has no nonzero nilpotent elements
...
)
4
...
Prove
that if for every nonzero element a of R we have aR 5 R, then R is
a field
...
Let A, B, and C be ideals of a ring R
...
(Compare this with Euclid’s
Lemma in Chapter 0
...
Show, by example, that the intersection of two prime ideals need
not be a prime ideal
...
Let R denote the ring of real numbers
...

What happens if R is replaced by any field F?
8
...

9
...
Prove that Zn has no nonzero
nilpotent elements
...
Let R be a commutative ring with unity
...
Show that a 1 b is a unit
...
)
11
...
If A # B < C, show that
A # B or A # C
...
For any element a in a ring R, define ͗a͘ to be the smallest ideal of
R that contains a
...
Show, by example, that if R is commutative but does not have a unity, then ͗a͘ and aR may be different
...
Let R be a ring with unity
...

14
...


14 | Supplementary Exercises for Chapters 12–14

277

15
...
If A > B 5 {0}, show that ab 5 0
when a [ A and b [ B
...
Show that the direct sum of two integral domains is not an integral
domain
...
Consider the ring R 5 {0, 2, 4, 6, 8, 10} under addition and multiplication modulo 12
...
What is the characteristic of Zm % Zn? Generalize
...
Let R be a commutative ring with unity
...
Show that R is a field
...
Suppose that I is an ideal of J and that J is an ideal of R
...
(Be careful not to assume that
the unity of I is the unity of R
...
)
21
...
Find a nontrivial idempotent (that is, not 0 and not 1) in
Q[x]/͗x4 1 x2͘
...
In a principal ideal domain, show that every nontrivial prime ideal
is a maximal ideal
...
Find an example of a commutative ring R with unity such that a,
b [ R, a 2 b, and an 5 bn and am 5 bm, where n and m are positive
integers that are relatively prime
...
)
3
24
...
[That is, Q( "2) is the subfield with the property that Q( "2)
3
3
contains Q and "2 and if F is any subfield containing Q and "2,
3
3
then F contains Q( "2)
...

25
...
If A is a
proper ideal of R, show that R/A has the same characteristic as R
...
Let F be a field of order pn
...

27
...

28
...
Prove that the additive group of
R/C(R) is not cyclic
...
Let
R5 ec

a b
d ` a, b, c, d [ Z 2 f
c d

278

Rings

with ordinary matrix addition and multiplication modulo 2
...

31
...


33
...

35
...

37
...

39
...


41
...

43
...


1 0
dr ` r [ Rf
0 0

is not an ideal of R
...
)
If R is an integral domain and A is a proper ideal of R, must R/A be
an integral domain?
Let A 5 {a 1 bi | a, b [ Z, a mod 2 5 b mod 2}
...

Suppose that R is a commutative ring with unity such that for each
a in R there is a positive integer n greater than 1 (n depends on a)
such that an 5 a
...

State a “finite subfield test”; that is, state conditions that guarantee
that a finite subset of a field is a subfield
...
Prove that the
sum of all of the elements of F is 0
...

Use this fact to find a zero-divisor in Z13[i]
...
Show that b is a unity for R
...

Show that the characteristic of Z[i]/͗a 1 bi͘ divides a2 1 b2
...

For any commutative ring R, R[x, y] is the ring of polynomials in x
and y with coefficients in R (that is, R[x, y] consists of all finite sums
of terms of the form axiyj, where a [ R and i and j are nonnegative
integers)
...
) Prove that ͗x, y͘
is a prime ideal in Z[x, y] but not a maximal ideal in Z[x, y]
...

Prove that ͗2, x, y͘ is a maximal ideal in Z[x, y]
...
Prove that (a, b) is nilpotent in R % S if and
only if both a and b are nilpotent
...
Prove that (a, b) is a zero-divisor
in R % S if and only if a or b is a zero-divisor or exactly one of a or
b is 0
...
Determine all idempotents in Zp k, where p is a prime
...
Let R be a commutative ring with unity 1
...

47
...
Define addition and multiplication as in Z["2], except that modulo n arithmetic is used to
combine the coefficients
...

48
...
Prove that every zero-divisor in Z pn is a nilpotent
element
...
If x is a nilpotent element in a commutative ring R, prove that rx is
nilpotent for all r in R
...

I
...
HERSTEIN, Topics in Algebra

Definition and Examples
In our work with groups, we saw that one way to discover information
about a group is to examine its interaction with other groups by way of
homomorphisms
...

Just as a group homomorphism preserves the group operation, a ring
homomorphism preserves the ring operations
...


A ring homomorphism that is both one-to-one and onto is called a
ring isomorphism
...

Again as with group theory, the roles of isomorphisms and homomorphisms are entirely distinct
...

A schematic representation of a ring homomorphism is given in
Figure 15
...
The dashed arrows indicate the results of performing the
ring operations
...
The reader
should supply the missing details
...
b

φ

281

φ (a)
...
1

EXAMPLE 1 For any positive integer n, the mapping k S k mod n is
a ring homomorphism from Z onto Zn (see Exercise 11 in Chapter 0)
...

EXAMPLE 2 The mapping a 1 bi S a 2 bi is a ring isomorphism
from the complex numbers onto the complex numbers (see Exercise 25
in Chapter 6)
...
The mapping f(x) S f(1) is a ring homomorphism from
R[x] onto R
...
Although showing that f(x 1 y) 5
f(x) 1 f(y) appears to be accomplished by the simple statement that
5(x 1 y) 5 5x 1 5y, we must bear in mind that the addition on the left is
done modulo 4, whereas the addition on the right and the multiplication
on both sides are done modulo 10
...
So, to verify that f preserves both operations, we write x 1 y 5 4q1 1 r1 and xy 5 4q2 1 r2, where 0 # r1 , 4
and 0 # r2 , 4
...
Similarly, using the fact that
5 ? 5 5 5 in Z10, we have f(xy) 5 f(r2) 5 5r2 5 5(xy 2 4q2) 5 5xy 2
20q2 5 (5 ? 5)xy 5 5x5y 5 f(x)f(y) in Z10
...

By Example 10 in Chapter 10, the only group homomorphisms from Z12
to Z30 are x S ax, where a 5 0, 15, 10, 20, 5, or 25
...
This requirement rules out 20 and 5
as possibilities for a
...


282

Rings

EXAMPLE 6 Let R be a commutative ring of characteristic 2
...

EXAMPLE 7 Although 2Z, the group of even integers under addition, is group-isomorphic to the group Z under addition, the ring 2Z is
not ring-isomorphic to the ring Z
...

EXAMPLE 8 (Test for Divisibility by 9)
An integer n with decimal representation akak21 ? ? ? a0 is divisible by 9
if and only if ak 1 ak21 1 ? ? ? 1 a0 is divisible by 9
...
Then, letting a denote
the natural homomorphism from Z to Z9 [in particular, a(10) 5 1], we
note that n is divisible by 9 if and only if
0 5 a(n) 5 a(ak)(a(10))k 1 a(ak21)(a(10))k21 1 ? ? ? 1 a(a0)
5 a(ak) 1 a(ak21) 1 ? ? ? 1 a(a0)
5 a(ak 1 ak21 1 ? ? ? 1 a0)
...

EXAMPLE 9 (Theorem of Gersonides)
Among the most important unsolved problems in number theory is the
so-called “abc conjecture
...
Gersonides proved that the only pairs of positive integers
that are powers of 2 and powers of 3 which differ by 1 are 1, 2; 2, 3; 3,
4; and 8, 9
...
To verify that this is so for 2m 5 3n 1 1, observe that
for all n we have 3n mod 8 5 3 or 1
...
On
the other hand, for m
...
To handle the case
where 2m 5 3n 2 1, we first note that for all n, 3n mod 16 5 3, 9, 11, or
1, depending on the value of n mod 4
...
Since 2m mod 16 5 0 for m $ 4, we have ruled out the cases where
n mod 4 5 1, 2, or 3
...
But the only values for 2m mod 5
are 2, 4, 3, and 1
...


15 | Ring Homomorphisms

283

Properties of Ring Homomorphisms
Theorem 15
...
Let A be a
subring of R and let B be an ideal of S
...
For any r [ R and any positive integer n, f(nr) 5 nf(r) and
f(r n) 5 (f(r))n
...
f(A) 5 {f(a) | a [ A} is a subring of S
...
If A is an ideal and f is onto S, then f(A) is an ideal
...
f21(B) 5 {r [ R | f(r) [ B} is an ideal of R
...
If R is commutative, then f(R) is commutative
...
If R has a unity 1, S 2 {0}, and f is onto, then f(1) is the unity
of S
...
f is an isomorphism if and only if f is onto and Ker f 5
{r [ R | f(r) 5 0} 5 {0}
...
If f is an isomorphism from R onto S, then f21 is an
isomorphism from S onto R
...
1 and 10
...

The student should learn the various properties of Theorem 15
...
Property 2 says that the homomorphic image of a subring is a subring
...

The next three theorems parallel results we had for groups
...

Theorem 15
...
Then Ker f
5 {r [ R | f(r) 5 0} is an ideal of R
...
3 First Isomorphism Theorem for Rings
Let f be a ring homomorphism from R to S
...
In
symbols, R/Ker f < f(R)
...
4 Ideals Are Kernels
Every ideal of a ring R is the kernel of a ring homomorphism of R
...


The homomorphism from R to R/A given in Theorem 15
...
Theorem 15
...

In Example 17 in Chapter 14 we gave a direct proof that ͗x͘ is a
prime ideal of Z[x] but not a maximal ideal
...

EXAMPLE 10 Since the mapping f from Z[x] onto Z given by
f( f(x)) 5 f(0) is a ring homomorphism with Ker f 5 ͗x͘ (see Exercise
29 in Chapter 14), we have, by Theorem 15
...
And because
Z is an integral domain but not a field, we know by Theorems 14
...
4 that the ideal ͗x͘ is prime but not maximal in Z[x]
...
5 Homomorphism from Z to a Ring with Unity
Let R be a ring with unity 1
...


PROOF Since the multiplicative group property am+n 5 aman translates to
(m 1 n)a 5 ma 1 na when the operation is addition, we have f(m 1 n) 5
(m 1 n) ? 1 5 m ? 1 1 n ? 1
...

That f also preserves multiplication follows from Exercise 15 in
Chapter 12, which says that (m ? a)(n ? b) 5 (mn) ? (ab) for all integers
m and n
...
So, f preserves multiplication as well
...
0, then
R contains a subring isomorphic to Zn
...


PROOF Let 1 be the unity of R and let S 5 {k ? 1 | k [ Z}
...
5
shows that the mapping f from Z to S given by f(k) 5 k ? 1 is a homomorphism, and by the First Isomorphism Theorem for rings, we have
Z/Ker f < S
...
3, n is also the characteristic of R
...
When R has characteristic 0, S <
Z/͗0͘ < Z
...


PROOF This follows directly from the statement of Theorem 15
...
(For example, in Z3, if
x 5 5, we have 5 ? 1 5 1 1 1 1 1 1 1 1 1 5 2
...
If F is a field of characteristic 0, then F contains
a subfield isomorphic to the rational numbers
...
In the latter case, let
T 5 {ab21 | a, b [ S, b 2 0}
...

Since the intersection of all subfields of a field is itself a subfield
(Exercise 11), every field has a smallest subfield (that is, a subfield
that is contained in every subfield)
...
It follows from Corollary 3 that the prime
subfield of a field of characteristic p is isomorphic to Zp, whereas the
prime subfield of a field of characteristic 0 is isomorphic to Q
...
)

The Field of Quotients
Although the integral domain Z is not a field, it is at least contained in a
field—the field of rational numbers
...
Can we mimic the
construction of the rationals from the integers for other integral domains? Yes
...
6 is called the field of
quotients of D
...
6, you should keep

286

Rings

in mind that we are using the construction of the rationals from the integers as a model for our construction of the field of quotients of D
...
6 Field of Quotients
Let D be an integral domain
...


PROOF Let S 5 {(a, b) | a, b [ D, b 2 0}
...
Now, let F be the set of equivalence classes of S under
the relation ; and denote the equivalence class that contains (x, y) by
x/y
...


(Notice that here we need the fact that D is an integral domain to ensure
that multiplication is closed; that is, bd 2 0 whenever b 2 0 and d 2 0
...
To do this, suppose that a/b 5 a9/b9
and c/d 5 c9/d9, so that ab9 5 a9b and cd9 5 c9d
...

Thus, by definition, we have
(ad 1 bc)/(bd) 5 (a9d9 1 b9c9)/(b9d9),
and, therefore, addition is well defined
...
That F is a
field is straightforward
...
Then 0/1 is the
additive identity of F
...
The remaining field
properties can be checked easily
...

EXAMPLE 11 Let D 5 Z[x]
...

When F is a field, the field of quotients of F[x] is traditionally denoted by F(x)
...
Then Zp(x) 5 {f(x)/g(x) | f(x), g(x) [
Zp[x], g(x) 2 0} is an infinite field of characteristic p
...

TITLE OF SONG BY JOHN LENNON AND
PAUL MCCARTNEY, December

1
...

3
...

5
...

7
...

9
...

11
...

13
...
1
...
2
...
3
...
4
...

Show that the correspondence x S 3x from Z4 to Z12 does not preserve multiplication
...
6 is a
ring homomorphism
...

Suppose that f is a ring homomorphism from Zm to Zn
...
Give an example to show that the converse
is false
...
Is the ring 2Z isomorphic to the ring 3Z?
b
...
(This exercise is referred to in this chapter
...
Show
that the field Z3[i] is ring-isomorphic to the field Z3[x]/͗x2 1 1͘
...

2b a
Show that f: C S S given by
a b
f(a 1 bi) 5 c
d
2b a
is a ring isomorphism
...
Let Z["2] 5 {a 1 b"2 | a, b [ Z}
...

b a

Show that Z["2] and H are isomorphic as rings
...
Consider the mapping from M2(Z ) into Z given by c

a
c

b
d S a
...

16
...

19
...

21
...

23
...

25
...


27
...

29
...

31
...
Prove or disprove that the mapc

a b
d S a is a ring homomorphism
...

Is the mapping from Z10 to Z10 given by x S 2x a ring homomorphism?
Describe the kernel of the homomorphism given in Example 3
...
Prove
that a ring homomorphism carries an idempotent to an idempotent
...
Determine all
ring homomorphisms from Z20 to Z30
...

Determine all ring homomorphisms from Z to Z
...
What
are the possibilities for f((1, 0))?
Determine all ring homomorphisms from Z % Z into Z % Z
...
Show that the group A/B is isomorphic to the group Z4 but that the ring A/B is not ring-isomorphic to
the ring Z4
...
Prove that S has a unity
...

Determine all ring homomorphisms from Z % Z to Z
...

Let m be a positive integer and let n be an integer obtained from m
by rearranging the digits of m in some way
...
) Show that m 2 n is divisible by 9
...


a
0

15 | Ring Homomorphisms

289

32
...
Prove that n is divisible by 11 if and only
if a0 2 a1 1 a2 2 ? ? ? (21)kak is divisible by 11
...
Show that the number 7,176,825,942,116,027,211 is divisible by 9
but not divisible by 11
...
Show that the number 9,897,654,527,609,805 is divisible by 99
...
(Test for divisibility by 3) Let n be an integer with decimal representation akak21 ? ? ? a1a0
...

36
...
Prove that n is divisible by 4 if and only
if a1a0 is divisible by 4
...
Show that no integer of the form 111,111,111,
...

38
...
Find values for a and b such that n is divisible by 99
...
Suppose n is a positive integer written in the form n 5 ak3k 1
ak213k21 1 ? ? ? 1 a13 1 a0, where each of the ai’s is 0, 1, or 2 (the
base 3 representative of n)
...

40
...

41
...

42
...

43
...
If f is a homomorphism from R onto S and the characteristic of R is nonzero, prove
that the characteristic of S divides the characteristic of R
...
Let R be a commutative ring of prime characteristic p
...

45
...
Show that a homomorphism from a field onto a ring with more
than one element must be an isomorphism
...
Suppose that R and S are commutative rings with unities
...

a
...

b
...


290

Rings

48
...
Show that the homomorphic image of a principal
ideal ring is a principal ideal ring
...
Let R and S be rings
...
Show that the mapping from R % S onto R given by (a, b) S a
is a ring homomorphism
...
Show that the mapping from R to R % S given by a S (a, 0) is a
one-to-one ring homomorphism
...
Show that R % S is ring-isomorphic to S % R
...
Show that if m and n are distinct positive integers, then mZ is not
ring-isomorphic to nZ
...
Prove or disprove that the field of real numbers is ring-isomorphic
to the field of complex numbers
...
Show that the only ring automorphism of the real numbers is the
identity mapping
...
Determine all ring homomorphisms from R to R
...
Suppose that n divides m and that a is an idempotent of Zn (that is,
a2 5 a)
...
Show that the same correspondence need not yield a
ring homomorphism if n does not divide m
...
Show that the operation of multiplication defined in the proof of
Theorem 15
...

56
...
Show that these two rings are not ring-isomorphic
...
Let Z[i] 5 {a 1 bi | a, b [ Z}
...
(This exercise
is referred to in Chapter 18
...
Let F be a field
...

59
...

Show that if E is any field that contains D, then E contains a
subfield that is ring-isomorphic to F
...
)
60
...
(Compare with Theorem 15
...
)
61
...
6 is an
equivalence relation
...
Give an example of a ring without unity that is contained in a field
...
Prove that the set T in the proof of Corollary 3 to Theorem 15
...


15 | Ring Homomorphisms

291

64
...
If R has a unity and S is an integral domain,
show that f carries the unity of R to the unity of S
...

65
...
If a 1 bi is a complex zero of f(x) (here i 5 "21),
show that a 2 bi is a zero of f(x)
...
)
66
...

b a
a
...

b
...

c
...

d
...
Is Ker f a maximal ideal?
67
...
(This exercise is referred to in
this chapter
...
Let n be a positive integer
...

69
...

takes c

Suggested Readings
J
...
Gallian and J
...

In this article, formulas are given for the number of group homomorphisms from Zm into Zn and the number of ring homomorphisms from
Zm into Zn
...
d
...
edu/
~jgallian/homs
...

In this article written by undergraduates, it is shown that R[x] is a
principal ideal ring if and only if R < R1 % R2 % ? ? ? % Rn, where
each Ri is a field
...

This article gives a formula for the number described in the title
...
d
...
edu/~jgallian/puzzle
This site has a math puzzle that is based on the ideas presented in this
chapter
...
Finally, another integer is created by using
all but one of the digits of the previous integer in any order
...


16 Polynomial Rings

Wit lies in recognizing the resemblance among things which differ and the
difference between things which are alike
...
In high school,
students study polynomials with integer coefficients, rational coefficients, real coefficients, and perhaps even complex coefficients
...
Notice that all of these
sets of polynomials are rings, and, in each case, the set of coefficients is
also a ring
...

Definition Ring of Polynomials over R

Let R be a commutative ring
...

Two elements
anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0
and
bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0
of R[x] are considered equal if and only if ai 5 bi for all nonnegative
integers i
...
n and bi 5 0 when i
...
)

293

294

Rings

In this definition, the symbols x, x2,
...
Rather, their purpose
is to serve as convenient placeholders that separate the ring elements
an, an21,
...
We could have avoided the x’s by defining a polynomial as an infinite sequence a0, a1, a2,
...
, but our
method takes advantage of the student’s experience in manipulating
polynomials where x does represent a variable
...
For example, in Z3[x], the polynomials f (x) 5 x3 and g(x) 5 x5 determine the same function from Z3
to Z3, since f(a) 5 g(a) for all a in Z3
...
Also, in the ring Zn[x], be careful to reduce only the
coefficients and not the exponents modulo n
...

To make R[x] into a ring, we define addition and multiplication in
the usual way
...
Then
f(x) 1 g(x) 5 (as 1 bs)xs 1 (as21 1 bs21)xs21
1 ? ? ? 1 (a1 1 b1)x 1 a0 1 b0,
where s is the maximum of m and n, ai 5 0 for i
...
m
...
, m 1 n
...
This substitution is a homomorphism from R[x] to R
...
So, just multiply polynomials over a
commutative ring R in the same way that polynomials are always multiplied
...

Consider f(x) 5 2x3 1 x2 1 2x 1 2 and g(x) 5 2x2 1 2x 1 1 in Z3[x]
...
Now, using
the definitions and remembering that addition and multiplication of the
coefficients are done modulo 3, we have
f(x) 1 g(x) 5 (2 1 0)x3 1 (1 1 2)x2 1 (2 1 2)x 1 (2 1 1)
5 2x3 1 0x2 1 1x 1 0
5 2x3 1 x
and
f(x) ? g(x) 5 (0 ? 1 1 0 ? 2 1 2 ? 2 1 1 ? 0 1 2 ? 0 1 2 ? 0)x5
1 (0 ? 1 1 2 ? 2 1 1 ? 2 1 2 ? 0 1 2 ? 0)x4
1 (2 ? 1 1 1 ? 2 1 2 ? 2 1 2 ? 0)x3
1 (1 ? 1 1 2 ? 2 1 2 ? 2)x2 1 (2 ? 1 1 2 ? 2)x 1 2 ? 1
5 x5 1 0x4 1 2x3 1 0x2 1 0x 1 2
5 x5 1 2x3 1 2
Our definitions for addition and multiplication of polynomials were
formulated so that they are commutative and associative, and so that
multiplication is distributive over addition
...

It is time to introduce some terminology for polynomials
...

The polynomial f(x) 5 0 has no degree
...
We often write deg f(x) 5 n to indicate
that f(x) has degree n
...

Very often properties of R carry over to R[x]
...


296

Rings

Theorem 16
...


PROOF Since we already know that D[x] is a ring, all we need
to show is that D[x] is commutative with a unity and has no zero-divisors
...
If 1 is the unity element of
D, it is obvious that f(x) 5 1 is the unity element of D[x]
...
Then, by definition, f(x)g(x) has leading coefficient anbm and, since D is an integral domain, anbm 2 0
...
The
next theorem is the analogous statement for polynomials over a field
...
2 Division Algorithm for F[x]
Let F be a field and let f(x) and g(x) [ F[x] with g(x) 2 0
...


PROOF We begin by showing the existence of q(x) and r(x)
...

So, we may assume that n 5 deg f(x) $ deg g(x) 5 m and let f(x) 5
anxn 1 ? ? ? 1 a0 and g(x) 5 bmxm 1 ? ? ? 1 b0
...
Thus, resorting to long division, we let f1(x) 5

16 | Polynomial Rings

297

21
f(x) 2 anbm xn2mg(x)
...

[Technically, we should get the induction started by proving the case
in which deg f(x) 5 0, but this is trivial
...

So, the polynomials q(x) 5 anbm21xn2m 1 q1(x) and r(x) 5 r1(x) have
the desired properties
...
Then, subtracting these two equations, we obtain
0 5 g(x)[q(x) 2 q(x)] 1 [r(x) 2 r(x)]
or
r(x) 2 r(x) 5 g(x)[q(x) 2 q(x)]
...
Since the latter is clearly impossible, we have r (x) 5 r(x) and
q(x) 5 q (x) as well
...
When the
ring of coefficients of a polynomial ring is a field, we can use the long
division process to determine the quotient and remainder
...
qan xn 1
...

m

n

f1 (x)

So,
f1(x) 5 (anxn 1 ? ? ?) 2 anbm21xn2m(bmxm 1 ? ? ?)

298

Rings

EXAMPLE 1 To find the quotient and remainder upon dividing
f(x) 5 3x4 1 x3 1 2x2 1 1 by g(x) 5 x2 1 4x 1 2, where f(x) and g(x)
belong to Z5[x], we may proceed by long division, provided we keep in
mind that addition and multiplication are done modulo 5
...
Therefore,
3x4 1 x3 1 2x2 1 1 5 (x2 1 4x 1 2)(3x2 1 4x) 1 2x 1 1
...
If f(x) and g(x) [ D[x], we say that g(x)
divides f(x) in D[x] [and write g(x) | f(x)] if there exists an h(x) [ D[x]
such that f(x) 5 g(x)h(x)
...

An element a is a zero (or a root) of a polynomial f(x) if f(a) 5 0
...
]
When F is a field, a [ F, and f(x) [ F[x], we say that a is a zero of
multiplicity k (k $ 1) if (x 2 a)k is a factor of f(x) but (x 2 a)k11 is not
a factor of f(x)
...
No doubt you have seen these
for the special case where F is the field of real numbers
...
Then f(a) is the remainder in
the division of f(x) by x 2 a
...

Corollary 2 The Factor Theorem
Let F be a field, a [ F, and f(x) [ F[x]
...


PROOF The proof of Corollary 2 is left as an exercise (Exercise 7)
...


PROOF We proceed by induction on n
...
Now suppose that f(x) is a polynomial of degree n over a field and a is a zero of f(x) of multiplicity k
...
If
f(x) has no zeros other than a, we are done
...
By the Second Principle of Mathematical Induction, we know
that q(x) has at most deg q(x) 5 n 2 k zeros, counting multiplicity
...

We remark that Corollary 3 is not true for arbitrary polynomial rings
...
(See
Exercise 3
...

EXAMPLE 2 The Complex Zeros of xn 2 1
We find all complex zeros of xn 2 1
...
It follows from DeMoivre’s Theorem (see Example 7
in Chapter 0) that vn 5 1 and vk 2 1 for 1 # k , n
...
, vn21 is a zero of xn 2 1 and, by Corollary 3, there are no
others
...

We conclude this chapter with an important theoretical application
of the division algorithm, but first an important definition
...


Theorem 16
...
Then F[x] is a principal ideal domain
...
1, we know that F[x] is an integral domain
...
If I 5 {0}, then I 5 ͗0͘
...
We will
show that I 5 ͗g(x)͘
...
Now
let f(x) [ I
...
Since r(x) 5 f(x) 2
g(x)q(x) [ I, the minimality of deg g(x) implies that the latter condition
cannot hold
...
This shows that
I # ͗g(x)͘
...
3 also establishes the following
...
4 Criterion for I 5 ͗g(x)͘
Let F be a field, I a nonzero ideal in F[x], and g(x) an element of
F[x]
...


As an application of the First Isomorphism Theorem for Rings
(Theorem 15
...
4, we verify the remark we made in
Example 12 in Chapter 14 that the ring R[x]/͗x2 1 1͘ is isomorphic to
the ring of complex numbers
...
Then
x2 1 1 [ Ker f and is clearly a polynomial of minimum degree in Ker f
...


Exercises
If I feel unhappy, I do mathematics to become happy
...

PAUL TURÁN

1
...
Compute f(x) 1 g(x) and f(x) ? g(x)
...
In Z3[x], show that the distinct polynomials x4 1 x and x2 1 x
determine the same function from Z3 to Z3
...
Show that x2 1 3x 1 2 has four zeros in Z6
...
)

16 | Polynomial Rings

301

4
...

5
...
2
...
List all the polynomials of degree 2 in Z2[x]
...
Prove Corollary 2 of Theorem 16
...

8
...
Show that R[x] has a subring isomorphic to R
...
If f: R → S is a ring homomorphism, define f:R[x] → S[x] by
(anxn 1 ? ? ? 1 a0) → f(an)xn 1 ? ? ? 1 f(a0)
...
(This exercise is referred to in Chapter 33
...
If the rings R and S are isomorphic, show that R[x] and S[x] are
isomorphic
...
Let f(x) 5 x3 1 2x 1 4 and g(x) 5 3x 1 2 in Z5[x]
...

12
...

Determine the quotient and remainder upon dividing f(x) by g(x)
...
Show that the polynomial 2x 1 1 in Z4[x] has a multiplicative inverse in Z4[x]
...
Are there any nonconstant polynomials in Z[x] that have multiplicative inverses? Explain your answer
...
Let p be a prime
...

16
...
2 is false for any commutative ring that has a zero divisor
...
(Degree Rule) Let D be an integral domain and f(x), g(x) [ D[x]
...
Show, by example, that for commutative ring R it is possible that deg f(x)g(x) ,
deg f(x) 1 deg g(x) where f(x) and g(x) are nonzero elements in
R[x]
...
)
18
...

19
...
Let a be a zero of f(x) of
multiplicity n, and write f(x) 5 (x 2 a)nq(x)
...
(This exercise is referred to in this chapter
...
Prove that for any positive integer n, a field F can have at most a
finite number of elements of order at most n
...
Let F be an infinite field and let f(x) [ F[x]
...


302

Rings

22
...
If f(a) 5 g(a) for
infinitely many elements a of F, show that f(x) 5 g(x)
...
Let F be a field and let p(x) [ F[x]
...
(This exercise is
referred to in Chapter 20
...
Prove that Z[x] is not a principal ideal domain
...
3
...
Find a polynomial with integer coefficients that has 1/2 and 21/3
as zeros
...
Let f(x) [ R[x]
...
Show that a is a zero of f(x) of multiplicity 1
...
Show that Corollary 2 of Theorem 16
...

28
...
2 is true for polynomials
over integral domains
...
Let F be a field and let
I 5 {anxn 1 an21xn21 1 ? ? ? 1 a0 | an, an21,
...

Show that I is an ideal of F[x] and find a generator for I
...
Let F be a field and let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ F[x]
...

31
...
For any integer a, let a denote
a mod m
...
(This exercise is referred to in Chapters
17 and 33
...
Find infinitely many polynomials f(x) in Z3[x] such that f(a) 5 0 for
all a in Z3
...
For every prime p, show that
xp21 2 1 5 (x 2 1)(x 2 2) ? ? ? [x 2 ( p 2 1)]
34
...

36
...


in Zp[x]
...
1, prove that (n 2 1)!
mod n 5 n 2 1 if and only if n is prime
...

Find the remainder upon dividing 98! by 101
...


16 | Polynomial Rings

303

38
...

39
...

40
...
If I is a prime ideal of R,
prove that I[x] is a prime ideal of R[x]
...
Let F be a field, and let f(x) and g(x) belong to F[x]
...
(This exercise is referred to in Chapter 20
...
Prove that Q[x]/͗x2 2 2͘ is ring-isomorphic to Q["2 ] 5 {a 1
b"2 | a, b [ Q}
...
Let f(x) [ R[x]
...

44
...

Prove that I is an ideal in F[x]
...
When F is finite, find a monic
polynomial g(x) such that I 5 ͗g(x)͘
...
Let g(x) and h(x) belong to Z[x] and let h(x) be monic
...
(This exercise is referred to in Chapter 33
...
For any field F, recall that F(x) denotes the field of quotients of the
ring F[x]
...

47
...
Show that there exist a, b [ F with the property
that x2 1 x 1 1 divides x43 1 ax 1 b
...
Let f(x) 5 am x m 1 am21x m21 1 ? ? ? 1 a0 and g(x) 5 bn xn 1 bn21x n21 1
? ? ? 1 b0 belong to Q[x] and suppose that f 8 g belongs to Z[x]
...

49
...
If a mod m 5 b mod m, prove that f (a)
mod m 5 f(b) mod m
...

50
...

51
...


Saunders Mac Lane
The 1986 Steele Prize for cumulative
influence is awarded to Saunders Mac Lane
for his many contributions to algebra and
algebraic topology, and in particular for his
pioneering work in homological and
categorical algebra
...
He was born on August 4, 1909, in
Norwich, Connecticut
...
After applying for employment
as a master at a private preparatory school for
boys, Mac Lane received a two-year instructorship at Harvard in 1934
...
In 1947, he went back to Chicago permanently
...
His book, Survey of Modern
Algebra, coauthored with Garrett Birkhoff,
influenced generations of mathematicians
and is now a classic
...
He was elected to the National
Academy of Sciences, received the National
Medal of Science and supervised 41 Ph
...

theses
...

To find more information about Mac
Lane, visit:
http://www-groups
...
st-and
...
uk/~history/

Factorization
17 of Polynomials
The value of a principle is the number of things it will explain
...
In this chapter, we consider the same problems in a
more abstract setting
...

Definition Irreducible Polynomial, Reducible Polynomial

Let D be an integral domain
...
A nonzero,
nonunit element of D[x] that is not irreducible over D is called
reducible over D
...

EXAMPLE 1 The polynomial f(x) 5 2x2 1 4 is irreducible over Q
but reducible over Z, since 2x2 1 4 5 2(x2 1 2) and neither 2 nor x2 1 2
is a unit in Z[x]
...

EXAMPLE 3 The polynomial x2 2 2 is irreducible over Q but reducible over R
...

In general, it is a difficult problem to decide whether or not a particular polynomial is reducible over an integral domain, but there are special cases when it is easy
...
It applies
to the three preceding examples
...
1 Reducibility Test for Degrees 2 and 3
Let F be a field
...


PROOF Suppose that f(x) 5 g(x)h(x), where both g(x) and h(x) belong
to F[x] and have degrees less than that of f(x)
...
Say g(x) 5 ax 1 b
...

Conversely, suppose that f(a) 5 0, where a [ F
...

Theorem 17
...
, p 2 1
...
On the other hand,
because neither 0, 1, nor 2 is a zero of x2 1 1 over Z3, x2 1 1 is irreducible over Z3
...
For example, in
Q[x], the polynomial x4 1 2x2 1 1 is equal to (x2 1 1)2, but has no
zeros in Q
...

To simplify the proof of the first of these, we introduce some terminology and isolate a portion of the argument in the form of a lemma
...
, a0
...


17 | Factorization of Polynomials

307

Gauss’s Lemma
The product of two primitive polynomials is primitive
...
Let p be a prime divisor of the content of
f(x)g(x), and let f (x), g(x), and f(x)g(x) be the polynomials obtained
from f(x), g(x), and f(x)g(x) by reducing the coefficients modulo p
...

Thus, f (x) 5 0 or g(x) 5 0
...
Hence, either f(x)
is not primitive or g(x) is not primitive
...

Remember that the question of reducibility depends on which ring of
coefficients one permits
...
In Chapter 20, we will prove that every polynomial of degree greater than 1 with coefficients from an integral
domain is reducible over some field
...
2 shows that in the
case of polynomials irreducible over Z, this field must be larger than
the field of rational numbers
...
2 Reducibility over Q Implies Reducibility Over Z
Let f(x) [ Z[x]
...


PROOF Suppose that f(x) 5 g(x)h(x), where g(x) and h(x) [ Q[x]
...
Let a be the least common
multiple of the denominators of the coefficients of g(x), and b the least
common multiple of the denominators of the coefficients of h(x)
...
Let c1 be the content of ag(x) and let c2 be the content of bh(x)
...
Since f(x) is primitive, the content of abf(x) is ab
...
Thus, ab 5 c1c2 and f(x) 5
g1(x)h1(x), where g1(x) and h1(x) [ Z[x] and deg g1(x) 5 deg g(x) and
deg h1(x) 5 deg h(x)
...
2 by tracing
through it for the polynomial f(x) 5 6x2 1 x 2 2 5 (3x 2 3/2)(2x 1
4/3) 5 g(x)h(x)
...


Irreducibility Tests
Theorem 17
...
The next theorem often allows us
to simplify the problem even further
...
3 Mod p Irreducibility Test
Let p be a prime and suppose that f(x) [ Z[x] with deg f(x) $ 1
...
If f (x) is irreducible over Zp and
deg f (x) 5 deg f(x), then f(x) is irreducible over Q
...
2 that if f(x) is reducible over Q, then f(x) 5 g(x)h(x) with g(x), h(x) [ Z[x], and both
g(x) and h(x) have degree less than that of f(x)
...
Since deg f(x) 5 deg f (x) , we have deg
g(x) # deg g(x) , deg f (x) and deg h(x) # deg h(x) , deg f (x)
...

EXAMPLE 6 Let f(x) 5 21x3 2 3x2 1 2x 1 9
...
Thus, f(x) is irreducible over Q
...
3
to conclude that f(x) is irreducible over Q
...
3
...
For example, consider f(x) 5 21x3 2 3x2 1 2x 1 8
...
But over Z5, f (x) has no zeros and
therefore is irreducible over Z5
...
Note that
this example shows that the Mod p Irreducibility Test may fail for
some p and work for others
...
However, this is not always possible, since f(x) 5 x4 1 1 is irreducible over Q but reducible
over Zp for every prime p
...
)
The Mod p Irreducibility Test can also be helpful in checking for
irreducibility of polynomials of degree greater than 3 and polynomials
with rational coefficients
...
We will
show that f(x) is irreducible over Q
...
Then f(x) is irreducible over Q if h(x) is irreducible
over Z
...
Clearly, h(x) has no zeros in Z2
...
[For if so, the factor would have
to be either x2 1 x 1 1 or x2 1 1
...
] Thus h(x) is irreducible over Z2[x]
...

EXAMPLE 8 Let f(x) 5 x 5 1 2x 1 4
...
1 nor the Mod 2 Irreducibility Test helps here
...

Substitution of 0, 1, and 2 into f (x) does not yield 0, so there are no linear
factors
...
If so, we may assume it has
the form x2 1 ax 1 b (see Exercise 5)
...
We can immediately rule out each of the nine that has a zero over
Z3, since f (x) does not have one
...
These are eliminated by long division
...
(Why is it unnecessary to check for cubic or fourth-degree factors?)
Another important irreducibility test is the following one, credited to
Ferdinand Eisenstein (1823–1852), a student of Gauss
...

Theorem 17
...

If there is a prime p such that p B an, p | an21,
...


310

Rings

PROOF If f(x) is reducible over Q, we know by Theorem 17
...
Say g(x) 5 br xr 1 ? ? ? 1 b0 and
h(x) 5 cs xs 1 ? ? ? 1 c0
...
Let us say p | b0
and p B c0
...
So, there is a
least integer t such that p B bt
...
By assumption, p divides at and, by choice of t, every summand
on the right after the first one is divisible by p
...
This is impossible, however, since p is prime and p
divides neither bt nor c0
...


PROOF Let
f(x)5Fp(x11)5

(x11) p 21
p
p
p
5x p21 1 a b x p22 1 a b x p23 1
...

(x11)21
1
2
1

Then, since every coefficient except that of xp21 is divisible by p and
the constant term is not divisible by p2, by Eisenstein’s Criterion, f(x) is
irreducible over Q
...
Since this is impossible, we conclude that Fp(x) is irreducible over Q
...

The principal reason for our interest in irreducible polynomials
stems from the fact that there is an intimate connection among them,
maximal ideals, and fields
...


17 | Factorization of Polynomials

311

Theorem 17
...
Then ͗p(x)͘ is a maximal ideal
in F[x] if and only if p(x) is irreducible over F
...
Clearly,
p(x) is neither the zero polynomial nor a unit in F[x], because neither
{0} nor F[x] is a maximal ideal in F[x]
...
Thus, ͗p(x)͘ 5 ͗g(x)͘
or F[x] 5 ͗g(x)͘
...
In
the second case, it follows that deg g(x) 5 0 and, consequently, deg h(x) 5
deg p(x)
...

Now, suppose that p(x) is irreducible over F
...
Because F[x] is a principal ideal domain, we know that I 5 ͗g(x)͘ for some g(x) in F[x]
...
Since p(x) is irreducible over F, it follows that either g(x) is a constant or h(x) is a constant
...
So, ͗p(x)͘ is maximal in F[x]
...
Then
F[x]/͗p(x)͘ is a field
...
5 and 14
...

The next corollary is a polynomial analog of Euclid’s Lemma for
primes (see Chapter 0)
...
If p(x) is irreducible
over F and p(x) | a(x)b(x), then p(x) | a(x) or p(x) | b(x)
...
From Theorem 14
...
Thus,
a(x) [ ͗p(x)͘ or b(x) [ ͗p(x)͘
...

The next two examples put the theory to work
...
By
Theorem 17
...
5, it suffices to find a
cubic polynomial over Z2 that has no zero in Z2
...
Thus, Z2[x]/͗x3 1 x 1 1͘ 5 {ax2 1 bx 1 c 1 ͗x3 1
x 1 1͘ | a, b, c [ Z2} is a field with eight elements
...
Since the sum of two polynomials of
the form ax2 1 bx 1 c is another one of the same form, addition is easy
...

On the other hand, multiplication of two coset representatives need not
yield one of the original eight coset representatives:
(x2 1 x 1 1 1 ͗x3 1 x 1 1͘) ? (x2 1 1 1 ͗x3 1 x 1 1͘)
5 x4 1 x3 1 x 1 1 1 ͗x3 1 x 1 1͘ 5 x4 1 ͗x3 1 x 1 1͘
(since the ideal absorbs the last three terms)
...

Another way is to observe that x3 1 x 1 1 1 ͗x3 1 x 1 1͘ 5 0 1
͗x3 1 x 1 1͘ implies x3 1 ͗x3 1 x 1 1͘ 5 x 1 1 1 ͗x3 1 x 1 1͘
...

Similarly,
(x2 1 x 1 kx3 1 x 1 1l) ? (x 1 kx3 1 x 1 1l)
5 x3 1 x2 1 kx3 1 x 1 1l
5 x2 1 x 1 1 1 kx3 1 x 1 1l
...
1
...

Table 17
...
Keep in mind that x3 can be replaced by
x 1 1 and x4 by x2 1 x
...
Thus, Z3[x]/͗x2 1 1͘ is a field
...
Thus, this field has
nine elements
...
1 by replacing i by x
...
In Chapter 18,
we will study this property in greater depth
...
6 was given by Gauss
...

Theorem 17
...
Furthermore, if
b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ? qn(x),
where the b’s and c’s are irreducible polynomials of degree 0, and the
p(x)’s and q(x)’s are irreducible polynomials of positive degree, then
s 5 t, m 5 n, and, after renumbering the c’s and q(x)’s, we have bi 5
6ci for i 5 1,
...
, m
...
If
deg f(x) 5 0, then f(x) is constant and the result follows from the
Fundamental Theorem of Arithmetic
...
0, let b denote the
content of f(x), and let b1b2 ? ? ? bs be the factorization of b as a product
of primes
...
Thus, to prove the existence portion
of the theorem, it suffices to show that a primitive polynomial f(x) of
positive degree can be written as a product of irreducible polynomials
of positive degree
...
If deg f(x) 5 1,
then f(x) is already irreducible and we are done
...
If f(x) is irreducible,
there is nothing to prove
...
Thus, by induction, both g(x) and h(x) can be written as a product of irreducibles of
positive degree
...

To prove the uniqueness portion of the theorem, suppose that
f(x) 5 b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ?
qn(x), where the b’s and c’s are irreducible polynomials of degree 0, and
the p(x)’s and q(x)’s are irreducible polynomials of positive degree
...
Since the p(x)’s and q(x)’s are
primitive, it follows from Gauss’s Lemma that p1(x)p2(x) ? ? ? pm(x) and
q1(x)q2(x) ? ? ? qn(x) are primitive
...

It then follows from the Fundamental Theorem of Arithmetic that s 5 t
and, after renumbering, bi 5 6ci for i 5 1, 2,
...
Thus, by canceling the constant terms in the two factorizations for f(x), we have
p1(x)p2(x) ? ? ? pm(x) 5 6q1(x) q2(x) ? ? ? qn(x)
...
5 and induction (see
Exercise 27) that p1(x) | qi(x) for some i
...
Then, since q1(x) is irreducible, we have q1(x) 5 (r/s)p1(x),
where r, s [ Z
...
So, q1(x) 5 6p1(x)
...
Now, we may repeat the argument
above with p2(x) in place of p1(x)
...

Clearly, this is impossible
...
n, after n steps we
would have 61 on the right and a nonconstant polynomial on the left—
another impossibility
...


Weird Dice: An Application
of Unique Factorization
EXAMPLE 12 Consider an ordinary pair of dice whose faces are
labeled 1 through 6
...
In a 1978 issue of

315

17 | Factorization of Polynomials

Scientific American [1], Martin Gardner remarked that if one were to
label the six faces of one cube with the integers 1, 2, 2, 3, 3, 4 and the six
faces of another cube with the integers 1, 3, 4, 5, 6, 8, then the probability of obtaining any particular sum with these dice (called Sicherman
dice) would be the same as the probability of rolling that sum with ordinary dice (that is, 1/36 for a 2, 2/36 for a 3, and so on)
...
1
...
To do so,
we utilize the fact that Z[x] has the unique factorization property
...
1

To begin, let us ask ourselves how we may obtain a sum of 6, say, with
an ordinary pair of dice
...
Next we consider the product of the
two polynomials created by using the ordinary dice labels as exponents:
(x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x)
...
Notice the correspondence between pairs of labels whose sums are 6 and pairs of terms
whose products are x6
...
So, let a1, a2, a3, a4, a5, a6 and
b1, b2, b3, b4, b5, b6 be any two lists of positive integer labels for the faces
of a pair of cubes with the property that the probability of rolling any
particular sum with these dice (let us call them weird dice) is the same as
the probability of rolling that sum with ordinary dice labeled 1 through
6
...

(1)

316

Rings

Now all we have to do is solve this equation for the a’s and b’s
...
The polynomial x6 1 x5 1
x4 1 x3 1 x2 1 x factors uniquely into irreducibles as
x(x 1 1)(x2 1 x 1 1)(x2 2 x 1 1)
so that the left-hand side of Equation (1) has the irreducible factorization
x2(x 1 1)2(x2 1 x 1 1)2(x2 2 x 1 1)2
...
6, this means that these factors are the only possible
irreducible factors of P(x) 5 xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6
...

To restrict further the possibilities for these four parameters, we evaluate P(1) in two ways
...
Clearly, this means that r 5 1 and t 5 1
...
On the other hand, if
q 5 2, the smallest possible sum one could roll with the corresponding
labels for dice would be 3
...
Let’s consider each of these possibilities in turn
...

When u 5 1, P(x) 5 x6 1 x5 1 x4 1 x3 1 x2 1 x, so the die labels
are 6, 5, 4, 3, 2, 1—an ordinary die
...

This proves that the Sicherman dice do give the same probabilities
as ordinary dice and that they are the only other pair of dice that have
this property
...

HOMER SIMPSON

1
...

2
...

If f(x) [ D[x] and f(x) is irreducible over F but reducible over D,
what can you say about the factorization of f(x) over D?

17 | Factorization of Polynomials

317

3
...
(This exercise is referred to in this chapter
...
Suppose that f(x) 5 xn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]
...

5
...

a
...

b
...

c
...

d
...

(This exercise is referred to in this chapter
...
Suppose that f(x) [ Zp[x] and is irreducible over Zp, where p is a
prime
...

7
...

8
...

9
...
Does this fact contradict the corollary to Theorem 17
...
Determine which of the polynomials below is (are) irreducible
over Q
...
x5 1 9x4 1 12x2 1 6
b
...
x4 1 3x2 1 3
d
...
(5/2)x5 1 (9/2)x4 1 15x3 1 (3/7)x2 1 6x 1 3/14
11
...
(This
exercise is referred to in this chapter
...
Show that x2 1 x 1 4 is irreducible over Z11
...
Let f(x) 5 x3 1 6 [ Z7[x]
...

14
...
Write f(x) as a product of irreducible polynomials over Z2
...
Let p be a prime
...
Show that the number of reducible polynomials over Zp of the
form x2 1 ax 1 b is p( p 1 1)/2
...
Determine the number of reducible quadratic polynomials over Zp
...
Let p be a prime
...
Determine the number of irreducible polynomials over Zp of the
form x2 1 ax 1 b
...
Determine the number of irreducible quadratic polynomials
over Zp
...
Show that for every prime p there exists a field of order p2
...
Prove that, for every positive integer n, there are infinitely many
polynomials of degree n in Z[x] that are irreducible over Q
...
Show that the field given in Example 11 in this chapter is isomorphic to the field given in Example 9 in Chapter 13
...
Let f(x) [ Zp[x]
...

21
...

22
...

23
...

24
...
Find
all zeros of f(x) by using the Quadratic Formula (2b 6 "b2 24ac) ?
(2a)21 (all calculations are done in Z7)
...
Try the Quadratic Formula on g(x)
...

25
...


27
...

29
...


and an 2 0
...

Let F be a field and f(x) [ F[x]
...
(This assumption is useful when one uses a computer
to check for irreducibility
...
, ak(x) [ F[x], where
p(x) is irreducible over F
...
(This exercise is referred to in the proof of
Theorem 17
...
)
Explain how the Mod p Irreducibility Test (Theorem 17
...

Show that x4 1 1 is reducible over Zp for every prime p
...
)
If p is a prime, prove that xp21 2 xp22 1 xp23 2 ? ? ? 2 x 1 1 is
irreducible over Q
...
Let F be a field and let p(x) be irreducible over F
...
(This exercise is referred to in
Chapter 20
...
Prove that the ideal ͗x2 1 1͘ is prime in Z[x] but not maximal in Z[x]
...
Let F be a field and let p(x) be irreducible over F
...
(This
exercise is referred to in Chapter 20
...
Suppose there is a real number r with the property that r 1 1/r is
an odd integer
...

35
...
Carry out the analysis given in Example 12 for a pair of tetrahedrons instead of a pair of cubes
...
)
37
...
2) ordinary dice
each labeled 1 through 6, instead of just two
...

38
...
Carry out an analysis similar to that given in Example
12 to derive these labels
...

