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GROUP THEORY

NOTES FOR THE COURSE ALGEBRA 3, MATH 370

MCGILL UNIVERSITY, FALL 2003,

VERSION: November 3, 2003

EYAL Z
...
Basic Concepts and Key Examples
1
...
1
...
2
...
Main examples
2
...
Z, Z/nZ and (Z/nZ)×
2
...
The dihedral group D2n
2
...
The symmetric group Sn
2
...
Matrix groups and the quaternions
2
...
Groups of small order
2
...
Direct product
3
...
Constructing subgroups
4
...
Commutator subgroup
4
...
Centralizer subgroup
4
...
Normalizer subgroup
5
...
Lagrange’s theorem
7
...
The Isomorphism Theorems
8
...
1
...
2
...
The first isomorphism theorem
10
...
The third isomorphism theorem
12
...
Group Actions on Sets
13
...
Basic properties
15
...
Cayley’s theorem
16
...
Applications to construction of normal subgroups
17
...
1
...
2
...
3
...
4
...
The Symmetric Group
18
...
The simplicity of An

34
34
35

Part 5
...
The class equation
21
...
1
...
Cauchy’s Theorem
23
...
1
...
Finitely Generated Abelian Groups, Semi-direct Products and Groups of
Low Order
24
...
Semi-direct products
25
...
Application to groups of order pq
...
Groups of low, or simple, order
26
...
Groups of prime order
26
...
Groups of order p2
26
...
Groups of order pq, p < q
27
...
Groups of order 8
29
...
Composition series, the Jordan-H¨
older theorem and solvable groups
30
...
1
...
2
...
The Jordan-H¨older theorem
32
...
Basic Concepts and Key Examples

Groups are among the most rudimentary forms of algebraic structures
...
For example, vector spaces, which have very
complex definition, are easy to classify; once the field and dimension are known, the vector space is
unique up to isomorphism
...

In the study of vector spaces the objects are well understood and so one focuses on the study of
maps between them
...
g
...
In contrast,
at least in the theory of finite groups on which this course focuses, there is no comparable theory of
maps
...

While we shall define such maps (called homomorphisms) between groups in general, there will be
a large set of so called simple groups1 for which there are essentially no such maps: the image of a
simple group under a homomorphism is for all practical purposes just the group itself
...
The classification of all simple groups was completed in the second
half of the 20-th century and has required thousands of pages of difficult math
...
We will occupy ourselves with understanding the structure of subgroups of a
finite group, with groups acting as symmetries of a given set and with special classes of groups (cyclic,
simple, abelian, solvable, etc
...
First definitions

1
...
Group
...
2
(2) (Identity) There is an element e ∈ G such that for all g ∈ G we have eg = ge = g
...

It follows quite easily from associativity that given any n elements g1 ,
...
For that reason we allow ourselves
to write simply g1 · · · gn (though the actual computation of such product is done by successively
multiplying two elements, e
...
(((g1 g2 )(g3 g4 ))g5 ) is a way to compute g1 g2 g3 g4 g5
...

2In fuller notation m(f, m(g, h)) = m(m(f, g), h)
...
1

GROUP THEORY

2

The identity element is unique: if e0 has the same property then e0 = ee0 = e
...
The element
h provided in axiom (3) is unique as well: if h0 has the same property then hg = e = h0 g and so
hgh = h0 gh, thus h = he = hgh = h0 gh = h0 e = h0
...
A useful identity is (f g)−1 = g −1 f −1
...

−1

We define by induction g n = g n−1 g for n > 0 and g n = (g −n )
definition
...


for n < 0
...
It is called abelian if it is commutative:
gh = hg for all g, h ∈ G
...
2
...

A subgroup H of a group G is a non-empty subset of G such that (i) e ∈ H, (ii) if g, h ∈ H then
gh ∈ H, and (iii) if g ∈ H then also g −1 ∈ H
...
One
checks that {e} and G are always subgroups, called the trivial subgroups
...
1, 2
...
4

Hto indicate that H is a subgroup of G
...
Note that
n
{h : n ∈ Z} is always a cyclic subgroup
...
The order of an element h ∈ G,
o(h), is defined to be the minimal positive integer n such that hn = e
...