RENÉ J
...
d
...
edu/~jgallian
1
...
Use it to
test the polynomials in the examples given in this chapter and the
polynomials given in Exercise 10 for irreducibility
...
Use software such as Mathematica, Maple, or GAP to express
xn 2 1 as a product of irreducible polynomials with integer coefficients for n 5 4, 8, 12, and 20
...

Test your conjecture for n 5 105
...
Use software such as Mathematica, Maple, or GAP to express x p 2 x
as a product of irreducibles over Zp for several choices of the prime
p and n
...


Reference
1
...


Suggested Readings
Duane Broline, “Renumbering the Faces of Dice,” Mathematics Magazine
52 (1979): 312–315
...

J
...
Gallian and D
...
Rusin, “Cyclotomic Polynomials and Nonstandard
Dice,” Discrete Mathematics 27 (1979): 245–259
...

Randall Swift and Brian Fowler, “Relabeling Dice,” Mathematics
Magazine 72 (1999): 204–208
...


Serge Lang
Lang’s Algebra changed the way graduate
algebra is taught
...

Citation for the Steele Prize

SERGE LANG was a prolific mathematician,
inspiring teacher, and political activist
...
His
family moved to Los Angeles when he was a
teenager
...
A
...
D
...
His
first permanent position was at Columbia
University in 1955, but in 1971 Lang resigned his position at Columbia as a protest
against Columbia’s handling of Vietnam antiwar protesters
...

Lang made significant contributions to
number theory, algebraic geometry, differential geometry, and analysis
...
His
most famous and influential book was his
graduate-level Algebra
...
Lang often got into
heated discussions about mathematics, the
arts, and politics
...

Among Lang’s honors were the Steele
Prize for Mathematical Exposition from the
American Mathematcial Society, the Cole
Prize in Algebra (see Chapter 25), and election to the National Academy of Sciences
...

For more information about Lang, visit:
http://wikipedia
...
You can keep your sterile truth for yourself
...
Several of those results—unique factorization in Z[x] and the division algorithm for F[x], for instance—are natural counterparts to theorems about the integers
...

Definition Associates, Irreducibles, Primes

Elements a and b of an integral domain D are called associates if
a 5 ub, where u is a unit of D
...
A nonzero element a of an
integral domain D is called a prime if a is not a unit and a | bc implies
a | b or a | c
...
Notice that an element a is a prime if and only if
͗a͘ is a prime ideal
...
The source of the confusion is that in the case of the integers,
the concepts of irreducibles and primes are equivalent, but in general, as
we will soon see, they are not
...
(These rings are of
fundamental importance in number theory
...
To do this, we define a function N, called the norm, from Z["d]
into the nonnegative integers by N(a 1 b"d) 5 |a2 2 db2|
...

EXAMPLE 1 We exhibit an irreducible in Z[" 3] that is not prime
...
Consider 1 1 " 3
...
Then N(xy) 5
N(x)N(y) 5 N(1 1 " 3) 5 4, and it follows that N(x) 5 2
...
Thus, x or y is a unit and
1 1 " 3 is an irreducible
...
On the
2
2
2
other hand, for integers a and b to exist so that 2 5 (1 1 " 3)(a 1
2
b" 3) 5 (a 2 3b) 1 (a 1 b)" 3, we must have a 2 3b 5 2 and a 1
2
2
b 5 0, which is impossible
...
1
...
The example also shows that the converse of the fourth
property above for the norm is not true
...

EXAMPLE 2 The element 7 is irreducible in the ring Z["5]
...
Then
49 5 N(7) 5 N(x) N(y), and since x is not a unit, we cannot have N(x) 5
1
...
Let x 5 a 1 b"5
...
This means that a2 2 5b2 5
67
...
But this means that
both a and b are divisible by 7, and this implies that |a2 2 5b2| 5 7 is
divisible by 49, which is false
...
The answer: no
...
1 Prime Implies Irreducible
In an integral domain, every prime is an irreducible
...

We must show that b or c is a unit
...
Say at 5 b
...
Thus, c is a unit
...
The next theorem reveals a circumstance
in which primes and irreducibles are equivalent
...
2 PID Implies Irreducible Equals Prime
In a principal ideal domain, an element is an irreducible if and only
if it is a prime
...
1 shows that primes are irreducibles
...
We must show that a | b or a | c
...
Since a [ I, we can write
a 5 dr, and because a is irreducible, d is a unit or r is a unit
...
Then c 5 acx 1 bcy,
and since a divides both terms on the right, a also divides c
...
Thus, a divides b
...
3)
...

EXAMPLE 3 We show that Z[x] is not a principal ideal domain
...
We claim that I is not
of the form ͗h(x)͘
...

By the degree rule (Exercise 17 in Chapter 16), 0 5 deg 2 5 deg h(x) 1
deg f(x), so that h(x) is a constant polynomial
...
Thus, h(1) 5 61 or 62
...
But then x 5 62g(x), which is
nonsense
...
It is natural to ask whether all integral
domains have this property
...


Historical Discussion
of Fermat’s Last Theorem
There are infinitely many nonzero integers x, y, z that satisfy the equation x2 1 y2 5 z2
...
The tremendous effort put forth by the likes of
Euler, Legendre, Abel, Gauss, Dirichlet, Cauchy, Kummer, Kronecker,
and Hilbert to prove that there are no solutions to this equation has
greatly influenced the development of ring theory
...
The problem lay dormant until 1637, when the French mathematician Pierre de Fermat
(1601–1665) wrote in the margin of a book, “
...
”
Because Fermat gave no proof, many mathematicians tried to prove
the result
...
The case where n 5 4 is elementary and was
done by Fermat himself
...

Since the validity of the case for a particular integer implies the validity for all multiples of that integer, the next case of interest was n 5 7
...
In 1847, Lamé stirred excitement by
announcing that he had completely solved the problem
...
Thus, his
factorization took place in the ring Z[a] 5 {a0 1 a1a 1 ? ? ? 1
ap21a p21 |ai [ Z}
...
In
fact, three years earlier, Ernst Eduard Kummer had proved that this is
not always the case
...
Within a few weeks of Lamé’s announcement, Kummer had shown that Fermat’s Last Theorem is true for all
primes of a special type
...

Kummer’s work has led to the theory of ideals as we know it today
...
The famous number theorist Edmund Landau received so many
of these that he had a form printed with “On page ____, lines ____ to
____, you will find there is a mistake
...

Recent discoveries tying Fermat’s Last Theorem closely to modern
mathematical theories gave hope that these theories might eventually
lead to a proof
...
1)
...
In
June 1993, excitement spread through the mathematics community
with the announcement that Andrew Wiles of Princeton University had
proved Fermat’s Last Theorem (see Figure 18
...
The Princeton mathematics department chairperson was quoted as saying, “When we
heard it, people started walking on air
...
This story does have a happy ending
...

In view of the fact that so many eminent mathematicians were unable to prove Fermat’s Last Theorem, despite the availability of the
vastly powerful theories, it seems highly improbable that Fermat had a
correct proof
...


18 | Divisibility in Integral Domains

Figure 18
...
2

Andrew Wiles

Unique Factorization Domains
We now have the necessary terminology to formalize the idea of
unique factorization
...
every nonzero element of D that is not a unit can be written as a
product of irreducibles of D, and
2
...


Another way to formulate part 2 of this definition is the following:
If p1 n 1 p 2 n 2 ? ? ? pr n r and q1 m 1 q 2m 2 ? ? ? q s m s are two factorizations of some
element as a product of irreducibles, where no two of the pi’s are associates and no two of the q j’s are associates, then r 5 s, and each p i is an
associate of one and only one q j and ni = mj
...
6
says that Z[x] is a unique factorization domain
...

Before proving our next theorem, we need the ascending chain condition for ideals
...


PROOF Let I1 , I2 , ? ? ? be a chain of strictly increasing ideals in
an integral domain D, and let I be the union of all the ideals in this chain
...

Then, since D is a principal ideal domain, there is an element a in D
such that I 5 ͗a͘
...
Clearly, then, for any member Ii of the
chain, we have Ii # I 5 ͗a͘ # In, so that In must be the last member of
the chain
...
3 PID Implies UFD
Every principal ideal domain is a unique factorization domain
...
We will show that a0 is a product of irreducibles (the
product might consist of only one factor)
...
If a0 is irreducible, we are done
...
If a1 is not irreducible, then we can write a1 5 b2a2,
where neither b2 nor a2 is a unit and a2 is nonzero
...
of elements that are not units
in D and a sequence a0, a1, a2,
...
Hence, ͗a0͘ , ͗a1͘ , ? ? ? is a strictly increasing
chain of ideals (see Exercise 5), which, by the preceding lemma, must
be finite, say, ͗a0͘ , ͗a1͘ , ? ? ? , ͗ar͘
...
This argument shows that every nonzero nonunit
in D has at least one irreducible factor
...
If c1
is not irreducible, then we can write c1 5 p2c2, where p2 is irreducible
and c2 is not a unit
...
Let us say that the sequence ends with ͗cs͘
...
This completes the proof that every nonzero nonunit of a principal ideal domain is a product of irreducibles
...
To do this, suppose that
some element a of D can be written
a 5 p1p2 ? ? ? pr 5 q1q2 ? ? ? qs,
where the p’s and q’s are irreducible and repetition is permitted
...
If r 5 1, then a is irreducible and, clearly, s 5 1 and p1 5 q1
...
Since p1 divides q1q2 ? ? ? qs, it must divide
some qi (see Exercise 29), say, p1 | q1
...
Since
up1p2 ? ? ? pr 5 uq1q2 ? ? ? qs 5 q1(uq2) ? ? ? qs
and
up1 5 q1,
we have, by cancellation,
p2 ? ? ? pr 5 (uq2) ? ? ? qs
...

Hence, the same is true about the two factorizations of a
...
3, the only
way we used the fact that the integral domain D is a principal ideal
domain was to say that D has the property that there is no infinite,
strictly increasing chain of ideals in D
...
Noetherian
domains are of the utmost importance in algebraic geometry
...

One such example is Z[x]
...

As an immediate corollary of Theorem 18
...


18 | Divisibility in Integral Domains

331

Corollary F[x] Is a UFD
Let F be a field
...


PROOF By Theorem 16
...
So, F[x]
is a unique factorization domain, as well
...
4)
...
, p | a0

and

p2 B a0
...
If f(x) is reducible over Q,
we know by Theorem 17
...
Let
f (x), g(x), and h(x) be the polynomials in Zp[x] obtained from f (x),
g(x), and h(x) by reducing all coefficients modulo p
...
Since Zp is a field, Zp[x] is a unique factorization domain
...
So, g(0) 5 h(0) 5 0 and, therefore, p | g(0) and p | h(0)
...


Euclidean Domains
Another important kind of integral domain is a Euclidean domain
...
d(a) # d(ab) for all nonzero a, b in D; and
2
...


EXAMPLE 5 The ring Z is a Euclidean domain with d(a) 5 |a| (the
absolute value of a)
...
Then F[x] is a Euclidean domain with
d( f(x)) 5 deg f(x) (see Theorem 16
...

Examples 5 and 6 illustrate just one of many similarities between the
rings Z and F[x]
...
1
...
1 Similarities Between Z and F[x]
Z

F[x]

Form of elements:
an10n 1 an2110n21 1 ? ? ? 1 a110 1 a0
Euclidean domain:
d(a) 5 |a|
Units:
a is a unit if and only if |a| 5 1
Division algorithm:
For a, b [ Z, b 2 0, there exist q, r [ Z
such that a 5 bq 1 r, 0 # r , |b|

↔

PID:
Every nonzero ideal I 5 ͗a͘, where
a 2 0 and |a| is minimum
Prime:
No nontrivial factors
UFD:
Every element is a “unique” product of
primes

↔

↔

↔

↔
↔

Form of elements:
anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0
Euclidean domain:
d( f(x)) 5 deg f(x)
Units:
f(x) is a unit if and only if deg f(x) 5 0
Division algorithm:
For f(x), g(x) [ F[x], g(x) 2 0, there
exist q(x), r(x) [ F[x] such that f(x)
5 g(x)q(x) 1 r(x), 0 # deg r(x) ,
deg g(x) or r(x) 5 0
PID:
Every nonzero ideal I 5 ͗f(x)͘, where
deg f(x) is minimum
Irreducible:
No nontrivial factors
UFD:
Every element is a “unique” product of
irreducibles

EXAMPLE 7 The ring of Gaussian integers
Z[i] 5 {a 1 bi | a, b [ Z}
is a Euclidean domain with d(a 1 bi) 5 a2 1 b2
...
That d(x) # d(xy) for x, y [ Z[i] follows
directly from the fact that d(xy) 5 d(x)d(y) (Exercise 7)
...
Say
xy21 5 s 1 ti, where s, t [ Q
...
(These integers may not be uniquely
determined, but that does not matter
...
Then
xy21 5 s 1 ti 5 (m 2 m 1 s) 1 (n 2 n 1 t)i
5 (m 1 ni) 1 [(s 2 m) 1 (t 2 n)i]
...

We claim that the division condition of the definition of a Euclidean
domain is satisfied with q 5 m 1 ni and
r 5 [(s 2 m) 1 (t 2 n)i]y
...
Finally,
d(r) 5 d([(s 2 m) 1 (t 2 n)i])d(y)
5 [(s 2 m)2 1 (t 2 n)2]d(y)
#a

1
1
1 b d(y) , d(y)
...
4 ED (Euclidean Domain) Implies PID
Every Euclidean domain is a principal ideal domain
...
Among
all the nonzero elements of I, let a be such that d(a) is a minimum
...
For, if b [ I, there are elements q and r such that b 5 aq 1 r,
where r 5 0 or d(r) , d(a)
...
Thus, r 5 0 and b [ ͗a͘
...

Although it is not easy to verify, we remark that there are principal
ideal domains that are not Euclidean domains
...
Motzkin in 1949
...

As an immediate consequence of Theorems 18
...
4, we have
the following important result
...


We may summarize our theorems and remarks as follows:
ED ⇒ PID ⇒ UFD
UFD ⇒ PID ⇒ ED
/
/

334

Rings

(You can remember these implications by listing the types alphabetically
...

Since Z is a unique factorization domain, the next theorem is a broad
generalization of this fact
...

Theorem 18
...


We conclude this chapter with an example of an integral domain that
is not a unique factorization domain
...
It is straightforward
2
that Z[" 5] is an integral domain (see Exercise 9 in Chapter 13)
...
Since N(xy) 5
N(x)N(y) and N(x) 5 1 if and only if x is a unit (see Exercise 1), it follows that the only units of Z[" 5] are 61
...

2
2

We claim that each of these four factors is irreducible over Z[" 5]
...

2
Then 4 5 N(2) 5 N(x)N(y) and, therefore, N(x) 5 N(y) 5 2, which is
impossible
...
Thus, there would be integers a and b such that a2 1 5b2 5
23
...
The same argument applies to 1 6
3" 5
...
The answer is only when d 5 21
or 22 (see [1, p
...
The case where d 5 21 was first proved, naturally
enough, by Gauss
...

THOMAS MANN, The Magic Mountain

1
...
(This exercise is referred to in this chapter
...
In an integral domain, show that a and b are associates if and only
if ͗a͘ 5 ͗b͘
...
Show that the union of a chain I1 , I2 , ? ? ? of ideals of a ring R is
an ideal of R
...
)
4
...

5
...
Show that ͗ab͘ is a proper subset of ͗b͘
...
)
6
...
Define a , b if a and b are associates
...

7
...

8
...
Prove that u is a unit
in D if and only if d(u) 5 d(1)
...
Let D be a Euclidean domain with measure d
...

10
...
Prove that ͗p͘ is a
maximal ideal in D if and only if p is irreducible
...
Trace through the argument given in Example 7 to find q and r in
Z[i] such that 3 2 4i 5 (2 1 5i)q 1 r and d(r) , d(2 1 5i)
...
Let D be a principal ideal domain
...

13
...

14
...

15
...
(Hint:
2
Factor 10 in two ways
...
Give an example of a unique factorization domain with a subdomain that does not have a unique factorization
...
In Z[i], show that 3 is irreducible but 2 and 5 are not
...
Prove that 7 is irreducible in Z["6], even though N(7) is not prime
...
Prove that if p is a prime in Z that can be written in the form a2 1 b2,
then a 1 bi is irreducible in Z[i]
...

20
...

2
21
...

2
2
22
...

23
...

24
...
Show that in F[x] a prime ideal is a maximal ideal
...
Let d be an integer less than 21 that is not divisible by the square
of a prime
...

26
...

27
...

28
...

29
...
If p | a1a2 ? ? ? an, prove that
p divides some ai
...
)
30
...
Explain why this does not contradict the corollary of Theorem 18
...

31
...

Prove that D/͗p͘ is a field
...
Show that an integral domain with the property that every strictly
decreasing chain of ideals I1
...
? ? ? must be finite in length is
a field
...
An ideal A of a commutative ring R with unity is said to be finitely
generated if there exist elements a1, a2,
...
, an͘
...
Show that an integral domain R
satisfies the ascending chain condition if and only if every ideal of
R is finitely generated
...
Prove or disprove that a subdomain of a Euclidean domain is a
Euclidean domain
...
Show that for any nontrivial ideal I of Z[i], Z[i]/I is finite
...
Find the inverse of 1 1 "2 in Z["2]
...
In Z[" 7], show that N(6 1 2" 7) 5 N(1 1 3" 7) but 6 1
2
2
2" 7 and 1 1 3" 7 are not associates
...
Let R 5 Z % Z % ? ? ? (the collection of all sequences of integers
under componentwise addition and multiplication)
...
with the property that I1 , I2 , I3 , ? ? ?
...
)
39
...

40
...
(That is, R is contained in every integral domain in F 3x4 that
contains x 2 and x 3
...

41
...


Computer Exercise
I never use a computer
...
d
...
edu/~jgallian
1
...
Run the program for several cases and
formulate a conjecture based on your data
...
H
...
Stark, An Introduction to Number Theory, Chicago, Ill
...

2
...
C
...


Suggested Readings
Oscar Campoli, “A Principal Ideal Domain That Is Not a Euclidean
Domain,” The American Mathematical Monthly 95 (1988): 868–871
...


338

Rings

Gina Kolata, “At Last, Shout of ‘Eureka!’ in Age-Old Math Mystery,” The
New York Times, June 24, 1993
...

C
...

The demise of Miyaoka’s proof of Fermat’s Last Theorem is charmingly lamented
...

This article gives a short proof that Z["2 n] 5 {a 1 b"2 n | a, b [ Z} is
an integral domain that is not Euclidean when n
...

Simon Singh and Kenneth Ribet, “Fermat’s Last Stand,” Scientific American 277 (1997): 68–73
...


Suggested Video
The Proof, Nova, http://shop
...
org/product/show/7827
This documentary film shown on PBS’s NOVA program in 1997
chronicles the seven-year effort of Andrew Wiles to prove Fermat’s Last
Theorem
...
youtube
...


Suggested Websites
http://www
...
umn
...

http://en
...
org/wiki/Fermat%27s_Last_Theorem
This website provides a concise history of the efforts to prove Fermat’s Last
Theorem
...


Sophie Germain
One of the very few women to overcome
the prejudice and discrimination that
tended to exclude women from the pursuit
of higher mathematics in her time was
Sophie Germain
...
She educated herself by reading the
works of Newton and Euler in Latin and the
lecture notes of Lagrange
...

Gauss gave Germain’s results high praise and
a few years later, upon learning her true identity, wrote to her:
But how to describe to you my admiration
and astonishment at seeing my esteemed correspondent Mr
...
A taste for the abstract
sciences in general and above all the mysteries of numbers is excessively rare: it is not a
subject which strikes everyone; the enchanting charms of this sublime science reveal

themselves only to those who have the
courage to go deeply into it
...


Germain is best known for her work on
Fermat’s Last Theorem
...

For more information about Germain,
visit:
http://www-groups
...
st-and
...
uk/~history

339

Andrew Wiles
For spectacular contributions to number
theory and related fields, for major
advances on fundamental conjectures,
and for settling Fermat’s Last Theorem
...
00 each for public lectures on the proof
...
00 a ticket
for the sold-out event
...
By December, Wiles released a
Fermat’s Last Theorem
...
In September of 1994, a
paper by Wiles and Richard Taylor, a former
IN 1993, ANDREW WILES of Princeton electristudent of Wiles, circumvented the gap in
fied the mathematics community by announcthe original proof
...
His proof, which
errors
...
” In 1997,
and group theory
...

based on deep results that had already shed
Wiles was born in 1953 in Cambridge,
much light on the problem, many experts in
England
...
Wiles’s
University in 1980
...
The New York
Princeton in 1982, and he has received many
Times ran a front-page story on it, and one TV
prestigious awards
...

To find more information about Wiles,
Wiles even made People magazine’s list of the
visit:
25 most intriguing people of 1993! In San
Francisco a group of mathematicians rented a
http://www-groups
...
st-and
...
uk/~history/

340

341

18 | Supplementary Exercises for Chapters 15–18

Supplementary Exercises for Chapters 15–18
The intelligence is proved not by ease of learning, but by understanding
what we learn
...
d
...
edu/~jgallian/TF
1
...
Show that F is isomorphic to Zp for some prime p
...
Let Q["2] 5 {r 1 s"2 | r, s [ Q}
...

3
...
Show that A > B is an ideal of A and that
A/(A > B) is isomorphic to (A 1 B)/B
...
)
4
...
Show that A/B is an ideal of R/B and
(R/B)/(A/B) is isomorphic to R/A
...
Let f(x) and g(x) be irreducible polynomials over a field F
...

6
...
Explain why Exercise 5 is a
special case of this theorem
...
Prove that the set of all polynomials all of whose coefficients are
even is a prime ideal in Z[x]
...
Let R 5 Z[" 5] and let I 5 {a 1 b" 5 | a, b [ Z, a 2 b is
2
2
even}
...

9
...
Show that the mapping from R into itself given by x → axa21 is a ring automorphism
...
Let a 1 b" 5 belong to Z[" 5] with b 2 0
...

11
...
How many elements does it have?
12
...
In Z["2] 5 {a 1 b"2 | a, b [ Z}, show that every element of the
form (3 1 2"2)n is a unit, where n is a positive integer
...
Let p be a prime
...

15
...
Show that if 1 1 k is an
idempotent in Zn, then n 2 k is an idempotent in Zn
...
Show that Zn (where n
...
(The number is 2d, where d is the number of distinct prime
divisors of n
...
Show that the equation x2 1 y2 5 2003 has no solutions in the
integers
...
Prove that if both k and k 1 1 are idempotents in Zn and k 2 0, then
n 5 2k
...
Prove that x4 1 15x3 1 7 is irreducible over Q
...
For any integers m and n, prove that the polynomial x3 1 (5m 1 1)x 1
5n 1 1 is irreducible over Z
...
Prove that ͗"2͘ is a maximal ideal in Z["2]
...
Prove that Z[" 2] and Z["2] are unique factorization domains
...
)
23
...
Express both 13 and 5 1 i as products of irreducibles from Z[i]
...
Let R 5 {a/b | a, b [ Z, 3 B b}
...

Find its field of quotients
...
Give an example of a ring that contains a subring isomorphic to Z
and a subring isomorphic to Z3
...
Show that Z[i]/͗3͘ is not ring-isomorphic to Z3 % Z3
...
For any n
...

30
...

32
...


a 0
d ` a, b P Z n f is ring0 b

isomorphic to Zn % Zn
...
Prove
that R[x]/I[x] is isomorphic to (R/I)[x]
...

Find an ideal I of Z8[x] such that the factor ring Z8[x]/I is an integral
domain but not a field
...

Find an ideal I of Z[x] such that Z[x]/I is ring-isomorphic to Z3
...
hmco
...

J
...
R
...

Thus far we have covered groups and rings in some detail, and we have
touched on the notion of a field
...
In this chapter, we provide a concise review of this material
...

1
...

3
...


a(v 1 u) 5 av 1 au
(a 1 b)v 5 av 1 bv
a(bv) 5 (ab)v
1v 5 v

The members of a vector space are called vectors
...
The operation that combines a scalar a and
a vector v to form the vector av is called scalar multiplication
...


345

346

Fields

EXAMPLE 1 The set Rn 5 {(a1, a2,
...
Here the operations are the obvious ones
...
, an) 1 (b1, b2,
...
, an 1 bn)
and
b(a1, a2,
...
, ban)
...
The operations are
c

a1 a2
b b
a 1 b1 a2 1 b2
d 1 c 1 2d 5 c 1
d
a3 a4
b3 b4
a3 1 b3 a4 1 b4

and
bc

a1 a2
ba1 ba2
d
...

EXAMPLE 4 The set of complex numbers C 5 {a 1 bi | a, b [ R}
is a vector space over R
...

The next example is a generalization of Example 4
...

EXAMPLE 5 Let E be a field and let F be a subfield of E
...
The operations are the operations of E
...

Definition Subspace

Let V be a vector space over a field F and let U be a subset of V
...


EXAMPLE 6 The set {a2x2 1 a1x 1 a0 | a0, a1, a2 [ R} is a subspace of the vector space of all polynomials with real coefficients
over R
...
, vn be
(not necessarily distinct) elements of V
...
, vn͘ 5 {a1v1 1 a2v2 1 ? ? ? 1 anvn | a1, a2,
...
, vn
...
, vn
...
, vn͘ 5 V, we say that {v1, v2,
...


Linear Independence
The next definition is the heart of the theory
...
, vn from S and elements a1, a2,
...
A set of vectors
that is not linearly dependent over F is called linearly independent
over F
...
To verify this, assume that there are real
numbers a, b, and c such that a(1, 0, 0) 1 b(1, 0, 1) 1 c(1, 1, 1) 5
(0, 0, 0)
...
From this we see that
a 5 b 5 c 5 0
...

Definition Basis

Let V be a vector space over F
...


The motivation for this definition is twofold
...
Second, with every vector
space spanned by finitely many vectors, we can use the notion of basis
to associate a unique integer that tells us much about the vector space
...
)
a
a1b
d ` a, b P R f
a1b
b
is a vector space over R (see Exercise 17)
...
To prove that the set
1 0
1 1
B is linearly independent, suppose that there are real numbers a and b
such that
B5 ec

a c
This gives c

1 1
0 1
0 0
d 1b c
d 5 c
d
...
On the other
0

hand, since every member of V has the form
c

a
a1b
1 1
0 1
d 5 ac
d 1 bc
d,
a1b
b
1 0
1 1

we see that B spans V
...

Theorem 19
...
, um} and {w1, w2,
...


PROOF Suppose that m 2 n
...

Consider the set {w1, u1, u2 ,
...
Since the u’s span V, we
know that w1 is a linear combination of the u’s, say, w1 5 a1u1 1
a2u2 1 ? ? ? 1 amum, where the a’s belong to F
...
For convenience, say a1 2 0
...
, um} spans V (see
Exercise 21)
...
, um}
...
, um, say, w2 5 b1w1 1 b2u2 1
...
Then at least one of b2,
...
Let us say
b2 2 0
...
, um span V
...
, wm} spans V
...
, wm and, therefore, the set {w1,
...
This contradiction finishes the proof
...
1 shows that any two finite bases for a vector space have
the same size
...

However, there is no vector space that has a finite basis and an infinite
basis (see Exercise 25)
...
For completeness, the trivial vector space {0} is
said to be spanned by the empty set and to have dimension 0
...
A vector
space that has a finite basis is called finite dimensional; otherwise, it is
called infinite dimensional
...

THE LAW OF THUMB

1
...
Find a basis for each of the vector spaces in Examples 1–4
...
(Subspace Test) Prove that a nonempty subset U of a vector space
V over a field F is a subspace of V if, for every u and u9 in U and
every a in F, u 1 u9 [ U and au [ U
...
Verify that the set in Example 6 is a subspace
...
Is {x2 1 x 1 1, x 1 5, 3} a basis?
4
...
, vn͘ defined in Example 7 is a subspace
...
Determine whether or not the set {(2, 21, 0), (1, 2, 5), (7, 21, 5)} is
linearly independent over R
...
Determine whether or not the set
ec

2 1
0 1
1 1
d, c
d, c
df
1 0
1 2
1 1

is linearly independent over Z5
...
If {u, v, w} is a linearly independent subset of a vector space, show
that {u, u 1 v, u 1 v 1 w} is also linearly independent
...
If S is a linearly dependent set of vectors, prove that one of the vectors in S is a linear combination of the others
...
(Every finite spanning collection contains a basis
...
, vn}
spans a vector space V, prove that some subset of the v’s is a basis
for V
...
(Every independent set is contained in a basis
...
, vn} be a linearly

350

Fields

independent subset of V
...
, wm
such that {v1, v2,
...
, wm} is a basis for V
...
If V is a vector space over F of dimension 5 and U and W are subspaces of V of dimension 3, prove that U > W 2 {0}
...

12
...

13
...
Is V a vector space over Q?
14
...
Is W a subspace of V? If so, what is its dimension?
15
...
Is W a subspace
of V? If so, what is its dimension?
a b
d ` a, b, c P Q f
...

Verify that the set V in Example 9 is a vector space over R
...
Prove that P is a subspace of R3
...
Give a geometric description of P
...
Show that B is a basis for V
if and only if every member of V is a unique linear combination of
the elements of B
...
)
If U is a proper subspace of a finite-dimensional vector space V,
show that the dimension of U is less than the dimension of V
...
1, prove that {w1, u2,
...

If V is a vector space of dimension n over the field Zp, how many
elements are in V?
Let S 5 {(a, b, c, d) | a, b, c, d [ R, a 5 c, d 5 a 1 b}
...

Let U and W be subspaces of a vector space V
...


16
...

18
...


20
...

22
...

24
...
If a vector space has one basis that contains infinitely many elements, prove that every basis contains infinitely many elements
...
)
26
...
Since 1 u 2
2
2
1
3 v 2 6 w 5 (0, 0, 0), can we conclude that the set {u, v, w} is linearly dependent over Z7?
27
...
Such a mapping is called a linear transformation
...

28
...
Prove that the image
of V under T is a subspace of W
...
Let T be a linear transformation of a vector space V
...

30
...
If {v1, v2,
...
, T(vn)} spans W
...
If V is a vector space over F of dimension n, prove that V is isomorphic as a vector space to Fn 5 {(a1, a2,
...
(This
exercise is referred to in this chapter
...
Let V be a vector space over an infinite field
...


Emil Artin
For Artin, to be a mathematician meant to
participate in a great common effort, to
continue work begun thousands of years
ago, to shed new light on old discoveries,
to seek new ways to prepare the developments of the future
...

RICHARD BRAUER,
Bulletin of the American
Mathematical Society

EMIL ARTIN was one of the leading mathematicians of the 20th century and a major
contributor to linear algebra and abstract algebra
...
He received a Ph
...
in 1921 from the University
of Leipzig
...
After one year at
Notre Dame, Artin went to Indiana
University
...
The last four
years of his career were spent where it
began, at Hamburg
...
He made contributions to number theory, group theory, ring theory, field theory,
Galois theory, geometric algebra, algebraic

352

topology, and the theory of braids—a field he
invented
...

Artin was an outstanding teacher of
mathematics at all levels, from freshman
calculus to seminars for colleagues
...
D
...
Through his research, teaching, and
books, Artin exerted great influence among
his contemporaries
...

For more information about Artin, visit:
http://www-groups
...
st-and
...
uk/~history/

Olga Taussky-Todd
“Olga Taussky-Todd was a distinguished
and prolific mathematician who wrote
about 300 papers
...
Taussky-Todd received her doctoral
degree in 1930 from the University of Vienna
...
She also
edited lecture notes of Emil Artin and assisted Richard Courant
...

In 1937, she taught at the University of
London
...
In 1957, she became the first
woman to teach at the California Institute of
Technology as well as the first woman to
receive tenure and a full professorship in
mathematics, physics, or astronomy there
...
D
...
D
...


In addition to her influential contributions to linear algebra, Taussky-Todd did
important work in number theory
...
She was elected a Fellow of the
American Association for the Advancement
of Science and vice president of the American
Mathematical Society
...

Taussky-Todd died on October 7, 1995, at the
age of 89
...
dcs

...
ac
...
edu/lriddle/women/women
...

RICHARD A
...
Indeed, we saw that Z3[x]/͗x2 1 1͘ is a field of order 9,
whereas R[x]/͗x2 1 1͘ is a field isomorphic to the complex numbers
...

Definition Extension Field

A field E is an extension field of a field F if F # E and the operations
of F are those of E restricted to F
...

Theorem 20
...

Then there is an extension field E of F in which f(x) has a zero
...
Clearly, it suffices to construct an extension
field E of F in which p(x) has a zero
...

We already know that this is a field from Corollary 1 of Theorem 17
...

Also, since the mapping of f: F S E given by f(a) 5 a 1 ͗ p(x)͘ is
one-to-one and preserves both operations, E has a subfield isomorphic
354

20 | Extension Fields

355

to F
...

Finally, to show that p(x) has a zero in E, write
p(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0
...

EXAMPLE 1 Let f(x) 5 x2 1 1 [ Q[x]
...

Of course, the polynomial x2 1 1 has the complex number " as a
21
zero, but the point we wish to emphasize here is that we have constructed a field that contains the rational numbers and a zero for the
polynomial x2 1 1 by using only the rational numbers
...
Our method utilizes only the field we
are given
...
Then, the irreducible factorization of f(x) over Z3 is (x2 1 1)(x3 1 2x 1 2)
...

Since every integral domain is contained in its field of quotients
(Theorem 15
...
The next example shows that this is not
true for commutative rings in general
...
Then f(x) has no zero in any
ring containing Z4 as a subring, because if b were a zero in such a ring,

356

Fields

then 0 5 2b 1 1, and therefore 0 5 2(2b 1 1) 5 2(2b) 1 2 5
(2 ? 2)b 1 2 5 0 ? b 1 2 5 2
...


Splitting Fields
To motivate the next definition and theorem, let’s return to Example 1 for
a moment
...
Then, since a and 2a are both zeros of x2 1 1 in (Q[x]/
͗x2 1 1͘)[x], it should be the case that x2 1 1 5 (x 2 a)(x 1 a)
...
First note that
(x 2 a)(x 1 a) 5 x2 2 a2 5 x2 2 (x2 1 ͗x2 1 1͘)
...

This shows that x2 1 1 can be written as a product of linear factors in
some extension of Q
...

The polynomial given in Example 2 presents a greater challenge
...
But first a definition
...
We say that f(x)
splits in E if f(x) can be factored as a product of linear factors in E[x]
...


Note that a splitting field of a polynomial over a field depends not
only on the polynomial but on the field as well
...

The next example illustrates how a splitting field of a polynomial f (x)
over field F depends on F
...
Since

2
2
x2 1 1 5 (x 1 " 1)(x 2 " 1), we see that f(x) splits in C, but a splitting

field over Q is Q(i) 5 {r 1 si | r, s [ Q}
...
Likewise, x2 2 2 [ Q[x] splits in R, but a splitting field over Q is
Q("2) 5 {r 1 s "2 | r, s [ Q}
...
Just as it makes no sense to
say “f(x) is irreducible,” it makes no sense to say “E is a splitting field for
f(x)
...
”
The following notation is convenient
...
, an be elements of some extension E of F
...
, an) to denote the smallest subfield of E that contains F
and the set {a1, a2,
...
It is an easy exercise to show that F(a1,
a2,
...
, an}
...
, an) is a splitting field for
f(x) over F in E
...
For example, we denoted the set {a 1 b"2 |
a, b [ Z} by Z["2] and the set {a 1 b"2 | a, b [ Q} by Q("2 )
...
In
general, parentheses are used when one wishes to indicate that the set is
a field, although no harm would be done by using, say, Q["2] to denote
{a 1 b"2 | a, b [ Q} if we were concerned with its ring properties
only
...

Theorem 20
...
Then
there exists a splitting field E for f(x) over F
...
If deg f (x) 5 1, then
f (x) is linear
...
By Theorem 20
...
Then we
may write f (x) 5 (x 2 a1)g(x), where g(x) [ E[x]
...
, an
...
, an)
...
Obviously, the zeros of f (x) in C are 6"2 and 6i
...


EXAMPLE 6 Consider f(x) 5 x2 1 x 1 2 over Z3
...

At the same time, we know by the proof of Kronecker’s Theorem that
the element x 1 ͗x2 1 x 1 2͘ of
F 5 Z3[x]/͗x2 1 x 1 2͘
is a zero of f(x)
...
2) that the other zero of f(x) must
also be in F
...
But how do we factor f(x) in F? Factoring f(x) in F is confusing because we are using the symbol x in two distinct ways: It is used as a
placeholder to write the polynomial f(x), and it is used to create the coset
representatives of the elements of F
...
With this identification, the field
Z3[x]/͗x2 1 x 1 2͘ can be represented as {0, 1, 2, b, 2b, b 1 1, 2b 1 1,
b 1 2, 2b 1 2}
...

For example, (2b 1 1)(b 1 2) 5 2b2 1 5b 1 2 5 2(2b 1 1) 1 5b 1
2 5 9b 1 4 5 1
...

So, x2 1 x 1 2 5 (x 2 b)(x 1 b 1 1)
...


20 | Extension Fields

359

The next theorem shows how the fields F(a) and F[x]/͗ p(x)͘ are
related in the case where p(x) is irreducible over F and a is a zero of
p(x) in some extension of F
...
3 F(a) < F[x]/͗p(x)͘
Let F be a field and let p(x) [ F[x] be irreducible over F
...
Furthermore, if deg p(x) 5 n, then every member of F(a)
can be uniquely expressed in the form
cn21an21 1 cn22an22 1 ? ? ? 1 c1a 1 c0,
where c0 , c1,
...


PROOF Consider the function f from F[x] to F(a) given by f(f(x)) 5
f(a)
...
We claim that Ker f 5 ͗p(x)͘
...
) Since p(a) 5 0, we have ͗p(x)͘ #
Ker f
...
5 that ͗p(x)͘ is a
maximal ideal in F[x]
...
At this
point it follows from the First Isomorphism Theorem for Rings and
Corollary 1 of Theorem 17
...
Noting
that f(F[x]) contains both F and a and recalling that F(a) is the smallest such field, we have F[x]/͗p(x)͘ < f(F[x]) 5 F(a)
...
, cn21 [ F (see Exercise 23 in Chapter 16) and the
natural isomorphism from F[x]/͗ p(x)͘ to F(a) carries ckxk 1 ͗ p(x)͘
to ckak
...
3, we have the following
attractive result
...
If a is a
zero of p(x) in some extension E of F and b is a zero of p(x) in some
extension E9 of F, then the fields F(a) and F(b) are isomorphic
...
3, we have
F(a) < F[x]/͗p(x)͘ < F(b)
...
, vn with the property that every member
of the vector space can be expressed uniquely in the form a1v1 1
a2v2 1 ? ? ? 1 anvn, where the a’s belong to F (Exercise 19 in Chapter 19)
...
3 says
that if a is a zero of an irreducible polynomial over F of degree n, then the
set {1, a,
...

Theorem 20
...

EXAMPLE 7 Consider the irreducible polynomial f(x) 5 x6 2 2
6
over Q
...
3 that the
6
set {1, 21/6, 22/6, 23/6, 24/6, 25/6} is a basis for Q(" 2) over Q
...

6

This field is isomorphic to Q[x]/͗x6 2 2͘
...
Because of this important result, Theorem 20
...

In Example 6, we produced two splitting fields for the polynomial
x2 1 x 1 2 over Z3
...
But are these different-looking splitting fields
algebraically different? Not really
...

To make it easier to apply induction, we will prove a more general result
...
Since this mapping agrees with f on F, it is convenient and natural to use f to denote this mapping as well
...
If f is a field isomorphism from
F to F9 and b is a zero of f(p(x)) in some extension of F9, then there
is an isomorphism from F(a) to F9(b) that agrees with f on F and
carries a to b
...
It is straightforward to check that the mapping
from F[x]/͗p(x)͘ to F9[x]/͗f(p(x))͘ given by
f(x) 1 ͗p(x)͘ S f(f(x)) 1 ͗f(p(x))͘
is a field isomorphism
...
(If you object, put a bar over the f
...
3, we know that there is an isomorphism a from F(a) to
F[x]/͗p(x)͘ that is the identity on F and carries a to x 1 ͗p(x)͘
...
Thus, bfa is the desired
mapping from F(a) to F9(b)
...
1
...
1

Theorem 20
...
If E is a splitting field for f(x) over F and E9 is a
splitting field for f(f (x)) over F9, then there is an isomorphism
from E to E9 that agrees with f on F
...
If deg f (x) 5 1, then E 5 F and E9 5
F9, so that f itself is the desired mapping
...
1, let p(x) be
an irreducible factor of f (x), let a be a zero of p(x) in E, and let b be a
zero of f(p(x)) in E9
...

Now write f (x) 5 (x 2 a)g(x), where g(x) [ F(a)[x]
...
Since deg g(x) , deg f (x), there is an isomorphism from E
to E9 that agrees with a on F(a) and therefore with f on F
...
Then any two splitting fields
of f(x) over F are isomorphic
...
The
result follows immediately from Theorem 20
...

In light of the corollary above, we may refer to “the” splitting field
of a polynomial over F without ambiguity
...

The splitting field is easy to obtain, however
...
Then each of
a1/n, va1/n, v2a1/n,
...

n

Zeros of an Irreducible Polynomial
Now that we know that every nonconstant polynomial over a field
splits in some extension, we ask whether irreducible polynomials must
split in some special way
...
To discover how, we borrow
something whose origins are in calculus
...
The
derivative of f(x), denoted by f9(x), is the polynomial nanx n21 1
(n 2 1)an21x n22 1 ? ? ? 1 a1 in F[x]
...
The
standard rules for handling sums and products of functions in calculus
carry over to arbitrary fields as well
...
Then
1
...
(af(x))9 5 af 9(x)
3
...


20 | Extension Fields

363

PROOF Parts 1 and 2 follow from straightforward applications of the
definition
...
This also follows directly from
the definition
...
Such zeros are called multiple zeros
...
5 Criterion for Multiple Zeros
A polynomial f(x) over a field F has a multiple zero in some
extension E if and only if f(x) and f9(x) have a common factor of
positive degree in F[x]
...
Since f 9(x) 5
(x 2 a)2g9(x) 1 2(x 2 a)g(x), we see that f 9(a) 5 0
...
Now if f(x) and f 9(x)
have no common divisor of positive degree in F[x], there are polynomials
h(x) and k(x) in F[x] such that f(x)h(x) 1 f 9(x)k(x) 5 1 (see Exercise 41
in Chapter 16)
...
Since this is nonsense, f(x) and f 9(x)
must have a common divisor of positive degree in F[x]
...
Let a be a zero of the common factor
...
Since a is a zero of f (x), there is a polynomial q(x) such that
f (x) 5 (x 2 a)q(x)
...
Thus, x 2 a is a factor of q(x) and a is a multiple zero of f(x)
...
6 Zeros of an Irreducible
Let f(x) be an irreducible polynomial over a field F
...
If F has characteristic p 2 0, then f(x) has a multiple zero only if it is of the
form f(x) 5 g(x p) for some g(x) in F[x]
...
5, f (x) and
f 9(x) have a common divisor of positive degree in F[x]
...
Because a polynomial over a field cannot
divide a polynomial of smaller degree, we must have f 9(x) 5 0
...
Thus, f 9(x) 5 0 only when kak 5 0 for k 5 1,
...

So, when char F 5 0, we have f(x) 5 a0, which is not an irreducible
polynomial
...
Thus, f(x) has no multiple zeros
...
Thus,
the only powers of x that appear in the sum anx n 1 ? ? ? 1 a1x 1 a0 are
those of the form x pj 5 (x p) j
...
[For example, if f(x) 5 x4p 1 3x2p 1 x p 1 1, then g(x) 5
x4 1 3x2 1 x 1 1
...
6 shows that an irreducible polynomial over a field of
characteristic 0 cannot have multiple zeros
...

Definition

A field F is called perfect if F has characteristic 0 or if F has
characteristic p and Fp 5 {ap | a [ F} 5 F
...

Theorem 20
...


PROOF Let F be a finite field of characteristic p
...
We claim that
f is a field automorphism
...
Moreover, f(a 1 b) 5 (a 1 b)p 5 ap 1 a b ap21b 1
1
p p22 2
p
a b a b 1???1 a
b ab p21 1 b p 5 a p 1 b p, since each
2
p21
p
a b is divisible by p
...

i
Thus, f is one-to-one and, since F is finite, f is onto
...


365

20 | Extension Fields

Theorem 20
...


PROOF The case where F has characteristic 0 has been done
...
From Theorem 20
...
Since Fp 5 F, each ai in F can be written
in the form bip for some bi in F
...
But then f(x) is not irreducible
...

Theorem 20
...
Then all the zeros of f(x) in E have the
same multiplicity
...
If a has multiplicity m,
then in E[x] we may write f(x) 5 (x 2 a)mg(x)
...
4 and from Theorem 20
...
Thus,
f(x) 5 f( f(x)) 5 (x 2 b)mf(g(x))
and we see that the multiplicity of b is greater than or equal to the multiplicity of a
...
So, we
have proved that a and b have the same multiplicity
...
9 we have the following
appealing result
...
Then f(x) has the form
a(x 2 a1)n(x 2 a2)n ? ? ? (x 2 at)n
where a1, a2,
...


We conclude this chapter by giving an example of an irreducible
polynomial over a field that does have a multiple zero
...

EXAMPLE 9 Let F 5 Z2(t) be the field of quotients of the ring Z2[t]
of polynomials in the indeterminate t with coefficients from Z2
...
) Consider f(x) 5 x2 2 t [
F[x]
...
Well, suppose that h(t)/k(t) is a zero of f(x)
...
Since h(t), k(t) [ Z2[t],
we then have h(t2) 5 tk(t2) (see Exercise 45 in Chapter 13)
...
So, f(x) is irreducible over F
...
So, by
Theorem 20
...
(Indeed, it
has a single zero of multiplicity 2 in K 5 F[x]/͗x2 2 t͘
...

PAUL ANDERSON,

New Scientist

1
...

2
...

3
...
Express your answer in
the form Q(a)
...
Find the splitting field of x4 1 1 over Q
...
Find the splitting field of
3

x4 1 x2 1 1 5 (x2 1 x 1 1)(x2 2 x 1 1)
over Q
...
Let a, b [ R with b 2 0
...

_
7
...

8
...
Suppose that a is a
zero of f(x) in some extension of F
...
Write out a
complete multiplication table for F(a)
...
Let F(a) be the field described in Exercise 8
...

10
...
Show that a2 and a2 1 a
are zeros of x3 1 x 1 1
...
Describe the elements in Q(p)
...
Let F 5 Q(p3)
...

13
...
Do the same for
x10 2 x
...
Find all ring automorphisms of Q( " 5)
...
Let F be a field of characteristic p and let f (x) 5 x p 2 a [ F[x]
...

16
...
Write f(x) as a product of linear factors in E[x]
...
Find a, b, c in Q such that
(1 1 "4)/(2 2 "2) 5 a 1 b "2 1 c " 4
...

3

3

3

3

3

18
...

19
...

2
20
...
If c
belongs to some extension of F, prove that F(c) 5 F(ac 1 b)
...
)
21
...
Show that f(x) and f(x 1 a) have the
same splitting field over F
...
Recall that two polynomials f (x) and g(x) from F[x] are said to be
relatively prime if there is no polynomial of positive degree in F[x]
that divides both f (x) and g(x)
...

23
...


368

Fields

24
...
Prove that
F(a, b) 5 F(a)(b) 5 F(b)(a)
...
Write x3 1 2x 1 1 as a product of linear polynomials over some
field extension of Z3
...
Express x8 2 x as a product of irreducibles over Z2
...
Prove or disprove that Q("3) and Q(" 3) are ring-isomorphic
...
For any prime p, find a field of characteristic p that is not perfect
...
If b is a zero of x2 1 x 1 2 over Z5, find the other zero
...
Show that x4 1 x 1 1 over Z2 does not have any multiple zeros in
any field extension of Z2
...
Show that x21 1 2x8 1 1 does not have multiple zeros in any
extension of Z3
...
Show that x21 1 2x9 1 1 has multiple zeros in some extension of Z3
...
Let F be a field of characteristic p 2 0
...

34
...
Write f(x) as a product of linear factors
...
Let F, K, and L be fields with F # K # L
...

36
...

37
...
Prove that every nonidentity element is a generator of the
cyclic group (Z2 3x4>͗f(x) ͘) *
...

E
...
BELL

LEOPOLD KRONECKER was born on December
7, 1823, in Leignitz, Prussia
...
Kronecker entered the
University of Berlin in 1841 and completed
his Ph
...
dissertation in 1845 on the units in a
certain ring
...
Thereafter, being well-off financially,
he spent most of his time doing research in algebra and number theory
...
He innovatively applied rings and
fields in his investigations of algebraic numbers, established the Fundamental Theorem of
Finite Abelian Groups, and was the first mathematician to master Galois’s theory of fields
...
He believed
that all mathematics should be based on relationships among integers
...
His most famous remark on the
matter was “God made the integers, all the
rest is the work of man
...

Kronecker died on December 29, 1891,
at the age of 68
...
dcs

...
ac
...
”
S
...
ULAM, Adventures of a Mathematician

Characterization of Extensions
In Chapter 20, we saw that every element in the field Q("2) has the
particularly simple form a 1 b"2, where a and b are rational
...
The fields of the first type have a
great deal of structure
...

Definition Types of Extensions

Let E be an extension field of a field F and let a [ E
...
If
a is not algebraic over F, it is called transcendental over F
...
If E is not an algebraic extension of F, it is called a
transcendental extension of F
...


Leonhard Euler used the term transcendental for numbers that are
not algebraic because “they transcended the power of algebraic methods
...
Charles Hermite proved that e is transcendental over
Q in 1873, and Lindemann showed that p is transcendental over Q in
1882
...

With a precise definition of “almost all,” it can be shown that almost all
real numbers are transcendental over Q
...
1 shows why we make the distinction between elements
that are algebraic over a field and elements that are transcendental over
a field
...

Theorem 21
...
If a is
transcendental over F, then F(a) < F(x)
...
Moreover, p(x) is irreducible over F
...
If a is transcendental over F, then Ker f 5 {0}, and so
we may extend f to an isomorphism f:F(x) S F(a) by defining
f ( f(x)/g(x)) 5 f(a)/g(a)
...
4,
there is a polynomial p(x) in F[x] such that Ker f 5 ͗p(x)͘ and p(x) has
minimum degree among all nonzero elements of Ker f
...

The proof of Theorem 21
...
The details are left to the reader (see Exercise 1)
...
2 Uniqueness Property
If a is algebraic over a field F, then there is a unique monic irreducible polynomial p(x) in F[x] such that p(a) 5 0
...
2 is
called the minimal polynomial for a over F
...
3 Divisibility Property
Let a be algebraic over F, and let p(x) be the minimal polynomial for
a over F
...


If E is an extension field of F, we may view E as a vector space over F
(that is, the elements of E are the vectors and the elements of F are the
scalars)
...


372

Fields

Finite Extensions
Definition Degree of an Extension

Let E be an extension field of a field F
...
If [E:F] is finite, E is called a finite extension of F; otherwise,
we say that E is an infinite extension of F
...
1 illustrates a convenient method of depicting the degree
of a field extension over a field
...
1

EXAMPLE 1 The field of complex numbers has degree 2 over the
reals, since {1, i} is a basis
...

EXAMPLE 2 If a is algebraic over F and its minimal polynomial
over F has degree n, then, by Theorem 20
...
,
an21} is a basis for F(a) over F; and, therefore, [F(a):F] 5 n
...

Theorem 21
...


PROOF Suppose that [E:F] 5 n and a [ E
...
, an}
is linearly dependent over F; that is, there are elements c0, c1,
...

Clearly, then, a is a zero of the nonzero polynomial
f(x) 5 cnx n 1 cn21x n21 1 ? ? ? 1 c1x 1 c0
...
4 is not true, for otherwise, the degrees of the elements of every algebraic extension of E over F would
3
4
be bounded
...
) is an algebraic extension of Q
that contains elements of every degree over Q (see Exercise 3)
...
Like all counting theorems, it has far-reaching
consequences
...
5 [K:F] 5 [K:E][E:F]
Let K be a finite extension field of the field E and let E be a finite
extension field of the field F
...


PROOF Let X 5 {x1, x2,
...
, ym} be a basis for E over F
...
To do this, let a [ K
...
, bn [ E such that
a 5 b1x1 1 b2x2 1 ? ? ? 1 bnxn
...
, n, there are elements ci1, ci2,
...

Thus,

n

n

m

i51

i51

j51

a 5 a bi xi 5 a a a cij yj b xi 5 a cij (yj xi )
...

Now suppose there are elements cij in F such that

0 5 g c ij (yj x i) 5 g A g (c ij yj)Bx i
...
But each cij [ F and Y is a basis for E over F, so each cij 5 0
...

Using the fact that for any field extension L of a field J, [L:J] 5 n if
and only if L is isomorphic to Jn as vector spaces (see Exercise 29), we
may give a concise conceptual proof of Theorem 21
...
Let

374

Fields
K
n
nm

E
m
F

[K:F ] = [K:E ][E:F ]

Figure 21
...
Then K < En and E < Fm, so that K < En <
(Fm)n < Fmn
...

The content of Theorem 21
...
2
...
5 is often utilized
...
5 shows that {1, "3, "5, "15} is a basis for Q("3, "5)
over Q
...
3
...
Then [Q(" 2, " 3) : Q] 5 12
...

3
4
3
4
3
Thus, [Q(" 2, " 3) : Q] $ 12
...
Therefore,
3
4
3
4
3
3
[Q(" 2, " 3) : Q] 5 [Q(" 2, " 3 ): Q(" 2)][Q(" 2) : Q] # 4 ? 3 5 12
...
4
...
3

3

3

Q(√2)

4

Q(√3)

12
3

4
Q

Figure 21
...
5 can sometimes be used to show that a field does not
contain a particular element
...
Let b be a zero of h(x) in some extension of Q
...
For, if so, then Q , Q(" 2) #
3
3
Q(b) and 4 5 [Q(b) : Q] 5 [Q(b) : Q(" 2)][Q(" 2) : Q] implies that
3 divides 4
...

EXAMPLE 6 Consider Q("3, "5)
...
The inclusion Q("3 1 "5) # Q("3, "5) is clear
...
It
2
follows that [("3 1 "5) 1 ("3 2 "5)]/2 5 "3 and [("3 1 "5 )2
("3 2 "5)]/2 5 "5 both belong to Q("3 1 "5), and therefore
Q("3, "5) # Q("3 1 "5 )
...
3 that the
minimal polynomial for "3 1 "5 over Q has degree 4
...
Then x2 5 3 1
2"15 1 5
...
Thus, "3 1 "5 is a zero of x4 2 16x 1 4
...

Example 6 shows that an extension obtained by adjoining two elements to a field can sometimes be obtained by adjoining a single
element to the field
...

Theorem 21
...


PROOF Let p(x) and q(x) be the minimal polynomials over F for a and
b, respectively
...
, am and b1, b2,

...
Among the infinitely many elements of F, choose an
element d not equal to (ai 2 a)/(b 2 bj) for all i $ 1 and all j
...
In
particular, ai 2 a 1 d(b 2 bj) for j
...

We shall show that c 5 a 1 db has the property that F(a, b) 5 F(c)
...
To verify that F(a, b) # F(c), it suffices to
prove that b [ F(c), for then b, c, and d belong to F(c) and a 5 c 2 bd
...
Since both
q(b) 5 0 and r(b) 5 p(c 2 db) 5 p(a) 5 0, both q(x) and r(x) are divisible by the minimal polynomial s(x) for b over F(c) (see Theorem 21
...

Because s(x) [ F(c)[x], we may complete the proof by proving that
s(x) 5 x 2 b
...
But r(bj) 5 p(c 2 dbj) 5 p(a 1db 2 dbj) 5 p(a 1 d(b 2 bj)) and
d was chosen such that a 1 d(b 2 bj) 2 ai for j
...
It follows that b
is the only zero of s(x) in K[x] and, therefore, s(x) 5 (x 2 b)u
...
6 guarantees
that u 5 1
...
6
and induction that any finite extension of a field of characteristic 0 is a
simple extension
...


Properties of Algebraic Extensions
Theorem 21
...


PROOF Let a [ K
...
Since a is algebraic over E, we know that a is the zero
of some irreducible polynomial in E[x], say, p(x) 5 bnx n 1 ? ? ? 1 b0
...
, Fn 5 Fn21(bn)
...
, bn),

21 | Algebraic Extensions

377

K
E

Fn(a)
Fn
F 21
n
F1
F0
F

Figure 21
...
Thus, [Fn(a):Fn] 5 n; and, because each bi is algebraic over F, we know that each [Fi11:Fi] is finite
...
(See Figure 21
...
)
Corollary Subfield of Algebraic Elements
Let E be an extension field of the field F
...


PROOF Suppose that a, b [ E are algebraic over F and b 2 0
...
But note that
[F(a, b):F] 5 [F(a, b):F(b)][F(b):F]
...

Thus, both [F(a, b):F(b)] and [F(b):F] are finite
...

One might wonder if there is such a thing as a maximal algebraic
extension of a field F—that is, whether there is an algebraic extension E
of F that has no proper algebraic extensions
...
Otherwise, it follows
from Kronecker’s Theorem that E would have a proper algebraic extension
...
If every member of E[x] splits in E,
and K is an algebraic extension of E, then every member of K is a zero of

378

Fields

some element of E[x]
...
A field
that has no proper algebraic extension is called algebraically closed
...
This field is
called the algebraic closure of F
...

In 1799, Gauss, at the age of 22, proved that C is algebraically
closed
...
” Over a 50-year period,
Gauss found three additional proofs of the Fundamental Theorem
...
In view of the ascendancy of abstract
algebra in the 20th century, a more appropriate phrase for Gauss’s result
would be “The Fundamental Theorem of Classical Algebra
...

EDGAR GUEST

1
...
2 and Theorem 21
...

2
...
Show that every polynomial in
F[x] splits in E
...
Prove that Q("2, " 2, " 2,
...
(This exercise is referred to in this
chapter
...
Let E be an algebraic extension of F
...

5
...
Show that F is algebraically closed
...
Suppose that f (x) and g(x) are irreducible over F and that deg f (x)
and deg g(x) are relatively prime
...

7
...
Show that Q("a) 5 Q("b) if
and only if there exists some c [ Q such that a 5 bc2
...
Find the degree and a basis for Q("3 1 "5) over Q("15)
...

9
...
Show that, for
every a in E, F(a) 5 F or F(a) 5 E
...
Let a be a complex number that is algebraic over Q
...

11
...
If a is algebraic
over F of degree m, and b is algebraic over F of degree n, where m
and n are relatively prime, show that [F(a, b):F] 5 mn
...
Find an example of a field F and elements a and b from some
extension field such that F(a, b) 2 F(a), F(a, b) 2 F(b), and
[F(a, b):F] , [F(a):F][F(b):F]
...
Let K be a field extension of F and let a [ K
...
Find examples to illustrate that [F(a):F(a3)] can
be 1, 2, or 3
...
Find the minimal polynomial for " 3 1 "2 over Q
...
Let K be an extension of F
...
If [E1:F] and [E2:F] are both prime,
show that E1 5 E2 or E1 > E2 5 F
...
Find the minimal polynomial for " 2 1 " 4 over Q
...
Let E be a finite extension of R
...

18
...
Show that there is an integer d such that
E 5 Q("d) where d is not divisible by the square of any prime
...
Suppose that p(x) [ F[x] and E is a finite extension of F
...

20
...
Show that [E:F] is finite if and only
if E 5 F(a1, a2,
...
, an are algebraic over F
...
If a and b are real numbers and a and b are transcendental over Q,
show that either ab or a 1 b is also transcendental over Q
...
Let f (x) be a nonconstant element of F[x]
...

23
...
Find a primitive element for the
splitting field for f(x) over Q
...
Find the splitting field for x4 2 x2 2 2 over Z3
...
Let f (x) [ F[x]
...

26
...
Prove that
Q("a) 5 Q(a)
...
If F is a field and the multiplicative group of nonzero elements of
F is cyclic, prove that F is finite
...
Let a be a complex number that is algebraic over Q and let r be a
rational number
...

29
...
(See Exercise 27 in
Chapter 19 for the appropriate definition
...
)
30
...

Prove or disprove that [Q(a1/n):Q] 5 n
...
Let a and b belong to some extension of F and let b be algebraic
over F
...

32
...
If a is a zero of f (x) and
b is a zero of g(x), show that f (x) is irreducible over F(b) if and
only if g(x) is irreducible over F(a)
...
Let b be a zero of f(x) 5 x5 1 2x 1 4 (see Example 8 in Chapter 17)
...

3
6
34
...

35
...
Show that Q("a, "b) 5
Q("a1"b)
...
Let F, K and L be fields with F 8 K 8 L
...

37
...
Show that K is a finite extension of F
...
Prove that C is not the splitting field of any polynomial in Q3x4
...
Prove that "2 is not an element of Q(p)
...
Let a 5 cos 2p 1 i sin 2p and b 5 cos 2p 1 i sin 2p
...

41
...
Show that a and 1 1 a21
have the same degree over F
...
L
...

In this paper, it is proved that if p1, p2,
...
, " pn):Q] 5 2n
...
Yale, “Automorphisms of the Complex Numbers,” Mathematics
Magazine 39 (1966): 135–141
...


Irving Kaplansky
He got to the top of the heap
by being a first-rate doer and
expositor of algebra
...
HALMOS, I Have a

Photographic Memory

IRVING KAPLANSKY was born on March 22,
1917, in Toronto, Canada, a few years after
his parents emigrated from Poland
...

As an undergraduate at the University of
Toronto, Kaplansky was a member of the
winning team in the first William Lowell
Putnam competition, a mathematical contest
for United States and Canadian college students
...
A
...
A
...
In
1939, he entered Harvard University to earn
his doctorate as the first recipient of a Putnam
Fellowship
...
D
...
After
one year at Columbia University, he went to
the University of Chicago, where he remained
until his retirement in 1984
...


Kaplansky’s interests were broad, including areas such as ring theory, group theory,
field theory, Galois theory, ergodic theory,
algebras, metric spaces, number theory, statistics, and probability
...
The Steele Prize citation says, in part, “
...
”
Kaplansky died on June 25, 2006 at the age
of 89
...
dcs

...
ac
...

RICHARD A
...
Finite
fields were first introduced by Galois in 1830 in his proof of the unsolvability of the general quintic equation
...
To this day, matrix groups over finite fields are among the
most important classes of groups
...
But, besides the many uses of
finite fields in pure and applied mathematics, there is yet another good
reason for studying them
...
We have already seen that every finite field
has prime-power order (Exercise 47 in Chapter 13)
...

Theorem 22
...


PROOF Consider the splitting field E of f(x) 5 xpn 2 x over Zp
...
Since f(x) splits in E, we know that f(x) has exactly pn zeros in E, counting multiplicity
...
5,
every zero of f(x) has multiplicity 1
...

382

22 | Finite Fields

383

On the other hand, the set of zeros of f(x) in E is closed under addition,
subtraction, multiplication, and division by nonzero elements (see
Exercise 35), so that the set of zeros of f(x) is itself a field extension of
Zp in which f(x) splits
...

To show that there is a unique field for each prime-power, suppose
that K is any field of order pn
...
So, K must be a splitting field for f(x) over Zp
...
4, there is only one such field up to
isomorphism
...
1 appeared in the works of
Galois and Gauss in the first third of the 19th century
...
The uniqueness portion of the theorem was
proved by E
...
Moore in an 1893 paper concerning finite groups
...
T
...

Because there is only one field for each prime-power pn, we may unambiguously denote it by GF( pn), in honor of Galois, and call it the
Galois field of order pn
...

Theorem 22
...

n factors
As a group under multiplication, the set of nonzero elements of
GF( pn) is isomorphic to Zpn21 (and is, therefore, cyclic)
...
3), every
nonzero element of GF( pn) has additive order p
...


384

Fields

To see that the multiplicative group GF( pn)* of nonzero elements of
GF(pn) is cyclic, we first note that by the Fundamental Theorem of
Abelian Groups (Theorem 11
...
If the orders of these components are pairwise relatively prime then it follows from Corollary 1
of Theorem 8
...
Hence we may assume that
there is an integer d
...
From the Fundamental Theorem of Cyclic Groups (Theorem
4
...

This means that GF(pn)* has two distinct subgroups of order d, call
them H and K
...
2)
...
, 0), (0, 1, 0,
...
, (0, 0,
...

Corollary 1
[GF(p n):GF(p)] 5 n

Corollary 2 GF(pn) Contains an Element of Degree n
Let a be a generator of the group of nonzero elements of GF( pn)
under multiplication
...


PROOF Observe that [GF( p)(a):GF( p)] 5 [GF( pn):GF( p)] 5 n
...
Since x4 1
x 1 1 is irreducible over Z2, we know that
GF(16) < {ax3 1 bx2 1 cx 1 d 1 ͗x4 1 x 1 1͘ | a, b, c, d [ Z2}
...
For example,
(x3 1 x2 1 x 1 1)(x3 1 x) 5 x3 1 x2,

22 | Finite Fields

385

since the remainder upon dividing
(x3 1 x2 1 x 1 1)(x3 1 x) 5 x6 1 x5 1 x2 1 x
by x4 1 x 1 1 in Z2[x] is x3 1 x2
...

Thus,
x6 1 x5 1 x2 1 x 5 (x3 1 x2) 1 (x2 1 x) 1 x2 1 x 5 x3 1 x2
...
To take advantage of this, we must first find a generator of this
group
...
Obviously, x has these properties
...
, x14}, where x15 5 1
...
For example, x10 ? x7 5 x17 5 x2, but what is x10 1 x7? So, we face
a dilemma
...
On the
other hand, if we write the elements of F* in the multiplicative form xi,
then multiplication is easy and addition is hard
...
All we need to do is use the relation x4 5 x 1 1 to
make a two-way conversion table, as in Table 22
...

So, we see from Table 22
...

Don’t be misled by the preceding example into believing that the
element x is always a generator for the cyclic multiplicative group
of nonzero elements
...
(See Exercise 17
...
1 Conversion Table for Addition and Multiplication in GF(16)
Multiplicative
Form to
Additive Form
1
x
x2
x3
x4
x5
x6
x7
x8
x9
x10
x11
x12
x13
x14

Additive Form to
Multiplicative
Form

1
x
x2
x3
x11
x2 1 x
x3 1 x2
x3 1 x 1 1
x2 1 1
x3 1 x
x2 1 x 1 1
x3 1 x2 1 x
x3 1 x2 1 x 1 1
x3 1 x2 1 1
x3 1 1

1
x
x11
x2
x2 1 x
x2 1 1
x2 1 x 1 1
x3
x3 1 x2
x3 1 x
x3 1 1
x3 1 x2 1 x
x3 1 x2 1 1
x3 1 x 1 1
x3 1 x2 1 x 1 1

1
x
x4
x2
x5
x8
x10
x3
x6
x9
x14
x11
x13
x7
x12

irreducible polynomials of the same degree over Zp[x] yield isomorphic
fields, some are better than others for computational purposes
...
We will show how
to write f(x) as the product of linear factors
...
Then |F| 5 8 and |F*| 5 7
...
1, we know that |a| 5 7
...
3,
F 5 {0, 1, a, a2, a3, a4, a5, a6}
5 {0, 1, a, a 1 1, a2, a2 1 a 1 1, a2 1 1, a2 1 a}
...
We can simplify the calculations by using
the fact that a3 1 a2 1 1 5 0 to make a conversion table for the two
forms of writing the elements of F
...
Then,
a4 5 a3 1 a 5 (a2 1 1) 1 a 5 a2 1 a 1 1,
a5 5 a3 1 a2 1 a 5 (a2 1 1) 1 a2 1 a 5 a 1 1,
a6 5 a2 1 a,
a7 5 1
...

f(a2) 5 (a2)3 1 (a2)2 1 1 5 a6 1 a4 1 1
5 (a2 1 a) 1 (a2 1 a 1 1) 1 1 5 0
...
Next we try a3
...

Now a4
...

So, a4 is our remaining zero
...


Subfields of a Finite Field
Theorem 22
...
The
following theorem gives us a complete description of all the subfields
of a finite field
...
3, which describes all the subgroups of a finite cyclic group
...
3 Subfields of a Finite Field
For each divisor m of n, GF( pn) has a unique subfield of order pm
...


PROOF To show the existence portion of the theorem, suppose that
m divides n
...
For simplicity, write pn 2 1 5
m
(pm 2 1)t
...
We leave it as an easy exercise for the reader to show that K is a subfield of GF(pn)
...

m
Since the polynomial x p 2 x has at most pm zeros in GF(pn), we have
|K| # pm
...
Then |at| 5 pm 2 1 and since (a t) p m21 5 1,
it follows that at [ K
...

The uniqueness portion of the theorem follows from the observation
that if GF(pn) had two distinct subfields of order pm, then the polynom
mial x p 2 x would have more than pm zeros in GF(pn)
...
2
...
Then F is isomorphic
to GF( pm) for some m and, by Theorem 21
...

Thus, m divides n
...
2 and 22
...
3, make the task
of finding the subfields of a finite field a simple exercise in arithmetic
...
Then
there are exactly three subfields of F, and their orders are 2, 4, and 16
...
To find the subfield of order 4, we merely observe that the
three nonzero elements of this subfield must be the cyclic subgroup of
F* 5 ͗x͘ of order 3
...

EXAMPLE 4 If F is a field of order 36 5 729 and a is a generator of
F*, then the subfields of F are
1
...
GF(9) 5 {0} < ͗a91͘
3
...
GF(729) 5 {0} < ͗a͘
...

MARY CASE

1
...

3
...

5
...

7
...

9
...

11
...

13
...

15
...


17
...

19
...

If m divides n, show that [GF( pn):GF( pm)] 5 n/m
...

Let a be a zero of x3 1 x2 1 1 in some extension of Z2
...

Let a be a zero of f(x) 5 x2 1 2x 1 1 in some extension of Z3
...

Let a be a zero of f(x) 5 x3 1 x 1 1 in some extension of Z2
...

Let K be a finite extension field of a finite field F
...

How many elements of the cyclic group GF(81)* are generators?
Let f(x) be a cubic irreducible over Z2
...

Prove that the rings Z3[x]/͗x2 1 x 1 2͘ and Z3[x]/͗x2 1 2x 1 2͘ are
isomorphic
...
(This exercise is referred to in Chapter 32
...

Prove that the degree of any irreducible factor of x 8 2 x over Z2 is
1 or 3
...

Show that x is a generator of the cyclic group (Z3[x]/͗x3 1 2x 1 1͘)*
...
Prove that x is a generator of the cyclic group (Z2 3x4>
͗f(x) ͘) *
...
Find one such generator
...

Prove the uniqueness portion of Theorem 22
...


390

Fields

20
...
Show that ab is a generator of GF(81)*
...
Construct a field of order 9 and carry out the analysis as in Example 1, including the conversion table
...
Show that any finite subgroup of the multiplicative group of a field
is cyclic
...
Show that the set K in the proof of Theorem 22
...

n
24
...

25
...
(This exercise is referred to in the proof of Theorem 22
...
)
26
...

27
...
If p(x) is a polynomial in Zp[x] with no multiple zeros, show that
p(x) divides x pn 2 x for some n
...
Suppose that p is a prime and p 2 2
...

Show that a is a nonsquare in GF( pn) if n is odd and that a is a
square in GF( pn) if n is even
...
Let f (x) be a cubic irreducible over Zp, where p is a prime
...

31
...

32
...
List the elements of each subfield of F
...
Suppose that F is a field of order 125 and F* 5 ͗a͘
...

34
...

n
35
...
Show that the
set of zeros of f (x) in E is closed under addition, subtraction, multiplication, and division (by nonzero elements)
...
1
...
Suppose that L and K are subfields of GF( pn)
...
Give an example to show that the mapping a S ap need not be an
automorphism for arbitrary fields of prime characteristic p
...
In the field GF(pn ) , show that for every positive divisor d of n,
n
xp 2 x has an irreducible factor over GF(p) of degree d
...

JEFF PESIS

Software for the computer exercise in this chapter is available at the
website:
http://www
...
umn
...
This software tests cubic polynomials over Zp for irreducibility
...
1
...
Use the table to
write (x2 1 x) (x2 1 x 1 1) (x2 1 1) as a power of x
...

2
...
When a polynomial f(x) is irreducible, the
software finds a generator for the cyclic group of nonzero elements of the field Zp 3x4> ͗ f(x)͘ and creates a conversion table
for addition and multiplication similar to Table 22
...

Run the program for p 5 2 and x4 1 x 1 1
...
Use the table to
write x23 in additive form
...
Smith and J
...
Gallian, “Factoring Finite Factor Rings,”
Mathematics Magazine 58 (1985): 93–95
...


L
...
Dickson
One of the books [written by L
...
Dickson]
is his major, three-volume History of the
Theory of Numbers which would be a life’s
work by itself for a more ordinary man
...
A
...

In 1896, he received the first Ph
...
to be
awarded in mathematics at the University of
Chicago
...

Dickson was one of the most prolific
mathematicians of the 20th century, writing
267 research papers and 18 books
...

Dickson had a disdainful attitude toward
applicable mathematics; he would often say,
“Thank God that number theory is unsullied
by any applications
...
Dickson would often mention his
honeymoon: “It was a great success,” he
said, “except that I only got two research
papers written
...
He was the first to be awarded the
prize from the American Association for the
Advancement of Science for the most notable
contribution to the advancement of science,
and the first to receive the Cole Prize in algebra from the American Mathematical Society
...
Dickson died
on January 17, 1954
...
dcs

...
ac
...
This was one of the great events
of my life, as dazzling as first love
...
They were
especially interested in constructions that could be achieved using only a
straightedge without markings and a compass
...
But
they did not know how to trisect every angle or how to construct a regular seven-sided polygon (heptagon)
...
Legend has it that
the ancient Athenians were told by the oracle at Delos that a plague
would end if they constructed a new altar to Apollo in the shape of a cube
with double the volume of the old altar, which was also a cube
...
They knew
how to solve all these problems using other means, such as a compass
and a straightedge with two marks, or an unmarked straightedge and a
spiral, but they could not achieve any of the constructions with a compass
and an unmarked straightedge alone
...
The resolution of these perplexities was made
possible when they were transferred from questions of geometry to questions of algebra in the 19th century
...
It had been known since Euclid that
regular polygons with a number of sides of the form 2k, 2k ? 3, 2k ? 5, and
2k ? 3 ? 5 could be constructed, and it was believed that no others were
393

394

Fields

possible
...
In 1801, Gauss asserted that a
regular polygon of n sides is constructible if and only if n hass the form
2kp1 p2 ? ? ? pi, where the p’s are distinct primes of the form 22 1 1
...
5
...
How these constructions can be effected is another matter
...

Gauss’s result on the constructibility of regular n-gons eliminated
another of the famous unsolved problems because the ability to trisect
a 60° angle enables one to construct a regular 9-gon
...
In 1837, Wantzel proved that it was not possible to double
the cube
...


Constructible Numbers
With the field theory we now have, it is an easy matter to solve the following
problem: Given an unmarked straightedge, a compass, and a unit length, what
other lengths can be constructed? To begin, we call a real number a constructible if, by means of an unmarked straightedge, a compass, and a line
segment of length 1, we can construct a line segment of length |a| in a finite
number of steps
...
(See the
exercises for hints
...
What we desire is an algebraic characterization
of this field
...
Call the subset {(x, y)[ R2| x, y [ F} of the real plane the plane of F,
call any line joining two points in the plane of F a line in F, and call any circle
whose center is in the plane of F and whose radius is in F a circle in F
...


In particular, note that to find the point of intersection of a pair of lines
in F or the points of intersection of a line in F and a circle in F, one

23 | Geometric Constructions

395

need only solve a linear or quadratic equation in F
...
Starting with points in the plane of some field F,
which points in the real plane can be obtained with an unmarked
straightedge and a compass? Well, there are only three ways to construct points, starting with points in the plane of F
...
Intersect two lines in F
...
Intersect a circle in F and a line in F
...
Intersect two circles in F
...
In case 2, the point of intersection is
the solution to either a linear equation in F or a quadratic equation in
F
...
In case 3, no new points are obtained, because,
if the two circles are given by x2 1 y2 1 ax 1 by 1 c 5 0 and
x2 1 y2 1 a9x 1 b9y 1 c9 5 0, then we have (a 2 a9)x 1 (b 2 b9)y 1
(c 2 c9) 5 0, which is a line in F
...

It follows, then, that the only points in the real plane that can be
constructed from the plane of a field F are those whose coordinates
lie in fields of the form F("a), where a [ F and a is positive
...
Continuing in this fashion, we see that a real
number c is constructible if and only if there is a series of fields Q 5
F1 # F2 # ? ? ? # Fn # R such that Fi11 5 Fi ("ai), where ai [ Fi
and c [ Fn
...
5 that if
c is constructible, then [Q(c):Q] 5 2k for some nonnegative integer k
...
Consider doubling the cube of volume 1
...
But [Q(" 2):Q] 5 3, so such a cube cannot be constructed
...
If it were possible to trisect an angle of 60°, then cos 20° would be constructible
...
1
...
Now, using
the trigonometric identity cos 3u 5 4 cos3 u 2 3 cos u, with u 5 20°, we

(cos 20°, sin 20°)

(0, 0)

(1, 0)

Figure 23
...
But, since 8x3 2 6x 2 1 is irreducible over Q (see
Exercise 13), we must also have [Q(cos 20°):Q] 5 3
...

The remaining problems are relegated to the reader as Exercises 14,
15, and 17
...
In 1775, the Paris
Academy, so overwhelmed with these claims, passed a resolution to no
longer examine these claims or claims of machines purported to exhibit
perpetual motion
...

Most of these people have heard that this is impossible but have refused
to believe it
...
158]
...
80]
...
73]
...
D
...
127]
...
Two men
who did this in 1961 succeeded in having their accomplishment noted in
the Congressional Record [2, p
...
Occasionally, newspapers and
magazines have run stories about “doing the impossible,” often giving
the impression that the construction may be valid
...
One had his printed in four
languages! There are two delightful books written by mathematicians
about their encounters with these people
...


Exercises
Only prove to me that it is impossible, and I will set about it this very
evening
...


1
...
0, give a geometric proof that a 1 b and a 2 b are constructible
...
If a and b are constructible, give a geometric proof that ab is constructible
...
Notice that all segments in the figure can be made with an unmarked straightedge and a
compass
...
Prove that if c is a constructible number, then so is "|c|
...
) (This exercise is referred to in Chapter 33
...
If a and b (b 2 0) are constructible numbers, give a geometric proof
that a/b is constructible
...
)
b

0

1
a

5
...

6
...

7
...

8
...

9
...

10
...

11
...


398

Fields

12
...
Give an example
of a circle and a line in the plane of Q whose points of intersection
are not in the plane of Q
...
Prove that 8x3 2 6x 2 1 is irreducible over Q
...
Use the fact that 8 cos3(2p/7) 1 4 cos2(2p/7) 2 4 cos(2p/7) 2 1 5 0
to prove that a regular seven-sided polygon is not constructible with
an unmarked straightedge and a compass
...
Show that a regular 9-gon cannot be constructed with an unmarked
straightedge and a compass
...
Show that if a regular n-gon is constructible, then so is a regular
2n-gon
...
(Squaring the Circle) Show that it is impossible to construct, with
an unmarked straightedge and a compass, a square whose area
equals that of a circle of radius 1
...

18
...

19
...
Can the cube be “quadrupled”?
21
...
If a, b, and c are constructible, show that the real roots of ax2 1
bx 1 c are constructible
...
Augustus De Morgan, A Budget of Paradoxes, 2nd ed
...
H
...

2
...


Suggested Website
http://en
...
org/wiki/Squaring_the_circle
This website provides an excellent account of efforts to square the
circle, and links for articles about trisecting the angle and doubling
the cube
...

SENECA

True/false questions for Chapters 19–23 are available on the Web at
http://www
...
umn
...

1
...
Suppose that p(x) is a quadratic polynomial with rational coefficients and is irreducible over Q
...

3
...
If n divides q 2 1, prove that the equation x n 5 a has either no
solutions in F or n distinct solutions in F
...
Without using the Primitive Element Theorem, prove that if [K:F]
is prime, then K has a primitive element
...
Let a be a zero of x2 1 x 1 1
...

4
6
...

7
...
Does your argument apply equally
well if a3 is replaced with a2 and a4?
8
...

9
...
Show that
a21 is also algebraic over F of degree n
...
Prove that p2 2 1 is algebraic over Q(p3)
...
If ab is algebraic over F and b 2 0, prove that a is algebraic over F(b)
...
Let E be an algebraic extension of a field F
...

13
...

14
...
Show that a finite extension of a finite field is a simple extension
...
Let R be an integral domain that contains a field F as a subring
...

x50

400

Fields

17
...
1) matrices such that each pair of elements in the basis
commutes under multiplication
...
Let Pn 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | each ai is a real
number}
...
Find a basis for the vector space { f [ P3|f (0) 5 0}
...
)
20
...
, f (n)} is a basis for Pn
...
)
21
...
Let L 5 {a [ K|a p [ F for some nonnegative integer n}
...

22
...
hmco
...
” All of them are of fundamental importance
...

G
...
MILLER, Theory and Application

of Finite Groups

Conjugacy Classes
In this chapter, we derive several important arithmetic relationships
between a group and certain of its subgroups
...
Another fruitful method of partitioning the
elements of a group is by way of conjugacy classes
...
We say that a and b are
conjugate in G (and call b a conjugate of a) if xax21 5 b for some x
in G
...


We leave it to the reader (Exercise 1) to prove that conjugacy is an
equivalence relation on G, and that the conjugacy class of a is the equivalence class of a under conjugacy
...
Let’s look at one example
...

Similarly, one may verify that
cl(R0) 5 {R0},
cl(R90) 5 {R90, R270} 5 cl(R270),
cl(R180) 5 {R180},
cl(V) 5 {V, H} 5 cl(H),
cl(D) 5 {D, D9} 5 cl(D9)
...
1 gives an arithmetic relationship between the size of
the conjugacy class of a and the size of C(a), the centralizer of a
...
1 The Number of Conjugates of a
Let G be a finite group and let a be an element of G
...


PROOF Consider the function T that sends the coset xC(a) to the
conjugate xax21 of a
...
Thus, the number of conjugates of a is the index of the centralizer of a
...


The Class Equation
Since the conjugacy classes partition a group, the following important
counting principle is a corollary to Theorem 24
...

Corollary 2 The Class Equation
For any finite group G,

|G| 5 S |G:C(a)|,
where the sum runs over one element a from each conjugacy class of G
...
† Theorem 24
...


†“Never

underestimate a theorem that counts something
...


24 | Sylow Theorems

405

Theorem 24
...

Then Z(G) has more than one element
...
Thus, by culling out these elements, we may write the
class equation in the form
|G| 5 |Z(G)| 1 S|G:C(a)|,
where the sum runs over representatives of all conjugacy classes with
more than one element (this set may be empty)
...
Hence,
|G| 2 S|G:C(a)| 5 |Z(G)|,
where each term on the left is divisible by p
...

Corollary Groups of Order p2 Are Abelian
If |G| 5 p2, where p is prime, then G is Abelian
...
2 and Lagrange’s Theorem, |Z(G)| 5 p or p2
...
If |Z(G)| 5 p, then
|G/Z(G)| 5 p, so that G/Z(G) is cyclic
...
3, G is
Abelian
...
1 and the class equation
...
) Suppose we select two elements
at random (with replacement) from a finite group
...
Then the probability Pr(G) that two elements
selected at random from G commute is |K|/n2, where K 5 {(x, y) [
G % G | xy 5 yx}
...
Thus,
0 K 0 5 a 0 C(x) 0
...
1 that if x and y are in the same
conjugacy class, then |C(x)| 5 |C(y)| (see Exercise 53)
...
, at}, then
|C(a1)| 1 |C(a2)| 1 ? ? ? 1 |C(at)| 5 t|C(a)|
5 |G:C(a)| |C(a)| 5 |G| 5 n
...
, xm, we have
m

0 K 0 5 a 0 C(x) 0 5 a 0 G:C(xi ) 0 0 C(xi ) 0 5 m # n
...

Obviously, when G is non-Abelian, Pr(G) is less than 1
...
Consequently, Pr(G) is large when the sizes of the conjugacy classes
are small
...

Since G is non-Abelian, it follows from Theorem 9
...
Thus, in the extreme case, we would have
|Z(G)| 5 |G|/4, and the remaining (3/4)|G| elements would be distributed
in conjugacy classes with two elements each
...
The dihedral group D4 is an example of a group that has probability equal to 5/8
...
Recall that the converse of Lagrange’s
Theorem is false; that is, if G is a group of order m and n divides m,
G need not have a subgroup of order n
...
It, as well as Theorem 24
...

Sylow’s Theorem and Lagrange’s Theorem are the two most important
results in finite group theory
...

Theorem 24
...
If p k divides |G|, then G
has at least one subgroup of order p k
...
If |G| 5 1, Theorem 24
...
Now assume that the statement is true for all groups of
order less than |G|
...
Thus, we may henceforth assume that pk does not
divide the order of any proper subgroup of G
...
Since pk divides |G| 5 |G:C(a)| |C(a)| and pk does not divide
|C(a)|, we know that p must divide |G:C(a)| for all a o Z(G)
...
The Fundamental
Theorem of Finite Abelian Groups (Theorem 11
...
5, then
guarantees that Z(G) contains an element of order p, say x
...
Now observe that pk21 divides |G/͗x͘|
...
Finally, note that |H/͗x͘| 5 pk21 and |͗x͘| 5 p imply that
|H| 5 pk, and this completes the proof
...
Say we have a group G of order 23 ? 32 ? 54 ? 7
...
On the
other hand, Sylow’s First Theorem tells us nothing about the possible
existence of subgroups of order 6, 10, 15, 30, or any other divisor of
|G| that has two or more distinct prime factors
...

Definition Sylow p-Subgroup

Let G be a finite group and let p be a prime divisor of |G|
...


So, returning to our group G of order 23 ? 32 ? 54 ? 7, we call any subgroup of order 8 a Sylow 2-subgroup of G, any subgroup of order 625 a Sylow 5-subgroup of G, and so on
...


408

Special Topics

Since any subgroup of order p is cyclic, we have the following generalization of Theorem 9
...
His proof
ran nine pages!
Corollary Cauchy’s Theorem
Let G be a finite group and let p be a prime that divides the order
of G
...


Sylow’s First Theorem is so fundamental to finite group theory that
many different proofs of it have been published over the years [our proof
is essentially the one given by Georg Frobenius (1849–1917) in 1895]
...

Observe that the corollary to the Fundamental Theorem of Finite
Abelian Groups and Sylow’s First Theorem show that the converse of
Lagrange’s Theorem is true for all finite Abelian groups and all finite
groups of prime-power order
...
But first we introduce a new term
...
We say that H and K are
conjugate in G if there is an element g in G such that H 5 gKg21
...

Theorem 24
...


PROOF Let K be a Sylow p-subgroup of G and let C 5 {K1, K2,
...
Since conjugation is an
automorphism, each element of C is a Sylow p-subgroup of G
...
For each g [ G, define
fg:C S C by fg(Ki) 5 gKig21
...

Now define a mapping T:G S SC by T(g) 5 fg
...


24 | Sylow Theorems

409

Next consider T(H), the image of H under T
...
2)
...
3), for each i, |orbT(H)(Ki)| divides
|T(H)|, so that |orbT(H)(Ki)| is a power of p
...
But the only elements of N(Ki) that have orders that
are powers of p are those of Ki (see Exercise 9)
...

So, to complete the proof, all we need to do is show that for some i,
|orbT(H)(Ki)| 5 1
...
1, we have |C| 5 |G:N(K)|
(see Exercise 21)
...
Because the orbits partition C, |C| is the sum of
powers of p
...
Thus, there is an orbit
of size 1, and the proof is complete
...
5 Sylow’s Third Theorem
Let p be a prime and let G be a group of order pkm, where p does not
divide m
...
Furthermore, any two Sylow p-subgroups
of G are conjugate
...
, Kn} with K 5 K1 be the set of all conjugates of K in G
...

Let SC and T be as in the proof of Theorem 24
...
This time
we consider T(K), the image of K under T
...
Thus, |orbT(K)(K1)| 5 1 and |orbT(K)(Ki)| is a power of p
greater than 1 for all i 2 1
...

Next we show that every Sylow p-subgroup of G belongs to C
...
Let
SC and T be as in the proof of Theorem 24
...
As in the previous paragraph, |C| is the sum of the orbits’ sizes
under the action of T(H)
...
Thus, |C| is a sum of terms each divisible by p, so that, modulo p,
n 5 |C| 5 0
...

Finally, that n divides |G| follows directly from the fact that n 5
|G:N(K)| (see Exercise 21)
...
Observe that the first portion of Sylow’s Third Theorem is a
counting principle
...

Corollary A Unique Sylow p-Subgroup Is Normal
A Sylow p-subgroup of a finite group G is a normal subgroup of G if
and only if it is the only Sylow p-subgroup of G
...

EXAMPLE 1 Consider the Sylow 2-subgroups of S3
...
According to Sylow’s Third
Theorem, we should be able to obtain the latter two of these from the
first by conjugation
...

EXAMPLE 2 Consider the Sylow 3-subgroups of A4
...
(See the table on
page 107
...

Thus, the number of Sylow 3-subgroups is 1 modulo 3, and the four
Sylow 3-subgroups are conjugate
...
1 shows the subgroup lattices for S3 and A4
...
Notice that the three subgroups of
order 2 in A4 are contained in a Sylow 2-group, as required by Sylow’s
Second Theorem
...

In contrast to the two preceding examples, observe that the
dihedral group of order 12 has seven subgroups of order 2, but that
conjugating {R0, R180} does not yield any of the other six
...


you can, count
...
1 Lattices of subgroups for S3 and A4
...

Say G is a group of order 40
...
Similarly, G has either one or five subgroups
of order 8
...
If there
are five subgroups of order 8, none is normal and all five can be obtained by starting with any particular one, say H, and computing xHx21
for various x’s
...
(See
Exercise 7, Supplementary Exercises for Chapters 5–8
...

What about a group G of order 30? It must have either one or six
subgroups of order 5 and one or ten subgroups of order 3
...
Thus, G has one subgroup of order 3 and one of order 5, and at least one of these is normal

412

Special Topics

in G
...
[This, in turn, implies that both
the subgroup of order 3 and the subgroup of order 5 are normal in G
(Exercise 57 in Chapter 9)
...

Note that in these two examples we were able to deduce all of this information from knowing only the order of the group—so many conclusions from one assumption! This is the beauty of finite group theory
...
As a further illustration of the power of the Sylow
theorems, we next give a sufficient condition that guarantees that a
group of order pq, where p and q are primes, must be Zpq
...
6 Cyclic Groups of Order pq
If G is a group of order pq, where p and q are primes, p , q,
and p does not divide q 2 1, then G is cyclic
...


PROOF Let H be a Sylow p-subgroup of G and let K be a Sylow
q-subgroup of G
...
So 1 1 kp is equal
to 1, p, q, or pq
...

Similarly, there is only one Sylow q-subgroup of G
...
5, H and K are normal subgroups of G
...
To show that G is cyclic, it suffices to show that x
and y commute, for then |xy| 5 |x||y| 5 pq
...

Thus, xyx21y21 [ K > H 5 {e}, and hence xy 5 yx
...
6 demonstrates the power of the Sylow theorems in
classifying the finite groups whose orders have small numbers of prime
factors
...
6 exist for groups of orders p2q, p2q2, p3, and p4, where p and q are prime
...
2 The number of groups of a given order up to 100
...
2 lists the number of nonisomorphic
groups with orders at most 100
...
Also observe that, generally speaking, it is not the size
of the group that gives rise to a large number of groups of that size but the
number of prime factors involved
...
Contrast this with the fact reported in
1989 that there are 2328 groups of order 128 and 56,092 of order 256 [3]
...
csse
...
edu
...
Estimates put the
number of groups of order 512 at more than one million
...
In fact, our arguments serve as a good review of much of our work in group theory
...
Let H be a Sylow 3-subgroup
of G and let K be a Sylow 11-subgroup of G
...
Similarly,
H is normal in G
...
6, that elements from H and K commute, and therefore
G 5 H 3 K
...
Thus,
G is isomorphic to Z99 or Z3 % Z33
...
Let H be a Sylow 3-subgroup of
G and let K be a Sylow 11-subgroup of G
...
Thus, HK is a subgroup of G of order 33 (Exercise 55 in Chapter 9
and Exercise 7, Supplementary Exercises for Chapters 5–8)
...
6), we may write HK 5 ͗x͘
...
Since ͗x͘ has index 2 in G, we know it is
normal
...
Then, yx 5 xiy and,
since every member of G is of the form xsyt, the structure of G is completely determined by the value of i
...
To prove this, observe that |xi| 5 |x| (Exercise 5, Supplementary Exercises for Chapters 1–4)
...
But also, since y has order 2,
2

x 5 y21( yxy21)y 5 y21xiy 5 yxiy21 5 (yxy21)i 5 (xi)i 5 x i
...
From this it follows that
11 divides i 6 1, and therefore i 5 0 6 1, i 5 11 6 1, i 5 22 6 1, or
i 5 33 6 1
...
This proves that there are at
most four groups of order 66
...
For example, D11 % Z3 has 11 elements of order 2,
whereas D3 % Z11 has only three elements of order 2
...
)
EXAMPLE 5 The Only Group of Order 255 is Z255
Let G be a group of order 255 5 3 ? 5 ? 17, and let H be a Sylow 17-subgroup of G
...
By Example 15 in Chapter 10, |N(H)/C(H)| divides
|Aut(H)| 5 |Aut(Z17)|
...
5, |Aut(Z17)| 5 |U(17)| 5 16
...
Thus,
C(H) 5 G
...
Thus, 17 divides |Z(G)|, which in
turn divides 255
...
But the only groups of order 15, 5, 3, or 1 are the
cyclic ones, so we know that G/Z(G) is cyclic
...
3) shows that G is Abelian, and the Fundamental Theorem of
Finite Abelian Groups tells us that G is cyclic
...
That’s when I’ve really learned
...
Show that conjugacy is an equivalence relation on a group
...
Calculate all conjugacy classes for the quaternions (see Exercise 4,
Supplementary Exercises for Chapters 1–4)
...
Show that the function T defined in the proof of Theorem 24
...

4
...

5
...

6
...
Describe an isomorphism from
this group to D4
...
Suppose that G is a group of order 48
...

8
...

9
...
Prove that if x [
N(K) and the order of x is a power of p, then x [ K
...
)
10
...
Suppose that G is a group and ZGZ 5 p nm, where p is prime and
p
...
Prove that a Sylow p-subgroup of G must be normal in G
...
Let H be a Sylow p-subgroup of G
...

13
...
If G has more than one
Sylow 7-subgroup, exactly how many does it have?
14
...

15
...
Let G be a noncyclic group of order 21
...
Prove that a noncyclic group of order 21 must have 14 elements of
order 3
...
How many Sylow 5-subgroups of S5 are there? Exhibit two
...
How many Sylow 3-subgroups of S5 are there? Exhibit five
...
Prove that a group of order 175 is Abelian
...
Let H be a subgroup of a group G
...
Hint: Mimic the proof of Theorem 24
...

(This exercise is referred to in this chapter and in Chapter 25
...
Generalize the argument given in Example 3 to obtain a theorem
about groups of order p2q, where p and q are distinct primes
...
What is the smallest possible odd integer that can be the order of a
non-Abelian group?

416

Special Topics

24
...

25
...
6, prove that a group of order 15 is
cyclic
...
)
26
...

27
...

28
...
Show that G has exactly four elements of order 5 or exactly 24 elements of order 5
...
Show that the center of a group of order 60 cannot have order 4
...
Suppose that G is a group of order 60 and G has a normal subgroup N of order 2
...
G has normal subgroups of orders 6, 10, and 30
...
G has subgroups of orders 12 and 20
...
G has a cyclic subgroup of order 30
...
Let G be a group of order 60
...

32
...

33
...
Show that G has normal subgroups of order pk for all k between 1 and n (inclusive)
...
Suppose that G is a group of order pn, where p is prime, and G has
exactly one subgroup for each divisor of pn
...

35
...
If H is a proper subgroup of G,
prove that N(H)
...
(This exercise is referred to in Chapter 25
...
Suppose that G is a finite group and that all its Sylow subgroups are
normal
...

37
...
Show that a(H) 5 H for all automorphisms a of G
...
If H is a normal subgroup of a finite group G and |H| 5 pk
for some prime p, show that H is contained in every Sylow
p-subgroup of G
...
Let H and K denote a Sylow 3-subgroup and a Sylow 5-subgroup
of a group, respectively
...
If 3 divides |N(K)|, show that 5 divides |N(H)|
...
Let G be a group of order p2q2, where p and q are distinct primes,
q B p2 2 1, and p B q2 2 1
...
List three pairs
of primes that satisfy these conditions
...
Let H be a normal subgroup of a group G
...
Is this true when H
is not normal in G?
42
...
If the order of every element of a finite group G is
a power of p, prove that |G| is a power of p
...
)
43
...