Lemma 1
...
1
...

end of lecture 1
Proof
...
Since for every n we have hn+o(h) = hn ho(h) = hn we see that
< h >= {e, h, h2 ,
...
Thus, also ] < h > is finite and at most o(h)
...
Then the elements {e = h0 , h,
...
Therefore, hj−i = e and we
conclude that o(h) is finite and o(h) is at most ] < h >
...

¤
Corollary 1
...
2
...
, hn−1 } and consists of precisely n
elements (that is, there are no repetitions in this list
...
Let {gα : α ∈ I} be a set consisting of elements of G (here I is some index set)
...
It is clearly the intersection
of all subgroups of G containing {gα : α ∈ I}
...
2
...
The subgroup < {gα : α ∈ I} > is the set of all finite expressions h1 · · · ht where each
hi is some gα or gα−1
...
Clearly < {gα : α ∈ I} > contains each gα hence all the expressions h1 · · · ht where each hi
is some gα or gα−1
...
Clearly e (equal to the empty product, or to gα gα−1 if you
prefer) is in it
...
Finally, since
−1
(h1 · · · ht )−1 = h−1
¤
t · · · h1 it is also closed under taking inverses
...

The Cayley graph
...
We define an oriented graph taking as
vertices the elements of G and taking for every g ∈ G an oriented edge from g to ggα
...

Suppose that the set of generators consists of n elements
...
We see therefore that all Cayley
graphs are regular graphs
...

Suppose we take as group the symmetric group (see below) Sn and the transpositions as
generators
...
Thus, in the Cayley graph, the distance between a
permutation and the identity (the distance is defined as the minimal length of a path between
the two vertices) is the minimal way to write a permutation as a product of transpositions,
and could be thought of as a certain measure of the complexity of a transposition
...

It is a 3-regular oriented graph and a 6 regular graph
...
Main examples
2
...
Z, Z/nZ and (Z/nZ)×
...
, −2, −1, 0, 1, 2, 3,
...
It is cyclic; both 1 and −1 are generators
...
, n − 1}, with addition modulo n, is a finite
abelian group
...
In fact (see the section on cyclic
groups), an element x generates Z/nZ if and only if (x, n) = 1
...
It is a group whose order is
denoted by φ(n) (the function n 7→ φ(n) is call Euler’s phi function)
...
The congruence class 1 is the identity and the existence of inverse follows
from finiteness: given a ∈ Z/nZ× consider the function x 7→ ax
...
Since (a, n) = 1 we conclude that n|(x − y) that is, x = y in Z/nZ
...


Dummit & Foote
§§0
...
3

2
...
The dihedral group D2n
...
Consider the linear transformations of the plane that
take a regular polygon with n sides, symmetric about zero, unto itself
...
One concludes that every such symmetry is of the form y a xb for suitable
and unique a ∈ {0, 1}, b ∈ {1,
...
One finds that y 2 = e = xn and that yxy = x−1
...


Dummit & Foote
§1
...
1
...

The Dihedral group is thus a group of order 2n generated by a reflection y and a rotation x
satisfying y 2 = xn = xyxy = e
...


end of second lecture

GROUP THEORY

5

2
...
The symmetric group Sn
...
3

σ : {1, 2,
...
, n}
...

This makes Sn into a group, whose identity e is the identity function e(i) = i, ∀i
...
n

...
in
This defines a permutation σ by the rule σ(a) = ia
...
is ), where the ij are distinct elements of {1, 2,
...

This defines a permutation σ according to the following convention: σ(ia ) = ia+1 for 1 ≤ a < s,
σ(is ) = i1 , and for any other element x of {1, 2,
...
Such a permutation is called
a cycle
...