44
...
Prove that K > S is a Sylow psubgroup of K
...
If G is a group of odd order and x [ G, show that x21 is not in cl(x)
...
Determine the groups of order 45
...
Show that there are at most three nonisomorphic groups of order 21
...
Prove that if H is a normal subgroup of index p2 where p is prime,
then G9 # H (see Exercise 3 in the Supplementary Exercises for
Chapters 5–8 for a description of G9)
...
Show that Z2 is the only group that has exactly two conjugacy
classes
...
If H is a finite subgroup of a group G and x [ G, prove that
0N(H) 0 5 0N(xHx 21 ) 0
...
Let G be a group with ZGZ 5 595 5 5 ? 7 ? 17
...

52
...
Prove that if x and y are in the same conjugacy class of a group,
then |C(x)| 5 |C(y)|
...
)
54
...
Express the probability that
a randomly selected element from G commutes with a in terms of
orders of subgroups of G
...
Find Pr(D4), Pr(S3), and Pr(A4)
...
Prove that Pr(G % H) 5 Pr(G) ? Pr(H)
...
Let R be a finite noncommutative ring
...

8
[Hint: Mimic the group case and use the fact that the additive
group R/C(R) is not cyclic
...
Use the website http://people
...
uwa
...
au/gordon/remote/
cubcay/ to look up the number of groups of order 4p, where p is an
odd prime up to 37
...


References
1
...
H
...

2
...

3
...
A
...

4
...
Paley and P
...


Suggested Readings
J
...
Gallian and D
...

It is shown that Zn is the only group of order n if and only if n and f(n)
are relatively prime
...
d
...
edu/~jgallian/pq
...

W
...
Gustafson, “What Is the Probability That Two Group Elements
Commute?” The American Mathematical Monthly 80 (1973): 1031–1034
...
It is shown
that for all finite non-Abelian groups and certain infinite non-Abelian
groups, the probability that two elements from a group commute is at
most 5/8
...

Desmond MacHale, “Commutativity in Finite Rings,” The American
Mathematical Monthly 83 (1976): 30–32
...

Several properties of Pr(G) when G is a finite group are stated
...
Also, there is no group G such
that 7/16 , Pr(G) , 1/2
...
In the
course of a century this remarkable theorem has been the basis for the construction of numerous theories
...
A
...
While a student
at Christiania University, Sylow won a gold
medal for competitive problem solving
...
During the school year 1862–1863,
Sylow received a temporary appointment at
Christiania University and gave lectures
on Galois theory and permutation groups
...
From 1873 to 1881,
Sylow, with some help from Lie, prepared a
new edition of Abel’s works
...

Sylow’s great discovery, Sylow’s Theorem, came in 1872
...
Jordan called it “one of
the essential points in the theory of permutations
...

In 1869, Sylow was offered a professorship at Christiania University but turned it
down
...
Sylow held
this position until his death on September 7,
1918
...
dcs
...
ac
...
This will certainly be one of the great
achievements of mathematics of this century
...
Simple group theory is a vast and difficult subject; we call it
the El Dorado of group theory because of the enormous effort put forth
by hundreds of mathematicians over many years to discover and
classify all finite simple groups
...

Definition Simple Group

A group is simple if its only normal subgroups are the identity
subgroup and the group itself
...
The simplicity of A5, the group of even permutations on five
symbols, played a crucial role in his proof that there is not a solution by
radicals of the general fifth-degree polynomial (that is, there is no
“quintic formula”)
...
These building blocks may be determined in the following way
...
Then the factor group G0/G1 is simple, and we next
choose a proper normal subgroup G2 of G1 of largest order
...
The simple groups G0 /G1, G1/G2,
...
More than 100 years ago, Jordan and Hölder
420

25 | Finite Simple Groups

421

proved that these factors are independent of the choices of the normal
subgroups made in the process described
...

This and the fact that many questions about finite groups can be reduced
(by induction) to questions about simple groups make clear the importance of determining all finite simple groups
...
This follows directly from the
corollary in Chapter 11
...
The best we can do here is to give a few
examples and mention a few words about their discovery
...
The next
discoveries were made by Jordan in 1870, when he found four infinite
families of simple matrix groups over the field Zp, where p is prime
...
Between the years 1892 and 1905, the American
mathematician Leonard Dickson (see Chapter 22 for a biography) generalized Jordan’s results to arbitrary finite fields and discovered several new
infinite families of simple groups
...
A
...
N
...
Mathieu in 1861 were in fact simple groups
...
”
The next important discoveries came in the 1950s
...
The first step was Richard
Brauer’s observation that the centralizer of an element of order 2 was an
important tool for studying simple groups
...
D
...

In the early 1960s came the momentous Feit-Thompson Theorem,
which says that a non-Abelian simple group must have even order
...
The proof of the Feit-Thompson
Theorem filled an entire issue of a journal [1], 255 pages in all (see
Figure 25
...
This result provided the impetus to classify the finite simple groups—that is, a program to discover all finite simple groups and
prove that there are no more to be found
...

It seems that old Burnside their orders has
guessed
Except for the cyclic ones, even the rest
...

Groups made up with permutes will
produce some more:
For An is simple, if n exceeds 4
...

Still others have come on to study this thing
...

With matrices finite they made quite a list
The question is: Could there be others
they’ve missed?
Suzuki and Ree then maintained it’s the
case

That these methods had not reached the end
of the chase
...

That made up a simple group
...

A group, when the order won’t factor by two,
Is cyclic or solvable
...

Suzuki and Ree had caused eyebrows to raise,
But the theoreticians they just couldn’t faze
...

Still, some hardy souls felt a thorn in their
side
...
1

25 | Finite Simple Groups

For the five groups of Mathieu all reason
defied;
Not An, not twisted, and not Chevalley,
They called them sporadic and filed them
away
...

He found out that nobody wanted to know:
The masters had missed 1 7 5 5 6 0
...
)
By Janko and Conway and Fischer and Held
McLaughlin, Suzuki, and Higman, and Sims
...

Well, that is, quite simply, a sign of the time
...


The floodgates were opened! New groups
were the rage!

methods introduced in the Feit-Thompson proof were generalized and
improved with great success by many mathematicians
...

Despite many spectacular achievements, research in simple group theory in the 1960s was haphazard, and the decade ended with many people believing that the classification would never be completed
...
The anonymously written “song” in Figure 25
...
) Others, more optimistic, were predicting that it
would be accomplished in the 1990s
...
This honor is among
the highest forms of recognition that a mathematician can receive
(more information about the Fields Medal is given near the end of this
chapter)
...
First, Thompson published what is regarded as the single most important paper in simple group theory—the
N-group paper
...
Second, Daniel Gorenstein produced an elaborate outline for the classification, which he delivered in a series of lectures at
the University of Chicago in 1972
...
The army of researchers now had a battle plan and
a commander-in-chief
...
Thus came the third critical development: the involvement of
Michael Aschbacher
...
In fact, so much progress was made by Aschbacher
and others that by 1976, it was clear to nearly everyone involved that
enough techniques had been developed to complete the classification
...

The 1980s were ushered in with Aschbacher following in the footsteps of Feit and Thompson by winning the American Mathematical
Society’s Cole Prize in algebra (see the last section of this chapter)
...
Griess made the spectacular announcement
that he had constructed the “Monster
...
In fact, it has vastly more elements than there
are atoms on the earth! Its order is
808,017,424,794,512,875,886,459,904,961,710,757,005,754,
368,000,000,000
(hence, the name)
...
The Monster is a
group of rotations in 196,883 dimensions
...

At the annual meeting of the American Mathematical Society in 1981,
Gorenstein announced that the “Twenty-five Years’ War” to classify all the
finite simple groups was over
...
The proof was
spread out over hundreds of papers—both published and unpublished—
and ran more than 10,000 pages in length
...
By the end of the decade, group theorists
had concluded that there was indeed a gap in the unpublished work that
would be difficult to rectify
...
In 2004, at the annual meeting of the
American Mathematical Society, Aschbacher announced that he and Smith
had completed the classification
...
Aschbacher concluded his remarks by saying that he would not bet
his house that the proof is now error-free
...


†The

name was coined by John H
...


25 | Finite Simple Groups

425

Nonsimplicity Tests
In view of the fact that simple groups are the building blocks for all
groups, it is surprising how scarce the non-Abelian simple groups are
...
In this section, we give a few theorems
that are useful in proving that a particular integer is not the order of a
non-Abelian simple group
...

Theorem 25
...
If 1 is the only divisor of n that is equal to 1 modulo p,
then there does not exist a simple group of order n
...
If n is not a prime-power, then
every Sylow subgroup is proper, and, by Sylow’s Third Theorem, we
know that the number of Sylow p-subgroups of a group of order n is
equal to 1 modulo p and divides n
...

How good is this test? Well, applying this criterion to all the nonprime integers between 1 and 200 would leave only the following integers as possible orders of finite non-Abelian simple groups: 12, 24, 30,
36, 48, 56, 60, 72, 80, 90, 96, 105, 108, 112, 120, 132, 144, 150, 160,
168, 180, and 192
...
See [2]
for more on this
...

Theorem 25
...


PROOF Let G be a group of order 2n, where n is odd and greater
than 1
...
1)

426

Special Topics

that the mapping g S Tg is an isomorphism from G to a permutation
group on the elements of G [where Tg(x) 5 gx for all x in G]
...
Then, when the permutation Tg is written in disjoint
cycle form, each cycle must have length 1 or 2; otherwise, |g| 2 2
...
Thus, in cycle form, Tg consists of exactly n
transpositions, where n is odd
...

This means that the set of even permutations in the image of G is a
normal subgroup of index 2
...
) Hence, G is not simple
...

We will make heavy use of its two corollaries
...
3 Generalized Cayley Theorem
Let G be a group and let H be a subgroup of G
...
Then there is a
homomorphism from G into S whose kernel lies in H and contains
every normal subgroup of G that is contained in H
...
As in the proof of Cayley’s Theorem, it is easy to
verify that the mapping of a:g S Tg is a homomorphism from G into S
...
Thus, Ker a # H
...
Thus,
Tk(xH) 5 kxH 5 xk9H 5 xH
and, therefore, Tk is the identity permutation
...

We have proved, then, that every normal subgroup of G contained in H
is also contained in Ker a
...
3, we obtain the following very
powerful arithmetic test for nonsimplicity
...
In particular, G is not simple
...
3
...
Thus, |G/Ker a| 5 |G|/|Ker a| divides
|S| 5 |G:H|!
...

Corollary 2 Embedding Theorem
If a finite non-Abelian simple group G has a subgroup of index n,
then G is isomorphic to a subgroup of An
...
By the Generalized
Cayley Theorem, there is a nontrivial homomorphism from G into Sn
...
Recall from Exercise 19
in Chapter 5 that any subgroup of Sn consists of even permutations only
or half even and half odd
...
Thus, G is isomorphic to a subgroup of An
...
For example, let G be any group of order 80 5 16 ? 5
...
Since 80 is not a divisor of 5!,
there is no simple group of order 80
...
Let’s consider these orders
...
Consider 56
...
Now, any
two Sylow p-subgroups that have order p must intersect in only the
identity
...
But there are only 56 elements in
all
...

An analogous argument also eliminates the integers 105 and 132
...
Of these, 60 and

428

Special Topics

168 do correspond to simple groups
...

The easiest case to handle is 112 5 24 ? 7
...
A Sylow 2-subgroup of G must have index 7
...

But 112 does not divide |A7|, which is a contradiction
...
Recall from Exercise 21 in Chapter 24 that the number of Sylow p-subgroups of G is np 5 |G: N(H)|,
where H is any Sylow p-subgroup of G
...
So
n3 5 4
...

Next consider the possibility of a simple group G of order 144 5 9 ? 16
...
The
Index Theorem rules out the case where n3 5 4, so we know that there are
16 Sylow 3-subgroups
...
So, let H and H9 be a pair of Sylow
3-subgroups whose intersection has order 3
...
2 (or by Exercise 35
in Chapter 24), we see that N(H > H9) must contain both H and H9 and,
therefore, the set HH9
...
) Thus,
|N(H > H9)| $ |HH9| 5

0 H 0 0 Hr 0
9
...

5
0 H d Hr 0
3

Now, we have three arithmetic conditions on k 5 |N(H > H9)|
...
Clearly, then, k $ 36,
and so |G : N(H > H9)| # 4
...

Finally, suppose that G is a non-Abelian simple group of order 180 5
22 ? 32 ? 5
...
First, assume that
n5 5 36
...
Now, if each pair
of the Sylow 3-subgroups intersects in only the identity, then there are 80
more elements in the group, which is a contradiction
...
Then, as was the case for order 144, we have
9
...

3
Thus,
|N(L3 > L 3 5 9 ? k,
9)|

25 | Finite Simple Groups

429

where k $ 3 and k divides 20
...

9)|
The Index Theorem now gives us another contradiction
...
In this case, we let H be the normalizer of a
Sylow 5-subgroup of G
...
In Chapter 24, we proved that every group of
order 30 has an element of order 15
...
But A6 has no element of order 15
...
)
Unfortunately, the argument for 120 is fairly long and complicated
...
We leave this as an
exercise (Exercise 17)
...


The Simplicity of A5
Once 120 has been disposed of, we will have shown that the only integers between 1 and 200 that can be the orders of non-Abelian simple
groups are 60 and 168
...
A similar argument can be used
to show that the factor group SL(2, Z7)/Z(SL(2, Z7)) is a simple group
of order 168
...
]
If A5 had a nontrivial proper normal subgroup H, then |H| is equal to
2, 3, 4, 5, 6, 10, 12, 15, 20, or 30
...
Now, if |H| is equal to 3, 6, 12, or 15, then |A5/H| is relatively
prime to 3, and by Exercise 59 in Chapter 9, H would have to contain
all 20 elements of order 3
...
If |H| 5 30, then |An /H| is relatively prime to
5
both 3 and 5, and so H would have to contain all the elements of orders
3 and 5
...

But we know from our results in Chapter 24 that any group of order 30
or 15 has an element of order 15
...
This proves that A5 is simple
...
A few years later, Felix
Klein showed that the group of rotations of a regular icosahedron is

430

Special Topics

simple and, therefore, isomorphic to A5 (see Exercise 27)
...
Klein was the first to
prove that there is a simple group of order 168
...
His arguments for the integers 144 and 180
alone used up 10 pages
...
See [3] for a detailed account of this endeavor
...


The Fields Medal
Among the highest awards for mathematical achievement is the Fields
Medal
...
Although the Fields Medal is considered by many mathematicians to be the equivalent of the Nobel Prize, there are great differences
between these awards
...
† This tradition stems from John
Charles Fields’s stipulation, in his will establishing the medal, that the
awards should be “an encouragement for further achievement
...

More details about the Fields Medal can be found at www

...
com
...
The prize was founded in honor of Frank Nelson Cole on the
occasion of his retirement as secretary of the American Mathematical
Society
...
The effective, moving, vitalizing work of the world is done between the ages of twenty-five and
forty
...
I, chap
...


25 | Finite Simple Groups

431

recipients of the prize—Dickson, Chevalley, Brauer, Feit, Thompson,
and Aschbacher—have made fundamental contributions to simple
group theory at some time in their careers
...

HERBERT V
...

2
...

4
...

6
...

8
...

10
...

12
...


14
...

16
...

18
...


20
...

Prove that there is no simple group of order 280 5 23 ? 5 ? 7
...

Prove that there is no simple group of order 300 5 22 ? 3 ? 52
...

Prove that there is no simple group of order 540 5 22 ? 33 ? 5
...

Prove that there is no simple group of order 315 5 32 ? 5 ? 7
...

Prove that there is no simple group of order n, where 201 #
n # 235
...

Without using the “2 ? odd” test, prove that there is no simple
group of order 210
...
Choose
three odd integers between 200 and 1000
...

Show that there is no simple group of order pqr, where p, q, and r
are primes ( p, q, and r need not be distinct)
...

Show that S5 does not contain a subgroup of order 40 or 30
...
(This
exercise is referred to in this chapter
...

Suppose that H is a subgroup of a finite group G and that |H| and
(|G:H| 2 1)! are relatively prime
...
What
does this tell you about a subgroup of index 2 in a finite group?
Suppose that p is the smallest prime that divides |G|
...


432

Special Topics

21
...

(This exercise is referred to in Chapter 32
...
Prove that a simple group of order 60 has a subgroup of order 6
and a subgroup of order 10
...
Show that PSL(2, Z7) 5 SL(2, Z7)/Z(SL(2, Z7)), which has order
168, is a simple group
...
)
24
...

25
...

Show that H 5 S5
...
)
26
...
Show that |H| 5 |K|
...
Show that (up to isomorphism) A5 is the only simple group of
order 60
...
)
28
...

29
...
p
...
If a simple group G has a subgroup K that is a normal subgroup of
two distinct maximal subgroups, prove that K 5 5e6
...
Show that a finite group of even order that has a cyclic Sylow 2subgroup is not simple
...

JANET RENO

Software for the computer exercises in this chapter is available at the
website:
http://www
...
umn
...
This software uses a counter M to keep track of how many integers
Theorem 25
...
Run the program
for the following intervals: 1–100; 501–600; 5001–5100;
10,001–10,100
...
This software uses a counter M to keep track of how many integers
the Index Theorem eliminates in any given interval of integers
...
How does
M seem to behave as the sizes of the integers grow?

25 | Finite Simple Groups

433

References
1
...
Feit and J
...
Thompson, “Solvability of Groups of Odd Order,” Pacific Journal of Mathematics 13 (1963): 775–1029
...
J
...
Gallian, “Computers in Group Theory,” Mathematics Magazine 49
(1976): 69–73
...
J
...
Gallian, “The Search for Finite Simple Groups,” Mathematics Magazine 49 (1976): 163–179
...
Cornell, N
...
Wage, “Simple Groups of Orders Less Than
1000,” Journal of Undergraduate Research 5 (1973): 77–86
...
All but the last one are orders of simple groups
...

K
...

This note gives an elementary proof that A5 is simple using commutators
...
A
...
d
...
edu/~jgallian/simple
...

A historical account is given of the search for finite simple groups
...

This article gives an elementary introduction to groups and a discussion of simple groups, including the “Monster
...

You won’t find an article on a complex subject better written for the
layperson than this one
...

Sandra M
...

The author shows that the group SL(2, Z7)/Z(SL(2, Z7)) of order 168 is
simple using a counting argument
...
In rapid
succession he proved one astonishing
theorem after another
...

DANIEL GORENSTEIN, Scientific American

MICHAEL ASCHBACHER was born on April 8,
1944, in Little Rock, Arkansas
...
When he was nine years old, his family moved to East Lansing, Michigan; six
years later, they moved to Los Angeles
...
In addition to his schoolwork, he passed the first
four actuary exams and was employed for a
few years as an actuary, full-time in the summers and part-time during the academic year
...
In his senior year, Aschbacher took
abstract algebra but showed little interest
in the course
...

In 1966, Aschbacher went to the University of Wisconsin for a Ph
...
degree
...

Aschbacher’s dissertation work in the
area of combinatorial geometries had led
him to consider certain group-theoretic
questions
...
The
1980 Cole Prize Selection Committee said
of one of his papers, “[It] lifted the subject
to a new plateau and brought the classification within reach
...


Daniel Gorenstein
Gorenstein was one of the most influential
mathematicians of the last few decades
...
Upon graduating from
Harvard in 1943 during World War II,
Gorenstein was offered an instructorship at
Harvard to teach mathematics to army personnel
...
He received his Ph
...
degree in 1951, working in algebraic geometry
under Oscar Zariski
...
In 1951, Gorenstein
took a position at Clark University in
Worcester, Massachusetts, where he stayed
until moving to Northeastern University in
1964
...

In 1957, Gorenstein switched from algebraic geometry to finite groups, learning the
basic material from I
...
Herstein while collaborating with him over the next few years
...


It was there that Gorenstein, assimilating the
revolutionary techniques then being developed
by John Thompson, began his fundamental
work that contributed to the classification of
finite simple groups
...

Among the honors received by Gorenstein
are the Steele Prize from the American
Mathematical Society and election to membership in the National Academy of Sciences
and the American Academy of Arts and
Sciences
...
dcs
...
ac
...

DANIEL GORENSTEIN

JOHN G
...
In 1951, he entered
Yale University as a divinity student but
switched to mathematics in his sophomore
year
...
D
...
After one year on the
faculty at Harvard, Thompson returned to
Chicago
...
In 1993, Thompson accepted an appointment at the University of Florida
...

In his dissertation, he verified a 50-year-old
conjecture about finite groups possessing a
certain kind of automorphism
...
) The novel methods Thompson
used in his dissertation foreshadowed the
revolutionary ideas he would later introduce
in the Feit-Thompson paper and the classification of minimal simple groups (simple
groups that contain no proper non-Abelian
simple subgroups)
...

In the late 1970s, Thompson made significant contributions to coding theory, the
theory of finite projective planes, and the
theory of modular functions
...

Among Thompson’s many honors are the
Cole Prize in algebra and the Fields Medal
...
In 2000,
President Clinton presented Thompson the
National Medal of Science
...

To find more information about Thompson,
visit:
http://www-groups
...
st-and

...
uk/~history/

Generators
26
and Relations
One cannot escape the feeling that these mathematical formulae have an
independent existence and an intelligence of their own, that they are
wiser than we are, wiser even than their discoverers, that we get more
out of them than we originally put into them
...
Simply put, we begin with a set of elements
that we want to generate the group, and a set of equations (called relations) that specify the conditions that these generators are to satisfy
...
This will uniquely determine the group up to isomorphism
...
Consider D4, the group of symmetries of a square
...

Observe that R and H are related in the following ways:
R4 5 H 2 5 (RH)2 5 R0

(the identity)
...
For
example, (RH)2 5 R0 yields HR 5 R21H21, and R4 5 H 2 5 R0 yields
R21 5 R3 and H21 5 H
...
In fact, every relation between
R and H can be derived from those given in Equation (1)
...
This last
stipulation is necessary because the subgroup {R0, R180, H, V} of D4 is
generated by the two elements a 5 R180 and b 5 H that satisfy the relations a4 5 b2 5 (ab)2 5 e
...
It
is natural to ask whether this description of D4 applies to some other group
437

438

Special Topics

as well
...
Any other group generated by two elements a
and b satisfying only the relations a4 5 b2 5 (ab)2 5 e, and those that can
be derived from these relations, is isomorphic to D4
...
The purpose of
this chapter is to show that this procedure can be reversed; that is, we
can begin with any set of generators and relations among the generators
and construct a group that is uniquely described by these generators and
relations, subject to the stipulation that all other relations among the
generators can be derived from the original ones
...
For any set S 5 {a, b, c,
...
} by replacing each x in S by x21
...

The elements of W(S) are called words from S
...
This word is called the empty word and
is denoted by e
...
Observe that this operation is associative and the
empty word is the identity
...

At this stage we have everything we need to make a group out of
W(S) except inverses
...
But abb21a21 is
not the empty word! You may recall that we faced a similar obstacle
long ago when we carried out the construction of the field of quotients
of an integral domain
...
But their product, ab/(ba),
was a formal symbol that was not the same as the formal symbol 1/1,
the identity
...

Definition Equivalence Classes of Words

For any pair of elements u and v of W(S), we say that u is related to v if v
can be obtained from u by a finite sequence of insertions or deletions of
words of the form xx21 or x21x, where x [ S
...
(See Exercise 1
...
Then acc21b is equivalent to ab;
aab21bbaccc21 is equivalent to aabac; the word a21aabb21a21 is
equivalent to the empty word; and the word ca21b is equivalent to
cc21caa21a21bbca21ac21b21
...


Free Group
Theorem 26
...
For any word u in W(S), let u
denote the set of all words in W(S) equivalent to u (that is, u is the
equivalence class containing u)
...


PROOF This proof is left to the reader
...
1 is called a free group on S
...
2 shows why free groups are important
...
2 The Universal Mapping Property
Every group is a homomorphic image of a free group
...
(Such
a set exists, because we may take S to be G itself
...
Unfortunately, since our notation for any word in W(S) also
denotes an element of G, we have created a notational problem for ourselves
...
As before, x1x2
...
Notice that x1x2
...

Now consider the mapping from F into G given by
f(x1x2
...


440

Special Topics

[All we are doing is taking a product in F and viewing it as a product in
G
...
]
Clearly, f is well defined, for inserting or deleting expressions of the
form xx21 or x21x in elements of W(S) corresponds to inserting or deleting the identity in G
...
xn ) (y1y2
...
xny1y2
...

Finally, f is onto G because S generates G
...
2
and the First Isomorphism Theorem for Groups
...


Generators and Relations
We have now laid the foundation for defining a group by way of generators and relations
...

EXAMPLE 2 Let F be the free group on the set {a, b} and let N be
the smallest normal subgroup of F containing the set {a4, b2, (ab)2}
...
We begin by observing that the
mapping f from F onto D4, which takes a to R90 and b to H (horizontal
reflection), defines a homomorphism whose kernel contains N
...
On the other hand, we claim that the set
K 5 {N, aN, a2N, a3N, bN, abN, a2bN, a3bN}
of left cosets of N is F/N itself
...
So, it suffices
to show that K is closed under multiplication on the left by a and b
...
For b, we will
do only one of the eight cases
...
Consider b(aN)
...
Upon completion of the other cases (Exercise 3), we
know that F/N has at most eight elements
...
Since F/Ker f is a factor
group of F/N [indeed, F/Ker f < (F/N)/(Ker f/N)], it follows that F/N
also has eight elements and F/N 5 F/Ker f < D4
...
, an} and let
F be the free group on A
...
, wt} be a subset of F and
let N be the smallest normal subgroup of F containing W
...
, an and the relations w1 5 w2 5 ? ? ? 5
wt 5 e if there is an isomorphism from F/N onto G that carries ai N to ai
...
, an | w1 5 w2 5 ? ? ? 5 wt 5 e͘
...
This restriction is
not necessary, however
...
For example, the relation a21b23ab 5 e is
often written as ab 5 b3a
...
Instead, one
just manipulates the generators and treats anything in N as the identity, as our notation suggests
...
, an | w1 5 w2 5 ? ? ? 5 wt 5 e͘,
many authors prefer to say that G has the presentation
͗a1, a2,
...

Notice that a free group is “free” of relations; that is, the equivalence
class containing the empty word is the only relation
...

Free groups are of fundamental importance in a branch of algebra
known as combinatorial group theory
...

EXAMPLE 4 The group of integers is the free group on one letter; that
is, Z < ͗a| ͘
...
)

442

Special Topics

The next theorem formalizes the argument used in Example 2 to
prove that the group defined there has eight elements
...
3 (Dyck, 1882)
Let

G 5 ͗a1, a2,
...
, an | w1 5 w2 5 ? ? ? 5 wt 5
wt11 5 ? ? ? 5 wt1k 5 e͘
...


PROOF See Exercise 5
...
3 says that if you start with generators and relations for a group G and create a group G by imposing additional
relations, then G is a homomorphic image of G
...


PROOF See Exercise 5
...
What does G look like? Formally, of course, G is isomorphic to F/N, where F is free on {a, b} and N is the smallest normal subgroup of F containing b22a2 and (ab)22a2
...
Then, just as in Example 2, it follows that S is closed under
multiplication by a and b from the left
...
Thus, we can determine the elements of G once we know
exactly how many elements there are in H
...
) To do this, first observe that b2 5 (ab)2 5 abab implies
b 5 aba
...
Hence, H has at most four elements, and therefore G has at most
eight—namely, e, b, b2, b3, a, ab, ab2, and ab3
...
For example, Z2 % Z2
satisfies the defining relations and has only four elements
...
How can we show that the eight
elements listed above are distinct? Well, consider the group G generated
by the matrices
A5 c

0 1
d
21 0

and

B5 c

0 i ,
d
i 0

where i 5 "
...
So, it follows from the corollary to Dyck’s Theorem
that G is isomorphic to G and therefore G has order 8
...

EXAMPLE 6 Let
G 5 ͗a, b | a3 5 b9 5 e, a21ba 5 b21͘
...
Thus,
G 5 {aib j | 0 # i # 2, 0 # j # 8},
and therefore G has at most 27 elements
...

Hence,
b 5 ebe 5 a23ba3 5 a22(a21ba)a2 5 a22b21a2
5 a21(a21b21a)a 5 a21ba 5 b21
...
But
b2 5 e 5 b9 further implies b 5 e
...
But Z3 satisfies the defining relations with a 5 1 and b 5 0
...

We hope Example 6 convinces you of the fact that, once a list of the
elements of the group given by a set of generators and relations has
been obtained, one must further verify that this list has no duplications
...
Obviously, experience plays a role here
...


444

Special Topics

EXAMPLE 7 Let G be the group with the 26 letters of the alphabet as
generators
...
For
example, buy 5 by 5 bye, hour 5 our, lead 5 led, whole 5 hole
...
With these examples in mind, we ask, What is the
group given by these generators and relations? Surprisingly, the answer
is the infinite cyclic group generated by v
...
The former can easily
be done with common words
...
From too 5 to
/
we have o 5 0
...
In
contrast, the reference Handbook of Homophones by W
...
Townsend (available at the website http://members
...
org/~jeremy/
dictionaryclassic/chapters/homophones
...
Of course, including these makes the group trivial
...

Theorem 26
...


PROOF The Fundamental Theorem of Finite Abelian Groups takes
care of the Abelian cases
...
Also, let G1 5 ͗a, b | a4 5 b2 5 (ab)2 5 e͘ and let G2 5 ͗a, b | a2 5
b2 5 (ab)2͘
...
Thus, it suffices to
show that G must satisfy the defining relations for G1 or G2
...
Then, if b is any element of G not in ͗a͘, we
know that
G 5 ͗a͘ < ͗a͘b 5 {e, a, a2, a3, b, ab, a2b, a3b}
...
Which of the eight elements of G can it
be? Not b, ab, a2b, or a3b, by cancellation
...
Not a3, for the same reason
...

Suppose b2 5 e
...
From this and the fact that |bab21| 5 |a|, we then conclude
that bab21 5 a or bab21 5 a21
...
But then, since b2 5 e, we have
(ab)2 5 e, and therefore G satisfies the defining relations for G1
...

The classification of the groups of order 8, together with our results
on groups of order p2, 2p, and pq from Chapter 24, allow us to classify
the groups of order up to 15, with the exception of those of order 12
...

An argument along the lines of Theorem 26
...
This group, called the dicyclic
group of order 12 and denoted by Q6, has presentation ͗a, b | a6 5 e,
a3 5 b2, b21ab 5 a21͘
...
1 lists the groups of order at most 15
...

Table 26
...
For n $ 3, we have used Dn to denote the group of
symmetries of a regular n-gon
...
By analogy, these
generators and relations serve to define D1 and D2 also
...
) Finally, we define the infinite dihedral group
D` as ͗a, b | a2 5 b2 5 e͘
...

Theorem 26
...


PROOF Let G be a group generated by a pair of distinct elements of
order 2, say, a and b
...
If |ab| 5 `, then G is
infinite and satisfies the relations of D`
...
By Dyck’s Theorem, G is isomorphic to some factor group
of D`, say, D`/H
...
Since every element
of D` has one of the forms (ab)i, (ba)i, (ab)ia, or (ba)ib, by symmetry,
we may assume that h 5 (ab)i or h 5 (ab)ia
...
Since (ab)i is
in H, we have
H 5 (ab)iH 5 (abH)i,
so that (abH)21 5 (abH)i21
...

Thus,
D`/H 5 ͗aH, bH͘ 5 ͗aH, abH͘
(see Exercise 7), and D`/H satisfies the defining relations for Di (use
Exercise 9 with x 5 aH and y 5 abH)
...

If h 5 (ab)ia, then
H 5 (ab)iaH 5 (ab)iHaH,

26 | Generators and Relations

447

and therefore
(abH)i 5 (ab)iH 5 (aH)21 5 a21H 5 aH
...

However,
(abH)2i 5 (aH)2 5 a2H 5 H,
so that D`/H is again finite
...

Finally, suppose that |ab| 5 n
...
But (ab)21 5 b21a21 5 ba,
since a and b have order 2
...
1
...

A
aba ab

FF

a

e

FF

B
b ba

FF

bab baba

FF

Figure 26
...

For example, if we place a pair of mirrors facing each other at a 45°
angle, we obtain the group D4
...
2, the effect of
reflecting an object in mirror A, then mirror B, is a rotation of twice the
angle between the two mirrors (that is, 90°)
...
2 The group D4—reflections in mirrors at a 45° angle

In Figure 26
...
The corresponding group is D180
...
As n becomes larger and larger, the mirrors approach a
parallel position
...

Β

Α
ab

a

e

b ba

bab ba
ba

FF FF FF FF

aba

Figure 26
...
The principal advantage is that in many situations—particularly in knot theory,
algebraic topology, and geometry—groups defined by way of generators and relations arise in a natural way
...
Among the disadvantages of defining a group by
generators and relations is the fact that it is often difficult to decide
whether or not the group is finite, or even whether or not a particular
element is the identity
...
Nowadays,
these questions are frequently tackled with the aid of a computer
...

Title of a Song by RINGO STARR, May 1971

1
...
Show that the relation defined
on W(S) in this chapter is an equivalence relation
...
Let n be an even integer
...

3
...

4
...

5
...
3 and its corollary
...
Let G be the group {61, 6i, 6j, 6k} with multiplication defined
as in Exercise 52 in Chapter 9
...
(Hence, the name “quaternions
...
In any group, show that ͗a, b͘ 5 ͗a, ab͘
...
5
...
Let a 5 (12)(34) and b 5 (24)
...

9
...

(This exercise is referred to in the proof of Theorem 26
...
)
10
...

11
...
Show that xy 5 yx
...
Let G 5 ͗a, b | a2 5 b4 5 e, ab 5 b3a͘
...
Express a3b2abab3 in the form bia j where 0 # i # 1 and
0 # j # 3
...
Express b3abab3a in the form bia j where 0 # i # 1 and 0 # j # 3
...
Let G 5 ͗a, b | a2 5 b2 5 (ab)2͘
...
Express b2abab3 in the form bia j
...
Express b3abab3a in the form bia j
...
Let G be the group defined by the following table
...

1
1
2
3
4
5
6
:
2n

2

3

4

5

6

???

2n

1
2
3
4
5
6
:
2n

2
1
4
3
6
5
:
2n 2 1

3
2n
5
2
7
4
:
2n 2 2

4
2n 2 1
6
1
8
3
:
2n 2 3

5
2n 2 2
7
2n
9
2
:
2n 2 4

6
2n 2 3
8
2n 2 1
10
1
:
2n 2 5

???
???
???
???
???
???
:
???

2n
3
2
5
4
7
:
1

450

Special Topics

15
...
Show that |G| # 16
...

16
...
1 of all groups of
orders 1 to 11
...
Let G be defined by some set of generators and relations
...

18
...
Show that the permutations (23) and (13)
satisfy the defining relations of G
...

19
...

20
...
Show that Z(G) 5
{e, xn}
...
(The group G is called the dicyclic group of order 4n
...
Let G 5 ͗a, b | a6 5 b3 5 e, b21ab 5 a3͘
...
Let G 5 ͗x, y | x4 5 y4 5 e, xyxy21 5 e͘
...
Assuming that |G| 5 16, find the center of G and show that G/͗ y2͘ is
isomorphic to D4
...
Determine the orders of the elements of D`
...
Let G5•£0 1 c § † a, b, c P Z2 ¶
...

25
...
Determine |G|
...
Let G 5 ͗a, b|a2 5 e, b2 5 e, aba 5 bab͘
...
Let G 5 ͗a, b|a3 5 e, b2 5 e, aba21b21 5 e͘
...
Given an example of a non-Abelian group that has exactly three
elements of finite order
...
Referring to Example 7 in this chapter, show as many letters as you
can that are equivalent to ~
...
Suppose that a group of order 8 has exactly five elements of order 2
...


References
1
...
-F
...
Schoof, L
...
Zagier, “Quotient homophones des groupes libres [Homophonic Quotients of Free Groups],” Experimental Mathematics 2 (1993): 153–155
...
H
...
Whitford, A Dictionary of American Homophones and Homographs,
New York: Teachers College Press, 1962
...
Fran, Jr
...
J
...

This book is replete with the group-theoretic aspects of the Magic
Cube
...
The book has numerous challenging exercises stated in group-theoretic terms
...

This article shows how a group can be associated with a knotted string
...
The theory of knots—a branch of topology—
seeks to classify and analyze the different ways of embedding such a
curve
...
Among other knots, Neuwirth describes the construction of the
knot group for the trefoil knot pictured
...


The trefoil knot

David Peifer, “An Introduction to Combinatorial Group Theory and the
Word Problem,” Mathematics Magazine 70 (1997): 3–10
...


Marshall Hall, Jr
...

HANS ZASSENHAUS, Notices of
the American Mathematical Society

MARSHALL HALL, JR
...
Louis, Missouri
...
He completed a B
...
degree in 1932 at
Yale
...
H
...
D
...
At the
outbreak of World War II, he joined Naval
Intelligence and had significant success in deciphering both the Japanese codes and the
German Enigma messages
...
After the
war, Hall had faculty appointments at the
Ohio State University, Caltech, and Emory
University
...

Hall’s highly regarded books on group
theory and combinatorial theory are classics
...
His 1943 paper on
projective planes ranks among the most cited
papers in mathematics
...
One of Hall’s
most celebrated results is his solution to the
“Burnside Problem” for exponent 6—that is,
a finitely generated group in which the order
of every element divides 6 must be finite
...
It was Hall who
suggested Thompson’s Ph
...
dissertation
problem
...
D
...

To find more information about Hall,
visit:
http://www–groups
...
st–and

...
uk/~history/

27 Symmetry Groups

I’m not good at math, but I do know that the universe is formed with
mathematical principles whether I understand them or not, and I am
going to let that guide me
...
In this chapter and the next, we examine this fundamentally
important concept in some detail
...

Definition Isometry

An isometry of n-dimensional space Rn is a function from Rn onto Rn
that preserves distance
...
With this definition, we may now
make precise the definition of the symmetry group of an n-dimensional figure
...
The symmetry group of F in Rn is the set
of all isometries of Rn that carry F onto itself
...


It is important to realize that the symmetry group of an object depends not only on the object, but also on the space in which we view it
...

453

454

Special Topics

Although we have formulated our definitions for all finite dimensions, our chief interest will be the two-dimensional case
...
46])
...
A reflection
across a line L is that transformation that leaves every point of L fixed
and takes every point Q, not on L, to the point Q9 so that L is the perpendicular bisector of the line segment from Q to Q9 (see Figure 27
...

The line L is called the axis of reflection
...
Some authors call an axis of reflective symmetry L a mirror because L acts like
a two-sided mirror; that is, the image of a point Q in a mirror placed on
the line L is, in fact, the image of Q under the reflection across the line
L
...
For example, the reflected image of a clockwise spiral is a
counterclockwise spiral
...
(See Figure 27
...
)
Q
Q'
L
Axis of reflection

Axis of reflection

Figure 27
...
For example, if p and q are points in a plane
and T is a translation, then the two directed line segments joining p to
T( p) and q to T(q) have the same length and direction
...
This line is called the glide-axis
...
2, the arrow gives the direction and length of the translation,
and is contained in the axis of reflection
...
Successive footprints in wet sand are related by a
glide-reflection
...
2 Glide-reflection

27 | Symmetry Groups

455

Classification of Finite Plane
Symmetry Groups
Our first goal in this chapter is to classify all finite plane symmetry
groups
...
(For convenience, call
D2 the plane symmetry group of a nonsquare rectangle and D1 the
plane symmetry group of the letter “V
...
) The cyclic groups Zn are easily seen to be plane symmetry
groups also
...
3 is an illustration of an organism whose plane
symmetry group consists of four rotations and is isomorphic to Z4
...
The famous mathematician
Hermann Weyl attributes the following theorem to Leonardo da Vinci
(1452–1519)
...
3 Aurelia Insulinda, an organism
whose plane symmetry group is Z4

Theorem 27
...


PROOF Let G be a finite plane symmetry group of some figure
...
Now observing that the
composition of two reflections preserves orientation, we know that
such a composition is a translation or rotation
...
Thus, every two reflections in G have reflection axes
that intersect in some point
...
Then because f f9 preserves orientation, we know that f f9
is a rotation
...
This means that their
axes of reflection must intersect at point P
...

For convenience, let us denote a rotation about P of s degrees
by Rs
...
(Such an angle exists, since G is finite and R360 belongs to G
...

To see this, suppose that Rs is in G
...

Then, b # s and there is some integer t such that tb # s ,
(t 1 1)b
...

Since b represents the smallest positive angle of rotation among the
elements of G, we must have s 2 tb 5 0, and therefore, Rs 5 (Rb)t
...

For convenience, let us say that |Rb| 5 n
...
If G has at least one reflection, say f, then
f, fRb, f(Rb)2,
...
Furthermore, this is the entire set of reflections of G
...
Thus, g 5 f21(Rb)k 5 f(Rb)k
...
, (Rb)n21, f, fRb, a(Rb)2,
...
Hence, by our
characterization of the dihedral groups (Theorem 26
...


Classification of Finite Groups
of Rotations in R3
One might think that the set of all possible finite symmetry groups in
three dimensions would be much more diverse than is the case for two
dimensions
...
For example, moving to
three dimensions introduces only three new groups of rotations
...


27 | Symmetry Groups

457

Theorem 27
...


Theorem 27
...
3), makes easy work of determining the group of rotations of an
object in R3
...
4, which is composed of six congruent squares and eight congruent equilateral triangles
...
Obviously, there are four rotations that map this square to itself,
and the designated square can be rotated to the location of any of the
other five
...
3), the rotation group has order 4 ? 6 5 24
...
2, G is one of Z24, D12,
and S4
...
So, G is isomorphic to S4
...
4

The group of rotations of a tetrahedron (the tetrahedral group) is isomorphic to A4; the group of rotations of a cube or an octahedron (the
octahedral group) is isomorphic to S4; the group of rotations of a dodecahedron or an icosahedron (the icosahedral group) is isomorphic to
A5
...
271–273] specifies which portions of the polyhedra
are being permuted in each case
...
5
...
5 The five regular solids as depicted by Johannes Kepler
in Harmonices Mundi, Book II (1619)

Exercises
Perhaps the most valuable result of all education is the ability to make
yourself do the thing you have to do, when it ought to be done, whether you
like it or not
...

2
...

4
...

Show that the translations of Rn form a group
...

Show that the group of rotations in R3 of a 3-prism (that is, a prism
with equilateral ends, as in the following figure) is isomorphic to D3
...
What is the order of the (entire) symmetry group in R3 of a 3-prism?
6
...
What is the order of the symmetry group in R3 of an n-prism?

27 | Symmetry Groups

459

8
...

9
...
a subset of R1,
b
...
a subset of R3
...
)
10
...
) The letters of the alphabet can be sorted into the following
categories:
1
...
BCDEK
3
...
HIOX
What defines the categories?
11
...
Why is inversion [that is, f(x, y) 5 (2x, 2y) ] not listed as one of
the four kinds of isometries in R2 ?
13
...

14
...
Describe a rotation that has
this same effect
...
In R2, a rotation fixes a point; in R3, a rotation fixes a line
...

16
...

17
...

18
...
Which isometry is it?
19
...
What
type of isometry must it be?
20
...
What is A followed by B? How is the composite motion related to the points a and b?

460

Special Topics

References
1
...
S
...
Coxeter, Introduction to Geometry, 2nd ed
...

2
...
, New York: W
...
Freeman, 1993
...

It is shown that the group of rotations of a dodecahedron and the
group of rotations of an icosahedron are both A5
...
Rosen, Symmetry Discovered, Cambridge: Cambridge University Press,
1975
...
It includes sections on group theory, spatial symmetry, temporal
symmetry, and color symmetry, and chapters on symmetry in nature
and the uses of symmetry in science
...

This article discusses how chemists use group theory to understand
molecular structure and how physicists use it to study the fundamental
forces and particles
...
wikipedia
...
Included are
essays, photos, links, and references
...


Frieze Groups
28 and Crystallographic
Groups
Symmetry, considered as a law of regular composition of structural objects,
is similar to harmony
...
In our opinion the whole
esthetics of scientific and artistic creativity lies in the ability to feel this
where others fail to perceive it
...
V
...
A
...
There are two
types of such groups
...
These kinds of designs are the ones used for decorative strips and
for patterns on jewelry, as illustrated in Figure 28
...
In mathematics,
familiar examples include the graphs of y 5 sin x, y 5 tan x, y 5 |sin x|,
and |y| 5 sin x
...

In previous chapters, it was our custom to view two isomorphic
groups as the same group, since we could not distinguish between them
algebraically
...
To emphasize this
difference, we will treat them separately
...

A proof that there are exactly seven types of frieze patterns is given in
the appendix to [6]
...
1 Frieze patterns

The symmetry group of pattern I (Figure 28
...
Letting x denote a translation to the right of one unit (that
is, the distance between two consecutive R’s), we may write the symmetry group of pattern I as
F1 5 {xn | n [ Z}
...
2 Pattern I

R

28 | Frieze Groups and Crystallographic Groups

463

The group for pattern II (Figure 28
...
Letting x denote a glide-reflection, we may write the
symmetry group of pattern II as
F2 5 {xn | n [ Z}
...
3 Pattern II

Notice that the translation subgroup of pattern II is just ͗x2͘
...
4) is generated by a
translation x and a reflection y across the dashed vertical line
...
Any one will
do
...


RR

RR

RR

RR

RR

Figure 28
...
Thus, by Theorem 26
...
A geometric fact about pattern III
worth mentioning is that the distance between consecutive pairs of vertical reflection axes is half the length of the smallest translation vector
...
5), the symmetry group F4 is generated by a
translation x and a rotation y of 180° about a point p midway between
consecutive R’s (such a rotation is often called a half-turn)
...
(Another rotation point lies between a
top and bottom R
...
) Therefore,
F4 5 {xnym | n [ Z, m 5 0 or m 5 1}
...
5 Pattern IV

R

R

R

R

Special Topics

RR

RR
p

RR
RR

RR

RR

Figure 28
...
6) is yet another
infinite dihedral group generated by a glide-reflection x and a rotation y
of 180° about the point p
...
The rotation points are midway between the vertical reflection axes
...

The symmetry group F6 for pattern VI (Figure 28
...
The group is
F6 5 {xnym | n [ Z, m 5 0 or m 5 1}
...
In fact, F6
is isomorphic to Z % Z2
...
(Conversely, a glidereflection is nontrivial if its glide-axis is not an axis of reflective symmetry for the pattern
...
7 Pattern VI

The symmetry group F7 of pattern VII (Figure 28
...
It is
isomorphic to the direct product of the infinite dihedral group and Z2
...
Therefore,
F7 5 {x ny mz k | n [ Z, m 5 0 or m 5 1, k 5 0 or k 5 1}
...
8 Pattern VII

The preceding discussion is summarized in Figure 28
...
Figure 28
...

In describing the seven frieze groups, we have not explicitly said
how multiplication is done algebraically
...
9 The seven frieze patterns and their groups of symmetries

composition
...

For example, we know that every element of F7 can be written in the
form x ny mz k
...
We may do this simply by
looking at the effect that g has on pattern VII
...
To distinguish this R, we enclose it in a shaded box
...
See
Figure 28
...

Now, comparing the starting position of the shaded R with its final
position, we see that x21yzxz 5 x22y
...


466
Special Topics

Is there a vertical reflection?
yes

no

Is there a horizontal reflection?

Is there a horizontal reflection
or glide-reflection?
no

Is there a
horizontal
reflection?

R R

no

III
R R

no

II
R R

R R

I

R

R
R

yes

R

yes

VI
R

R

R

IV
R
R

R R

R

yes

R
R

V

R

R

R R

Figure 28
...

Adapted from [6, p
...


R
R

R R

R R

no

Is there a
half-turn?

R

R R

no

R

Is there a
half-turn?

VII

R R

R R

yes

R

yes

R

R R

28 | Frieze Groups and Crystallographic Groups

RR

RR

RR

RR

RR

RR

RR

RR

467

RR

RR

z

RR

RR

RR

RR

RR

RR

RR

RR

x

RR

RR

RR

RR

RR

RR

RR

RR

z

RR

RR

RR

RR

RR
RR

RR

RR

RR

RR

y

RR

RR

RR

RR
xϪ1

RR

RR

RR

RR

RR

RR

RR
RR

Figure 28
...
However,
there are 17 additional kinds of discrete plane symmetry groups that
arise from infinitely repeating designs in a plane
...
Consequently, the patterns are invariant under
linear combinations of two linearly independent translations
...
Another term occasionally used for these groups is wallpaper groups
...
It
is adapted from the excellent article by Schattschneider [5] and the
monograph by Crowe [1]
...
We begin with some examples
...
12 Study of Regular Division of the Plane with Fish and Birds, 1938
...
The arrows are translation vectors
...
In Figure 28
...
The crystallographic notation for it is p1
...
)
The symmetry group of the pattern in Figure 28
...
This group has no (nonzero) rotational or
reflective symmetry
...

Figure 28
...
The notation for this
group is p3
...
15 and 28
...

Figure 28
...


28 | Frieze Groups and Crystallographic Groups

469

Figure 28
...
L
...
The
solid arrow is a translation
vector
...


Figure 28
...
Escher
drawing with
symmetry p3
(disregarding
shading)
...


470

Special Topics

Figure 28
...
15 and 28
...
The designs for p3 and pg are based on elements of
Chinese lattice designs found in [2]; the design for pm is based on a
weaving pattern from the Sandwich Islands, found in [3]
...
16 The plane symmetry groups

471

472

Special Topics

Figure 28
...
18
...

These three pieces of information will narrow the list of candidates to at
most two
...

For example, consider the two patterns in Figure 28
...
Both patterns have a smallest positive rotational symmetry of 120°; both have reflectional and nontrivial glide-reflectional
symmetry
...
18, these patterns must be of
type p3m1 or p31m
...
Thus, the left pattern is p3m1, and the right pattern is p31m
...
1 (reproduced from [5, p
...
A lattice of points of a pattern is a set of images of any
particular point acted on by the translation group of the pattern
...
The possible lattices for periodic patterns in a
plane, together with lattice units, are shown in Figure 28
...
A generating region (or fundamental region) of a periodic pattern is the smallest
portion of the lattice unit whose images under the full symmetry group of
the pattern cover the plane
...
12, 28
...
14 are given in
Figure 28
...
In Figure 28
...
The only symmetry pattern in which the lattice unit and the generating region coincide is the p1 pattern illustrated in
Figure 28
...
Table 28
...

Notice that Table 28
...
This fact is commonly
called the crystallographic restriction
...
Barlow
...
1 can also
be used in reverse to create patterns with a specific symmetry group
...
19 were made in this way
...
18 Identification flowchart for symmetries of plane periodic patterns

yes

no

pmm

cmm

28 | Frieze Groups and Crystallographic Groups

475

p31m

p3m1

Figure 28
...
20 Possible lattices for plane periodic patterns

In sharp contrast to the situation for finite symmetry groups, the transition from two-dimensional crystallographic groups to three-dimensional
crystallographic groups introduces a great many more possibilities, since
the motif is repeated indefinitely by three independent translations
...
These were independently determined by Fedorov, Schönflies, and
Barlow in the 1890s
...
1 Identification Chart for Plane Periodic Patternsa

Type

Lattice

Highest
Order of
Rotation

Reflections

Nontrivial
GlideGenerating
Reflections
Region

p1
p2
pm
pg
cm
pmm
pmg

Parallelogram
Parallelogram
Rectangular
Rectangular
Rhombic
Rectangular
Rectangular

1
2
1
1
1
2
2

No
No
Yes
No
Yes
Yes
Yes

No
No
No
Yes
Yes
No
Yes

1 unit
1
2 unit
1
2 unit
1
2 unit
1
2 unit
1
4 unit
1
4 unit

pgg
cmm

Rectangular
Rhombic

2
2

No
Yes

Yes
Yes

1
4
1
4

unit
unit

p4
p4m

Square
Square

4
4

No
Yes

No
Yes

1
4
1
8

unit
unit

p4g

Square

4

Yes

Yes

1
8

unit

p3
p3m1

Hexagonal
Hexagonal

3
3

No
Yes

No
Yes

1
3
1
6

unit
unit

p31m

Hexagonal

3

Yes

Yes

1
6

unit

p6
p6m

Hexagonal
Hexagonal

6
6

No
Yes

No
Yes

Helpful
Distinguishing
Properties

1
6 unit
1
12 unit

Parallel reflection
axes
Perpendicular
reflection axes
Fourfold centers
on reflection
axes
Fourfold centers
not on
reflection axes
All threefold
centers on
reflection axes
Not all threefold
centers on
reflection axes

aA

rotation through an angle of 360°/n is said to have order n
...


famous lecture in 1900 at the International Congress of Mathematicians in
Paris
...
This was answered affirmatively by L
...
We mention in passing that
in four dimensions, there are 4783 symmetry groups for infinitely repeating patterns
...
In fact, a crystal is defined as a rigid

28 | Frieze Groups and Crystallographic Groups

477

Figure 28
...
12, 28
...
14
...


body in which the component particles are arranged in a pattern that repeats in three directions (the repetition is caused by the chemical bonding)
...
In crystalline materials, the motif units are atoms, ions, ionic
groups, clusters of ions, or molecules
...
In 1912, Max von Laue, a
young German physicist, hypothesized that a narrow beam of x-rays directed onto a crystal with a photographic film behind it would be

478

Special Topics

deflected (the technical term is “diffracted”) by the unit cell (made up of
atoms or ions) and would show up on the film as spots
...
3
...
This discovery marked the birth of modern mineralogy
...
Von Laue was awarded the Nobel
Prize in physics in 1914, and the Braggs were jointly awarded the
Nobel Prize in physics in 1915
...
One of these was a graduate student named
Francis Crick; another was an x-ray crystallographer, Rosalind Franklin
...
At this point, we let Horace
Judson [4, pp
...

Crick saw in Franklin’s words and numbers something just as important,
indeed eventually just as visualizable
...
Crystallographers distinguish 230 different space groups, of which the face-centered monoclinic
cell with its curious properties of symmetry is only one—though in biological substances a fairly common one
...
So Crick saw at
once the symmetry that neither Franklin nor Wilkins had comprehended,
that Perutz, for that matter, hadn’t noticed, that had escaped the theoretical
crystallographer in Wilkins’ lab, Alexander Stokes—namely, that the
molecule of DNA, rotated a half turn, came back to congruence with itself
...


This was a crucial fact
...
In 1962, Watson, Crick, and Maurice
Wilkins received the Nobel Prize in medicine and physiology for their
discovery
...
172]
...

YOGI BERRA

1
...

3
...

5
...

7
...

9
...

11
...

How many nonisomorphic frieze groups are there?
In the frieze group F7, write x2yzxz in the form x ny mz k
...

In the frieze group F7, show that yz 5 zy and xy 5 yx
...

Use the results of Exercises 5 and 6 to do Exercises 3 and 4
through symbol manipulation only (that is, without referring to the
pattern)
...
)
Prove that in F7 the cyclic subgroup generated by x is a normal
subgroup
...

Look up the word frieze in an ordinary dictionary
...

Determine which of the seven frieze groups is the symmetry group
of each of the following patterns
...


b
...


d
...


f
...
Determine the frieze group corresponding to each of the following
patterns:
a
...
y 5 |sin x|,
c
...
y 5 tan x,
e
...

13
...


14
...
17
...
Determine which of the 17 crystallographic groups is the symmetry group of each of the following patterns
...


b
...


d
...
In the following figure, there is a point labeled 1
...
The image of 1 under the
composition of a and b is labeled ab
...


β
1

αβ
α

17
...
An extensive study of automobile tires revealed that only
five of the seven frieze patterns occur
...

18
...
16
...
Determine which of the frieze groups is the symmetry group of
each of the following patterns
...
? ? ? D D D D ? ? ?
b
...
? ? ? L L L L ? ? ?
d
...
? ? ? N N N N ? ? ?
f
...
? ? ? L L ? ? ?
20
...
17
...
Donald Crowe, Symmetry, Rigid Motions, and Patterns, Arlington, Va
...

2
...
Dye, A Grammar of Chinese Lattice, Harvard-Yenching Institute
Monograph Series, vol
...
: Harvard University Press,
1937
...
)

482

Special Topics

3
...
(Reproduction of the same title, first published in 1856 and
reprinted in 1910 and 1928
...
Horace Freeland Judson, The Eighth Day of Creation, New York: Simon
and Schuster, 1979
...
D
...

6
...
K
...
W
...


Suggested Readings
S
...
, For All Practical Purposes, 7th ed
...
H
...

This book has a well-written, richly illustrated chapter on symmetry in
art and nature
...
G
...

This article uses automobile tires as a tool for introducing and explaining the symmetry terms and concepts important in chemistry
...

D
...

A loving, lavish, encyclopedic book on the drawings of M
...
Escher
...
von Baeyer, “Impossible Crystals,” Discover 11(2) (1990): 69–78
...
The x-ray diffraction patterns of quasicrystals exhibit fivefold symmetry—something that had been thought to
be impossible
...
mcescher
...
C
...
It features many of
his prints and most of his 136 symmetry drawings
...
disted
...
bc
...
htm
This spectacular website on symmetry and tessellations has numerous activities and links to many other sites on related topics
...


M
...
Escher
I never got a pass mark in math
...

M
...
ESCHER

M
...
ESCHER was born on June 17, 1898, in
the Netherlands
...
Gradually, he became less and
less interested in the visible world and became increasingly absorbed in an inventive
approach to space
...
He also
studied the mathematician George Pólya’s
paper on the 17 plane crystallographic
groups
...

Escher was fond of incorporating various
mathematical ideas into his works
...

Although Escher originals are now quite
expensive, it was not until 1951 that he derived a significant portion of his income
from his prints
...

His prints have been used to illustrate ideas
in hundreds of scientific works
...
Escher died on
March 27, 1972, in Holland
...
C
...
mcescher
...

A toast from FRANK HARARY on the
occasion of Pólya’s 90th birthday

GEORGE PÓLYA was born in Budapest,
Hungary, on December 13, 1887
...

In 1912, he was awarded a Ph
...
in mathematics
...

After two years at Brown University, he went
to Stanford University, where he remained
until his death in 1985 at the age of 97
...
B
...
Escher, a geologist, sent a
copy of the paper to his artist brother, M
...

Escher, who used Pólya’s black-and-white
geometric patterns as a guide for making his
own interlocking colored patterns featuring
birds, reptiles, and fish
...
Pólya is also famous for his
books on problem solving and for his teaching
...
The Society for Industrial
and Applied Mathematics, the London Mathematical Society, and the Mathematical Association of America have prizes named after
Pólya
...
He never learned to
drive a car and took his first plane trip at
age 75
...

For more information about Pólya, visit:
http://www-groups
...
st-and
...
uk/~history/

485

John H
...
”
RONALD GRAHAM Speaking of John H
...
CONWAY ranks among the most
original and versatile contemporary mathematicians
...
As a youngster,
he was often beaten up by older boys and
did not do well in high school
...

A pattern that uses repeated shapes to
cover a flat surface without gaps or overlaps
is called a tiling
...

Unlike patterns whose symmetry group is
one of the 17 plane crystallographic groups,
Penrose patterns can be neither translated nor
rotated to coincide with themselves
...
In 1993,

486

Conway discovered a new prism that can be
used to fill three-dimensional space without
gaps or overlaps
...
Among his most important discoveries are three simple groups, which are now
named after him
...
) Conway
is fascinated by games and puzzles
...

Conway has received numerous prestigious
honors
...

For more information about Conway,
visit:
http://www-groups
...
st-and
...
uk/~history/

Symmetry
29 and Counting
Let us pause to slake our thirst one last time at symmetry’s bubbling spring
...
Consider, for example, the task of
coloring the six vertices of a regular hexagon so that three are black
6
and three are white
...
1 shows the a b 5 20 possibilities
...
1(a) as different, since all six designs shown there can be obtained
from one of them by rotating
...
) In this case, we say that the designs in Figure 29
...
Similarly, the
designs in Figure 29
...
1(c) and those in Figure 29
...
And, since
no design from Figure 29
...
However, the designs in Figure 29
...
1(b)
can be reflected to yield the designs in Figure 29
...
For example, for
purposes of arranging three black beads and three white beads to form a
necklace, the designs shown in Figure 29
...

In general, we say that two designs (arrangements of beads) A and B
are equivalent under a group G of permutations of the arrangements if
there is an element f in G such that f(A) 5 B
...
It follows, then,

487

488

Special Topics

(a)

(c)

(d)

(b)

Figure 29
...
(The set being permuted is the set of all
possible designs or arrangements
...
1 divide into four orbits under
the group of rotations but only three orbits under the group D6, since
the designs in Figure 29
...
Thus,
we could obtain all 20 tile designs from just four tiles, but we could
obtain all 20 necklaces from just three of them
...
Such an approach was provided by Georg Frobenius
in 1887
...
By an accident of history, Frobenius’s theorem has come to
be known as “Burnside’s Theorem
...
If G is a group of permutations on a set S and i [ S, then
stabG(i) 5 {f [ G | f(i) 5 i} and orbG(i) 5 {f(i) |
f [ G}
...

Definition Elements Fixed by f

For any group G of permutations on a set S and any f in G, we let
fix(f) 5 {i [ S | f(i) 5 i}
...


Theorem 29
...

0 G 0 f[G
PROOF Let n denote the number of pairs (f, i), with f [ G, i [ S,
and f(i) 5 i
...
First, for
each particular f in G, the number of such pairs is exactly |fix(f)|
...

fPG

(1)

Second, for each particular i in S, observe that |stabG(i)| is exactly the
number of pairs (f, i) for which f(i) 5 i
...

iPS

(2)

It follows from Exercise 33 in Chapter 7 that if s and t are in the same
orbit of G, then orbG(s) 5 orbG(t), and thus by the Orbit-Stabilizer Theorem (Theorem 7
...
So, if we choose s [ S and sum over orbG(s), we have
a

t[orbG(s)

0stabG (t) 0 5 0orbG (s) 0 0stabG (s) 0 5 0G 0
...


490

Special Topics

Applications
To illustrate how to apply Burnside’s Theorem, let us return to the ceramic
tile and necklace problems
...
Obviously, the identity
fixes all 20 designs
...
1 that rotations of 60°, 180°,
or 300° fix none of the 20 designs
...
2 shows fix(f) for
the rotations of 120° and 240°
...
1
...
2 Tile designs fixed by 120°
rotation and 240° rotation
Table 29
...

6
Now let’s use Burnside’s Theorem to count the number of necklace
arrangements consisting of three black beads and three white beads
...
) For this problem, two arrangements are equivalent if they
are in the same orbit under D6
...
3 shows the arrangements fixed

Figure 29
...
2

Type of Element
Identity
Rotation of order 2 (180°)
Rotation of order 3 (120° or 240°)
Rotation of order 6 (60° or 300°)
Reflection across diagonal
Reflection across side bisector

Number of
Elements
of This
Type

Number of
Arrangements
Fixed by Type
of Element

1
1
2
2
3
3

20
0
2
0
4
0

by a reflection across a diagonal
...
2 summarizes the information
needed to apply Burnside’s Theorem
...

Now that we have gotten our feet wet on a few easy problems, let’s
try a more difficult one
...
1)
...
How shall we decide when two colorings
of the tetrahedron are nonequivalent? Certainly, if we were to pick up a
tetrahedron colored in a certain manner, rotate it, and put it back down,
we would think of the tetrahedron as being positioned differently rather
than as being colored differently (just as if we picked up a die labeled in
the usual way and rolled it, we would not say that the die is now differently labeled)
...
1 and is isomorphic to A4
...
) Every rotation permutes the 729 colorings, and to apply Burnside’s Theorem we must determine the size of
fix(f) for each of the 12 rotations of the group
...
Next, consider the element (234) of order 3, shown in the bottom row, second from the left in
Figure 5
...
Suppose that a specific coloring is fixed by this element

492

Special Topics

Table 29
...
Since (234) carries edge 12 to edge 13, edge 13 to edge
14, and edge 14 to edge 12, these three edges must agree in color (edge
ij is the edge joining vertex i and vertex j)
...
So,
|fix(234)| 5 32, since there are three choices for each of these two sets
of three edges
...
3 show the possible colorings of the two sets of three edges
...

Now consider the rotation (12)(34) of order 2
...
1
...
This yields 3 ? 3 choices for those two edges
...
Similarly,
edges 23 and 14 must agree
...
This
means that we have 3 ? 3 ? 3 ? 3 ways to color the tetrahedron that will
be equivalent under (12)(34)
...
4 gives the complete list of 81
colorings
...

Now that we have analyzed the three types of group elements, we
can apply Burnside’s Theorem
...
4 81 Colorings Fixed by (12)(34) (X and Y can be any of R, W, and B)

Edge
12
34
13
24
23
14

Colorings
X
Y
R
R
R
R

X
Y
R
R
W
W

X
Y
R
R
B
B

X
Y
W
W
W
W

X
Y
W
W
R
R

X
Y
W
W
B
B

X
Y
B
B
B
B

X
Y
B
B
R
R

X
Y
B
B
W
W

29 | Symmetry and Counting

493

colorings of the edges of a tetrahedron with 3 colors is
1
(1 ? 36 1 8 ? 32 1 3 ? 34) 5 87
...

Just as surely, you are wondering who besides mathematicians are interested in counting problems such as the ones we have discussed
...
Indeed, one set of benzene derivatives can be viewed as
six carbon atoms arranged in a hexagon with one of the three radicals
NH2, COOH, or OH attached at each carbon atom
...
4 for
one example
...
4 A benzene derivative

So Burnside’s Theorem enables a chemist to determine the number of
benzene molecules (see Exercise 4)
...
Again,
the number of such molecules can be easily counted using Burnside’s
Theorem
...
If G is a group
and S is a set of objects, we say that G acts on S if there is a homomorphism g from G to sym(S), the group of all permutations on S
...
) For convenience, we denote the image of g under g as gg
...
Notice that when g is one-to-one, the elements of G may be regarded as permutations on S
...
Thus, a group acting on a set is a natural generalization
of the permutation group concept
...
Then gR0, gR180,
gD, gD9 are the identity; gR90, gR270, gH, gV interchange the two diagonals;
and the mapping g S gg from D4 to sym(S) is a group homomorphism
...
In this case, the mapping g S gg from GL(n, F) to sym(S) is a
one-to-one homomorphism
...
The proof of Cayley’s Theorem ( Theorem 6
...
4 and 24
...
3) has G acting on the left cosets of a
subgroup H
...

EPICURUS

1
...
(It is permissible to use a
single color on all four corners
...
Determine the number of different necklaces that can be made using 13 white beads and three black beads
...
Determine the number of ways in which the vertices of an equilateral triangle can be colored with five colors so that at least two colors are used
...
A benzene molecule can be modeled as six carbon atoms arranged
in a regular hexagon in a plane
...
How many such
compounds are possible?

29 | Symmetry and Counting

495

5
...
How many compounds are possible?
6
...

7
...

8
...

9
...

10
...

11
...
How many ways can
we color the cake with n colors assuming that each piece receives
one color?
12
...
Let G be a finite group and let sym(G) be the group of all permutations on G
...
Show that G acts on itself under
the action g S fg
...

14
...
For each g in G, let gg denote the element of sym(S) defined by gg(xH) 5 gxH
...

15
...
Show that D4 acts on {L1, L2} and determine the kernel of the mapping g S gg
...

Chapter 20 of this book presents a more detailed treatment of the
subject of symmetry and counting
...

This article discusses a combinatorial problem concerning generating
periodic patterns that the artist M
...
Escher posed and solved in an
algorithmic way
...
The article can be downloaded free from the website
http://www
...
org/

William Burnside
In one of the most abstract domains of
thought, he [Burnside] has systematized
and amplified its range so that, there, his
work stands as a landmark in the widening
expanse of knowledge
...

A
...
FORSYTH

WILLIAM BURNSIDE was born on July 2,
1852, in London
...
He then accepted a position at the Royal Naval College at Greenwich and spent the rest of his
career in that post
...
He is best remembered, however, for his pioneering work in
group theory and his classic book Theory of
Groups, which first appeared in 1897
...

One mark of greatness in a mathematician is the ability to pose important and
challenging problems—problems that open
up new areas of research for future generations
...
It was he

who first conjectured that a group G of odd
order has a series of normal subgroups,
G 5 G0 $ G1 $ G2 $ ? ? ? $ Gn 5 {e},
such that Gi /Gi11 is Abelian
...
In 1994, Efim Zelmanov
received the Fields Medal for his work on a
variation of one of Burnside’s conjectures
...
He served as president of the Council of the London Mathematical Society and
received its De Morgan medal
...

To find more information about Burnside, visit:
http://www-groups
...
st-and
...
uk/~history/

497

Cayley Diagraphs
30 of Groups
The important thing in science is not so much to obtain new facts as to
discover new ways of thinking about them
...
The idea was originated by Cayley
in 1878
...
Although there is a rich and important general theory of directed graphs with many applications, we are interested only in those
that arise from groups
...
We define
a digraph Cay(S:G), called the Cayley digraph of G with generating set
S, as follows
...
Each element of G is a vertex of Cay(S:G)
...
For x and y in G, there is an arc from x to y if and only if xs 5 y for
some s [ S
...
He called the
498

499

30 | Cayley Diagraphs of Groups

resulting figure the color graph of the group
...
Rather than use colors to distinguish the different generators,
we will use solid arrows, dashed arrows, and dotted arrows
...
An arrow
emanating from x and pointing to y indicates that there is an arc from x to y
...
Note that
there are several ways to draw the digraph of a group given by a particular generating set
...
These
connections are uniquely determined by the generating set
...
(In the digraphs below, a headless arrow joining two vertices x
and y indicates that there is an arc from x to y and an arc from y to x
...
For example, a generator of order 2 is its own inverse
...

1
5

0

0

1
4

2

1

3
4
3

2

5

Cay ({1}: Z 6 )

Cay ({1}: Z 6 )

EXAMPLE 2 Z3 % Z2 5 ͗(1, 0), (0, 1)͘
...

H

R90
R0

R0

H

R90
H

R270 H

R90H

R180 H

R90

R180 H

R90H

R180

R270 H

R270

R270

R180

Cay({R90, H}: D4 )

Cay({R90, H}: D 4)

EXAMPLE 4 S3 5 ͗(12), (123)͘
...

(12)
(1)

(13)

(13)
(12)

(132)

(123)

(23)

Cay({(12), (13)}: S 3 )

501

30 | Cayley Diagraphs of Groups

EXAMPLE 6 A4 5 ͗(12)(34), (123)͘
...

b

a

b

e

ab

a

a 2b

a2

a3b

a3
Cay({a, b}: Q4)

EXAMPLE 8 D` 5 ͗a, b | a2 5 b2 5 e͘
...
Consider, for
example, the product ab3ab22 from the group given in Example 7
...
Of course, b21 means traverse the b arc
in reverse
...
) Tracing the product
through, we obtain b
...


Hamiltonian Circuits and Paths
Now that we have these directed graphs, what is it that we care to know
about them? One question about directed graphs that has been the object
of much research was popularized by the Irish mathematician
Sir William Hamilton in 1859, when he invented a puzzle called
“Around the World
...
One solves this puzzle
by starting at any particular city (vertex) and traveling “around the
world,” moving along the arcs in such a way that each other city is
visited exactly once before returning to the original starting point
...
1, where the vertices are
visited in the order indicated
...
1 Around the World

16

30 | Cayley Diagraphs of Groups

503

arcs in such a way that each vertex is visited exactly once before
returning to the starting vertex
...
) Such a sequence of arcs is called a Hamiltonian circuit in
the digraph
...
In the rest of this chapter, we concern ourselves with the existence
of Hamiltonian circuits and paths in Cayley digraphs
...
2 and 30
...

Is there a Hamiltonian circuit in
Cay({(1, 0), (0, 1)}: Z3 % Z2)?
More generally, let us investigate the existence of Hamiltonian circuits in
Cay({(1, 0), (0, 1)}: Zm % Zn),
where m and n are relatively prime and both are greater than 1
...
2 Hamiltonian Path in Cay({(1, 0), (0, 1)}: Z3 % Z2)
from (0, 0) to (2, 1)
...
3 Hamiltonian Circuit in Cay({a, b}: Q4)
...
4
...
(Clearly,
such a vertex exists
...
5
...
, which is just the coset (a, b) 1 ͗(21, 1)͘
...
Obviously,
there cannot be a Hamiltonian circuit consisting entirely of horizontal
moves
...


(0, 0)

(0, 1)

(0, 2)

(0, n –1)

(1, 0)

(1, 1)

(1, 2)

(1, n –1)
(0, 1)
(1, 0)

(m –1, 0)

(m –1, 1)

(m –1, 2)

(m –1, n –1)

Figure 30
...

(a 2 1, b 1 1)

(a, b)

Figure 30
...
1 A Necessary Condition
Cay({(1, 0), (0, 1)}: Zm % Zn ) does not have a Hamiltonian circuit
when m and n are relatively prime and greater than 1
...


505

30 | Cayley Diagraphs of Groups

Theorem 30
...

(0, 0 )

(1, 0 )

(0, 1)

(0, 2 )

(1, 1)

(1, 2 )

First 3 3 3 block

(0, 1)

(2, 0 )

(3, 0)

(2, 1)

(2, 2 )

(3, 1)

(1, 0)

(3, 2 )
Repeat path used
in first block

(4, 0)

(5, 0)

(4, 1)

(4, 2 )

(5, 1)

(5, 2 )

Repeat path used
in first block
kth 3 3 3 block

(3k – 1, 0)

(3k – 1, 1)

(3k – 1, 2)

Figure 30
...
Then we may think of Zm % Zn as k blocks of
size n 3 n
...
6 for an example
...
Use the generator (0, 1) to move
horizontally across the first row to the end
...
Keep this process up until the
point (n 2 1, 0)—the lower left-hand corner of the first block—has
been reached
...
Keep this up until the bottom block is
covered
...

Notice that the circuit given in the proof of Theorem 30
...
A much
more convenient way to describe a Hamiltonian path or circuit is to
specify the starting vertex and the sequence of generators in the order
in which they are to be applied
...
In Example 3, we may start at R0 and successively apply R90, R90,
R90, H, R90, R90, R90, H
...
, c is
a sequence of group elements, we use k p (a, b,
...
, c)
...
With this notation, we may conveniently denote the Hamiltonian circuit given in Theorem 30
...

We leave it as an exercise (Exercise 11) to show that if x1, x2,
...

From Theorem 30
...
But Theorem
30
...
There are some Cayley digraphs for non-Abelian groups that do not
even have Hamiltonian paths, but we will not discuss them here
...
3 Abelian Groups Have Hamiltonian Paths
Let G be a finite Abelian group, and let S be any (nonempty†) generating set for G
...


S is the empty set, it is customary to define ͗S͘ as the identity group
...


†If

507

30 | Cayley Diagraphs of Groups

PROOF We induct on |S|
...
, am21, where |a| 5 m
...
Now assume that |S|
...

Choose some s [ S
...
, sn21͘ where S 5 {s1, s2,
...

(Notice that H may be equal to G
...
, ak) in
Cay(T:H)
...
, ak, s, a1, a2,
...
, a1, a2,
...
, ak),
where a1, a2,
...

Because S 5 T < {s} and T generates H, the coset Hs generates the
factor group G/H
...
) Hence, the
cosets of H are H, Hs, Hs2,
...
Starting
from the identity element of G, the path given by (a1, a2,
...
, ak) is a
Hamiltonian path in Cay(T:H)]
...
Starting from there, the path (a1, a2,
...
Then, s moves us to the coset
Hs2, and we visit each element of this coset exactly once
...
, Hsn, visiting each
vertex in each of these cosets exactly once
...
Thus we have a Hamiltonian path
...

EXAMPLE 9 Let
D3 5 ͗r, f | r 3 5 f 2 5 e, rf 5 fr 2͘
...
7
...
7

508

Special Topics

Although it is not easy to prove, it is true that
Cay({(r, 0), ( f, 0), (e, 1)}: Dn % Zm)
has a Hamiltonian circuit for all n and m
...
) Example 10 shows
the circuit for this digraph when m is even
...

Then a Hamiltonian circuit in
Cay({(r, 0), ( f, 0), (e, 1)}: Dn % Zm)
with m even is traced in Figure 30
...
The sequence of generators that
traces the circuit is
m p [(n 2 1) p (r, 0), ( f, 0), (n 2 1) p (r, 0), (e, 1)]
...
iterate
(r2 f,

0)

(r n – 1f, 0)

(r 2,

(r 2,

0)

(r n –1, 0)

(r n –1f, 1)

1)

(r n–1, 1)

Figure 30
...
One particular Cayley graph that
is used to design and analyze interconnection networks of parallel machines is the symmetric group Sn with the set of all transpositions as the
generating set
...
A Hamiltonian path in a Cayley
digraph of a group is simply an ordered listing of the group elements
without repetition
...
In 1948, R
...
Rankin
used these ideas (although not the terminology) to prove that certain bellringing exercises could not be done by the traditional methods employed
by bell ringers
...
22] for the group-theoretic aspects of bell
ringing
...
This program can produce repeating
hyperbolic patterns in color from among various infinite classes of symmetry groups
...
The 2003 Mathematics
Awareness Month poster featured one such image (see http://www
...
org/mam/03/index
...
Two Escher drawings and their
computer-drawn counterparts are given in Figures 30
...
12
...
It is also possible to associate a group—called
the automorphism group—with every directed graph
...


Figure 30
...
C
...
10 A computer duplication of the pattern of M
...
Escher’s Circle
Limit I [2]
...


Figure 30
...
C
...
12 A computer drawing inspired by the pattern of
M
...
Escher’s Circle Limit IV [2]
...


Exercises
A mathematician is a machine for turning coffee into theorems
...
Find a Hamiltonian circuit in the digraph given in Example 7 different from the one in Figure 30
...

2
...

3
...

4
...

5
...


512

Special Topics

6
...
Explain why two
paths from x to y in the digraph yield a group relation
...
am 5 b1b2
...

7
...

8
...


9
...
Verify that
6 p [3 p (r, 0), ( f, 0), 3 p (r, 0), (e, 1)]
is a Hamiltonian circuit in
Cay({(r, 0), ( f, 0), (e, 1)}: D4 % Z6)
...
Draw a picture of Cay({2, 5}: Z8)
...
If s1, s2,
...

(This exercise is referred to in this chapter
...
Show that the Cayley digraph given in Example 7 has a
Hamiltonian path from e to a
...
Show that there is no Hamiltonian path in
Cay({(1, 0), (0, 1)}: Z3 % Z2)
from (0, 0) to (2, 0)
...
Draw Cay({2, 3}: Z6)
...
a
...
Show that a sequence s1, s2,
...

b
...


30 | Cayley Diagraphs of Groups

513

16
...
Draw Cay({a, b}: D4)
...
Let Dn be as in Example 10
...

18
...
Find a Hamiltonian
circuit in Cay({a, b}: Q8)
...
Let Q8 be as in Exercise 18
...

20
...
Does it have a Hamiltonian path?
21
...

Does this circuit generalize to the case Dn11 % Zn for all n $ 3?
22
...
Find a Hamiltonian circuit in
Cay({(a, 0), (b, 0), (e, 1)}: Q8 % Zm) for all even m
...
Find a Hamiltonian circuit in
Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Z3)
...
Find a Hamiltonian circuit in
Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Zm) for all odd m $ 3
...
Write the sequence of generators that describes the Hamiltonian
circuit in Example 9
...
Let Dn be as in Example 10
...

27
...


29
...


Does your circuit generalize to the case Dn % Zn11 for all n $ 4?
Prove that Cay({(0, 1), (1, 1)}: Zm % Zn) has a Hamiltonian circuit
for all m and n greater than 1
...

Show that the circuit exits from every member of the coset (a, b)
1 ͗(1, 21)͘ vertically
...
Find a Hamiltonian circuit
in Cay({(r, 0), ( f, 0), (e, 1)}: D2 % Z3)
...
Find a Hamiltonian circuit in Cay({(a, 0),
(b, 0), (e, 1)}: Q8 % Z3)
...
In Cay({(1, 0), (0, 1)}: Z4 % Z5) find a sequence of generators that
visits exactly one vertex twice and all others exactly once and returns to the starting vertex
...
In Cay({(1, 0), (0, 1)}: Z4 % Z5) find a sequence of generators that
visits exactly two vertices twice and all others exactly once and returns to the starting vertex
...
Find a Hamiltonian circuit in Cay({(1, 0), (0, 1)}: Z4 % Z6)
...
(Factor Group Lemma) Let S be a generating set for a group G, let
N be a cyclic normal subgroup of G, and let
S 5 {sN | s [ S}
...
, ar N) is a Hamiltonian circuit in Cay(S:G/N) and the
product a1 ? ? ? ar generates N, prove that
|N| p (a1,
...

35
...

(One non-Abelian example is Q4
...
3 can be
generalized to include all Hamiltonian groups
...
F
...
Budden, The Fascination of Groups, Cambridge: Cambridge University Press, 1972
...
Douglas Dunham, John Lindgren, and David Witte, “Creating Repeating
Hyperbolic Patterns,” Computer Graphics 15 (1981): 215–223
...
David Witte, Gail Letzter, and Joseph A
...


Suggested Readings
Frank Budden, “Cayley Graphs for Some Well-Known Groups,” The
Mathematical Gazette 69 (1985): 271–278
...

E
...
Burrows and M
...
Clark, “Pictures of Point Groups,” Journal of
Chemical Education 51 (1974): 87–90
...
It gives a
comprehensive collection of the Cayley digraphs of groups important
to chemists
...

In this beautifully illustrated paper, a process for creating repeating
patterns of the hyperbolic plane is described
...

Joseph A
...

This article surveys research done on variations of the themes discussed in this chapter
...
Gallian and David Witte, “Hamiltonian Checkerboards,” Mathematics Magazine 57 (1984): 291–294
...
d
...
edu/~jgallian/checker
...

Paul Hoffman, “The Man Who Loves Only Numbers,” The Atlantic
Monthly 260 (1987): 60–74
...

Henry Levinson, “Cayley Diagrams,” in Mathematical Vistas: Papers
from the Mathematics Section, New York Academy of Sciences, J
...
McCarthy, eds
...

This richly illustrated article presents Cayley digraphs of many of the
groups that appear in this text
...
T
...

This article analyzes the practice of bell ringing by way of Cayley
digraphs
...

In this documentary, Erdös discusses politics, death, and mathematics
...
It is available for purchase at www
...
com
...
This free software is available at http://
sourceforge
...

SIR EDMUND WHITTAKER,

Scientific American

WILLIAM ROWAN HAMILTON was born on
August 3, 1805, in Dublin, Ireland
...
At five, he read and translated Latin,
Greek, and Hebrew; at 14, he had mastered
14 languages, including Arabic, Sanskrit,
Hindustani, Malay, and Bengali
...
In
1843, he made what he considered his greatest discovery—the algebra of quaternions
...


This stamp featuring the quaternions was
issued in 1983
...
Of
greater significance, however, is the fact that
the quaternions are noncommutative under
multiplication
...
The essential idea for the
quaternions suddenly came to Hamilton after
15 years of fruitless thought!
Today Hamilton’s name is attached to several concepts, such as the Hamiltonian function, which represents the total energy in a
physical system; the Hamilton-Jacobi differential equations; and the Cayley-Hamilton
Theorem from linear algebra
...

In his later years, Hamilton was plagued
by alcoholism
...

For more information about Hamilton,
visit:
http://www-groups
...
st-and
...
uk/~history/

Paul Erdös
Paul Erdös is a socially helpless Hungarian
who has thought about more mathematical problems than anyone else in history
...

Unlike most of his contemporaries, who
have concentrated on theory building, Erdös
focused on problem solving and problem
posing
...

Erdös was born on March 26, 1913, in
Hungary
...
Erdös, a Jew,
left Hungary in 1934 at the age of 21 because of the rapid rise of anti-Semitism in
Europe
...
He had little property and
no fixed address
...
His motto was, “Another roof, another proof
...

Erdös wrote more than 1500 research papers
...
These people are said to have
Erdös number 1
...
Erdös died of a heart attack on September 20, 1996, in Warsaw, Poland
...
dcs
...
ac
...
oakland
...
html

517

Introduction
31 to Algebraic
Coding Theory
Damn it, if the machine can detect an error, why can’t it locate the position
of the error and correct it?
RICHARD W
...
This theory,
which originated in the late 1940s, was created in response to practical
communication problems
...
) Algebraic codes are now used in compact disk and DVD
players, fax machines, digital televisions, and bar code scanners, and
are essential to computer maintenance
...
If the proposed
landing site appears unfavorable, we will command the craft to orbit
the planet; otherwise, we will command the craft to land
...
But it is possible that some sort of interference (called noise) could cause an incorrect message to be received
...
For example, if we
wish the craft to orbit Mars, we could send five 0s
...
So, if 00000 is sent and 10001 is received, the computer decides that 0 was the intended message
...
If we assume that errors occur
independently, it is less likely that three errors will occur than that two
or fewer errors will occur
...
Our particular situation is illustrated in Figure 31
...
The general coding procedure is illustrated in Figure 31
...

518

519

31 | Introduction to Algebraic Coding Theory

Original
message
0
earth

Encoded
message
00000
encoder

Received
message
10001

Noise
transmitter

spacecraft

Decoded
message
0

decoder

Figure 31
...
2 General encoding-decoding

In practice, the means of transmission are telephone, radiowave,
microwave, or even a magnetic disk
...
Throughout this chapter, we assume that errors in transmission occur independently
...

Now, let’s consider a more complicated situation
...
Further,
suppose that the probability that an error will be made in the transmission of any particular digit is
...
If we send this message directly, without any redundancy, the probability that it will be received error free is
(
...
0066
...
For example, the sequence 1011 is encoded as 111000111111
...
Now, what is the probability that our 500-digit message will be error free? Well, if a 1, say, is
sent, it will be decoded as a 0 if and only if the block received is 001,
010, 100, or 000
...
01)(
...
99) 1 (
...
99)(
...
99)(
...
01) 1 (
...
01)(
...
01)2[3(
...
01]
5
...
0003
...
9997, and it follows that the probability that the entire 500-digit message will be decoded correctly is
greater than (
...
86—a dramatic improvement
over
...


520

Special Topics

This example illustrates the three basic features of a code
...
The encoding procedure builds some
redundancy into the original messages; the decoding procedure corrects
or detects certain prescribed errors
...
In a
fivefold repetition code, 80% of all transmitted information is redundant
...

Before plunging into the formal theory, it is instructive to look at a
sophisticated example
...
Encoding will be done by viewing these messages as 1 3 4 matrices with entries from Z2 and multiplying each of the 16 messages on the right by
the matrix
1
0
G5≥
0
0

0
1
0
0

0
0
1
0

0
0
0
1

1
1
1
0

1
0
1
1

0
1
¥
...
) The resulting 7-tuples are called
code words
...
1
...
1
Message

Encoder G

Code Word

0000
0001
0010
0100
1000
1100
1010
1001

→
→
→
→
→
→
→
→

0000000
0001011
0010111
0100101
1000110
1100011
1010001
1001101

Message
0110
0101
0011
1110
1101
1011
0111
1111

Encoder G

Code Word

→
→
→
→
→
→
→
→

0110010
0101110
0011100
1110100
1101000
1011010
0111001
1111111

31 | Introduction to Algebraic Coding Theory

521

Notice that the first four digits of each code word constitute just the
original message corresponding to the code word
...
For this code, we
use the nearest-neighbor decoding method (which, in the case that the
errors occur independently, is the same as the maximum-likelihood decoding procedure)
...
If the choice of v9 is not unique, we can decide not to decode or
arbitrarily choose one of the code words closest to v
...
)
Once we have decoded the received word, we can obtain the message
by deleting the last three digits of v9
...
It would be encoded and transmitted as u 5
1000110
...
Similarly, the intended message 1111 would be encoded as
1111111
...

The code in Example 1 is one of an infinite class of important codes
discovered by Richard Hamming in 1948
...

The Hamming (7, 4) encoding scheme can be conveniently illustrated with the use of a Venn diagram, as shown in Figure 31
...
Begin
by placing the four message digits in the four overlapping regions I, II,
A

B

I

V

A

VI

1

1
0

III
IV

II

B

0

VII

1

1

C

0

C

Figure 31
...
For regions V, VI, and VII, assign 0 or 1 so that
the total number of 1s in each circle is even
...
This tells us that something is
wrong
...

Thus, the portion of the diagram that is in both A and B but not in C is
the source of the error
...
4
...

This is especially appropriate when it is easy to retransmit a message
...
For
example, computers are designed to use a parity check for numbers
...

If there is an even number of 1s in this representation, a 0 is attached to
the string; if there is an odd number of 1s in the representation, a 1
is attached to the string
...
Now, when the computer reads a
A

B

0

1

0

0
1

0
1

C

Figure 31
...
If the read number
has an odd number of 1s, the computer will know that an error has been
made, and it will reread the number
...

The methods of error detection introduced in Chapters 0 and 5 are
based on the same principle
...
If we find that
such a string does not satisfy that condition, we know that an error has
occurred
...

Definition Linear Code

An (n, k) linear code over a finite field F is a k-dimensional subspace V
of the vector space

Fn 5 F % F % ? ? ? % F
n copies
over F
...
When F is Z2, the
code is called binary
...
Note that an (n, k) linear code over a finite field F of order q has
q k code words, since every member of the code is uniquely expressible
as a linear combination of the k basis vectors with coefficients from F
...
Also, since errors in transmission may occur
in any of the n positions, there are qn possible vectors that can be
received
...
, an) more simply as a1a2 ? ? ? an, as we
did in Example 1
...


524

Special Topics

EXAMPLE 3 The set {0000, 0101, 1010, 1111} is a (4, 2) binary
code
...

EXAMPLE 4 The set
{0000, 0121, 0212, 1022, 1110, 1201, 2011, 2102, 2220}
is a (4, 2) linear code over Z3
...

To facilitate our discussion of the error-correcting and errordetecting capability of a code, we introduce the following terminology
...
The Hamming weight of a vector is the
number of nonzero components of the vector
...


We will use d(u, v) to denote the Hamming distance between the
vectors u and v, and wt(u) for the Hamming weight of the vector u
...
Then, d(s, t) 5 4, d(s, u) 5 4, d(s, v) 5 5, d(u, v) 5 1; and
wt(s) 5 4, wt(t) 5 4, wt(u) 5 4, wt(v) 5 5
...

Theorem 31
...


PROOF To prove that d(u, v) 5 wt(u 2 v), simply observe that both
d(u, v) and wt(u 2 v) equal the number of positions in which u and v
differ
...


31 | Introduction to Algebraic Coding Theory

525

With the preceding definitions and Theorem 31
...

Theorem 31
...
Alternatively, the same code
can detect any 2t or fewer errors
...

(If there is more than one such v9, we do not decode
...
Then, by the definition of distance between u and v, we have d(u, v) # t
...
Thus, by assumption,
2t 1 1 # wt(w 2 u) 5 d(w, u) # d(w, v) 1 d(v, u) # d(w, v) 1 t,
and it follows that t 1 1 # d(w, v)
...

To show that the code can detect 2t errors, we suppose that a transmitted code word u is received as the vector v and that at least one
error, but no more than 2t errors, was made in transmission
...
But v cannot be a code word, since
d(v, u) # 2t, whereas we know that the minimum distance between distinct code words is at least 2t 1 1
...
2 is often misinterpreted to mean that a linear code with
Hamming weight 2t 1 1 can correct any t errors and detect any 2t or
fewer errors simultaneously
...
The user must choose
one or the other role for the code
...
1
...
To understand why we can’t do
both, consider the received word 0001010
...

But there is no way for us to know which of these possibilities occurred
...
If our choice
were error detection, we simply would not decode
...
)
On the other hand, if we write the Hamming weight of a linear code
in the form 2t 1 s 1 1, we can correct any t errors and detect any t 1 s
or fewer errors
...
Detect any four errors (t 5 0, s 5 4)
...
Correct any one error and detect any two or three errors (t 5 1,
s 5 2)
...
Correct any two errors (t 5 2, s 5 0)
...

It is natural to wonder how the matrix G used to produce the Hamming code in Example 1 was chosen
...
A matrix of this form is called the standard
generator matrix (or standard encoding matrix) for the resulting code
...
An (n, k) linear code in which the k information
digits occur at the beginning of each code word is called a systematic
code
...

EXAMPLE 7 From the set of messages
{000, 001, 010, 100, 110, 101, 011, 111},
we may construct a (6, 3) linear code over Z2 with the standard generator matrix
1 0 0 1 1 0
G 5 £ 0 1 0 1 0 1 §
...
2
...

Table 31
...

0 1 2 2

The resulting code words are given in Table 31
...
Since the minimum
weight of the code is 3, it will correct any single error or detect any
double error
...
3
Message

Encoder G

Code Word

00
01
02
10
11
12
20
21
22

→
→
→
→
→
→
→
→
→

0000
0122
0211
1021
1110
1202
2012
2101
2220

Parity-Check Matrix Decoding
Now that we can conveniently encode messages with a standard generator matrix, we need a convenient method for decoding the received
messages
...
(When more than one error occurs in a
code word, our decoding method fails
...

Then, the n 3 (n 2 k) matrix
H5c

2A
d,
In2k

where 2A is the negative of A and In2k is the (n 2 k) 3 (n 2 k) identity matrix, is called the parity-check matrix for V
...
) The decoding procedure is:
1
...

2
...

3
...
s
...
If
there is more than one such instance, do not decode
...
When the code is binary, category 3 reduces to the following
...
If wH is more than one row of H,
do not decode
...
If wH does not fit into either category 2 or category 3, we know that
at least two errors occurred in transmission and we do not decode
...
The generator matrix is
1
0
G5≥
0
0

0
1
0
0

0
0
1
0

0
0
0
1

1
1
1
0

1
0
1
1

0
1
¥,
1
1

and the corresponding parity-check matrix is
1
1
1
H 5 G0
1
0
0

1
0
1
1
0
1
0

0
1
1
1W
...
Since this
is the first row of H and no other row, we assume that an error has been
made in the first position of v
...
Similarly, if w 5 1011111 is the received word, then wH 5 101,
and we assume that an error has been made in the second position
...
If the encoded message 1001101 is received as z 5 1001011 (with
errors in the fifth and sixth positions), we find zH 5 110
...
On the other hand, nearestneighbor decoding would yield the same incorrect result
...
We will soon see under what conditions this is true, but first we
need an important fact relating a code given by a generator matrix and
its parity-check matrix
...
Then, for any vector v
in F n, we have vH 5 0 (the zero vector) if and only if v belongs to C
...
Therefore, it follows from
the dimension theorem for linear transformations that n 5
n 2 k 1 dim (Ker H), so that Ker H has dimension k
...
)
Then, since the dimension of C is also k, it suffices to show that
2A
d
...
To do this, let G 5 [Ik | A], so that H 5 c
In2k
GH 5 [Ik | A] c

2
A
d 5 2A 1 A 5 [0]
In2k

(the zero matrix)
...
Thus, vH 5 (mG)H 5 m(GH) 5 m[0] 5 0 (the zero
vector)
...

But it will do more
...
3 Parity-Check Matrix Decoding
Parity-check matrix decoding will correct any single error if and only
if the rows of the parity-check matrix are nonzero and no one row is
a scalar multiple of any other row
...
In this
special situation, the condition on the rows is that they are nonzero and
distinct
...
Suppose that the transmitted code word w
was received with only one error, and that this error occurred in the ith
position
...
Now, using the
Orthogonality Lemma, we obtain
(w 1 ei)H 5 wH 1 eiH 5 0 1 eiH 5 eiH
...
Thus, if there was exactly one error in transmission, we can use the rows of the parity-check

31 | Introduction to Algebraic Coding Theory

531

matrix to identify the location of the error, provided that these rows are
distinct
...
)
Conversely, suppose that the parity-check matrix method correctly
decodes all received words in which at most one error has been made
in transmission
...
Thus, no row of H is the zero vector
...

Then, if some code word w is transmitted and the received word is
w 1 ei (that is, there is a single error in the ith position), we find
(w 1 ei)H 5 wH 1 eiH 5 ith row of H 5 jth row of H
...
This contradicts
our assumption that the method correctly decodes all received words in
which at most one error has been made
...
This method was devised by David Slepian in 1956
and is called coset decoding (or standard decoding)
...

The first row of the table is the set C of code words, beginning in column 1 with the identity 0 ? ? ? 0
...
Among all the
elements of the coset v 1 C, choose one of minimum weight, say, v9
...
Continue this process until all the
vectors in V have been listed in the table
...
] The words
in the first column are called the coset leaders
...

EXAMPLE 10 Consider the (6, 3) binary linear code
C 5 {000000, 100110, 010101, 001011, 110011, 101101, 011110, 111000}
...
Obviously,
100000 is not in C and has minimum weight among the elements of
100000 1 C, so it can be used to lead the second row
...
4 is the
completed table
...
4 A Standard Array for a (6, 3) Linear Code
Words
Coset
Leaders
000000
100000
010000
001000
000100
000010
000001
100001

100110
000110
110110
101110
100010
100100
100111
000111

010101
110101
000101
011101
010001
010111
010100
110100

001011
101011
011011
000011
001111
001001
001010
101010

110011
010011
100011
111011
110111
110001
110010
010010

101101
001101
111101
100101
101001
101111
101100
001100

011110
111110
001110
010110
011010
011100
011111
111111

111000
011000
101000
110000
111100
111010
111001
011001

If the word 101001 is received, it is decoded as 101101, since
101001 lies in the column headed by 101101
...

Recall that the first method of decoding that we introduced was the
nearest-neighbor method; that is, any received word w is decoded as
the code word c such that d(w, c) is a minimum, provided that there is
only one code word c such that d(w, c) is a minimum
...

Theorem 31
...