Every permutation is a product of disjoint cycles (uniquely up to permuting the cycles)
...
is ) is s
...
, σt are disjoint cycles of orders r1 ,
...
, rt
...


2
...
1
...

Lemma 2
...
1
...
Let Sn be the group of permutations of {1, 2,
...
There exists a
surjective homomorphism3 of groups
sgn : Sn −→ {±1}
(called the ‘sign’)
...

Proof
...
, xn ) =

Y

(xi − xj )
...

i
3That means sgn(στ ) = sgn(σ)sgn(τ )
4For n = 2 we get x − x
...

1
2
1
2
1
3
2
3

Dummit & Foote
§3
...

Thus, for a suitable choice of sign sgn(σ) ∈ {±1}, we have5
Y
Y
(xσ(i) − xσ(j) ) = sgn(σ) (xi − xj )
...

This function satisfies sgn( (k`) ) = −1 (for k < `): Let σ = (k`) and consider the product
Y
Y
Y
Y
(xi − xk )
(xσ(i) − xσ(j) ) = (x` − xk )
(xi − xj )
(x` − xj )
i
ii6=k,j6=`

kj6=`

i<`
i6=k

Counting the number of signs that change we find that
Y
Y
Y
(xσ(i) − xσ(j) ) = (−1)(−1)]{j:k ...
We first make the innocuous observation that
for any variables y1 ,
...

i
i
Let τ be a permutation
...
We get
sgn(τ σ)p(x1 ,
...
, xτ σ(n) )
= p(yσ(1) ,
...
, yn )
= sgn(σ)p(xτ (1) ,
...
, xn )
...

¤
Calculating sgn in practice
...
a` )(b1
...
(f1
...

Claim: sgn(a1
...

Corollary: sgn(σ) = (−1)] even length cycles
...
We write
(a1
...
(a1 a3 )(a1 a2 )
...

5For example, if n = 3 and σ is the cycle (123) we have

(xσ(1) − xσ(2) )(xσ(1) − xσ(3) )(xσ(2) − xσ(3) ) = (x2 − x3 )(x2 − x1 )(x3 − x1 ) = (x1 − x2 )(x1 − x3 )(x2 − x3 )
...


¤

GROUP THEORY

A Numerical example
...

9

Then
σ = (1 2 5)(3 4)(6 7 8 10 9)
...


We conclude that sgn(σ) = −1
...
Let F be any field
...
There is a unique linear
transformation
Tσ : Fn −→ Fn ,
such that
T (ei ) = eσ(i) ,

i = 1,
...
, en are the standard basis of Fn
...


...
) Since for every i we have Tσ Tτ (ei ) = Tσ eτ (i) = eστ (i) = Tστ ei , we have the
relation
Tσ Tτ = Tστ
...
For example, for n = 4
the matrices representing the permutations (12)(34) and (1 2 3 4) are, respectively




0 1 0 0
0 0 0 1
1 0 0 0
1 0 0 0




0 0 0 1 , 0 1 0 0
...
| eσ(n) =  ——– 
...


...
Because σ 7→ Tσ is a group homomorphism we may conclude that
Tσ−1 = Tσt
...


Dummit & Foote
p
...
| eσ(n)
¡
¢
= det e1 | e2 |
...


Recall that sgn(σ) ∈ {±1}
...


2
...
2
...
Let 1 ≤ i < j ≤ n and let σ = (ij)
...
Let T be the set of all transpositions (T has n(n − 1)/2 elements)
...
In fact, also the transpositions (12), (23),
...


Dummit & Foote
§3
...
3

2
...
3
...
Consider the set An of all permutations in Sn whose sign is 1
...
We see that e ∈ An and
that if σ, τ ∈ An also στ and σ −1
...

Thus, An is a group
...
It has n!/2 elements (use multiplication by
(12) to create a bijection between the odd and even permutations)
...
4
...
Let R be a commutative ring with 1
...

Proposition 2
...
1
...