PROOF Let C be a linear code, and let w be any received word
...
Then, w 1 C 5 v 1 C, so
w 5 v 1 c for some c in C
...
Now, if c9 is any code word, then w 2 c9 [ w 1 C 5 v 1 C, so
that wt(w 2 c9) $ wt(v), since the coset leader v was chosen as a vector
of minimum weight among the members of v 1 C
...

So, using coset decoding, w is decoded as a code word c such that
d(w, c) is a minimum
...
In the case of coset decoding, the
decoded value of v is always uniquely determined by the coset leader
of the row containing the received word
...

When we know a parity-check matrix for a linear code, coset decoding can be considerably simplified
...


The importance of syndromes stems from the following property
...
5 Same Coset—Same Syndrome
Let C be an (n, k) linear code over F with a parity-check matrix H
...


PROOF Two vectors u and v are in the same coset of C if and only if
u 2 v is in C
...

We may now use syndromes for decoding any received word w:
1
...

2
...

3
...

With this method, we can decode any received word with a table that
has only two rows—one row of coset leaders and another row with the
corresponding syndromes
...
The paritycheck matrix for this code is
1
1
0
H5F
1
0
0
†This

1
0
1
0
1
0

0
1
1
V
...
Hagelbarger in 1959
...

Since the coset leader v 5 000100 has 100 as its syndrome, we assume that
w 2 000100 5 101101 was sent
...
Notice that
these answers are in agreement with those obtained by using the standardarray method of Example 10
...
In medicine, it is used to
designate a collection of symptoms that typify a disorder
...

In this chapter, we have presented algebraic coding theory in
its simplest form
...
For example, Gorenstein (see Chapter 25 for a biography) and
Zierler, in 1961, made use of the fact that the multiplicative subgroup
of a finite field is cyclic
...

In some instances, two error-correcting codes are employed
...
One code checked for
independently occurring errors, and another—a so-called ReedSolomon code—checked for bursts of errors
...
999999
...
They can correct thousands of consecutive errors
...


We conclude this chapter with an adapted version of an article by Barry A
...
It was the first in a series of articles called “Mathematics
That Counts” in SIAM News, the news journal of the Society for Industrial and Applied
Mathematics
...

In this “Age of Information,” no one need be
reminded of the importance not only of
speed but also of accuracy in the storage, retrieval, and transmission of data
...
Just as architects design buildings that
will remain standing even through an earthquake, their computer counterparts have
come up with sophisticated techniques capable of counteracting digital disasters
...

“When you talk about CD players and digital audio tape and now digital television, and
various other digital imaging systems that are
coming—all of those need Reed-Solomon
[codes] as an integral part of the system,” says
Robert McEliece, a coding theorist in the
electrical engineering department at Caltech
...
Voyager II,
for example, was transmitting data at incredibly low power—barely a whisper—
over tens of millions of miles
...

Careful engineering can reduce the error
rate to what may sound like a negligible
level—the industry standard for hard disk
drives is 1 in 10 billion—but given the volume of information processing done these
days, that “negligible” level is an invitation
to daily disaster
...

In 1960, the theory of error-correcting
codes was only about a decade old
...
At the same time, Richard
Hamming introduced an elegant approach to
single-error correction and double-error
detection
...
But with
their SIAM journal paper, McEliece says,
Reed and Solomon “hit the jackpot
...
That feature makes
Reed-Solomon codes particularly good at
dealing with “bursts” of errors: six consecutive bit errors, for example, can affect at
most two bytes
...

Mathematically, Reed-Solomon codes
are based on the arithmetic of finite fields
...
” Starting from a “message” (a 0, a1,

...
, P(gN21)), where N is
the number of elements in K, g is a generator of the (cyclic) group of nonzero elements in K, and P(x) is the polynomial a0 1
a1x 1 ? ? ? 1 am21x m21
...
e
...

In today’s byte-sized world, for example,
it might make sense to let K be the field of
order 28, so that each element of K corresponds to a single byte (in computerese, there
are four bits to a nibble and two nibbles to a
byte)
...
, P(g255)
...

Despite their advantages, Reed-Solomon
codes did not go into use immediately—
they had to wait for the hardware technology to catch up
...

The Reed-Solomon paper “suggested some
nice ways to process data, but nobody knew
if it was practical or not, and in 1960 it
probably wasn’t practical
...
Many other bells
and whistles (some of fundamental theoretic significance) have also been added
...

Reed was among the first to recognize
the significance of abstract algebra as the
basis for error-correcting codes
...

However, he added, “coding theory was not
a subject when we published that paper
...

Three decades later, the impact is clear
...
“It’s clear they’re practical,
because everybody’s using them now,” says
Elwyn Berkekamp
...
In short, says McEliece, “it’s been an
extraordinarily influential paper
...

“But let your communication be yea, yea; nay, nay: for whatsoever is
more than these cometh of evil
...
Find the Hamming weight of each code word in Table 31
...

2
...

3
...

4
...
d(u, v) 5 d(v, u) (symmetry)
b
...
d(u, v) 5 d(u 1 w, v 1 w) (translation invariance)
5
...

0 0 1 1 1 0
6
...

7
...
Suppose that C is a linear code with Hamming weight 3 and that
C9 is one with Hamming weight 4
...
Let C be a binary linear code
...
(A subcode of a code is a subset of
the code that is itself a code
...
Let
C 5 {0000000, 1110100, 0111010, 0011101, 1001110,
0100111, 1010011, 1101001}
...
Suppose that the parity-check matrix of a binary linear code is
1
0
H 5 E1
1
0

0
1
1U
...
Use the generator matrix
G5c

1 0 1 1
d
0 1 2 1

to construct a (4, 2) ternary linear code
...

13
...

0
1

Find the parity-check matrix of this code
...
Show that in a binary linear code, either all the code words end with
0, or exactly half end with 0
...
Suppose that a code word v is received as the vector u
...

16
...

Construct a standard array for C
...
If the received word 11101 has exactly one error, can we determine the intended code word? If the
received word 01100 has exactly one error, can we determine the
intended code word?

31 | Introduction to Algebraic Coding Theory

539

17
...

0 0 1 1 0 1
Decode each of the received words
001001, 011000, 000110, 100001
by the following methods:
a
...
parity-check matrix method,
c
...
coset decoding using the syndrome method
...
Suppose that the minimum weight of any nonzero code word in a
linear code is 6
...

19
...

20
...

Determine the code
...
How many code words are there in a (6, 4) ternary linear code?
How many possible received words are there for this code?
22
...
Suppose that the parity-check matrix for a ternary code is
2
2
E1
1
0

1
2
2U
...

24
...
2 is true
...
Can a (6, 3) binary linear code be double-error-correcting using the
nearest-neighbor method? Do not assume that the code is systematic
...
Prove that there is no 2 3 5 standard generator matrix G that will
produce a (5, 2) linear code over Z3 capable of detecting all possible triple errors
...
Why can’t the nearest-neighbor method with a (4, 2) binary linear
code correct all single errors?
28
...

Determine the row that contains 100001
...
Use the field F 5 Z2[x]/͗x2 1 x 1 1͘ to construct a (5, 2) linear
code that will correct any single error
...
Find the standard generator matrix for a (4, 2) linear code over Z3
that encodes 20 as 2012 and 11 as 1100
...
Will the code correct all
single errors?
31
...
, n, the code C has at least one vector with a 1 in the
ith position
...

32
...
Show that not every vector of
weight t 1 1 in Fn can occur as a coset leader
...
Let C be an (n, k) binary linear code over F 5 Z 2
...

34
...
Show that either every member of C
has even weight or exactly half the members of C have even
weight
...
)
35
...
For each i with 1 # i # n, let Ci 5
{v [ C | the ith component of v is 0}
...


31 | Introduction to Algebraic Coding Theory

541

Reference
1
...
Cipra, “The Ubiquitous Reed-Solomon Codes,” SIAM News
26 (January 1993): 1, 11
...
H
...

The eminent mathematician G
...
Hardy insisted that “real” mathematics was almost wholly useless
...
Levinson uses the finite field of
order 16 to construct a linear code that can correct any three errors
...
M
...
Washington, D
...
: The Mathematical Association of
America, 1983
...


Richard W
...

Citation for the Piore
Award, 1979

RICHARD W
...
He graduated
from the University of Chicago with a B
...
degree in mathematics
...
A
...
D
...

During the latter part of World War II,
Hamming was at Los Alamos, where he was
involved in computing atomic-bomb designs
...

In 1950, Hamming published his famous
paper on error-detecting and error-correcting
codes
...
The Hamming codes are used in
many modern computers
...


542

Hamming received numerous prestigious awards, including the Turing Prize
from the Association for Computing Machinery, the Piore Award from the Institute of
Electrical and Electronics Engineers
(IEEE), and the Oender Award from the
University of Pennsylvania
...
Hamming Medal “for exceptional contributions to information sciences, systems
and technology” and named Hamming as
its first recipient
...

To find more information about Hamming, visit:
http://www-groups
...
st-and

...
uk/~history/

Jessie MacWilliams
She was a mathematician who was instrumental in developing the mathematical
theory of error-correcting codes from its
early development and whose Ph
...
thesis
includes one of the most powerful theorems in coding theory
...
She was born in
1917 in England
...
After one year at Johns
Hopkins, she went to Harvard for a year
...
Although she made a
major discovery about codes while a programmer, she could not obtain a promotion
to a math research position without a Ph
...

degree
...
D
...

She then returned to Harvard for a year
(1961–1962), where she finished her degree
...

MacWilliams returned to Bell Labs,
where she remained until her retirement in
1983
...
One of her results of great theoretical
importance is known as the MacWilliams
Identity
...

To find more information about
Mac Williams, visit:
http://www
...
org/
noetherbrochure/
MacWilliams80
...


VERA PLESS was born on March 5, 1931, to
Russian immigrants on the West Side of
Chicago
...
The program at Chicago emphasized great literature
but paid little attention to physics and mathematics
...
After
passing her master’s exam, she took a job as a
research associate at Northwestern University while pursuing a Ph
...
there
...

Over the next several years, Pless stayed
at home to raise her children while teaching

544

part-time at Boston University
...
One person told her outright, “I would never hire a woman
...
Although she had never even heard of
coding theory, she was hired because of her
background in algebra
...
In 1975,
she went to the University of Illinois–Chicago,
where she remained until her retirement
...
D
...


An Introduction to
32
Galois Theory
Galois theory is a showpiece of mathematical unification, bringing together
several different branches of the subject and creating a powerful machine
for the study of problems of considerable historical and mathematical
importance
...
Look at Figures 32
...
2
...
1
depicts the lattice of subgroups of the group of field automorphisms of
4
Q( " 2 , i)
...
Figure 32
...
The integer along an upward line from a field
K1 to a field K2 is the degree of K2 over K1
...
2 is the lattice of Figure 32
...
This is
only one of many relationships between these two lattices
...
1 Lattice of subgroups of the group of field automorphisms of
4
4
4
4
4
Q(" 2 , i), where a : i S i and " 2 S 2i " 2 , b : i S 2i, and " 2 S " 2
545

546

Special Topics
4

Q(√2, i)
2
4

Q(√2)
2

2

2

4

2

2

2

2

Q(√2)

2

4

4

Q(√2, i)

Q(i√2)

Q((11 i) √2)

Q((12 i)√2)
2

2

Q(i√2)

Q(i)
2

2

2

2

Q

Figure 32
...
This relationship was discovered in the process of attempting to solve a polynomial equation f(x) 5 0 by radicals
...

Definitions Automorphism, Galois Group, Fixed Field of H

Let E be an extension field of the field F
...
The Galois group of E over F,
Gal(E/F), is the set of all automorphisms of E that take every element
of F to itself
...


It is easy to show that the set of automorphisms of E forms a group
under composition
...
Be careful not to
misinterpret Gal(E/F) as something that has to do with factor rings or
factor groups
...

The following examples will help you assimilate these definitions
...

We leave to the reader the verifications that the mappings are indeed
automorphisms
...
Since
Q("2) 5 {a 1 b"2 | a, b [ Q}

32 | An Introduction to Galois Theory

547

and any automorphism of a field containing Q must act as the identity
on Q (Exercise 1), an automorphism f of Q("2 ) is completely determined by f("2 )
...
This proves that the group Gal(Q("2)/Q)
has two elements, the identity mapping and the mapping that sends a 1
b"2 to a 2 b "2
...
An automor3
3
phism f of Q(" 2 ) is completely determined by f(" 2 )
...
Since Q(" 2 ) is a subset of the real numbers and " 2
3
3
is the only real cube root of 2, we must have f(" 2 ) 5 " 2
...
Obviously, the fixed field of Gal(Q(" 2 )/Q) is Q (" 2 )
...
Any auto4
morphism f of Q(" 2 , i) fixing Q(i) is completely determined by
4
f(" 2 )
...
Thus, there are at most
4
four possible automorphisms of Q(" 2 , i) fixing Q(i)
...
Thus, Gal(Q(" 2 , i)/Q(i)) is a
2} (where e is the identity
cyclic group of order 4
...
The lattice of subgroups of Gal(Q(" 2 , i)/
4
Q(i)) and the lattice of subfields of Q(" 2 , i) containing Q(i) are shown
in Figure 32
...
As in Figures 32
...
2, the integers along the lines
4

4

{e, a, a 2, a 3}

Q(√2, i)

2

2

{e, a 2}

Q(√2, i)

2

2

{e}

Q(i)

Figure 32
...

EXAMPLE 4 Consider the extension Q("3 , "5 ) of Q
...
This time there are four automorphisms:
e

a

"3 S "3
"5 S "5

b

"3 S 2"3
"5 S "5

ab

"3 S "3
"5 S 2"5

"3 S 2"3
"5 S 2"5

Obviously, Gal(Q("3, "5)/Q) is isomorphic to Z2 % Z2
...
The lattice of subgroups of
Gal(Q("3, "5)/Q) and the lattice of subfields of Q("3, "5) are shown
in Figure 32
...

{e, a, b, ab}

2

Q(√3,√5 )

{e, b }

{e, a }

2

2

2

2

{e, ab }

2

Q(√5 )

2

2

2

Q (√3 )

2

2

{e}

Q(√3√5 )

2

Q

Figure 32
...
In
particular, the automorphism group is non-Abelian
...
Now, consider the
3
extension Q(v, " 2 ) of Q
...
There are six in all:
e

a

b

b2

ab

ab2

vSv
v S v2
vSv
vSv
v S v2
v S v2
3
3
3
3
3
3
3
3
3
3
3
2" 2 " 2 S v2" 2 " 2 S v" 2
" 2 S " 2 " 2 S " 2 " 2 S v" 2 " 2 S v
3

549

32 | An Introduction to Galois Theory

Since ab 2 ba, we know that Gal(Q(v, " 2 )/Q) is isomorphic to S3
...
2
...
5
...
5 Lattice of subgroups of Gal(Q(v, " 2 )/Q) and lattice
3
of subfields of Q(v, " 2 ), where v 5 21/2 1 i"3/2
...
5 have been arranged so that each nontrivial
proper field occupying the same position as some group is the fixed field
3
of that group
...

The preceding examples show that, in certain cases, there is an intimate connection between the lattice of subfields between E and F and
the lattice of subgroups of Gal(E/F)
...
Thus, we
may define a mapping g: ^ S & by g(K) 5 Gal(E/K) and a mapping
f :& S ^ by f (H) 5 EH
...
Similarly, if G and H belong to & and
G # H, then f (G) $ f (H)
...
We leave it to the reader to show that for any K
in ^, we have (fg)(K) $ K, and for any G in &, we have (gf )(G) $ G
...

However, when E is a suitably chosen extension of F, the Fundamental
Theorem of Galois Theory, Theorem 32
...
In particular, f and g
are inclusion-reversing isomorphisms between the lattices ^ and &
...
1 is true, but our theorem illustrates the fundamental principles involved
...
285] for additional details and proofs
...
1 Fundamental Theorem of Galois Theory
Let F be a field of characteristic 0 or a finite field
...
Furthermore, for any subfield K of E containing F,

1
...
[The
index of Gal(E/K) in Gal(E/F) equals the degree of K over F
...
If K is the splitting field of some polynomial in F[x],

then Gal(E/K) is a normal subgroup of Gal(E/F) and Gal(K/F)
is isomorphic to Gal(E/F)/Gal(E/K)
...
K 5 EGal(E/K)
...
]
4
...
[The
automorphism group of E fixing EH is H
...
For example, it is usually quite difficult to determine, directly, how many subfields a given field has, and it
is often difficult to decide whether or not two field extensions are the
same
...
Hence, the Fundamental Theorem of Galois Theory can
be a great labor-saving device
...
[Recall from
Chapter 20 that if f (x) [ F[x] and the zeros of f (x) in some extension
of F are a1, a2,
...
, an) is the splitting field of
f (x) over F
...
How many subfields does it have and what are
they? First, observe that Q(v) is the splitting field of x7 2 1 over Q, so
that we may apply the Fundamental Theorem of Galois Theory
...
Thus,
[Q(v):Q] 5 |Gal(Q(v)/Q)| $ 6
...

Thus, Gal(Q(v)/Q) is a cyclic group of order 6
...
See Figure 32
...

kφ l
2

3

kφ 2 l

kφ 3 l
3

2
{e}

Figure 32
...
To find the fixed field of
͗f3͘, we must find a member of Q(v) that is not in Q and that is fixed
by f3
...
Since [Q(v)͗f 3͘ : Q] 5 3 and [Q(v 1 v21) : Q]
divides [Q(v)͗f 3͘ : Q], we see that Q(v 1 v21) 5 Q(v)͗f 3͘
...
Thus,
we have found all subfields of Q(v)
...
Let
us determine Gal(E/F)
...
2, E has the form
F(b) for some b where b is the zero of an irreducible polynomial p(x) of
the form x n 1 an21x n21 1 ? ? ? 1 a1x 1 a0, where an21, an22,
...
Since any field automorphism f of E must take 1 to itself, it
follows that f acts as the identity on F
...
And because p(x) has at most n zeros, we know that there
are at most n possibilities for f(b)
...
2) that the group ͗s͘ has order n (see Exercise 11 in
Chapter 22)
...


552

Special Topics

Solvability of Polynomials by Radicals
For Galois, the elegant correspondence between groups and fields
given by Theorem 32
...
Galois sought to
solve a problem that had stymied mathematicians for centuries
...
In the 16th century, Italian mathematicians developed formulas for solving any third- or
fourth-degree equation
...
For example, the equation
x3 1 bx 1 c 5 0
has the three solutions
A 1 B,
2(A 1 B)/2 1 (A 2 B)" /2,
23
2(A 1 B)/2 2 (A 2 B)" /2,
23
where
A5

2c
b3
c2
1
1
Ç 2
Å 27
4
3

and

B5

2c
b3
c2
2
1
...

Both Abel and Galois proved that there is no general solution of a
fifth-degree equation by radicals
...
” Before discussing Galois’s method, which provided a grouptheoretic criterion for the solution of an equation by radicals and led to
the modern-day Galois theory, we need a few definitions
...
We say that f (x) is solvable by radicals over F if f (x) splits in some extension F(a1, a2,
...
, kn such that a 1k1 [ F and ai ki [
F(a1,
...
, n
...
In other words,

32 | An Introduction to Galois Theory

553

each zero of the polynomial can be written as an expression (usually a
messy one) involving elements of F combined by the operations of addition, subtraction, multiplication, division, and extraction of roots
...
Then
8
8
8
8
x8 2 3 splits in Q(v, " 3 ), v8 [ Q, and (" 3 )8 [ Q , Q(v)
...
Although the zeros of x8 2 3
8
8
8
are most conveniently written in the form " 3 , " 3 v, " 3 v2,
...

2

8

Thus, the problem of solving a polynomial equation for its zeros can
be transformed into a problem about field extensions
...
This is exactly how Galois showed that there are fifth-degree polynomials that
cannot be solved by radicals, and this is exactly how we will do it
...

Definition Solvable Group

We say that a group G is solvable if G has a series of subgroups
{e} 5 H0 , H1 , H2 , ? ? ? , Hk 5 G,
where, for each 0 # i , k, Hi is normal in Hi11 and Hi11/Hi is Abelian
...
So are the dihedral groups
and any group whose order has the form pn, where p is a prime (see
Exercises 22 and 23)
...
In a certain
sense, solvable groups are almost Abelian
...
In particular, A5 is not solvable
...
Our goal is to connect the notion of solvability of polynomials by radicals to that of
solvable groups
...


554

Special Topics

Theorem 32
...
If E is the splitting
field of xn 2 a over F, then the Galois group Gal(E/F) is solvable
...
Let b be a zero of xn 2 a in E
...
, vn21b, and therefore E 5 F(b)
...
To see this, observe that
any automorphism in Gal(E/F) is completely determined by its action
on b
...
That is, any element
of Gal(E/F) takes b to vib for some i
...
Then, since v [ F, f and s fix v and f(b) 5 v jb and
s(b) 5 v kb for some j and k
...
This shows
that sf 5 fs, and therefore Gal(E/F) is Abelian
...

Let v be a primitive nth root of unity and let b be a zero of xn 2 a in E
...
Since
vb is also a zero of xn 2 a, we know that both v and vb belong to E,
and therefore v 5 vb/b is in E as well
...
Analogously to the
case above, for any automorphisms f and s in Gal(F(v)/F) we have
f(v) 5 v j for some j and s (v) 5 v k for some k
...

Since elements of Gal(F(v)/F) are completely determined by their action on v, this shows that Gal(F(v)/F) is Abelian
...
1,
the series
{e} # Gal(E/F(v)) # Gal(E/F)

32 | An Introduction to Galois Theory

555

is a normal series
...

To reach our main result about polynomials that are solvable by radicals, we need two important facts about solvable groups
...
3 Factor Group of a Solvable Group Is Solvable
A factor group of a solvable group is solvable
...

If N is any normal subgroup of G, then
{e} 5 H0N/N , H1N/N , H2N/N , ? ? ? , HkN/N 5 G/N
is the requisite series of subgroups that guarantees that G/N is solvable
...
)
Theorem 32
...
If both N and G/N are
solvable, then G is solvable
...

Then the series
N0 , N1 , ? ? ? , Nt 5 H0 , H1 , ? ? ? , Hs 5 G
has Abelian factors (see Exercise 27)
...


556

Special Topics

Theorem 32
...
Suppose that
f(x) splits in F(a1, a2,
...
,
ai21) for i 5 2,
...
Let E be the splitting field for f(x) over F in
F(a1, a2,
...
Then the Galois group Gal(E/F) is solvable
...
For the case t 5 1, we have F # E # F(a1)
...
Then F #
E # L, and both E and L are splitting fields of polynomials over F
...
1, Gal(E/F) < Gal(L/F)/Gal(L/E)
...
2 that Gal(L/F) is solvable, and from Theorem 32
...
Thus, Gal(E/F) is solvable
...
1
...

Then L is a splitting field of (xn1 2 a) f (x) over F, and L is a splitting
field of f (x) over K
...
, at), so the induction hypothesis implies that Gal(L/K) is
solvable
...
2 asserts that Gal(K/F) is solvable, which,
from Theorem 32
...
Hence,
Theorem 32
...
So, by Part 2 of Theorem 32
...
3, we know that the factor group
Gal(L/F)/Gal(L/E) < Gal(E/F) is solvable
...
3 is true also;
that is, if E is the splitting field of a polynomial f (x) over a field F of
characteristic 0 and Gal(E/F) is solvable, then f (x) is solvable by radicals over F
...

However, one of the major unsolved problems in algebra, first posed
by Emmy Noether, is determining which finite groups can occur as
Galois groups over Q
...
” It is known that every solvable group is a Galois group over Q
...
The article by
Ian Stewart listed among this chapter’s suggested readings provides
more information on this topic
...


557

32 | An Introduction to Galois Theory

Consider g(x) 5 3x5 2 15x 1 5
...
4), g(x) is irreducible over Q
...
A similar analysis shows that g(x) also has real zeros between
0 and 1 and between 1 and 2
...
6
...
So, for g(x) to have four real zeros, g9(x)
would have to have three real zeros, and it does not
...

(See Exercise 65 in Chapter 15
...
Since any
automorphism of K 5 Q(a1, a2, a3, a4, a5) is completely determined by its
action on the a’s and must permute the a’s, we know that Gal(K/Q) is isomorphic to a subgroup of S5, the symmetric group on five symbols
...
Thus, the Fundamental
Theorem of Galois Theory tells us that 5 also divides |Gal(K/Q)|
...
3), we may conclude that
Gal(K/Q) has an element of order 5
...
The
mapping from C to C, sending a 1 bi to a 2 bi, is also an element of
Gal(K/Q)
...

But, the only subgroup of S5 that contains both a 5-cycle and a 2-cycle is
S5
...
) So, Gal(K/Q) is isomorphic to S5
...


Exercises
Seeing much, suffering much, and studying much are the three pillars
of learning
...
Let E be an extension field of Q
...
(This exercise is referred to in this chapter
...
Determine the group of field automorphisms of GF(4)
...
Let E be a field extension of the field F
...
(This exercise is referred to in this chapter
...
Given that the automorphism group of Q("2, "5, "7) is isomorphic to Z2 % Z2 % Z2, determine the number of subfields of Q("2,
"5, "7) that have degree 4 over Q
...
Let E be a field extension of a field F and let H be a subgroup of
Gal(E/F)
...
(This
exercise is referred to in this chapter
...
Let E be the splitting field of x4 1 1 over Q
...
Find
all subfields of E
...
Is there an automorphism of E
2
whose fixed field is Q?
7
...
, an
...
, an), show that Gal(K/F) is isomorphic to a group of
permutations of the ai’s
...
]
8
...

9
...

Show that v 1 v21 is fixed by f3 and v3 1 v5 1 v6 is fixed by f2
...
)
10
...
What is the order of the group Gal(E/Q)?
What is the order of Gal(Q("10)/Q)?
11
...
If Gal(E/F) is isomorphic to A4,
show that there is no subfield K of E such that [K:F] 5 2
...
Determine the Galois group of x3 2 1 over Q and x3 2 2 over Q
...
)
13
...
If [K:F] 5 p2q, where p and q are distinct
primes, show that K has subfields L1, L2, and L3 such that [K:L1] 5
p, [K:L2] 5 p2, and [K:L3] 5 q
...
Suppose that E is the splitting field of some polynomial over a field F
of characteristic 0
...

15
...
Show, by means of an example, that K
need not be the splitting field of some polynomial in F[x]
...
Suppose that E is the splitting field of some polynomial over a field
F of characteristic 0
...

17
...
If Gal(E/F) is an Abelian group of order 10,
draw the subfield lattice for the fields between E and F
...
Let v be a nonreal complex number such that v5 5 1
...

19
...

20
...

Exercises 21, 22, and 23 are referred to in this chapter
...

22
...

24
...


Show that S5 is not solvable
...

Show that a group of order pn, where p is prime, is solvable
...

Complete the proof of Theorem 32
...

26
...

27
...
Prove that K is a normal subgroup of G
...
)
28
...


Reference
1
...
Ehrlich, Fundamental Concepts of Abstract Algebra, Boston:
PWS-Kent, 1991
...

This article gives an elementary discussion of Galois’s proof that the general fifth-degree equation cannot be solved by radicals
...
In this regard,
Rothman refutes several accounts given by other Galois biographers
...

This nontechnical article discusses recent work of John Thompson pertaining to the question of “which groups can occur as Galois groups
...
dcs
...
ac
...

J
...
ROSEBLADE

PHILIP HALL was born on April 11, 1904, in
London
...
He demonstrated academic
prowess early by winning a scholarship to
Christ’s Hospital, where he had several outstanding mathematics teachers
...

Although abstract algebra was a field neglected at King’s College, Hall studied
Burnside’s book Theory of Groups and some
of Burnside’s later papers
...
That same year, Hall discovered a
major “Sylow-like” theorem about solvable
groups: If a solvable group has order mn,
where m and n are relatively prime, then
every subgroup whose order divides m is contained in a group of order m and all subgroups
of order m are conjugate
...
In the 1930s, Hall
also developed a general theory of groups of
prime-power order that has become a foundation of modern finite group theory
...

Among the concepts that have Hall’s name
attached to them are Hall subgroups, Hall
algebras, Hall-Littlewood polynomials, Hall
divisors, the marriage theorem from graph
theory, and the Hall commutator collecting
process
...
No fewer than
one dozen have become eminent mathematicians in their own right
...

To find more information about Hall,
visit:
http://www-groups
...
st-and

...
uk/~history/

33 Cyclotomic Extensions

“
...
”
ALBERT EINSTEIN

Motivation
For the culminating chapter of this book, it is fitting to choose a topic
that ties together results about groups, rings, fields, geometric constructions, and the history of mathematics
...
We begin with the history
...
And,
given a construction of a regular n-gon, it is easy to construct a regular
2n-gon
...
) but failed
...
Incredibly, Gauss, at age 19, showed that a regular 17-gon is
constructible, and shortly thereafter he completely solved the problem
of exactly which n-gons are constructible
...
Gauss was so proud of
this accomplishment that he requested that a regular 17-sided polygon
be engraved on his tombstone
...
In this chapter, we examine the factors of x n 2 1 and show how
Galois theory can be used to determine which regular n-gons are constructible with a straightedge and compass
...


561

562

Special Topics

Cyclotomic Polynomials
Recall from Example 2 in Chapter 16 that the complex zeros of xn 2 1
are 1, v 5 cos(2p ) 1 i sin(2p ), v2, v3,
...
Thus, the splitting
n
n
field of x n 2 1 over Q is Q(v)
...

Since v 5 cos(2p ) 1 i sin(2p ) generates a cyclic group of order
n
n
n under multiplication, we know from Corollary 3 of Theorem 4
...
These generators are called the primitive nth roots
of unity
...
The polynomials whose zeros are the f(n) primitive nth roots
of unity have a special name
...
, vf(n) denote the primitive nth
roots of unity
...


In particular, note that Fn(x) is monic and has degree f(n)
...
2 we will prove that Fn(x) has integer coefficients, and in
Theorem 33
...

EXAMPLE 1 F1(x) 5 x 2 1, since 1 is the only zero of x 2 1
...
F3(x) 5 (x 2 v)(x 2 v2), where v 5 cos(2p) 1 i sin(2p) 5 (21 1
3
3
i"3)/2, and direct calculations show that F3(x) 5 x2 1 x 1 1
...

In practice, one does not use the definition of Fn(x) to compute it
...

Theorem 33
...


33 | Cyclotomic Extensions

563

Before proving this theorem, let us be sure that the statement
is clear
...

PROOF Since both polynomials in the statement are monic, it suffices
to show that they have the same zeros and that all zeros have multiplicity 1
...
Then ͗v͘ is a cyclic group of
n
n
order n, and ͗v͘ contains all the nth roots of unity
...
3 we
know that for each j, |v j| divides n so that (x 2 v j) appears as a factor in
F|v j |(x)
...
Thus, x 2 a is a factor
of x n 2 1
...

Before we illustrate how Theorem 33
...

Theorem 33
...


PROOF The case n 5 1 is trivial
...
From Theorem 33
...
Thus,
Fn(x) [ Z[x]
...
If p is a prime, we have from Theorem 33
...
From Theorem 33
...
So, by long
division, F6(x) 5 x2 2 x 1 1
...

The exercises provide shortcuts that often make long division unnecessary
...
1
...
Judging from Table 33
...
1 The Cyclotomic Polynomials Fn(x) up to n 5 15
n

Fn(x)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

x21
x11
x2 1 x 1 1
x2 1 1
x4 1 x3 1 x2 1 x 1 1
x2 2 x 1 1
x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1
x4 1 1
x6 1 x3 1 1
x4 2 x3 1 x2 2 x 1 1
x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1
x4 2 x2 1 1
x12 1 x11 1 x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1
x6 2 x5 1 x4 2 x3 1 x2 2 x 1 1
x8 2 x7 1 x5 2 x4 1 x3 2 x 1 1

cyclotomic polynomials
...

The next theorem reveals why the cyclotomic polynomials are
important
...
3 (Gauss)
The cyclotomic polynomials Fn(x) are irreducible over Z
...

Because Fn(x) is monic and has no multiple zeros, it suffices to show
that every zero of Fn(x) is a zero of f(x)
...
Let v be a primitive nth root of unity that is a zero
of f (x)
...
Let p be any
prime that does not divide n
...
2,
v p is also a primitive nth root of unity, and therefore 0 5 (v p)n 2 1 5
f (v p)g(v p), and so f (v p) 5 0 or g(v p) 5 0
...
Then
g(v p) 5 0, and so v is a zero of g(x p)
...
3, f (x)
divides g(x p) in Q[x]
...
Say g(x p) 5 f (x)h(x), where
h(x) [ Z[x]
...
Since this reduction process is a ring homomorphism from Z[x] to Z p[x] (see Exercise 9 in Chapter 16), we

33 | Cyclotomic Extensions

565

have g(x p) 5 f (x)h(x) in Z p[x]
...
1, we then have ( g(x)) p 5 g(x p) 5 f (x)h(x),
and since Z p[x] is a unique factorization domain, it follows that g(x)
and f (x) have an irreducible factor in Zp[x] in common; call it m(x)
...
Then, viewing xn 2 1 as a member of Zp[x],
we have xn 2 1 5 f (x)g(x) 5 k1(x)k2(x)(m(x))2
...
But because p does not
divide n, the derivative nx n21 of x n 21 is not 0, and so nx n21 and x n 2 1
do not have a common factor of positive degree in Zp[x]
...
5, we must have f (v p) 5 0
...
Now let k be any integer between 1 and n that is relatively prime to n
...
It follows then that each of v, v p1, (v p1) p2,
...
Since every zero of Fn(x) has the
form v k, where k is between 1 and n and is relatively prime to n, we
have proved that every zero of Fn(x) is a zero of f (x)
...

Of course, Theorems 33
...
1 give us the factorization of
x n 2 1 as a product of irreducible polynomials over Q
...
1
is also useful for finding the irreducible factorization of x n 2 1 over Zp
...
Irreducible factors of x n 2 1
over Zp are used to construct error-correcting codes
...
From Table 33
...
Taking all the coefficients on both sides mod 2, we
obtain the same expression, but we must check that these factors are irreducible over Z2
...
1)
...
Over Z3,
we again start with the factorization x6 2 1 5 (x 2 1)(x 1 1)(x 2 1
x 1 1)(x 2 2 x 1 1) over Z and view the coefficients mod 3
...
Similarly, x 2 2 x 1 1 5 (x 2 2)(x 1 1) 5
(x 1 1)2
...

We next determine the Galois group of the cyclotomic extensions of Q
...
4
Let v be a primitive nth root of unity
...


PROOF Since 1, v, v2,
...
For each k in U(n), v k is a primitive nth root of unity, and by the lemma preceding Theorem 20
...
Moreover, these are all the automorphisms of Q(v), since any automorphism must map a primitive nth
root of unity to a primitive nth root of unity
...

This shows that the mapping from U(n) onto Gal(Q(v)/Q) given by
k S f k is a group homomorphism
...

The next example uses Theorem 33
...

EXAMPLE 3 Let a 5 cos(2p) 1 i sin(2p) and let b 5 cos(2p) 1
9
9
15
i sin(2p)
...


The Constructible Regular n-gons
As an application of the theory of cyclotomic extensions and Galois theory, we determine exactly which regular n-gons are constructible with a
straightedge and compass
...

Lemma
Let n be a positive integer and let v 5 cos A 2p B 1 i sin A 2p B
...

n

PROOF Observe that from (cos(2p) 1 i sin(2p))(cos(2p) 2 i sin(2p)) 5
n
n
n
n
cos2(2p) 1 sin2(2p) 5 1, we have cos(2p) 2 i sin(2p) 5 1/v
...
Thus, cos(2p ) [ Q(v)
...
5 (Gauss, 1796)
It is possible to construct the regular n-gon with a straightedge and
compass if and only if n has the form 2kp1 p2 ? ? ? pt, where k $ 0 and
the pi’s are distinct primes of the form 2m 1 1
...
By the results of
n
Chapter 23, we know that cos(2p) is constructible only if [Q(cos(2p)) : Q]
n
n
is a power of 2
...

Let v 5 cos(2p) 1 i sin(2p)
...
By the lemma on the preceding page, Q(cos(2p)) # Q(v), and by
n
Theorem 32
...

n
Recall that the elements s of Gal(Q(v)/Q) have the property that
s(v) 5 vk for 1 # k # n
...
If such a s belongs to Gal(Q(v)/Q(cos(2p ))), then we must
n
n
have cos(2pk ) 5 cos(2p )
...
So, |Gal(Q(v)/Q(cos(2p)))| 5 2, and therefore [Q(cos(2p)) : Q] 5
n
n
f(n)/2
...
Of course, this implies that f(n) is a power of 2
...
Then f(n)5|U(n)| 5 |U(2k)||U( p1n1)|
|U( p2n2)|? ? ? |U( pt n t)| 5 2k21p1n 121( p1 2 1) p2n221 ( p2 2 1) ? ? ?
pt n t21( pt 2 1) must be a power of 2
...
This completes the proof that the
condition in the statement is necessary
...
5 is also sufficient,
suppose that n has the form 2k p1p2 ? ? ? pt , where the pi’s are distinct
odd primes of the form 2m 1 1, and let v 5 cos(2p ) 1 i sin(2p )
...
3, Q(v) is a splitting field of an irreducible polynomial
over Q, and therefore, by the Fundamental Theorem of Galois Theory,
f(n) 5 [Q(v) : Q] 5 |Gal(Q(v)/Q)|
...
, t 2 1
...
So, for each i, we may choose bi [ EH such
i21
i
i21
that EH 5 EH (bi)
...

i
i
i21
i
Thus, it follows from Exercise 3 in Chapter 23 that every element of
Q(cos(2p )) is constructible
...
In fact, Gauss gave his proof 15 years before Galois was born
...
They should spur us to greater exertion
...
C
...
Determine the minimal polynomial for cos( p ) 1 i sin( p ) over Q
...
Factor x12 2 1 as a product of irreducible polynomials over Z
...
Factor x 8 2 1 as a product of irreducible polynomials over Z2, Z3,
and Z5
...
For any n
...

5
...
1, prove that the product of the nth roots of unity is
(21)n11
...
Let v be a primitive 12th root of unity over Q
...

7
...
Prove that there are only a finite
number of roots of unity in F
...
For any n
...

9
...

(Primes of the form 2n 1 1 are called Fermat primes
...
Prove that Fn(0) 5 1 for all n
...

11
...

12
...
Prove that the
splitting field of x mn 2 1 over Q is the same as the splitting field of
(x m 2 1)(x n 2 1) over Q
...
Prove that F2 n(x) 5 Fn(2x) for all odd integers n
...

14
...
Use this to find F8(x) and F27(x)
...
Prove the assertion made in the proof of Theorem 33
...
, t 2 1
...
)
16
...

(See Exercise 7 in Chapter 32 for the definition
...
Let p be a prime that does not divide n
...

18
...

19
...
Show that there is a
unique field K with the property that Q , K , E
...
Let E be the splitting field of x 6 2 1 over Q
...

21
...
Find the three elements of Gal(Q(v)/Q)
15
15
of order 2
...
When you try to
prove a theorem, you don’t just list the hypotheses, and then start to reason
...

PAUL HALMOS, I Want to Be a Mathematician
...
d
...
edu/~jgallian
1
...
Enter several
choices for n of the form pq and p2q, where p and q are distinct
primes
...

2
...
On the basis of this
information, make a conjecture about the nature of coefficients of
the irreducible factors of x n 2 1 for all n
...


Carl Friedrich Gauss
He [Gauss] lives everywhere in
mathematics
...
T
...
While still a teenager, he made
many fundamental discoveries
...
It commemorates Gauss’s construction of a regular 17-sided polygon with a
straightedge and compass
...
In his Ph
...
dissertation in 1799, he proved the Fundamental
Theorem of Algebra
...
Young
mathematicians who sought encouragement
from him were usually rebuffed
...

Gauss died in Göttingen at the age of 77 on
February 23, 1855
...
Appropriately, the base is in the
shape of a 17-point star
...

To find more information about Gauss,
visit:
http://www-groups
...
st-and
...
uk/~history/

Manjul Bhargava
We are watching him [Bhargava] very
closely
...

He's amazingly mature mathematically
...

PETER SARNAK

MANJUL BHARGAVA was born in Canada on
August 8, 1974, and grew up in Long Island,
New York
...
D
...
Bhargava investigated
a “composition law” first formulated by
Gauss in 1801 for combining two quadratic
equations (equations in a form such as
x2 1 3xy 16y2 5 0) in a way that was very
different from normal addition and revealed
a lot of information about number systems
...
He not only broke new
ground in that area but also discovered 13
more composition laws and developed a
coherent mathematical framework to explain
them
...
What
made Bhargava’s work especially remarkable is that he was able to explain all his
revolutionary ideas using only elementary
mathematics
...
”
Despite his youth, Bhargava already has
won many awards, including a Clay Research
Fellowship, the Clay Research Award, the
Blumenthal Award for the Advancement of
Research in Pure Mathematics, the SASTRA
Ramanujan Prize, and the 2008 Cole Prize in
number theory (see page 430)
...
In 2003,
Bhargava accepted a full professorship with
tenure at Princeton at the age of 28
...
He performs
extensively in the New York and Boston
areas
...
npr
...
php?storyId=4111253
To find more information about Bhargava,
visit
www
...
org and
www
...
umn
...
pdf

571

572

Special Topics

Supplementary Exercises for Chapters 24–33

Text not available due to copyright restrictions

True/false questions for Chapters 24–33 are available on the Web at
http://www
...
umn
...
Let G 5 ͗x, y | x 5 (xy)3, y 5 (xy)4͘
...
Let G 5 ͗z | z6 5 1͘ and H 5 ͗x, y | x2 5 y3 5 1, xy 5 yx͘
...

3
...

4
...
q
...

5
...
Assume that |H| 5 49, |K| 5 5, and K is a subgroup of N(H)
...

6
...

7
...
Prove that K is a normal subgroup of G
...
)
8
...

9
...
Prove that HK is a subgroup of G
if H # N(K)
...
Suppose that H is a subgroup of a finite group G and that H contains N(P), where P is some Sylow p-subgroup of G
...

11
...

12
...


33 | Supplementary Exercises for Chapters 24–33

573

13
...

14
...
Prove that every Sylow
subgroup of Dn is cyclic
...
Let G be the digraph obtained from Cay({(1, 0), (0, 1)}: Z3 % Z5)
by deleting the vertex (0, 0)
...
] Prove that G has a Hamiltonian circuit
...
Prove that the digraph obtained from Cay({(1, 0), (0, 1)}: Z4 % Z7)
by deleting the vertex (0, 0) has a Hamiltonian circuit
...
Let G be a finite group generated by a and b
...
, sn be
the arcs of a Hamiltonian circuit in the digraph Cay({a, b}: G)
...
Show that if
a vertex x travels by a, then every vertex in the coset x͗ab21͘ travels by a
...
Recall that the dot product u ? v of two vectors u 5 (u1, u2,
...
, vn) from F n is
u1v1 1 u2v2 1 ? ? ? 1 unvn
(where the addition and multiplication are those of F)
...
Show that
C > 5 {v [ Fn | v ? u 5 0 for all u [ C}
19
...

21
...

23
...
This code is called the dual of C
...
{00, 11},
b
...
{0000, 1111},
d
...

Let C be a binary linear code such that C # C >
...

Let C be an (n, k) binary linear code
...

Suppose that C is an (n, k) binary linear code and the vector
11 ? ? ? 1 [ C >
...

Suppose that C is an (n, k) binary linear code and C 5 C >
...
) Prove that n is even
...


574

Special Topics

24
...


The End
...
In these
cases, the student should supply the complete proof
...

B
...
FORBES

1
...
12, 2, 2, 10, 1, 0, 4, 5
...
1942, June 18; 1953, December 13
...
By using 0 as an exponent if necessary, we may write a 5 p1m1
...
pknk,
where the p’s are distinct primes and the m’s and n’s are nonnegative
...
pksk,
where si 5 max(mi, ni), and gcd(a, b) 5 p1t1
...
Then lcm(a, b) ?
gcd(a, b) 5 p1m11n1
...

9
...
We may assume that r1 $ r2
...
If a mod n 5 b mod n, then r1 5 r2 and n
divides a 2 b
...

11
...

13
...
2)
...
Let p be a prime greater than 3
...
Now observe that 6n, 6n 1 2, 6n 1 3, and 6n 1 4 are not prime
...
Since st divides a 2 b, both s and t divide a 2 b
...

19
...

21
...

23
...
By induction, S has 2n subsets that do not
contain a
...
So, there are 2 ? 2n 5 2n11 subsets in all
...
Consider n 5 200! 1 2
...
Say p1p2
...
qs, where the p’s and q’s are primes
...
Then p1 5 q1 and p2
...
qs
...
, pr 5 qr and r 5 s
...
Suppose that S is a set that contains a and whenever n $ a belongs to S, then n 1 1 [ S
...
Let T be the set of all integers greater
than a that are not in S and suppose that T is not empty
...
Then b 2 1 [ S, and therefore b 5 (b 2 1) 1 1 [ S
...
The statement is true for any divisor of 84 2 4 5 4092
...
6 P
...

35
...
a1a0 is a9 ? 109 1 a8 ? 108 1
...
Then use Exercise 11 and the fact that ai10i mod 9 5 ai mod 9 to deduce that the
check digit is (a9 1 a8 1
...

37
...
If a transposition involving the check digit c 5 (a1 1 a2 1
...
1 a 9 1 c) mod 9
...
1 a9 ) mod 9 5 0
...
1 a 9) 5 a1 1 a2 1
...
It follows that c 5 a10
...

39
...
a1a0 5 a8 ? 108 1 a7 ? 107 1
...
Then the error is
undetected if and only if (ai10i 2 ai910i) mod 7 5 0
...

41
...
Cases where (2a 2 b 2 c) mod 11 5 0 are undetected
...
The check digit would be the same
...
4302311568
51
...
Since b is one-to-one, b(a (a1)) 5 b(a(a2)) implies that a(a1) 5 a(a2) and since a is one-toone, a1 5 a2
...
Let c [ C
...
Thus, (ba)(a)
5 b(a(a)) 5 b(b) 5 c
...
Since a is one-to-one and onto we may define a21(x) 5 y if and only if a( y) 5 x
...

53
...
(1, 0) [ R and (0, 21) [ R, but (1, 21) o R
...
a belongs to the same subset as a
...
If a and b belong to the subset A and b and c belong to the subset B, then A 5 B, since the
distinct subsets of P are disjoint
...

57
...


Chapter 1
Think of what you’re saying, you can get it wrong and still think that it’s all right
...
Three rotations: 08, 1208, 2408, and three reflections across lines from vertices to midpoints of
opposite sides
...
no
5
...
, n 2 1
...

When n is odd, the axes of reflection are the lines from the vertices to the midpoints of the opposite
sides
...

7
...
However, a reflection fixes a line
...
Observe that 1 ? 1 5 1; 1(21) 5 21; (21)1 5 21; (21)(21) 5 1
...
”
11
...

13
...
See answer for Exercise 13
...
In each case, the group is D6
...
cyclic
21
...

23
...


Selected Answers

A3

Chapter 2
The noblest pleasure is the joy of understanding
...
Does not contain the identity; closure fails
...
Under modulo 4, 2 does not have an inverse
...

9 9
d
10 8
a
...
22a 1 2(2b 1 c); c
...

29
(ab)n need not equal anbn in a non-Abelian group
...

For the case n
...
For n , 0, note that e 5 (a21ba)n(a21ba)2n 5 (a21ba)n
(a21b2na) and solve for (a21ba)n
...
Say x 5 ab and x 5 ac; then cancellation yields
b 5 c
...

Use Exercise 23
...
So, nonidentity solutions come in pairs
...
So solutions to x2 2 e come in pairs
...
So, RFR 5 F21 5 F
...

Since a2 5 b2 5 (ab)2 5 e, we have aabb 5 abab
...

If n is not prime, the set is not closed under multiplication modulo n
...

1>2 1>2
Closure follows from the definition of multiplication
...
The inverse of
1>2 1>2
a a
1> (4a) 1> (4a)
c
d is c
d
...
c
7
...

11
...

15
...

19
...

23
...

27
...

31
...

35
...

39
...

ELEANOR DOAN

1
...

In U(10), |1| 5 1; |3| 5 |7| 5 4; |9| 5 2
...

In U(20), |1| 5 1; |3| 5 |7| 5 |13| 5 |17| 5 4; |9| 5 |11| 5 |19| 5 2
...

In each case, notice that the order of the element divides the order of the group
...
In Q, |0| 5 1 and all other elements have infinite order
...

5
...

7
...
Then e 5 ana 2m 5 an2m
...


A4

Selected Answers

9
...
are all distinct and belong to G, so G is infinite
...
, an 2 1 are distinct and belong to G
...
By brute force, show that k 4 5 1 for all k
...
For any integer n $ 3, Dn contains elements a and b of order 2 with |ab| 5 n
...

15
...

17
...
Uk(n) is closed
because (ab) mod k 5 (a mod k)(b mod k) 5 1 ? 1 5 1
...

19
...
If x [ > C(a), then xa 5 ax for all a in G,
a[G
a[G
so x [ Z(G)
...
The case that k 5 0 is trivial
...
If k is positive, then by induction on k, xak11 5 xaak1
axak 5 aakx 5 ak11x
...

23
...
C(5) 5 G; C(7) 5 {1, 3, 5, 7}
b
...
|2| 5 2; |3| 5 4
...

25
...
5
...
No
...

29
...
For the second part, use D4
...
Since the only elements of finite order in R* are 1 and 21, the only finite subgroups are {1} and
{1, 21}
...
2
35
...
Moreover, there is no positive integer
t , n>d such that (ad ) t 5 adt 5 e, for otherwise 0a 0 ? n
...
Note that c
d 5 c
d
...
For any positive integer n, a rotation of 3608/n has order n
...

41
...
(Note that ͗R90͘ 5 ͗R270͘)
...

43
...
So, there must be some element a of even order,
say |a| 5 2m
...

45
...
Then gr 5 gm2nq 5 gm(gn)2q 5 (gn)2q belongs to H
...

47
...
Let a, b [ H
...
2 can be
replaced by any positive integer
...
|͗3͘| 5 4
a b
a9 b9
51
...
It suffices to show that a 2 a9 1 b 2 b9 1 c 2 c9 1 d 2
c d
c9 d9
d9 5 0
...
If 0 is replaced by 1, H is
not a subgroup
...
If 2a and 2b [ K, then 2a(2b)21 5 2a2b [ K, since a 2 b [ H
...
c
d 5 c 2 1 d is not in H
...
If a 1 bi and c 1 di [ H, then (a 1 bi)(c 1 di)21 5 (ac 1 bd) 1 (bc 2 ad)i and (ac 1 bd)2 1
(bc 2 ad)2 5 1, so that H is a subgroup
...


a1b a
d 0 ab 1 b2 2 a2; a, b [ R f
a
b
a b 2
b
...
e c
d |a 2 0; a [ R f
0 a

59
...
e c

61
...
2
...

JOHN LENNON AND PAUL MCCARTNEY, “Let It Be,” single

1
...

3
...
͗ 3 ͘ 5 {3, 9, 7, 1}; ͗ 7 ͘ 5 {7, 9, 3, 1}
7
...

9
...
Six subgroups; generators are ak, where k is a
divisor of 20
...
By definition, a21 [ ͗a͘
...
By definition, a 5 (a 21 ) 21 [ ͗a 21 ͘
...

13
...
In the general case ͗am ͘ > ͗an ͘ 5 ͗ak ͘, where
k 5 1cm (m, n) mod 24
...
|g| divides 12 is equivalent to g12 5 e
...
The general result is given in Exercise 29 of Chapter 3
...
is odd or infinite
19
...
a
...
b
...
c
...
3, |a| 5 1, 2, 3, 4, 6, 8, 12, or 24
...
A similar argument eliminates all other possibilities except 24
...
Yes, by Theorem 4
...
The subgroups of Z are of the form ͗n͘ 5 {0, 6n, 62n, 63n,
...
The subgroups of ͗a͘ are of the form ͗an ͘ for n 5 0, 1, 2, 3,
...
For the first part, use Theorem 4
...

27
...

29
...
By Theorem 4
...
By Theorem 4
...

31
...
, ak}
...
Consider n 5 n1n2
...

33
...

35
...

37
...
Then there is some positive integer n such that (a/b)n 5 2
...
If n
...
But then
2 divides b as well
...

39
...
For n, use Z 2n21
...
Let t 5 lcm(m, n) and |ab| 5 s
...
Also, e 5
(ab)s 5 asbs, so that as 5 b2s, and therefore as and b2s belong to ͗a͘ > ͗b͘ 5 {e}
...
This proves that s 5 t
...

43
...
A finite cyclic group can have
only one subgroup for each divisor of its order
...
Another element of order p would give another subgroup of order p
...
1 ? 4, 3 ? 4, 7 ? 4, 9 ? 4; x 4,(x 4 ) 3, (x 4 ) 7, (x 4 ) 9
...
1 of order 1; 33 of order 2; 2 of order 3; 10 of order 11; 20 of order 33
49
...
Say a and b are distinct elements of order 2
...
If a and b do not commute, then aba is a third element of order 2
...
Use Exercise 18 of Chapter 3 and Theorem 4
...

55
...
In a cyclic group there are at most n solutions to the equation xn 5 e
...
12 or 60; 48

A6

Selected Answers

61
...
Thus 0a 0 is common divisor of 280 and 440, and therefore 0a 0
divides gcd(280, 440) 5 40
...
Say b is a generator of the group
...
2 that bp also generates the group
...

65
...

67
...
It is not cyclic because every nonzero element has order 3
...
Since m and n are relatively prime, it suffices to show both m and n divide k
...
1, it is enough to show that ak 5 e
...
Thus,
0 ͗a͘ > ͗b͘ 0 5 1
...
Observe that among the integers from 1 to pn the pn21 integers p, 2p, 3p,
...


Supplementary Exercises for Chapters 1–4
Four short words sum up what has lifted most successful individuals above the crowd: a little bit
more
...

A
...
a
...
Then (xh1x21)(xh2x21)21 5 xh1h221x21 [ xHx21 also
...
Let ͗h͘ 5 H
...
c
...
Suppose cl(a) > cl(b) 2 f
...
Then (y21x)a(y21x)21 5 b
...
This shows that cl(b) # cl(a)
...
Because a 5 eae21 [ cl(a), the union of the conjugacy classes is G
...
Observe that (xax21)k 5 xakx21
...

7
...

9
...

11
...

13
...
Let a 2 e belong to G
...
If |a| 5 3, then {e, a, a2} is a subgroup of G
...
Then the set {e, a, a2, b, ab, a2b}
consists of six different elements, a contradiction
...
Similarly, |a| 2 4
...
Pick a 2 e and b 2 e in G with a 2 b
...
Let c be the remaining element of G
...
It now follows that if a [ G
and a 2 e, then |a| 5 5
...
an(bn)21 5 (ab21)n, so Gn is a subgroup
...

17
...
Pick h [ H with h o K
...
Then, hk [ G, but hk o H
and hk o K
...

19
...

21
...
Thus, every member of the group can be
written in the form a i b j
...
D3 satisfies these conditions
...
xy 5 yx if and only if xyx21y21 5 e
...

25
...
Then x(gHg21)x21 5 gHg21
...

This means that g21xg [ N(H)
...
Reverse the argument to show gN(H)g21 #
N(gHg21)
...
Look at D11
...
Solution from Mathematics Magazine
...
Let a be an arbitrary element of S
...
} is finite, and therefore a m 5 a n for some m, n with m
...
By cancellation we

†Mathematics

Magazine 63 (April 1990): 136
...

33
...

37
...


41
...

45
...


A7

have ar(a) 5 a, where r(a) 5 m 2 n 1 1
...
If x is any element of S, then aar(a)21x 5 ar(a)x 5 ax,
and this implies that ar(a)21x 5 x
...
The identity element is unique, for if e9 is another identity, then e 5 ee9 5 e9
...
2
then ar(a)22 is an inverse of a, and if r(a) 5 2 then a2 5 a 5 e is its own inverse
...
”
11 is rational so H 2 f
...
Then (ab21)mn 5 (am)n/(bn)m is rational
...
H is not a subgroup when det A is an integer,
since det A21 need not be an integer
...
Then G 5 ͗x͘ < ͗y͘
...
To prove that |G| 5 pq or p3, use Theorem 4
...

If T and U are not closed, then there are elements x and y in T and w and z in U such that xy is not
in T and wz is not in U
...
Then xywz 5 (xy)wz [ U and xywz 5
xy(wz) [ T, a contradiction
...
Let Hk be the subgroup of polynomials of degree at most k together with the zero polynomial (the zero polynomial
does not have a degree)
...

Let S 5 {s1, s2, s3,
...
Then the set {gs121, gs221, gs321,
...
Say gsi21 5 sj
...

Let K 5 {x [ G | |x| divides d}
...
Let x [ H
...
3, |x| divides d
...
Let y [ K, |y| 5 t, and d 5 tq
...
3, H has a subgroup
of order t and G has only one subgroup of order t
...

To check associativity, note (a * b) * c 5 ((a 1 b) 2 1) * c 5 a 1 b 2 1 1 c 2 1 5 a 1 b 1
c 2 2 and a * (b * c) 5 a * (b 1 c 2 1) 5 a 1 (b 1 c 2 1) 2 1 5 a 1 b 1 c 2 2
...
Thus 1 is the identity (it is
obvious that the operation is commutative)
...
To find a generator, observe that for any positive
integer k, ak 5 ka 2 (k 2 1)
...
One can also check
that 2k 5 k 1 1 when k 5 0 or negative
...


Chapter 5
Mistakes are often the best teachers
...
FROUDE

1
...

5
...

9
...

13
...
2 b
...
5
a
...
12 c
...
6 e
...
2
12
For S6, the possible orders are 1, 2, 3, 4, 5, 6; for A6, 1, 2, 3, 4, 5; for A7, 1, 2, 3, 4, 5, 6, 7
...
even b
...
even d
...
even
even; odd
An even number of 2-cycles followed by an even number of 2-cycles gives an even number of
2-cycles in all
...

15
...
Then ab can be written as a product of m 1 n 2-cycles
...

1 2 3 4 5 6
17
...
a21 5 c
d
2 1 3 5 4 6
1 2 3 4 5 6
b
...
ab 5 c
d
6 2 1 5 3 4

A8

Selected Answers

19
...
Imitate the proof of Theorem 5
...

21
...

23
...
C(a3) 5 {a1, a2, a3, a4}; b
...
180; 75
27
...
(124586739), (142568793), (214856379)
...
Let a, b [ stab(a)
...
Also, a(a) 5 a implies a21(a(a)) 5
a21(a) or a 5 a21(a)
...
m is a multiple of 6 but not a multiple of 30
...
6!/5 5 144
37
...
Let a 5 (123) and b 5 (145)
...
(123)(12) 2 (12)(123) in Sn (n $ 3)
...
Cycle decomposition shows that any nonidentity element of A5 is a 5-cycle, a 3-cycle, or a product of
a pair of disjoint 2-cycles
...

45
...

47
...

49
...
To prove the general case,
observe that si(a) * s i11(b) 2 s i(b) * s i11(a) can be written in the form si(a) * s(si(b)) 2
si(b) * s(si(a)), which is the case already done
...
* si(ai) * si11(ai11) *
...
* si(ai11) * si11(ai) *
...

51
...
an) 5 (1a1)(1an)(1an21)
...

53
...

55
...
For n $ 5, observe that (12)(34)
and (12)(35) belong to H but their product does not
...
The product of an element of Z(A4) of order 2 and an element of A4 of order 3 would have order 6
...

59
...
Thus, we need only find the smallest subgroup of S4 containing a and g
...
This implies that the set is closed under multiplication and is therefore a group
...


Chapter 6
Think and you won’t sink
...
C
...

3
...

7
...

f(xy) 5 "xy 5 "x "y 5 f(x)f(y)
Try 1 → 1, 3 → 5, 5 → 7, 7 → 11
...


Selected Answers

A9

9
...
For the second part, observe that Tg 8 (Tg)21 5 Te 5
Tgg21 5 Tg 8 Tg21 and cancel
...
For any x in the group, we have (fgfh)(x) 5 fg(fh(x)) 5 fg(hxh21) 5 ghxh21g21 5 (gh)x(gh)21 5
fgh(x)
...
fR and fR disagree on H; fR and fH disagree on R90; fR and fD disagree on R90
...

15
...
We show that a21 is operation-preserving: a21(xy) 5 a21(x)a21(y) if and only
if a(a21(xy)) 5 a(a21(x)a21(y)), that is, if and only if xy 5 a(a21(x))a(a21(y)) 5 xy
...
That Inn(G) is a group follows from the equation fgfh 5 fgh
...
That a is one-to-one follows from the fact that r21 exists modulo n
...

19
...
2
...
The inverse of a one-to-one function is one-to-one
...
Then f21(ab) 5 f21(a) f21(b) if and only if ab 5 f(f21(a))f(f21(b)) 5 ab [we obtained the first equality by applying f to both sides of f21(ab) 5 f21(a)f21(b)]
...

Then f21(f(g)) 5 g, so that f21 is onto
...
Tg(x) 5 Tg(y) if and only if gx 5 gy or x 5 y
...
Let y [
G
...

25
...

27
...

a 2b
29
...

b
a
31
...

33
...
So,
fg 5 fzg
...
fg 5 fh implies gxg21 5 hxh21 for all x
...

37
...
Then f a n(x) 5 anxa2n 5 x, so that f an is the identity
...

39
...

41
...

43
...

45
...
0, then a 5 "a"a
...

47
...
It follows that the integer r maps to ar
and the rational r/s maps to ar/s
...


Chapter 7
Use missteps as stepping stones to deeper understanding and greater achievement
...

3
...

7
...

11
...
yes b
...
no
8/2 5 4, so there are four cosets
...
The cosets are H, 7H, 13H, 19H
...

Suppose that h [ H and h , 0
...
But hR1 is the set of all negative real
numbers
...

13
...
Use Lagrange’s Theorem (Theorem 7
...

17
...
So, using mod 7, we have 515 5 56 ? 56 ? 52 ? 5 5 1 ? 1 ?
4 ? 5 5 6; 713 mod 11 5 2
...
Use Corollary 4 of Lagrange’s Theorem (Theorem 7
...
2
...
By closure (234)(12) 5 (1342) belongs to H so that |H| is divisible by 3 and 4 and divides 24
...

23
...
Thus, for each x 2 e, we know that x 2 x21
...
an a n21 5 e
...
Let H be the subgroup of order p and K be the subgroup of order q
...
Let a be any element in G that is not in H < K
...
But |a| 2 p, for if so, then ͗a͘ 5 H
...

27
...
If |x| 5 33, then |x11| 5 3
...

29
...
Observe that by Lagrange’s Theorem, the elements of a group of order 55 must have orders
1, 5, 11, or 55; then use Theorem 4
...

31
...

33
...
Now suppose that c [ orbG(a) > orbG(b)
...
So, if x [ orbG(b), then x 5 g(b) 5 (gb21a)(a) for
some g
...
By symmetry, orbG(a) # orbG(b)
...
a
...
stabG(3) 5 {(1), (24)(56)}; orbG(3) 5 {3, 4, 1, 2}
c
...
Suppose that |Z(G)| 5 pn21 and let a be an element of G not in Z(G)
...
By Lagrange’s Theorem, we must have C(a) 5 G
...

39
...
Consider the mapping from G to G defined by f(x) 5 x2 and let |G| 5 2k 1 1
...

43
...
Then det(A21B) 5 1, so that A21B [ H and therefore B [
AH
...

45
...

47
...

49
...

GEORGE MOORE, The Bending of The Bough

1
...
The identity in the product is the n-tuple with the identity in each component
...
, gn) is (g121, g221,
...

3
...

5
...

7
...
In general, G1 % G2
...
,Gn
...
Yes, by Theorem 8
...

11
...
4 that as long as d divides n, the number of elements of order d in a
cyclic group depends only on d
...
Similarly for Zm % Zn
...
Try a 1 bi → (a, b)
...
Use Exercise 3 and Theorem 4
...

17
...


A11

Selected Answers

Selected Answers

A11

19
...
This is equivalent to gcd( |g|,|h|) 5 1
...
| (a,b,c)| 5 lcm5 |a|, |b|, |c|6 5 3 , unless a 5 b 5 c 5 e
...
% Zp, where p is prime, is p
...
Map c
d to (a, b, c, d)
...
Then the
group of m 3 n matrices under addition is isomorphic to Rmn
...
(g, g)(h, h)21 5 (gh21, gh21) When G 5 R, G % G is the plane and H is the line y 5 x
...
͗(3, 0)͘, ͗(3, 1)͘, ͗(3, 2)͘, ͗(0, 1)͘
29
...
{0, 400} % {0, 50, 100, 150}
33
...

35
...
The mapping f(3m9n ) 5 (m, n)
is not well-defined since f(3290 ) 5 f(30, 91 )
...
D24 has elements of order 24, whereas D3 % D4 does not
...
12
41
...

43
...
, nk that are even
...
No
...

47
...

49
...
So, |(a, b)| 5 lcm(|a|,
|b|) divides lcm(m, n)
...
Z, Z3, Z4, Z6
53
...
So, there are exactly (p2 2 1)/(p 2 1) 5 p 1 1 subgroups of order p
...
Look at Z % Z2
...
U(165) < U(11) % U(15) < U(5) % U(33) < U(3) % U(55) < U(3) % U(5) % U(11)
59
...

61
...
They are both isomorphic to Z10 % Z4
...
That U(n)2 is a subgroup follows from Exercise 15 of Supplementary Exercises
for Chapters 1–4
...

67
...
U(117) < U(9) % U(13) < Z6 % Z12, which contains ͗(2, 0)͘ % ͗(0, 4)͘
...
Consider U(49)
...
Consider U(65)
...
no

Supplementary Exercises for Chapters 5–8
All things are difficult before they are easy
...
Consider the finite and infinite cases separately
...
Now
use Theorem 4
...
For the infinite case, use Exercise 2 in Chapter 6
...
Observe that f(x21y21xy) 5 (f(x))21(f(y))21f(x)f(y), so f carries the generators of G9 to the
generators of G9
...
All nonidentity elements of G and H have order 3
...


A12

Selected Answers

7
...
However, not all symbols need represent distinct group
elements
...
We must determine the
extent to which this happens
...
But hk 5 h9k9 implies t 5 h21h9 5 k(k9)21 [
H > K, so that h9 5 ht and k9 5 t21k
...
So, |HK| 5 |H| |K|/|H > K|
...
U(n), where n 5 4, 8, 3, 6, 12, 24
...
Hint: 3 1 2i 5 "13 a
1
ib
"13
"13
13
...
Let f(1) 5 x0
...

15
...
The corresponding equation x2 5 b in Q1 does
not have a solution for all b
...
Suppose x p22 5 1
...
So, by cancellation, x 5 1
...
͗3͘ % ͗4͘
21
...

23
...
an and b 5 b1
...
Then ab21 5 a1a2
...
b 21 is a finite number of cycles
...

27
...

31
...

35
...

39
...

43
...

47
...

51
...

55
...


59
...

Count elements of order 2
...
Conversely, if G is Abelian, fa is the identity
...

20; (8, 7, (3251))
Let H 5 5x [ Zp2 % Zp2| xp 5 (0, 0) 6
...

(12)(34)(56789)
1260
b 5 (17395)(286)
Say the points in H lie on the line y 5 mx
...
This
set is the line y 2 b 5 m(x 2 a)
...
So (a21b)21 5 b21a [ H
...

These steps are reversible
...
3, U(pq) ; U(p) % U(q), so an element xn in U(pq) corresponds to an element
(xn, xn ) [ U(p) % U(q)
...
1 that (xn, xn ) 5 (1, 1) , the
1 2
1 2
identity of U(p) % U(q)
...
2,1)(12)(123
...
Also, (1n)(123
...
n 2 1)
...
n) generate Sn21
...
Now note that (1k)(1n)(1k) 5 (kn), so all 2-cycles are generated
...
In disjoint cycle form, b is a product of transpositions, so there must be some i
missing from this product
...
Pick j such that b( j) 2 j
...
If b commutes with s, it commutes with st as well
...
This proves that
stb 2 bst
...

BURTON HILLIS

1
...
Say i , j and let h [ Hi > Hj
...
Hi
...

5
...

7
...
If x [ H, then xH 5 H 5 Hx
...

But Hx is also the set of elements in G, not in H
...
G/H < Z4
G/K < Z2 % Z2
11
...

13
...

15
...
H 5 {0 1 ͗20͘, 4 1 ͗20͘, 8 1 ͗20͘, 12 1 ͗20͘, 16 1 ͗20͘}
...

19
...
By Theorem 9
...

Then |ab| 5 33
...
`; no, (6, 3) 1 ͗(4, 2)͘ has order 2
...
Z8
27
...
Mimic the argument given in Example 13 in this chapter
...
Certainly, every nonzero real number is of the form 6r, where r is a positive real number
...

33
...
If G 5 H 3 K, then |g| 5 lcm(|h|, |k|) provided that |h| and |k| are finite
...

35
...
For the second question, observe that 12 5 32162
...
Say |g| 5 n
...
Now use Corollary 2 to Theorem 4
...

39
...
We must show that gxg21h 5 hgxg21
...
Then we have (gxg21)h(gxg21)21 5 gxg21hgx21g21 5
gg21hgg21 5 h
...

41
...

43
...

45
...
Thus, |G : H| 5 1 or |NH : N| 5 1
...

47
...
”
49
...
Now assume
H/N is normal in G/N
...
Thus, xhx21N 5 h9N for h9 [ H
...

51
...
Then (R*)n 5 {x n | x [ R*} # H
...
So, H 5 R* or
H 5 R1
...
Use Exercise 7 and observe that VK 2 KV
...
Suppose n1h1 and n2h2 [ NH
...
Also (n1h1)21 5 h121n121 5 nh121 [ NH
...
Let N 5 ͗a͘, H 5 ͗ak͘, and x [ G
...


A14

Selected Answers

59
...
But |xH| divides |G/H|
...

61
...
1
...
Use Theorems 9
...
3
...
Say |gH| 5 n
...
For the second part, consider Z/͗k͘
...
It is not a group table
...

69
...
3 and Theorem 7
...

71
...
Try (abc)
...
To see that H is normal, observe that xgpx21 5 (xgx21 ) p
...

75
...
It follows that for every a
in G we have (aH) 2 5 H
...
Thus, a is in H
...

GARRY MARSHALL

1
...

5
...

9
...

13
...

17
...

21
...

25
...

29
...

33
...

37
...

41
...

Note that ( f 1 g)9 5 f9 1 g9
...
Odd values of r yield an isomorphism
...

Ker f is a normal subgroup of Ker sf
...
The kernel is {(e, h)|h [ H}
...
3
...

3, 13, 23
Suppose f is such a homomorphism
...
3, Ker f 5 ͗(8, 1)͘, ͗(0, 1)͘ or ͗(8, 0)͘
...
So, (Z16 % Z2) / Ker f is not isomorphic to Z4
% Z4
...

͗5͘
a
...

b
...

4 onto; 10 to
For each k with 0 # k # n 2 1, the mapping 1 → k determines a homomorphism
...
3 and properties 5, 7, and 8 of Theorem 10
...

f21(7) 5 7 Ker f 5 {7, 17}
11 Ker f
f((a, b) 1 (c, d )) 5 f((a 1 c, b 1 d)) 5 (a 1 c) 2 (b 1 d) 5 a 2 b 1 c 2 d 5 f((a, b)) 1
f((c, d))
...
f21(3) 5 {(a 1 3, a) | a [ Z}
...
Ker f 5 ͗cos 60° 1 i sin 60°͘
...

For each divisor d of k there is a unique subgroup of Zk of order d, and this subgroup is generated by f(d) elements
...
Furthermore, the order of the image of 1 must divide n, so we need consider only those divisors d of k that also divide n
...

45
...

49
...

53
...

57
...

61
...

It is infinite
...
Let H be a subgroup of G/N and let
γ21(H ) 5 H
...

The mapping g → fg is a homomorphism with kernel Z(G)
...
The kernel is the set of elements in Z[x] whose graphs pass through the
point (3, 0)
...
Since f(g) belongs to Z2 % Z2 5 ͗1, 0͘ x ͗0, 1͘ x ͗1, 1͘, it follows that
G 5 f21(͗1, 0͘) x f21(͗0, 1͘) x f21(͗1, 1͘)
...

Use Exercise 54 in Chapter 9 and Exercise 39 above to prove the first assertion
...

Mimic Example 16
...
Then G/H has no nontrivial, proper subgroup
...
But then for every coset g 1 H we have p(g 1 H) 5 H
so that pg [ H for all g [ G
...
Both Q and R satisfy the hypothesis
...
LEC, Unkempt Thoughts

1
...
n 5 36
Z9 % Z4, Z3 % Z3 % Z4, Z9 % Z2 % Z2, Z3 % Z3 % Z2 % Z2
5
...
In the first group, |3| 5 15; in the
second one, |(1, 1, 1)| 5 15
...

7
...
Z4 % Z2 % Z3 % Z5
11
...
Now go across the direct product and, for each distinct prime
you have, pick off the largest factor of the prime-power
...
Let us call the order of this new
factor n1
...
Then n2 divides n1, since each prime-power divisor of n2 is also a primepower divisor of n1
...
Example: If
G < Z27 % Z3 % Z125 % Z25 % Z4 % Z2 % Z2,
then
G < Z27 ? 125 ? 4 % Z3 ? 25 ? 2 % Z2
...

13
...
a
...
1 c
...
1 e
...
There is a unique Abelian group of order n if and only if n is
not divisible by the square of any prime
...
Z2 % Z2
19
...
n is square-free (no prime factor of n occurs more than once)
...
Among the first 11 elements in the table, there are nine elements of order 4
...

25
...

27
...
Z4 % Z4
31
...
1, 8
...
3
...
|͗a͘K| 5 |a||K|/|͗a͘ > K| 5 |a||K| 5 |a||K |p 5 |G|p 5 |G|
...
By the Fundamental Theorem of Finite Abelian Groups, it suffices to show that every group of the
form Zp n1 % Zp n 2 %
...
Consider first a group of the form
1
2
k
Zp n1 % Zp n 2 ( p1 and p2 need not be distinct)
...
Then U(qr) 5 U(q) % U(r) <
Ztp n1 % Zsp n2, and this latter group contains a subgroup isomorphic to Zp n1 % Zp n 2
...

37
...


Supplementary Exercises for Chapters 9–11
You cannot have success without the failures
...
G
...
Say aH 5 Hb
...
Then Ha 5 Hhb 5 Hb 5 aH
...
Suppose diag(G) is normal
...
Thus b 5 aba21
...
The index of diag(G) is |G|
...
Let a [ Aut(G) and fa [ Inn(G)
...

7
...
)
9
...

1 0 b
Z(H) 5 • £ 0
0
b
...

13
...

17
...


21
...

c
...

d
...

b(a/b 1 Z) 5 a 1 Z 5 Z
Use Exercise 5 of the Supplementary Exercises for Chapters 1– 4
...
For the first example, consider Dp, where p is a prime
...

Observe that hkh21k21 5 (hkh21)k21 [ K and hkh21k21 5 h(kh21k21) [ H
...
3 and Exercise 7 of Chapter 9
...
But then (Z8 % Z2 % Z2)/Ker f has more than three elements of order 2, whereas Z4 % Z4
has only three
...
2 together with the fact that S4 has no element of order 6
...
The number is m in all cases
...
The mapping g → gn is a homomorphism from G onto Gn with kernel Gn
...
Let |H| 5 p
...
But xHx21 is also a subgroup of order p
...

29
...
Then na/b 5 m for some integer m
...

31
...

33
...
For each positive integer n, let
Bn denote the set of elements of order n and suppose that f is an isomorphism from Q/Z to itself
...
2, f(Bn) # Bn
...
Since it follows from Exercise 11 and Exercise 29 that Q/Z 5 < Bn, where the union is taken over all positive integers n, we
have f(Q/Z) 5 Q/Z
...
If the group is not Abelian, for any element a not in the center, the inner automorphism induced by
a is not the identity; if the group is Abelian and contains an element a with |a|
...
% Z2
...
, ak) to (a2, a1, a3,
...

37
...
G/H is not isomorphic to a subgroup of G since G has only one
element of order 2
...
Observe that c
dc
d 5 c
d,
0 1 0 1
0
1
1 x 21
1 2x
d 5 c
d,
0 1
0
1
which is in H
...

so H is closed
...

Since c

a 1
dc
b 0

x 1
dc
1 0

2ab21
1
21 d 5 c
b
0

b21x
d belongs to H,
1

41
...
Since gKg21 5 K , conjugation is an automorphism of K
...


Chapter 12
Mistakes are the postals of discovery
...
For any n
...
The set M2(2Z ) of 2 3 2 matrices with even integer entries is an infinite noncommutative
ring that does not have a unity
...
In R, consider 5n"2 0 n [ Z 6
...
The proofs given for a group apply to a ring as well
...
In Zp, nonzero elements have multiplicative inverses
...

9
...
Thus
a 2 b and ab belong to each member of the intersection
...

11
...

So, ab 5 (2a)(2b)
...

Part 5: Use part 2
...


A18

Selected Answers

13
...

15
...
1 a)(b 1 b 1
...
1 ab), where the last term has mn summands
...

17
...

19
...
Then (a 2 b)x 5 ax 2 bx 5 xa 2 xb 5 x(a 2 b)
...

21
...
, xn)(a1,
...
, xn) for all xi in Ri if and only if xiai 5 xi for all xi in Ri and
i 5 1,
...

23
...
f (x) 5 1 and g(x) 5 21
...
If a is a unit, then b 5 a(a21b)
...
Consider a21 2 a22b
...
Try the ring M2(Z)
...
Note that 2x 5 (2x)3 5 8x3 5 8x
...
For Z6 use n 5 3
...
Say m 5 p2t where p is a prime
...

37
...

39
...
(ara)(asa) 5 ara2sa 5 arsa
...

41
...

43
...

45
...
Also, (n ? 1)(m ? 1) 5 (nm) ? ((1)(1)) 5 (nm) ? 1
...
{m/2n | m [ Z, n [ Z1}
49
...

51
...
(infinitely many copies)
...

B
...
FORBES, Epigrams

1
...
For Example 7, note that
c

1
0

0 0
dc
0 0

0
0
d 5 c
1
0

0
...

3
...
Then ab 5 a ? 0, so b 5 0
...
Let k [ Zn
...
If gcd(k, n) 5 d
...
Then k(n/d) 5
sd(n/d) 5 sn 5 0
...
Let s [ R, s 2 0
...
If S 5 R, then sr 5 1 (the unity) for some r
...
In this case, s(r1 2 r2) 5 0
...

9
...
Thus the set is a ring
...

11
...

13
...
1 an21 ) 5 1 1 a 1 a2 1
...
2 an 5 1 2 an 5
1 2 0 5 1
...
Suppose a 2 0 and an 5 0 (where we take n to be as small as possible)
...


Selected Answers

17
...

21
...

25
...

29
...

33
...


37
...

41
...

45
...

49
...

53
...

57
...

61
...

65
...
The other cases are similar
...
By the previous exercise, a 5 0
...

a2 5 a implies a(a 2 1) 5 0
...

See Theorems 3
...
3
...
Thus 0 5 aba 2 a 5 a(ba 2 1)
...

A subdomain of an integral domain D is a subset of D that is an integral domain under the operations of D
...
Every
subdomain contains 1 and is closed under addition and subtraction, so every subdomain
contains P
...

Use Theorems 13
...
4, and 7
...

By Exercise 32, 1 is the only element of an integeral domain if and only if 1 5 21
...

a
...
Then a 5 b because we can cancel a5 from both sides (since a5 5 b5)
...
Use the fact that there exist integers s and t such that 1 5 sn 1 tm, but remember that you
cannot use negative exponents in a ring
...

Z8
Let S 5 {a1, a2,
...
First show that S 5 {a1a1, a1a2,
...
Thus, a1 5 a1ai for some i
...

Say |x| 5 n and |y| 5 m with n , m
...

a
...

b
...

c
...

Use Theorems 13
...
5 and Exercise 43
...

c d
0 0
Use Exercise 50
...
2 b
...
2, 3, 6, 11 d
...

Use Exercise 25 and part a of Exercise 45
...

f(x) 5 f(x
...
f(1) so f(1) 5 1
...

Since a field of order 27 has characteristic 3, we have 3a 5 0 for all a
...


Chapter 14
Not one student in a thousand breaks down from overwork
...
Let r1a and r2a belong to ͗a͘
...
If r [ R and r1a [ ͗a͘, then
r(r1a) 5 (rr1)a [ ͗a͘
...
Clearly, I is not empty
...
1 rnan) 2 (s1a1 1
...
1 (rn 2 sn)an [ I
...
1 rnan) 5 (rr1)a1 1
...

That I # J follows from closure under addition and multiplication by elements from R
...
Let a 1 bi, c 1 di [ S
...
Also,
(a 1 bi)(c 1 di) 5 ac 2 bd 1 (ad 1 cb)i and ad 1 cb is even
...


A20

Selected Answers

7
...
, 216, 28, 0, 8, 16,
...
If n is prime, use Euclid’s Lemma (Chapter 0)
...

11
...
a 5 1 b
...
a 5 gcd(m, n)
13
...
a 5 12
b
...
To see this, note that every element of ͗6͗͘8͘ has the form 6t18k1 1 6t28k2 1
...
So, ͗6͗͘8͘ # ͗48͘
...

c
...
Let r [ R
...

17
...
Then r 5 r(u21u) 5 (ru21)u [ I
...
Observe that ͗2͘ and ͗3͘ are the only nontrivial ideals of Z6, so both are maximal
...

21
...
Also, if b1, b2, b3, and b4 are even, then every
entry of c

a1 a2 b1 b2
dc
d is even
...
Use the observation that every member of R can be written in the form c
d
...

2q3 1 r3 2q4 1 r4
r3 r4
25
...

27
...

29
...
If f(x) [ I, then f(x) 5 anxn 1
...
1 a1) [ ͗x͘
...
Suppose f(x) 1 A 2 A
...
Thus,
1
1 A
...
Now use Theorem 14
...

33
...
Also, i 1 ͗3 1 i͘ 5 23 1 ͗3 1 i͘ 5 7 1
͗3 1 i͘
...
, 9}, since 1 1 ͗3 1 i͘ has additive order 10
...
Use Theorems 14
...
4
...
Since every f(x) in ͗x, 2͘ has the form f(x) 5 xg(x) 1 2h(x), we have f(0) 5 2h(0), so that f(x) [ I
...
I is prime
and maximal
...

39
...
Every ideal is a subgroup
...

43
...

45
...
Then (b 2 c)a 5 ba 2 ca 5 0 2 0 5 0
...

47
...
͗3͘ b
...
͗3͘
49
...
We must show that x [ N(͗0͘)
...
Then, for some m, (xn)m 5 0, and therefore x [ N(͗0͘)
...
The set Z2[x]/͗x2 1 x 1 1͘ has only four elements and each of the nonzero ones has a multiplicative inverse
...

53
...

55
...

1
f(x) 5 2 [ R and g(x) 5 2 [ A, but f(x)g(x) o A
...
Hint: Any ideal of R/I has the form A/I, where A is an ideal of R
...
Use the fact that R/I is an integral domain to show that R/I 5 {I, 1 1 I}
...
͗x͘ , ͗x, 2n͘ , ͗x, 2n21͘ ,
...
Taking r 5 1 and s 5 0 shows that a [ I
...
If J is any
ideal that contains a and b, then it contains I because of the closure conditions
...
Then quit
...

W
...
FIELDS

1
...
In Z20, they are 0, 1, 5, and 16
...

3
...
First show this for the case when n is a power of 2
...

5
...
Pick a [ A and b [ B such that a, b o C
...

7
...
The ideals of F % F are {0} % {0}, F % F,
F % {0}, and {0} % F
...
Suppose that am mod n 5 0
...
By Euclid’s
Lemma (Chapter 0), p divides a, and since n is square-free, if follows that n divides a
...
Suppose a1, a2 [ A but a1 o B and a2 o C
...

13
...
Now show that the right-hand side contains a and is an
ideal
...
Since A is an ideal, ab [ A
...
So ab [ A > B 5 {0}
...
6
19
...

21
...

23
...

25
...
Then char R/A 5 the additive order of 1 1 A
...

27
...
2, 14
...
4
...
Observe that A 5 e c
d ` a, b [ Z2 f but c
dc
d 5 c
d is not in A
...
Z[i]/A has two elements
...
See Theorem 14
...
)
33
...

35
...

37
...
The inverse is 2x 1 3
...
Observe that Z5[x, y]/͗x, y͘ < Z5 and use Theorem 14
...

43
...
Then an 5 0 and bn 5 0
...

45
...
Since a and a 2 1 are relatively prime, pk|a or pk|(a 2 1)
...

47
...
In Z7 3"24 , (1 1 2"2 )
(1 1 5"2 ) 5 0
...
If xn 5 0, then (rx)n 5 rnxn 5 0
...

H
...
MENCKEN

1
...
Let s [ S and f(r) 5 s
...

Part 4: Let a and b belong to f21(B) and r belong to R
...
So,
f(a) 2 f(b) 5 f(a) 1 f(2b) 5 f(a 2 b) [ B
...
Also, f(ra) 5 f(r)f(a) [ B
and f(ar) 5 f(a)f(r) [ B
...

3
...
Let F(x 1 Ker f) 5 f(x)
...

5
...

7
...

9
...

11
...
Thus,
by Exercise 25 in Chapter 13, the intersection is a subfield
...
Apply the definition
...
Multiplication is not preserved
...
yes
19
...

21
...
For Z20 to Z30, 1 S 0,
1 S 6, 1 S 15, and 1 S 21 each define a homomorphism
...
The zero map and the identity map
...
Use Exercise 24
...
Say 1 is the unity of R
...
Then f(1)s 5 f(1)f(r) 5 f(1r) 5
f(r) 5 s
...

29
...
It follows that (a, b) → a, (a, b) → b,
and (a, b) → 0 are the only ring homomorphisms
...
Say m 5 akak21
...
b1b0
...
1 (a1 2 b1)10 1 (a0 2 b0)
...

33
...

35
...

37
...

39
...

41
...

43
...
3 and Theorem 10
...

45
...
The kernel must be an ideal
...
a
...
Then f(a)f(b) [ A, so that a [ f21(A) or b [ f21(A)
...
Consider the natural homomorphism from R to S/A
...
3 and 14
...

49
...
f((a, b) 1 (a9, b9)) 5 f((a 1 a9, b 1 b9)) 5 a 1 a9 5 f((a, b)) 1 f((a9, b9)) so f preserves
addition
...

b
...
f(a 1 b) 5 (a 1 b, 0) 5
(a, 0) 1 (b, 0) 5 f(a) 1 f(b)
...

c
...

51
...

53
...

55
...
So, acb9d9 5 (ab9)(cd9) 5
(ba9)(dc9) 5 bda9c9
...

57
...
Then prove (a 1 bi)/(c 1 di) [ Q[i]
...
The subfield of E is {ab21 | a, b [ D, b 2 0}
...
Reflexive and symmetric properties follow from the commutativity of D
...
Then adf 5 (bc)f 5 b(cf ) 5 bde, and cancellation yields af 5 be
...
Try ab21 → a/b
...
The mapping a 1 bi → a 2 bi is a ring isomorphism of C
...
Certainly the unity 1 is contained in every subfield
...
, p 2 1} is contained in every subfield
...
This subfield is isomorphic to Q
[map (m ? 1)(n ? 1)21 to m/n]
...
The mapping f(x) 5 (x mod m, x mod n) from Zmn to Zm { Zn is a ring isomorphism
...
Apply them!
SHERLOCK HOLMES,
The Hound of the Baskervilles

1
...
1, 2, 4, 5
5
...
Since deg (x 2 a) 5 1, deg r(x) 5 0 or r(x) 5 0
...
Also, f(a) 5 r(a)
...
Use Corollary 1 of Theorem 16
...

9
...
By inserting terms with the coefficient 0, we may write
f(x) 5 anxn 1
...
1 b0
...
1 f(a0 1 b0)

5 (f(an) 1 f(bn))xn 1
...
1 f(a0)) 1 (f(bn)xn 1
...


11
...

15
...

19
...

23
...

27
...

31
...

35
...

Quotient, 2x2 1 2x 1 1; remainder, 2
It is its own inverse
...
See Exercise 17
...
1 a0 and g(x) 5 bmxm 1
...
1 a0b0
...
Then we may write f(x) 5 (x 2 a)n (x 2 b)m q9(x), where
q9(x) is in F[x] and q9(b) Z 0
...

If b is a zero of f(x) greater than m, then b is a zero of g(x) 5 f(x)/(x 2 b)m 5 (x 2 a)nq9(x)
...

Use Corollary 3 of Theorem 16
...

If f(x) 2 g(x), then deg[f(x) 2 g(x)] , deg p(x)
...

Start with (x 2 1/2)(x 1 1/3) and clear fractions
...

By Theorem 16
...

Use Corollary 2 of Theorem 15
...

For any a in U(p), ap21 5 1, so every member of U(p) is a zero of x p21 2 1
...

Use Exercise 34
...
Observe that, modulo 101, (50!)2 5 (50!)(21)(22)
...
(51) 5 100!
and use Exercise 34
...
Take R 5 Z and I 5 ͗2͘
...
Hint: F[x] is a PID
...
Thus a(x) divides both f(x) and
g(x)
...

43
...
Use the product rule to compute f9(x)
...
Say deg g(x) 5 m, deg h(x) 5 n, and g(x) has leading coefficient a
...

Then deg k(x) , deg g(x) and h(x) divides k(x) in Z[x] by induction
...

47
...

49
...
To prove the second statement, assume that there is some integer k such that f(k) 5 0
...

This shows that k is not even
...
This contradiction completes the proof
...
A solution to x25 2 1 5 0 in Z37 is a solution to x25 5 1 in U(37)
...
1, |x| divides 25
...


Chapter 17
Experience enables you to recognize a mistake when you make it again
...
JONES

1
...
1
...
If f(x) is not primitive, then f(x) 5 ag(x), where a is an integer greater than 1
...

5
...
If f(x) 5 g(x)h(x), then af(x) 5 ag(x)h(x)
...
If f(x) 5 g(x)h(x), then f(ax) 5 g(ax)h(ax)
...
If f(x) 5 g(x)h(x), then f(x 1 a) 5 g(x 1 a)h(x 1 a)
...
Try a 5 1
...
Find an irreducible polynomial p(x) of degree 2 over Z5
...

9
...
No, since 4 is not a prime
...
Let f(x) 5 x4 1 1 and g(x) 5 f(x 1 1) 5 x4 1 4x3 1 6x2 1 4x 1 2
...
Eisenstein’s Criterion shows that g(x) is irreducible over Q
...
Also note that the complex zeros of
x4 1 1 must have order 8 (when considered as an element of C)
...

Then Example 2 in Chapter 16 tells us that the complex zeros of x4 1 1 are v, v3, v5, and v7, so
x4 1 1 5 (x 2 v)(x 2 v3)(x 2 v5)(x 2 v7)
...

15
...

19
...

23
...

(x 1 3)(x 1 5)(x 1 6)
a
...

b
...

Use Exercise 16, and imitate Example 10
...
This is a ring homomorphism with kernel ͗ x2 1 1͘
...


Selected Answers

A25

25
...
1 a0 5 0
...
1 sna0 5 0
...
Now use Euclid’s Lemma and the fact that r and s are relatively prime
...
Use induction and Corollary 2 of Theorem 17
...

29
...

If there is an a in Zp such that a2 5 2, then x4 1 1 5 (x2 1 ax 1 1)(x2 2 ax 1 1)
...

To show that one of these three cases must occur, consider the group homomorphism from Z* to
p
itself given by x → x2
...
Suppose that neither 21 nor 2 belongs to H
...
Thus, H 5 (21H)(21H) 5 (21H)(2H) 5 22H, so that 22 is in H
...
Since ( f 1 g)(a) 5 f(a) 1 g(a) and ( f ? g)(a) 5 f(a)g(a), the mapping is a homomorphism
...
By Theorem 17
...

33
...

35
...
Thus
the probability of going to jail is different
...
For example, if
in jail one cannot land on Virginia by rolling a pair of 2’s with Sicherman dice, but one is twice
as likely to land on St
...

37
...
Now, just as when n 5 2, we have q 5 r 5
t 5 1, but this time 0 # u # n
...
2, P(x) 5 x(x 1 1) ? (x2 1 x 1 1)(x2 2 x 1 1)u
has (2u 1 2)x2u13 as one of its terms
...
Thus, u # 2, as before
...

GENERAL CARL SPAATZ, in Presidents
Who Have Known Me, GEORGE E
...
1
...
Thus a 5 0 5 b, since otherwise d 5 1 or d is divisible by
the square of a prime
...
N((a 1 b"d )(a9 1 b9"d )) 5 N(aa9 1 dbb9 1 (ab9 1 a9b)"d ) 5 |(aa9 1 dbb9)2 2
d(ab9 1 a9b)2| 5 |a2a92 1 d 2b2b92 2 da2b92 2 da92b2| 5 |a2 2 db2||a92 2 db92| 5
N(a 1 b"d ) N(a9 1 b9"d )
...
If xy 5 1, then 1 5 N(1) 5 N(xy) 5 N(x)N(y) and N(x) 5 1 5 N(y)
...

4
...

3
...
Let a, b [ I and r [ R
...
Thus a, b [ Ik,
where k 5 max{i, j}
...

5
...
If ͗ab͘ 5 ͗b͘, then b 5 rab, so that 1 5 ra and a is a unit
...
Say x 5 a 1 bi and y 5 c 1 di
...

So
d(xy) 5 (ac 2 bd )2 1 (bc 1 ad)2 5 (ac)2 1 (bd )2 1 (bc)2 1 (ad)2
...

9
...
Then d(b) # d(bu) 5 d(a)
...

11
...

13
...
Mimic Example 8 to show that these are irreducible
...
Observe that 10 5 2 ? 5 and 10 5 (2 2 "26) (2 1 "26) and mimic Example 8
...

17
...
Then 9 5 d(3) 5 d(a)d(b), so that
d(a) 5 3
...
Observe that 2 5 2i(1 1 i)2 and
5 5 (1 1 2i)(1 2 2i)
...
Use Exercise 1 with d 5 21
...

21
...

23
...
Now use Exercise 22
...
Use the fact that x is a unit if and only if N(x) 5 1
...
See Example 3
...
p|(a1a2
...
an21 or p|an
...

31
...
4
...
Suppose R satisfies the ascending chain condition and there is an ideal I of R that is not finitely
generated
...
Since I is not finitely generated, ͗a1͘ is a proper subset of I, so
we may choose a2 [ I but a2 o ͗a1͘
...
Continuing in this fashion, we obtain a chain of infinite length ͗a1͘ ,
͗a1, a2͘ , ͗a1, a2, a3͘ ,
...

Let I 5 ...
, an͘
...
Letting k 5 max {i9 | i 5 1,
...
Thus, I # Ik and the
chain has length at most k
...
Say I 5 ͗a 1 bi͘
...
For any
c, d [ Z, let c 5 q1(a2 1 b2) 1 r1 and d 5 q2(a2 1 b2) 1 r2, where 0 # r1, r2 , a2 1 b2
...

37
...
For the other part, use Exercise 25
...
Theorem 18
...
So, assume that a is an irreducible in a UFD R
and that a|bc in R
...
Since a|bc there is an element d in R such that
bc 5 ad
...

41
...


Supplementary Exercises for Chapters 15–18
Errors, like straws, upon the surface flow;
He who would search for pearls must dive below
...
Use Theorem 15
...
4, and
Example 13 in Chapter 14
...
To show the isomorphism, use the First Isomorphism Theorem
...
Use the First Isomorphism Theorem
...
Consider the obvious homomorphism from Z[x] onto Z2[x]
...
3
...
As in Example 7 in Chapter 6, the mapping is onto, is one-to-one, and preserves multiplication
...

11
...
Note that
5 1 ͗2 1 i͘ 5 (2 1 i)(2 2 i) 1 ͗2 1 i͘
5 0 1 ͗2 1 i͘
...
Observe that (3 1 2"2) (3 2 2"2) 5 1
...
In Zn we are given (k 1 1)2 5 k 1 1
...

Also, (n 2 k)2 5 n2 2 2nk 1 k2 5 k2, so (n 2 k)2 5 n 2 k
...

19
...

23
...

27
...

31
...


A27

Observe that for any integer a, a2 mod 4 5 0 or 1
...

Use Theorem 14
...
The factor ring has two elements
...
4
...
Then (ad 2 bc)/(bd) and ac/(bd) [ R by Euclid’s Lemma
...

Z[i]/͗3͘ is a field and Z3 % Z3 is not
...
1 a0 → (an 1 I)x n 1
...

Let I 5 ͗2͘[x]
...

͗x, 3͘
...

GEORGE BERNARD SHAW

1
...
, 0), (0, 1, 0,
...
, (0, 0,
...
}; C has basis {1, i}
...
(a2x2 1 a1x 1 a0) 1 (a29x2 1 a19x 1 a09) 5 (a2 1 a29)x2 1 (a1 1 a19)x 1 (a0 1 a0 and
9)
a(a2x2 1 a1x 1 a0) 5 aa2x2 1 aa1x 1 aa0
...
Yes
...
Linearly dependent, since 23(2, 21, 0) 2 (1, 2, 5) 1 (7, 21, 5) 5 (0, 0, 0)
...
Suppose au 1 b(u 1 v) 1 c(u 1 v 1 w) 5 0
...
Since {u, v, w}
are linearly independent, we obtain c 5 0, b 1 c 5 0, and a 1 b 1 c 5 0
...