Proof
...
If A, B ∈
GLn (R) then det(AB) = det(A) det(B) gives that det(AB) is a unit of R and so AB ∈ GLn (R)
...
Note that A−1 A = Id implies that det(A−1 ) = det(A)−1 , hence an invertible
element of R
...

¤
Proposition 2
...
2
...

Proof
...

The first vector v1 in a basis can be chosen to be any non-zero vector and there are q n −1 such vectors
...

The third vector v3 can be chosen to be any vector not in Span(v1 , v2 ); there are q n − q 2 such vectors
...

¤

Dummit & Foote
§§1
...
5

GROUP THEORY

9

Exercise 2
...
3
...
It is also called a Borel subgroup
...
It is also called a unipotent subgroup
...

end of 3-rd lecture
Consider the case R = C, the complex numbers, and the set of eight matrices
½ µ
¶
µ
¶
µ
¶
µ
¶¾
1 0
i 0
0 1
0 i
±
,±
,±
,±

...
One can use the notation
±1, ±i, ±j, ±k
for the matrices, respectively
...

2
...
Groups of small order
...
1)
the following is a complete list of groups for the given orders
...

order

abelian groups

non-abelian groups

1

{1}

2

Z/2Z

3

Z/3Z

4

Z/2Z × Z/2Z, Z/4Z

5

Z/5Z

6

Z/6Z

7

Z/7Z

S3

8

Z/2Z × Z/2Z × Z/2Z, Z/2Z × Z/4Z, Z/8Z D8 , Q

9

Z/3Z × Z/3Z, Z/9Z

10

Z/10Z

11

Z/11Z

12

Z/2Z × Z/6Z, Z/12Z

D10
D12 , A4 , T

In the following table we list for every n the number G(n) of subgroups of order n (this is taken
from J
...
6
...
Let G, H be two groups
...


Dummit & Foote
§1
...

One checks that G × H is abelian if and only if both G and H are abelian
...
It follows that if G, H are cyclic groups whose orders
are co-prime then G × H is also a cyclic group
...
6
...
If H1 < H, G1 < G are subgroups then H1 × G1 is a subgroup of H × G
...
For example, the subgroups of Z/2Z × Z/2Z are
{0} × {0}, {0} × Z/2Z, Z/2Z × {0}, Z/2Z × Z/2Z and the subgroup {(0, 0), (1, 1)} which is not a
product of subgroups
...
Cyclic groups
Let G be a finite cyclic group of order n, G =< g >
...
0
...
We have o(g a ) = n/gcd(a, n)
...
3

Proof
...
Corollary 1
...
2)
...
Clearly a · n/gcd(a, n) is divisible by n so the
order of g a is less or equal to n/gcd(a, n)
...

¤
Proposition 3
...
3
...
This subgroup is
cyclic
...
We first show that every subgroup is cyclic
...
Then there is
a minimal 0 < a < n such that g a ∈ H and hence H ⊇< g a >
...
We may assume that
r > 0
...
Note that g r−ka ∈ H
...
Thus, H =< g a >
...
Thus, g a−gcd(a,n) ∈ H
...
Thus, every subgroup is cyclic and of the form < g a >
for a|n
...
We conclude that for every b|n there is a unique subgroup of order b and it
is cyclic, generated by g n/b
...
0
...
Let G be a finite group of order n such that for h|n the group G has at most one
subgroup of order h then G is cyclic
...
We define Euler’s phi function as
φ(h) = ]{1 ≤ a ≤ h : gcd(a, h) = 1}
...
7
7This can be proved as follows
...
Now
=
calculate the unit groups of both sides
...
316

GROUP THEORY

• n=

P
d|n

11

φ(d)
...

Consider an element g ∈ G of order h
...
We conclude that every element of order h must belong to this subgroup (because there
is a unique subgroup of order h in G) and that there are exactly ϕ(h) elements of order h in G
...
elts
...
On the other hand n = d|n φ(d)
...
This element is a generator of G
...
0
...
Let F be a finite field then F× is a cyclic group
...
Let q be the number of element of F
...
That is, the h elements in that subgroup must be the h
solutions of xh − 1
...