9
...
If not, then delete one of the vectors that is a linear
combination of the others (see Exercise 8)
...
Repeat this process until you
obtain a linearly independent subset
...

11
...
Use the fact that u1, u2, u3, w1, w2,
w3 are linearly dependent over F
...
dim V, then U > W 2 {0}
...
no
15
...
c
a1b
b
a9 1 b9
b9
a 1 b 1 a9 1 b9
b 1 b9
a
a1b
ac
ac 1 bc
d 5 c
d
...
Then every member of V is some linear combination of elements of B
...
1 anvn 5 a19v1 1
...
1 (an 2 an 9)vn 5 0
1
and ai 2 a i 9 5 0 for all i
...
Also, if a1v1 1
...
1 anvn 5
0v1 1
...

Since w1 5 a1u1 1 a2u2 1
...
2 anun),
and therefore u1 [ ͗w1, u2,
...
Clearly, u2,
...
, un͘
...
, un is in ͗w1, u2,
...

{(1, 0, 1, 1), (0, 1, 0, 1)}
...
1
...


21
...

25
...
If V and W are vector spaces over F, then the mapping must preserve addition and scalar multiplication
...
A vector space isomorphism from V to W
is a one-to-one linear transformation from V onto W
...
Suppose v and u belong to the kernel and a is a scalar
...

31
...
, vn} be a basis for V
...
1 anvn to (a1, a2,
...


Chapter 20
Well here’s another clue for you all
...
Compare with Exercise 24 in the Supplementary Exercises for Chapters 12–14
...
Q( "23)

7
...

9
...
The set of all expressions of the form
(anpn 1 an21pn21 1
...
1 b0),
5
...

13
...

15
...

17
...
Use the fact that 1 1 i 5 2(4 2 i) 1 5 and 4 2 i 5 5 2 (1 1 i)
...
If the zeros of f(x) are a1, a2,
...
, an 2 a
...

23
...
Let F 5 Z3[x]/͗x3 1 2x 1 1͘ and denote the cosets x 1 ͗x3 1 2x 1 1͘ by b and 2 1
͗x3 1 2x 1 1͘ by 2
...

27
...
Since f(1) 5 1, we have f(23) 5 23
...

31
...

35
...


Then 23 5 f(23) 5 f( "23"23) 5 (f( "23))2
...

Use long division
...
5
...
5
...
(x 2 an), where the
coefficients of f(x) belong to F
...

Since |(Z2[x]/͗ f(x)͘)*| 5 31, every nonidentity is a generator
...

DON HERALD

1
...
1 that if p(x) and q(x) are both monic irreducible polynomials in
F[x] with p(a) 5 q(a) 5 0, then deg p(x) 5 deg q(x)
...
1
...
3,
use the Division Algorithm (Theorem 16
...


Selected Answers

A29

3
...
5
...
Use Exercise 4
...
Suppose Q("a) 5 Q("b)
...
If "b o Q,
then "a o Q
...
It follows that r 5 0 and a 5 bs2
...

9
...

11
...

13
...
For the second part, take F 5 Q, a 5 1; F 5 Q, a 5
3
(21 1 i"3) >2; F 5 Q, a 5 "2
...
Suppose E1 > E2 2 F
...
Similarly, E2 5 E1 > E2
...
E must be an algebraic extension of R, so that E # C
...

19
...
First note [E(a):E] # [F(a):F] 5 deg p(x)
...
This implies that deg p(x) divides [E(a):E],
so deg p(x) 5 [E(a):E]
...
Hint: If a 1 b and ab are algebraic, then so is " (a 1 b) 2 2 4ab
...
"b2 2 4ac
25
...

27
...
If char F 5 0, then the prime subfield of F is isomorphic to Q
...
If a [ Zp, we
are done
...
If k
...
If k , 0, then a2k 1
a12k 5 1 and we are done
...
If [K:F] 5 n, then there are elements v1, v2,
...

The mapping a1v1 1
...
, an) is a vector space isomorphism from K to F n
...
, 0), (0, 1,
...
,
(0, 0,
...

31
...

33
...

35
...

37
...
, an), where a1, a2,
...
Now
use Theorem 21
...

39
...
1 a0)/(bnp n 1 bn21p n21 1
...
So, if "2 [ Q(p) , we have an expression of the form
2(bnp n 1 bn21p n21 1
...
1 a0)2
...
But then we have m 5 n, and "2 is equal to the rational
n
m
number am>bn
...
Observe that F(a) 5 F(1 1 a21 )
...

JOHN LENNON AND PAUL MCCARTNEY,
“Helter Skelter,” The White Album

1
...
The lattice of subfields of GF(64) looks like Figure 21
...

5
...
Use Theorem 22
...

9
...
See Exercise 8 in Chapter 20 for the
first case
...

11
...
1)
...
Use the fact that if g(x) is an irreducible factor of x8 2 x over Z 2 and deg g(x) 5 m, then the field
Z 2[x]/͗g(x)͘ has order 2m and is a subfield of GF(8)
...
3
...
Direct calculations show that given x3 1 2x 1 1 5 0, we have x2 2 1 and x13 2 1
...
Direct calculations show that x13 5 1, whereas (2x)2 2 1 and (2x)13 2 1
...

19
...
Now use Theorem
22
...
3
...
Find a quadratic irreducible polynomial p(x) over Z 3; then Z 3[x]/͗p(x)͘ is a field of order 9
...
Let a, b [ K
...
Also,
m
m
m
(ab) p 5 a p b p 5 ab
...

pn21 2 1 and use Corollary 3 of Lagrange’s Theorem (Theorem 7
...

25
...
identical
29
...
Note that |GF(p)[x]/͗g(x)͘| 5 p2, so that g(x) has a zero in GF(p2)
...
3
...
Use Exercise 11
...
Since F* is a cyclic group of order 124, it has a unique element of order 2
...
Use Exercise 45 in Chapter 13
...
Consider the field of quotients of Zp[x]
...


Chapter 23
Why, sometimes I’ve believed as many as six impossible things before breakfast
...
To construct a 1 b, first construct a
...
To construct a 2 b, use the compass to mark off a length of
b from the right end point of a line of length a
...
Let y denote the length of the hypotenuse of the right triangle with base 1, and let x denote the
length of the hypotenuse of the right triangle with base |c|
...
So, 1 1 2|c| 1 |c|2 5 1 1 d 2 1 |c|2 1 d 2, which simplifies to |c| 5 d 2
...
Use sin2 u 1 cos2 u 5 1
...
Use cos 2u 5 2 cos2 u 21
...
Use sin(a 2 b) 5 sin a cos b 2 cos a sin b
...
Solving two linear equations with coefficients from F involves only the operations of F
...
Use Theorem 17
...

15
...
Now use Exercise 10
...
This amounts to showing that "p is not constructible
...

However, [Q(p) : Q] is infinite
...
No, since [Q( "3 ):Q] 5 3
...
No, since [Q( "p):Q] is infinite
...

RALPH WALDO EMERSON, Journals

1
...
5
...
Suppose b is one solution of xn 5 a
...
Then each member of ͗c͘ is a solution to the equation xn 5 1
...


Selected Answers

A31

5
...
Now observe that since a2 1 a 1 1 5 0, we know that
a(2a 2 1) 5 1, and so a21 5 2a 2 1
...

7
...
Since F(a) 5 F(a21), we have degree of a 5 [F(a):F] 5 [F(a21):F] 5 degree of a21
...
If ab is a zero of cnx n 1
...
1
c1bx 1 c0 [ F(b)[x]
...
Every element of F(a) can be written in the form f(a)/g(a), where f(x), g(x) [ F[x]
...
By clearing
fractions and collecting like powers of a, we obtain a polynomial in a with coefficients from F
equal to 0
...

15
...
2
...
If the basis elements commute, then so would any combination of basis elements
...

19
...
Use Exercise 45 in Chapter 13
...
By Theorem 20
...
, an
...
, n are all the nth roots of unity
...

WILLIAM CULLEN BRYANT

1
...

a 5 xycy
3
...
This proves that T is well defined and one-to-one
...

5
...
Then H and K are normal and Abelian (corollary to Theorem 24
...
2)
...

7
...

9
...

11
...
5, np, the number of Sylow p-subgroups has the form 1 1 kp and np divides |m|
...
Thus k 5 0
...
5
...
8
15
...
By Exercise 16, G has seven subgroups of order 3
...
10; ͗(123)͘, ͗(234)͘, ͗(134)͘, ͗(345)͘, ͗(245)͘
23
...
Sylow’s Third Theorem implies that the Sylow 3- and Sylow 5-subgroups are unique
...
Then |x| 5 15
...
By Sylow, n17 5 1 or 35
...
Then the union of the Sylow 17-subgroups has 561
elements
...
Thus, we may form a cyclic subgroup of order 85 (Exercise 55 in
Chapter 9 and Theorem 24
...
But then there are 64 elements of order 85
...

29
...
3)
...
Let H be the Sylow 3-subgroup and suppose that the Sylow 5-subgroups are not normal
...
, K6
...
Also, each of the cyclic subgroups HK1,
...

Thus, there are 48 elements of order 15 which results in more than 60 elements in the group
...
Mimic the proof of Sylow’s First Theorem
...
Pick x [ Z(G) such that |x| 5 p
...
H/͗x͘, say y͗x͘ [ N(H/͗x͘)
but not H/͗x͘
...
If x o H, then x [ N(H), so that N(H)
...

37
...

39
...
Now use the fact
that H1K is cyclic group of order 15 and Exercise 25 in the Supplementary Exercises for
Chapters 1– 4
...
Normality of H implies cl(h) # H for h in H
...
This is true only when
H is normal
...
The mapping from H to xHx21 given by h → xhx21 is an isomorphism
...
Say cl(x) 5 {x, g1xg121, g2 xg221 ,
...
If x21 5 gi xgi21, then for each gj xgj21 in cl(x), we
have (gjxgj21) 21 5 gj x21gj21 5 gj(gi xgi21)gj21 [ cl(x)
...
It follows that |cl(x)| is even
...

47
...

49
...
Then cl(a) has n 2 1
elements all of the same order, say m
...
But then cl(a) 5 {a} and so n 5 2
...
2, then cl(a) has at most n 2 2
elements since conjugation of a by e, a, and a2 each yields a
...
Let H be a Sylow 5-subgroup
...
17, the only possibility is 1
...
Then by the N/C Theorem (Example 15 of
Chapter 10), |G>C(H)| divides both 4 and |G|
...

53
...
1
...
Pr(D4) 5 5/8, Pr(S3) 5 1/2, Pr(A4) 5 1/3
57
...
, xn}, Pr(R) 5 |K|/n2, where
K 5 {(x, y)|xy 5 yx, x, y [ R}
...
1 |C(xn)|
...
Thus,
|R/C(R)| $ 4 and so |C(R)| # |R|/4
...

Then starting with the elements in the center and proceeding to the elements not in the center,
we have |K| # |R|/4 1 (1/2)(3/4)|R| 5 (5/8)|R|
...

THOMAS BLANDI

1
...

3
...

5
...
Let L7 be a Sylow 7-subgroup of G
...
Let L be a subgroup of N(L 7) of order 5
...
6), N(L) $ N(L 7), so that 35 divides |N(L)|
...
4), which is Abelian (see the Corollary to Theorem 24
...
Thus, 25
divides |N(L)| as well
...
The Index Theorem now yields a
contradiction
...
n11 5 12
...

9
...
Use the N/C
Theorem given in Example 15 of Chapter 10 to show that C(L11) has an element of order 33
whereas A12 does not
...
If we can find a pair of distinct Sylow 2-subgroups A and B such that |A > B| 5 8, then
N(A > B) $ AB, so that N(A > B) 5 G
...

Then 16 ? 16/|H > K| 5 |HK| # 112 (Supplementary Exercise 7 for Chapters 5–8), so that

Selected Answers

15
...


19
...


23
...

27
...


31
...
If |H > K| 5 8, we are done
...
Then N(H > K)
picks up at least 8 elements from H and at least 8 from K (see Exercise 35 in Chapter 24)
...
So, |N(H > K)| 5 16, 56, or 112
...
If N(H > K) 5 H, then |H > K| 5 8,
since N(H > K) contains at least 8 elements from K
...

Then, we may take A 5 N(H > K) and B 5 H
...

n5 5 6 and n3 5 10 or 40
...
Now use the Embedding Theorem
...
If n35 10, use the N/C Theorem to
show that there is an element of order 6 and then use the Embedding Theorem and observe that A6
has no element of order 6
...
Then Ker a # H and |G/Ker a|
divides |G:H|!
...
A subgroup of index 2 is normal
...
But H > A5 5 A5 implies
H 5 A5, whereas H > A5 5 {e} implies H 5 {e} or |H| 5 2
...
)
Now use Exercise 70 in Chapter 9 and Exercise 48 in Chapter 5
...
Hint:
1 4
d has order 3
...
Now argue as we did to show that A5 is
simple
...
But then, G has a normal subgroup of order 14 or 28, which were
already ruled out
...

Suppose there is a simple group of order 60 that is not isomorphic to A5
...
Thus, n2 5 15
...
Now mimic the argument used in showing that there is no simple group of order 144 to show that the normalizer of
this intersection has index 5, 3, or 1, but the Embedding Theorem and the Index Theorem rule
these out
...
Since the number of Sylow q-subgroups is 1 modulo q
2
2
2
and divides p , it must be p
...
These elements,
2
together with the p elements in one Sylow p-subgroup, account for all p2q elements in G
...
But then the Sylow p-subgroup is normal in G
...
Let g be a generator of the Sylow 2-subgroup and
suppose that |G| 5 2kn where n is odd
...
This means that there are exactly n such cycles
...
This
means that the set of even permutations in the regular representations has index 2 and is therefore normal
...


A34

Selected Answers

Chapter 26
If you make a mistake, make amends
...
u , u because u is obtained from itself by no insertions; if v can be obtained from u by inserting or deleting words of the form xx21 or x21x, then u can be obtained from v by reversing the
procedure; if u can be obtained from v and v can be obtained from w, then u can be obtained
from w by obtaining first v from w and then u from v
...
b(a2N) 5 b(aN)a 5 a3bNa 5 a3b(aN)
5 a3a3bN
5 a6bN 5 a6Nb 5 a2Nb 5 a2bN
b(a3N) 5 b(a2N)a 5 a2bNa 5 a2b(aN)
5 a2a3bN
5 a5bN 5 a5Nb 5 aNb 5 abN
b(bN) 5 b2N 5 N
b(abN) 5 baNb 5 a3bNb 5 a3b2N 5 a3N
b(a2bN) 5 ba2Nb 5 a2bNb 5 a2b2N 5 a2N
b(a3bN) 5 ba3Nb 5 abNb 5 ab2N 5 aN
5
...
, an}
...
, wt} and let M be the smallest normal subgroup containing {w1, w2,
...
, wt1k}
...
The homomorphism from F/N to F/M given by
aN → aM induces a homomorphism from G onto G
...

7
...
Now show that a and b belong to ͗a, ab͘
...
Show that |G| # 2n and that Dn satisfies the relations that define G
...
Since x2 5 y2 5 e, we have (xy)21 5 y21x21 5 yx
...

13
...
b6 b
...
Center is ͗x2͘
...

17
...

19
...
But
yxy21 5 yxy11 5 (yxy)y10 5 xy10
...
6; the given relations imply that a2 5 e
...

23
...
ab 5 c 1 abc21 5 e
cd 5 a 1 (abc21)cd 5 ae 1 bd 5 e 1 d 5 b21
da 5 b 1 bda 5 b2 1 ea 5 b2 1 a 5 b2
ab 5 c 1 b3 5 c
So G 5 ͗b͘
...
So |G| 5 1 or 5
...

27
...

BRADENTON, Florida Herald

1
...

3
...
5
...
12
7
...
a
...
Z2 % Z2
c
...

11
...
An inversion in R3 leaves only a single point fixed, whereas a rotation leaves a line fixed
...
In R4, a plane is fixed
...

17
...
Let T be an isometry; p, q, and r the three noncollinear
points; and s any other point in the plane
...
Thus, T(s) is uniquely determined by
T ( p), T(q), and T (r)
...
a rotation

Chapter 28
The thing that counts is not what we know but the ability to use what we know
...
SPEARS

1
...

5
...

9
...

13
...

17
...


Try xnym → (n, m)
...
9
...

a
...
I c
...
VI e
...
III
cmm
a
...
p3 c
...
p6m
The principal purpose of tire tread design is to carry water away from the tire
...
Thus these designs would not carry water away equally
on both halves of the tire
...
VI b
...
I d
...
IV f
...
IV

Chapter 29
With every mistake we must surely be learning
...

3
...

7
...


6
30
13
45
126
1
11
...
For the first part, see Exercise 11 in Chapter 6
...

15
...


A36

Selected Answers

Chapter 30
I am not bound to please thee with my answers
...

3
...

7
...


13
...


17
...


21
...

25
...

29
...

33
...


The Merchant of Venice

4 * (b, a)
(m/2) * {3 * [(a, 0), (b, 0)], (a, 0), (e, 1), 3 * (a, 0), (b, 0), 3 * (a, 0), (e, 1)}
a3b
Both yield paths from e to a3b
...
Then we know the vertices x, xs1, xs1s2,
...
sn21 are distinct and
x 5 xs1s2
...
So if we apply the same sequence beginning at y, then cancellation shows that
y, ys1, ys1s2,
...
sn21 are distinct and y 5 ys1s2
...

If there were a Hamiltonian path from (0, 0) to (2, 0), there would be a Hamiltonian circuit in the
digraph, since (2, 0) 1 (1, 0) 5 (0, 0)
...
If s1, s2,
...
sj 5 e, then the vertex s1s2
...
Conversely, if sisi11
...
, s1s2
...

b
...

The sequence traces the digraph in a clockwise fashion
...
A circuit is 4 * (4 * 1, a), 3 * a, b,
7 * a, 1, b, 3 * a, b, 6 * a, 1, a, b, 3 * a, b, 5 * a, 1, a, a, b, 3 * a, b, 4 * a, 1, 3 * a, b, 3 * a, b,
3 * a, b
...
A circuit is 3 * (R, 1, 1), H,
2 * (1, R, R), R, 1, R, R, 1, H, 1, 1
...
A circuit is 2 * (1, 1, a), a, b,
3 * a, 1, b, b, a, b, b, 1, 3 * a, b, a, a
...
Then the sequence is r, r, f, r, r, 1,
f, r, r, f, r, 1, r, f, r, r, f, 1, r, r, f, r, r, 1, f, r, r, f, r, 1, r, f, r, r, f, 1
...
A circuit is 1, r, 1, 1, f, r, 1, r, 1, r,
f, 1
...

In the proof of Theorem 30
...
Now, if we
assume only that G is Hamiltonian, then H also is Hamiltonian and G/H exists
...

THOMAS R
...

3
...

7
...


wt(0001011) 5 3; wt(0010111) 5 4; wt(0100101) 5 3; etc
...

Observe that a vector has even weight if and only if it can be written as a sum of an even number
of vectors of weight 1
...
No, by Theorem 31
...


Selected Answers

A37

13
...

15
...
Coset decoding means v is at the head of the column containing u
...
Now suppose
u 2 v is a coset leader and u is decoded as y
...

Since v is a code word, u 5 u 2 v 1 v is in the row containing u 2 v
...

17
...


21
...

25
...

29
...

011000 is decoded as 111000 by all four methods
...

Since there are no code words whose distance from 100001 is 1 and three whose distance is 2,
the nearest-neighbor method will not decode or will arbitrarily choose a code word; parity-check
matrix decoding does not decode 100001; the standard-array and syndrome methods decode
100001 as 000000, 110101, or 101011, depending on which of 100001, 010100, or 001010 is a
coset leader
...
But each of these eight possibilities satisfies condition 2 or the first portion of condition 39 of the decoding procedure, so
decoding assumes that no error was made or one error was made
...

No; row 3 is twice row 1
...
For if so, nonzero code words would be all words with weight at least 5
...

Use Exercise 24, together with the fact that the set of code words is closed under addition
...
The following generating matrix will produce the
desired code:
c

1 0 1
1
x
d
...
Use Exercise 14
...
Let c, c9 [ C
...


A38

Selected Answers

35
...


Chapter 32
Wisdom rises upon the ruins of folly
...
Note that f(1) 5 1
...
Also, 1 5 f(1) 5 f(nn21) 5 f(n)f(n21) 5 nf(n21), so that
1/n 5 f(n21)
...
If a and b are automorphisms of E fixing F, so are a21 and ab
...
If a and b are fixed by elements of H, so are a 1 b, a 2 b, a ? b, and a/b
...
It suffices to show that each member of Gal(K/F) defines a permutation on the ai’s
...
1 c0
5 cn(x 2 a1)(x 2 a2)
...


9
...

13
...

17
...

21
...

25
...


Then f(x) 5 a(f(x)) 5 cn(x 2 a(a1))(x 2 a(a2)) ? ? ? (x 2 a(an))
...

f6(v) 5 v729 5 v
...

f2(v3 1 v5 1 v6) 5 v27 1 v45 1 v54 5 v6 1 v3 1 v5
...
(See Example 13 in Chapter 9
...

3
3
3
Let v be a primitive cube root of 1
...

Use the lattice of Z10
...
)
See Exercise 21 in Chapter 25
...

Use Exercise 40 in Chapter 10
...
So, xkx21 5 k9n for some n [ N
...


Chapter 33
All wish to posses knowledge, but few, comparatively speaking, are willing to pay the price
...
x 2 2 x 1 1
...
Over Z, x8 2 1 5 (x 2 1)(x 1 1)(x2 1 1) (x4 1 1)
...
So, over Z2, x8 2 1 5 (x 1 1)8
...
So, x8 2 1 5 (x 2 1)(x 1 1) (x2 1 1)(x2 1
x 1 2)(x2 2 x 2 1)
...
So, x8 2 1 5 (x 2 1)(x 1 1) (x 2 2)(x 1 2)(x2 1 2)(x2 2 2)
...
Let v be a primitive nth root of unity
...
vn 5 (21)n11
...
vn 5 vn(n11)/2
...
When n is even,
(vn/2)n11 5 (21)n11 5 21
...
If [F : Q] 5 n and F has infinitely many roots of unity, then there is no finite bound on their
multiplicative orders
...
n
...
But F $ Q(v) $ Q implies [Q(v) : Q] # n
...
Let 2n 1 1 5 q
...
So, by
Lagrange’s Theorem, 2n divides |U(q)| 5 q 2 1 5 2n
...
Let v be a primitive nth root of unity
...
, 6vn21
...

13
...
Thus, it suffices to show that every zero of Fn(2x) is a zero of F2n(x)
...

2p
15
...
Then, by

...
Now observe that |Hi11/H1 : Hi /H1| 5 |Hi11/Hi|
...
Instead, prove that Fn(x)Fpn(x) 5 Fn(xp)
...
If v is a zero of Fn(x), then
|v| 5 n
...
2, |vp| 5 n also
...
If v is a zero of Fpn(x), then
|v| 5 pn and therefore |v p| 5 n
...
Use Theorem 33
...
1
...
v → v4, v → v21, v → v24

Supplementary Exercises for Chapters 24–33
For those who keep trying, failure is temporary
...
Z6
3
...
If H v G, then
HK 5 45
...

5
...
Now, use Sylow’s Third Theorem
...
Note that gKg21 # gHg21 5 H
...

9
...

11
...
But A8 does not
have an element of order 21
...
Let G be a non-Abelian group of order 105
...
3, G/Z(G) is not cyclic
...
This leaves only 1 or 5 for |Z(G)|
...
Now, counting shows that K v G or L v G
...
But, KL has 24 elements of order 35 (since
|U(Z35)| 5 24)
...
Now, |HK| 5 15 and
HK is a cyclic subgroup of G
...
This means that 105 divides
|C(K)|
...

15
...
It suffices to show that x travels by a implies xab21 travels by a (for we may successively replace
x by xab21)
...

19
...
{00, 11}
b
...
{0000, 1100, 1010, 1001, 0101, 0110, 0011, 1111}
d
...
The mapping Tv: Fn → {0, 1} given by Tv(u) 5 u ? v is an onto homomorphism
...

23
...

Thus, in this problem, k 5 n 2 k
...


Text Credits
Page xi: Copyright © 1966 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page 14: “Brain Boggler” by Maxwell Carver, Copyright © 1988 by Discover
Magazine
...

Page 52: GRE materials selected from the GRE Practice Exam, Question 9 by
Educational Testing Service
...

Page 70: “A Programmer’s Lament,” from The Compleat Computer, Second
Edition, 1983, by Dennie L
...
Reprinted by
permission of Dennie L
...

Page 113: Copyright © Piet Hein Grooks
...

Page 173: From Gina Kolata, “Hitting the High Spots of Computer Theory,”
New York Times, December 13, 1994
...

Page 226: Copyright © 1969 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page 269: Copyright © Piet Hein Grooks
...

Page 287: Copyright © 1965 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page 339: Quote from Math’s Hidden Woman, Nova Online, http://www
...
org/wgbh/nova/proof/germain
...

Page 459: Copyright © 1994
...

Page 467: Adaptation of figure from Dorothy K
...
Crowe
...
Copyright ©
1988 by the University of Washington Press
...

Page 535–536: Adapted with permission from “The Ubiquitous Reed-Solomon
Codes” by Barry A
...
January 1993
...
Reprinted with kind permission
from Piet Hein a/s, DK-5500 Middelfart, Denmark
...
© 1980 Impulsive Music
...

Page A2: Copyright © 1965 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page A4: Copyright © 1970 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page A23: Copyright © 1968 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page A24: Copyright © 1968 (Renewed) Stony/ATV Tunes LLC
...
All rights reserved
...

Page A29: Quote from “While My Guitar Gently Weeps” by George Harrison
...
Used by permission
...


Photo Credits
Page 33: Tibor Zoltai; page 36: From Snow Crystals, by W
...
Bentley & W
...

Humphreys © Dover Publications
...
Johns University Collegeville, MN;
page 261: American Mathematical Society; page 275: The Granger Collection,
New York; page 304: Courtesy of Osa Mac Lane; page 321: Bogdan Oporowski;
page 328: AP Photo/Charles Rex Arbogast; page 339: Stock Montage; page
340: Princeton University; page 352: Courtesy Michael Artin; page 353:
MaryAnn McLoughlin; page 381: American Mathematical Society; page 392:
American Mathematical Society; page 419: Matematisk Unistitutt/Universitete
I Oslo; page 422: With permission of the Pacific Journal of Mathematics; page
434: American Mathematical Society; page 435: American Mathematical
Society; page 436: Courtesy John Thompson; page 452: Courtesy Jonathan
Hall; page 455: From Symmetry in Science and Art by A
...
Shubnikov & V/A
...
V
...
Kopstik © 1974 Plenum Publishing
Company; page 468: M
...
Escher’s symmetry Drawing E22 © 2004 The M
...

Escher Company-Baarn Holland
...
; page 469: (top) Creative
Publications; page 469: (bottom) M
...
Escher’s symmetry Drawing E21 ©
2004 The M
...
Escher Company-Baarn Holland
...
; pages
470–471: From Symmetry in Science and Art by A
...
Shubnikov & V/A
...
V
...
Kopstik © 1974 Plenum
Publishing Company; page 484: The Granger Collection, New York; page 485:
Courtesy of G
...
Alexanderson/Santa Clara University; page 486: Princeton
University; page 497: By permission of the Masters & Fellows of Pembroke
College in the University of Oxford; page 509: M
...
Escher’s Circle Limit I ©
2004 The M
...
Escher Company-Baarn Holland
...
; page 510:
(both) M
...
Escher’s Circle Limit I © 2004 The M
...
Escher Company-Baarn
Holland
...
; page 511: M
...
Escher’s Circle Limit I © 2004
The M
...
Escher Company-Baarn Holland
...
; page 516: Stock
Montage; page 517: Photo by Seymour Schuster, Carleton College; page 535:
SIAM News (January 1993, Volume 25, Number 1, pages 1 and 11); page 542:
Courtesy of Richard W
...


A42

Index of Mathematicians

(Biographies appear on pages in boldface
...
, 162, 164, 165, 173–174
Allenby, R
...
T
...
, 476
Birkhoff, Garrett, 304
Boole, George, 245
Brauer, Richard, 421, 431
Burnside, William, 421, 488,
497, 560

Eisenstein, Ferdinard, 309, 570
Erdös, Paul, 517
Escher, M
...
, 484, 485, 509–511
Euclid, 6, 393
Euler, Leonhard, 39, 44, 154, 325,
339, 370, 517
Feit, Walter, 421, 424, 431, 436,
497, 553
Fermat, Pierre, 143, 325, 326
Fields, John, 430
Fischer, Bernd, 424
Fourier, Joseph, 199
Fraenkel, Abraham, 237
Frobenius, Georg, 174, 408, 488

Cauchy, Augustin-Louis, 98, 104,
121, 187, 199, 266, 325, 354, 408
Cayley, Arthur, 31, 89, 90, 95, 126,
137, 382, 444, 498
Chevalley, Claude, 431
Cole, Frank, 421, 430
Conway, John H
...
A
...
E
...
H
...
N
...
A
...
, 162, 164, 173
Ruffini, Paolo, 102, 142

Kaplansky, Irving, 381
Klein, Felix, 275, 429, 430
Kline, Morris, 274
Knuth, Donald, 434, 452
Kronecker, Leopold, 218, 325,
354, 369
Kummer, Ernst, 325, 326, 369, 570
Lagrange, Joseph-Louis, 39, 121,
141, 154, 199, 339
Lamé, Gabriel, 325, 326
Landau, Edmund, 326
Lange, Serge, 321
Laplace, Pierre-Simon, 121, 154
Larson, Loren, 108
Legendre, Adrien-Marie, 199, 325
Lie, Sophus, 419
Lindemann, Ferdinand, 360, 369,
370, 394
Liouville, Joseph, 370
Mac Lane, Saunders, 304
MacWilliams, Jessie, 543
Mathieu, Emile, 421
Maurolycus, Francisco, 12
McElicee, Robert, 452, 535, 536
Miller, G
...
, 138, 174, 421
Miyaoka, Yoichi, 326
Moore, E
...
, 131, 383
Motzkin, T
...
J
...
, 421, 423, 424,
431, 435, 436, 452, 453, 497,
556, 560
van der Waerden, B
...
, 217
Verhoeff, J
...
, 534
Zorn, Max, 248

Index of Terms

Abel Prize, 39, 436
Abelian group, 32, 41
Addition modulo n, 7
Additive group of integers modulo n, 42
Algebraic
closure, 377, 378
element, 370
extension, 370
Algebraically closed field, 378
Alternating group, 106
Annihilator, 272
Arc, 498
Ascending chain condition, 329, 336
Associates, 322
Associativity, 32, 41
Automorphism(s)
Frobenius, 389
group, 131, 509
group of E over F, 546
inner, 130
of a group, 130
Axioms
for a group, 41
for a ring, 237
for a vector space, 345
Basis for a vector space, 347
Binary
code, 523
operation, 40
strings, 161
Boolean ring, 245
Burnside’s Theorem, 489
Cancellation
property for groups, 48
property for integral domains, 250
Cauchy’s Theorem, 187, 408
Cayley digraph, 498
Cayley table, 31
Cayley’s Theorem, 126
generalized, 426

Center
of a group, 62
of a ring, 243
Centralizer
of an element, 64
of a subgroup, 66
Characteristic of a ring, 252
Characteristic subgroup, 174
Check digit, 7
Check-digit scheme, 110
Chinese Remainder Theorem
for Rings, 341
Circle in F, 394
Class equation, 404
Closure, 31, 40
Code
binary, 523
dual of, 573
Hamming, 520
(n,k) linear, 523
self-dual, 573
systematic, 526
ternary, 524
word, 520, 523
Cole Prize, 321, 352, 392, 424, 430,
434, 436, 571
Commutative diagram, 208
Commutative operation, 32
Commutator subgroup, 174
Composition of functions, 19
Composition factors, 420
Conjugacy class, 91, 403
Conjugate
elements, 403
subgroups, 91, 408
Conjugation, 126
Constant polynomial, 295
Constructible number, 394
Constructible regular n-gons, 566
Content of a polynomial, 306
Coset
decoding, 531

A45

A46

INDEX OF TERMS

leader, 531
left, 138
representative, 138
right, 138
Crystallographic groups, 467
Crystallographic restriction, 473
Cube, rotation group of, 147
Cycle
m-, 98
notation, 98
Cyclic
group, 72
rotation group, 34
subgroup, 61
Cyclotomic
extension, 562
polynomial, 310, 562
Decoding
coset, 531
maximum-likelihood, 518
nearest neighbor, 520
parity-check matrix, 528
Degree
of a over F, 372
of an extension, 372
of a polynomial, 295
rule, 301
DeMoivre’s Theorem, 13
Derivative, 362
Determinant, 43
Diagonal of G % G, 168
Digital signatures, 165
Dihedral groups, 31, 32
Dimension of a vector space, 349
Direct product of groups
external, 155
internal, 188, 190
Direct sum
of groups, 192
of rings, 239
Dirichlet’s Theorem, 228
Discrete frieze group, 461
Distance between vectors, 524
Divides, 238, 298
Division algorithm
for F[x], 296
for Z, 3
Divisor, 3

Domain
Euclidean, 331
integral, 249
Noetherian, 330
unique factorization, 328
Doubling the cube, 393, 395
Dual code, 573
Eisenstein’s criterion, 309
Element(s)
algebraic, 370
conjugate, 403
degree of, 372
fixed by f, 489
idempotent, 255
identity, 31, 41, 238
inverse, 31, 41
nilpotent, 255
order of, 57
primitive, 376
square, 195
transcendental, 370
Embedding Theorem, 427
Empty word, 438
Equivalence class, 16
Equivalence relation, 16
Equivalent under group action, 487
Euclidean domain, 331
Euclid’s Lemma, 6
generalization of, 23
Euler phi-function, 79
Even permutation, 105
Exponent of a group, 175
Extension
algebraic, 370
cyclotomic, 562
degree, 372
field, 354
finite, 372
infinite, 372
simple, 370
transcendental, 370
External direct product, 155
Factor
group, 180
of a ring element, 238
ring, 263
Factor Theorem, 298

INDEX OF TERMS

Feit-Thompson Theorem, 421, 423,
436, 497, 553
Fermat prime, 568
Fermat’s Last Theorem, 325–327
Fermat’s Little Theorem, 143
Field
algebraic closure of, 377, 378
algebrically closed, 378
definition of, 250
extension, 354
fixed, 546
Galois, 383
of quotients, 285
perfect, 364
splitting, 356
Fields Medal, 423, 430, 436, 497
Finite dimensional vector space, 349
Finite extension, 372
First Isomorphism Theorem
for groups, 207
for rings, 283
Fixed field, 546
Free group, 439
Frieze pattern, 461
Frobenius map, 289, 389
Function
composition, 19
definition of, 18
domain, 18
image under, 18
one-to-one, 19
onto, 20
range, 18
Fundamental region, 473
Fundamental Theorem
of Algebra, 378
of Arithmetic, 6
of Cyclic Groups, 77
of Field Theory, 354
of Finite Abelian Groups, 218
of Galois Theory, 550
of Group Homomorphisms, 207
of Ring Homomorphisms, 284
GAP, 109
G/Z Theorem, 186
Galois
field, 383
group, 546, 558
Gaussian integers, 241, 332

A47

Gauss’s Lemma, 307
Generating region of a pattern, 473
Generator(s)
of a cyclic group, 61, 72
of a group, 47
in a presentation, 441
Geometric constructions, 393
Glide-reflection, 454
nontrivial, 464
trivial, 464
Greatest common divisor, 5
Group
Abelian, 32, 41
action, 493
alternating, 106
automorphism, 131, 509
automorphism of, 130
center of, 62
color graph of a , 499
commutative, 32
composition factors, 420
crystallographic, 467
cyclic 34, 61, 72
definition, 41
dicyclic, 445, 450
dihedral, 31, 32
discrete frieze group, 461
factor, 180
finite, 57
free, 439
frieze, 461
Galois, 546, 558
general linear, 43
generator(s), 47, 61, 72, 441
Hamiltonian, 514
Heisenberg, 54
homomorphism of, 200
icosahedral, 430, 457
infinite dihedral, 446
inner automorphism, 131
integers mod n, 42
isomorphic, 123
isomorphism, 123
non-Abelian, 32, 41
octahedral, 457
order of, 57
p- 404, 417
permutation, 95
presentation, 441
quarternions, 91, 196, 442

A48

INDEX OF TERMS

quotient, 180
representation, 211
simple, 420
solvable, 553
space, 475
special linear, 45
symmetric, 97
symmetry, 33, 34, 453
tetrahedral, 457
of units, 243
wallpaper, 467
Half-turn, 463
Hamiltonian
circuit, 503
group, 514
path, 503
Hamming
code, 520
distance, 524
weight of a code, 524
weight of a vector, 524
Homomorphism(s)
Fundamental Theorem of, 207, 284
kernel of, 200
of a group, 200
natural, 210, 284
of a ring, 280
Ideal
annihilator, 272
definition of, 262
finitely generated, 336
generated by, 263
maximal, 267
nil radical of, 272
prime, 267
principal, 263
product of, 270
proper, 262
sum of, 270
test, 262
trivial, 263
Idempotent, 255
Identity element, 31, 41, 238
Index of a subgroup, 142
Index Theorem, 426
Induction
first principle of, 13
second principle of, 14
Inner automorphism, 130

Integral domain, 249
Internal direct product, 188, 190
International standard book
number, 24
Inverse element, 31, 41
Inverse image, 204
Inversion, 135
Irreducibility tests, 306, 308
Irreducible element, 322
Irreducible polynomial, 305
ISBN, 24
Isometry, 453
Isomorphism(s)
class, 218
First Theorem for groups, 207
First Theorem for rings, 283
of groups, 123
of rings, 280
Second Theorem for
groups, 214
Second Theorem for rings, 341
Third Theorem for groups, 214
Third Theorem for rings, 341
Kernel
of a homomorphism, 200
of a linear transformation, 351
Key, 162
Kronecker’s Theorem, 354
Lagrange’s Theorem 141
Latin square, 53
Lattice
diagram, 80
of points, 473
unit, 473
Leading coefficient, 295
Least common multiple, 6
Left regular representation, 127
Line in F, 394
Linear
code, 523
combination, 347
transformation, 351
Linearly dependent vectors, 347
Linearly independent vectors, 347
Mathematical induction
First Principle, 13
Second Principle, 14
Mapping, 18

INDEX OF TERMS

Matrix
addition, 42
determinant of, 43
multiplication, 43
standard generator, 526
Maximal,
ideal, 267
subgroup, 232
Maximum-likelihood decoding, 518
Measure, 331
Minimal polynomial, 371
Mirror, 454
Mod p Irreducibility Test, 308
Modular arithmetic, 7
Monic polynomial, 295
Monster, 424, 556
Multiple, 3
Multiple zeros, 363
Multiplication modulo n, 7
Multiplicity of a zero, 298
Natural homomorphism, 210, 281, 284
Natural mapping, 208
N/C Theorem, 209
Nearest-neighbor decoding, 520
Nilpotent element, 255
Nil radical, 272
Noetherian domain, 330
Norm, 323
Normal subgroup, 178
Normal Subgroup Test, 179
Normalizer, 91
Odd permutation, 105
Operation
associative, 41
binary, 40
commutative, 32
preserving mapping, 123
table, 31
Opposite isometry, 454
Orbit of a point, 145
Orbit-Stabilizer Theorem, 146
Order
or a group, 57
of an element, 57
Orthogonality Relation, 530
PID, 299
Parity-check matrix, 528

Partition
of a set, 17
of an integer, 219
Perfect field, 364
Permutation
definition of, 95
even, 105
group, 95
odd, 105
p-group 404, 417
Phi-function, Euler, 79
Plane of F, 394
Plane symmetry, 33
Polynomial(s)
alternating, 106
constant, 295
content of, 306
cyclotomic, 310, 562
degree of, 295
derivative of, 362
Galois group of, 558
irreducible, 305
leading coefficient of, 295
minimal, 371
monic, 295
primitive, 306
reducible, 305
relatively prime, 303
ring of, 293
splits, 356
symmetric, 106
zero of, 298
Prime
element of a domain, 322
ideal, 267
integer, 3
relatively, 5, 303
subfield, 285
Primitive
element, 376
Element Theorem, 375
nth root of unity 299, 562
polynomial, 306
Principal ideal domain
271, 299
Principal ideal ring, 290
Projection, 212
Proper ideal, 262
Proper subgroup, 58
Pullback, 204

A49

A50

INDEX OF TERMS

Quaternions, 91, 196, 442
Quotient, 4, 297
Quotient group, 180
Quotients, field of, 285
Range, 18
Rational Root Theorem, 318
Reducible polynomial, 305
Reflection, 34, 454
Relation
equivalence, 16
in a presentation, 441
Relatively prime, 5, 303
Remainder, 4, 297
Remainder Theorem, 298
Ring(s)
Boolean, 245
center of, 243
characteristic of, 252
commutative, 238
definition of, 237
direct sum of, 239
factor, 263
homomorphism of, 280
isomorphism of, 280
of polynomials, 293
with unity, 238
RSA public encryption, 164
Rubik’s cube, 110
Scalar, 345
Scalar multiplication, 345
Self dual code, 573
Sicherman dice, 315
Simple extension, 370
Simple group, 420
Socks-Shoes Property, 50
Solvable by radicals, 552
Solvable group, 553
Spanning set, 347
Splitting field, 356
Squaring the circle, 393, 396
Stabilizer of a point, 115, 145
Standard array, 531
Standard generator matrix, 526
Subcode, 537
Subfield Test, 256
Subgroup(s)
centralizer, 66
characteristic, 174

commutator, 174
conjugate, 91, 408
cyclic, 61
definition of, 58
diagonal, 168
Finite Test, 61
generated by a, 61
index of, 142
lattice, 80
maximal, 232
nontrivial, 58
normal, 178
One-Step Test, 59
proper, 58
Sylow p- , 407
torsion, 92
trivial, 58
Two-Step Test, 60
Subring
definition of, 240
Test, 240
Trivial, 241
Subspace, 346
Subspace spanned by vectors, 347
Subspace Test, 349
Sylow p-subgroup, 407
Sylow test for nonsimplicity, 425
Sylow Theorems, 406, 408, 409
Symmetric group, 97
Symmetries of a square, 29
Symmetry group, 33, 34, 453
Syndrome of a vector, 533
Systematic code, 526
Torsion subgroup, 92
Transcendental element, 370
Transcendental extension, 370
Translation, 45, 454
Transposition, 103
Trisecting an angle, 393, 396
UFD, 328
Unique factorization domain, 328
Unique factorization theorem
for a PID, 329
for D[x], 334
for F[x], 331
for Z, 6
for Z[x], 313
in a Euclidean domain, 333

INDEX OF TERMS

Unity, 238
Universal Factor Group Property,
440
Universal Mapping Property, 439
Universal Product Code, 9
Vector, 345
Vector space
basis of, 347
definition of, 345
dimension of, 349
finite dimensional, 349
infinite dimensional, 349
spanned by a set, 347
trivial, 349
Vertex of a graph, 498

Wallpaper groups, 467
Weight of a vector, 524
Weighting vector, 10
Weird dice, 315
Well-defined function, 201
Well-ordering principle, 3
Word
code, 520, 523
empty, 438
in a group, 438
Zero
multiple, 363
multiplicity of, 298
of a polynomial, 298
Zero-divisor, 249

A51

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Essential Theorems in Abstract Algebra
Theorem 3
...

Theorem 4
...
If |͗a͘| = n, then for each positive divisor k of n, ͗a͘ has
exactly one subgroup of order k and no others
...
1 Lagrange’s Theorem
In a finite group the order of a subgroup divides the order of the group
...
1 Normal Subgroup Test
A subgroup H of G is normal in G if and only if xHxϪ1 # H for all x in G
...
3 First Isomorphism Theorem
If f is a group homomorphism from G to a group, then G/Ker f Ϸ f(G)
...
1 Fundamental Theorem of Finite Abelian Groups
Every finite Abelian group is a direct product of cyclic groups of prime-power order
...
3 Subring Test
A nonempty subset S of a ring R is a subring if a Ϫ b and ab are in S whenever a and b are in S
...
4 Characteristic of an Integral Domain
The characteristic of an integral domain is 0 or prime
...
1 Ideal Test
A nonempty subset A of a ring R is an ideal if a Ϫ b [ A whenever a and b are in A; and ra and ar
are in A whenever a [ A and r [ R
...
4 R/A is a Field if and only if A is Maximal
Let R be a communitive ring with unity and let A be an ideal of R
...

Theorem 15
...

Corollary 1 of Theorem 17
...
Then F[x]ր͗p(x)͘ is a field
...
5 [K : F] ϭ [K : E][E : F]
If K is a finite extension field of the field E and E is a finite extension field of the field F, then
[K : F] ϭ [K : E][E : F]
...
2 Structure of Finite Fields
The set of nonzero elements of a finite field is a cyclic group under multiplication
...
3 Sylow’s First Theorem
Let G be a finite group and p a prime
...

Theorem 24
...
Furthermore, any
two Sylow p-subgroups of G are conjugate
...
)

SET THEORY

SPECIAL SETS

>i[ISi
[a]
|s|

Z
Q
Q1
F*
R
R1
C

FUNCTIONS
AND ARITHMETIC

ALGEBRAIC SYSTEMS

f 21
t|s
tBs
gcd(a, b)
lcm(a, b)
f(a)
f: A → B
gf, ab
D4
Dn
e
Zn
det A
U(n)

Rn
SL(2, F)
GL(2, F)
g21
2g
UGU
UgU
H#G
H,G
kal
Z(G)

intersection of sets Si , i [ I
union of sets Si , i [ I
{x [ S | x , a}, equivalence class of S containing a, 16
number of elements in the set of S
integers, additive groups of integers, ring of integers
rational numbers, field of rational numbers
multiplicative group of positive rational numbers
set of nonzero elements of F
real numbers, field of real numbers
multiplicative group of positive real numbers
complex numbers
the inverse of the function f
t divides s, 3
t does not divide s, 3
greatest common divisor of the integers a and b, 5
least common multiple of the integers a and b, 6
image of a under f, 18
mapping of A to B, 18
composite function, 19
group of symmetries of a square, dihedral group of
order 8, 31
dihedral group of order 2n, 32
identity element, 41
group {0, 1,
...
, an) U a1, a2,
...
, anl
R/A
A1B
AB
Ann(A)
N(A)
F(x)
R[x]

{g [ G U ga 5 ag}, the centralizer of a in G, 64
{x [ G U xh 5 hx for all h [ H}, the centralizer
of H, 66
Euler phi function of n, 79
{x [ G U xHx21 5 H} 5 {x [ G U Hx 5 xH}, the normalizer of H in G, 91
conjugacy class of a, 91
{gn U g [ G}, 92
group of one-to-one functions from
{1, 2, ? ? ? , n} to itself, 97
alternating group of degree n, 106
G and G are isomorphic, 123
mapping given by fa(x) 5 axa21 for all x, 130
group of automorphisms of the group G, 131
group of inner automorphisms of G, 131
{ah U h [ H}, 138
{aha21 | h [ H}, 138
the index of H in G, 142
{f [ G U f(i) 5 i}, the stabilizer of i under the permutation group G, 145
{f(i) U f [ G}, the orbit of i under the
permutation group G, 145
external direct product of groups G1, G2,
...
, Hn, 190
kernel of the homomorphism f, 200
inverse image of g9 under f, 202
inverse image of K under f, 203
ring of polynomials with integer coefficients, 238
ring of all 2 3 2 matrices with integer entries, 238
direct sum of rings, 239
ring of multiples of n, 241
ring of Gaussian integers, 241
group of units of the ring R, 243
ring of Gaussian integers modulo n, 245
characteristic of R, 252
principal ideal generated by a, 263
ideal generated by a1, a2,
...
, vnl
F(a1, a2,
...
, an U w1 5 w2 5
...
, c)
(n, k)
Fn
d(u, v)
wt(u)
Gal(E/F)
EH
Fn(x)
C⊥

degree of the polynomial, 295
pth cyclotomic polynomial, 310
ring of 2 3 2 matrices over Q, 346
subspace spanned by v1, v2,
...
, an, 357
the derivative of f (x), 362
degree of E over F, 372
Galois field of order pn, 383
nonzero elements of GF( pn), 384
{xax21 U x [ G}, the conjugacy class of a, 403
probability that two elements from G commute, 405
the number of Sylow p-subgroups of a group, 410
set of all words from S, 438
group with generators a1, a2,
...
5 wt , 441
quarternions, 445
dicyclic group of order 12, 445
infinite dihedral group, 446
{i [ S U f(i) 5 i}, elements fixed by f, 489
Cayley digraph of the group G with generating set S,
498
concatenation of k copies of (a, b,

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