¤
end of 4-th lecture

4
...
1
...
Let G be a group
...
An element of the form xyx−1 y −1 is called a
commutator
...
It is not true in general that every element in
G0 is a commutator, though every element is a product of commutators
...
90

Example 4
...
1
...
First, note that every commutator
is an even permutation, hence contained in A3
...
It follows
that S30 = A3
...
2
...
Let H be a subgroup of G
...
One checks that it is a subgroup of G called the centralizer of H in G
...
It is a subgroup of G called
the centralizer of h in G
...

Taking H = G, the subgroup CG (G) is the set of elements of G such that each of them commutes
with every other element of G
...

Example 4
...
1
...
We first make the following useful
observation: τ στ −1 is the permutation obtained from σ by changing its entries according to τ
...
elts
...
0
...


Dummit & Foote
§2
...

Using this, we see that the centralizer of (12) in S5 is just S2 × S3 (Here S2 are the permutations
of 1, 2 and S3 are the permutations of 3, 4, 5
...


4
...
Normalizer subgroup
...
Define the normalizer of H in G, NG (H),
to be the set {g ∈ G : gHg −1 = H}
...
Note that H ⊂ NG (H), CG (H) ⊂ NG (H)
and H ∩ CG (H) = Z(H)
...
2

5
...
A left coset of H in G is a subset S of G of the form

Dummit & Foote
pp
...
A right coset is a subset of G of the form
Hg := {hg : h ∈ H}
for some g ∈ G
...

Example 5
...
1
...
The following table lists
the left cosets of H
...

g
gH
Hg
1

{1, (12)}

{1, (12)}

(12)

{(12), 1}

{(12), 1}

(13)

{(13), (123)}

{(13), (132)}

(23)

{(23), (132))}

{(23), (123))}

(123)

{(123), (13)}

{(123), (23)}

(132)

{(132), (23)}

{(132), (13)}

The first observation is that the element g such that S = gH is not unique
...
The second observation is that two cosets are either equal or disjoint; this is a
consequence of the following lemma
...
0
...
Define a relation g ∼ k if ∃h ∈ H such that gh = k
...

Proof
...
If gh = k for some h ∈ H then kh−1 = g
and h−1 ∈ H
...
Finally, if g ∼ k ∼ ` then gh = k, kh0 = ` for some
h, h0 ∈ H and so g(hh0 ) = `
...
¤

end of 5-th lecture

GROUP THEORY

13

G

g1 H

g2 H

gt H

Figure 5
...
Cosets of a subgroup H of a group G
...
One should note that in general gH 6= Hg; The table above provides an example
...
A difficult theorem of P
...
, gd such that g1 H,
...
, Hgd are precisely the right cosets of H
...
Hall,
Combinatorial Theory,
Ch
...
Lagrange’s theorem
Dummit & Foote
§3
...
0
...
Let H < G
...
If G is of finite
order then the number of left cosets of H in G is |G|/|H|
...

Proof
...
Recall that different equivalence classes are disjoint
...
We next show that for every
x, y ∈ G the cosets xH, yH have the same number of elements
...

Note that f is well defined (xh = xh0 ⇒ h = h0 ), injective (f (xh) = yh = yh0 = f (xh0 ) ⇒ h = h0 ⇒
xh = xh0 ) and surjective as every element of yH has the form yh for some h ∈ H hence is the image
of xh
...

¤
Corollary 6
...
4
...

Remark 6
...
5
...
The group A4 , which is of order 6, does not have a
subgroup of order 6
...
0
...
If G is a finite group then o(g) | |G| for all g ∈ G
...
We saw that o(g) = | < g > |
...
0
...
The converse does not hold
...


7
...
We say that N is a normal subgroup if for all g ∈ G we have gN = N g; equivalently,
gN g −1 = N for all g ∈ G; equivalently, gN ⊂ N g for all g ∈ G; equivalently, gN g −1 ⊂ N for all
g ∈ G
...
Note that an
equivalent way to say that N C G is to say that N < G and NG (N ) = G
...
81-85
...
0
...
The group A3 is normal in S3
...
Thus, τ A3 τ −1 ⊂ A3
...
Use the table above to see that (13)H 6=
H(13)
...
Let G/N denote the set of left cosets of N in G
...

Given two cosets aN and bN we define
aN ? bN = abN
...

Now, we know that for a suitable α, β ∈ N we have a0 α = a, b0 β = b
...
Note that since N C G and α ∈ N also b−1 αb ∈ N and so ab(b−1 αb)N =
abN
...
(Note that
(gN )−1 - the inverse of the element gN in the group G/N is also the set {(gn)−1 : n ∈ N } = N g −1 =
g −1 N
...
0
...
A group is called simple if its only normal subgroups are the trivial ones {e} and
G
...
0
...
We shall later prove that An is a simple group for n ≥ 5
...
On the other hand A4 is not simple
...

end of 6-th lecture
Recall the definition of the commutator subgroup G0 of G from §4
...
In particular, the notation
[x, y] = xyx−1 y −1
...

Hence, also g[x, y]−1 g −1 = [gxg −1 , gyg −1 ]−1
...
0
...
The subgroup G0 is normal in G
...
Furthermore, if G/N is abelian then N ⊇ G0
...
We know that G0 = {[x1 , y1 ]²1 · · · [xr , yr ]²r : xi , yi ∈ G, ²i = ±1}
...


GROUP THEORY

15

For every x, y ∈ G we have xG0 · yG0 = xyG0 = xy(y −1 x−1 yx)G0 = yxG0 = yG0 · xG0
...
If G/N is abelian then for every x, y ∈ G we have xN · yN = yN · xN
...
Thus, for every x, y ∈ G we have xyx−1 y −1 ∈ N
...

¤
Lemma 7
...
12
...

(1) B ∩ N is a normal subgroup of B
...
Also, N B is a subgroup of G
...

(3) If BC G then BN C G and B ∩ N C G
...
The same holds for N B
...


(1) B ∩ N is a normal subgroup of B: First B ∩ N is a subgroup of G, hence of B
...
Then bnb−1 ∈ B because b, n ∈ B and bnb−1 ∈ N because N C G
...
If bn, b0 n0 ∈ BN
then bnb0 n0 = [bb0 ][{(b0 )−1 nb0 }n0 ] ∈ BN
...

Note that BN = ∪b∈B bN = ∪b∈B N b = N B
...
Let g ∈ G and bn ∈ BN then
gbng −1 = [gbg −1 ][gng −1 ] ∈ BN , using the normality of both B and N
...
Thus, gxg −1 ∈ B ∩ N , which
shows B ∩ N is a normal subgroup of G
...


to prove the assertion it is enough to prove that every fibre f −1 x, x ∈ BN , has cardinality
|B ∩ N |
...
This shows that
−1
f (x) ⊇ {(by, y −1 n) : y ∈ B ∩ N }, a set of |B ∩ N | elements
...
Let y = y1−1 then (by)(y −1 n) = b1 n1
...
9
¤
Remark 7
...
13
...
Indeed, consider the case of G = S3 , B = {1, (12)}, N = {1, (13)} then
BN = {1, (12), (13), (132)} which is not a subgroup of S3
...
We can deduce though that
| < B, N > | ≥

|B| · |N |

...
Suppose, for example, that (|B|, |N |) = 1 then |B ∩ N | = 1 because
B ∩ N is a subgroup of both B and N and so by Lagrange’s theorem: |B ∩ N | divides both |B| and
|N |
...
For example, and subgroup of order 3 of A4 generates
A4 together with the Klein group
...
In particular, we do not need to assume that B or N are
normal
...

A group G is called simple if it has no non-trivial normal subgroups
...
A group of odd order, which is not prime, is not simple (Theorem of Feit
and Thompson)
...
We shall later prove
that the alternating group An is a simple group for n ≥ 5
...
It’s a group
...
Let PSLn (F) = SLn (F)/T
...
(See Rotman,
op
...
, §8)
...

If N C G then we have a short exact sequence
1 −→ N −→ G −→ G/N −→ 1
...
) Thus, might hope that the knowledge of N and G/N
allows to find the properties of G
...
e
...
Then G is a semi-direct
product
...


16

Dummit & Foote
p
...
; pp
...
6

GROUP THEORY

17

Part 2
...
Homomorphisms

8
...
Basic definitions
...
A homomorphism f : G −→ H is a function
satisfying f (ab) = f (a)f (b)
...

A homomorphism is called an isomorphism if it is 1 : 1 and surjective
...
Thus, f is an isomorphism if and only
if there exists a homomorphism g : H −→ G such that h ◦ g = idG , g ◦ h = idH
...
We use
the notation G ∼
= H
...

Example 8
...
1
...

Example 8
...
2
...
The group G is isomorphic to
Z/nZ: Indeed, define a function f : G −→ Z/nZ by f (g a ) = a
...
It is a homomorphism: g a g b = g a+b
...

It is injective, because f (g a ) = 0 implies that n|a and so g a = g 0 = e in the group G
...

For example, the kernel of the sign homomorphism Sn −→ {±1} is the alternating group An
...
1
...
We have an isomorphism S3 ∼
= D6 coming from the fact that a symmetry of a
triangle (an element of D6 ) is completely determined by its action on the vertices
...
1
...
The Klein V -group {1, (12)(34), (13)(24), (14)(23)} is isomorphic to Z/2Z × Z/2Z
by (12)(34) 7→ (0, 1), (13)(24) 7→ (1, 0), (14)(23) 7→ (1, 1)
...
1
...
The set Ker(f ) is a normal subgroup of G; f is injective if and only if Ker(f ) = {e}
...

Proof
...
If x, y ∈ Ker(f ) then f (xy) = f (x)f (y) = ee = e so
xy ∈ Ker(f ) and f (x−1 ) = f (x)−1 = e−1 = e so x−1 ∈ Ker(f )
...

If g ∈ G, x ∈ Ker(f ) then f (gxg −1 ) = f (g)f (x)f (g −1 ) = f (g)ef (g)−1 = e
...

If f is injective then there is a unique element x such that f (x) = e
...
Suppose
that Ker(f ) = {e} and f (x) = f (y)
...
That is x = y
and f is injective
...
Thus, if
h ∈ H and f (x) = h then the fibre f −1 (h) is precisely xKer(f )
...
1
...
If N C G then the canonical map πN : G −→ G/N , given by πN (a) = aN , is a
surjective homomorphism with kernel N
...
6

end of 7-th lecture

GROUP THEORY

18

Proof
...
Since
every element of G/N is of the form aN for some a ∈ G, π is surjective
...

¤
Corollary 8
...
7
...


Dummit & Foote
§3
...
83

8
...
Behavior of subgroups under homomorphisms
...

Proposition 8
...
1
...
If B < H then f −1 (B) < G
...

Proof
...
Furthermore, the identities f (x)f (y) = f (xy), f (x)−1 = f (x−1 )
show that f (A) is closed under multiplication and inverses
...

Let B < H
...
Let x, y ∈ f −1 (B) then f (xy) = f (x)f (y) ∈ B
because both f (x) and f (y) are in B
...
Also, f (x−1 ) = f (x)−1 ∈ B and so
x−1 ∈ f −1 (B)
...

Suppose now that BC H
...
Then f (gxg −1 ) = f (g)f (x)f (g)−1
...
Thus, f −1 (B)C G
...
2
...
It is not necessarily true that if AC G then f (A)C H
...


9
...
3

Theorem 9
...
3
...

There is an injective homomorphism f 0 : G/Ker(f ) −→ H such that the following diagram commutes:
f
/

...

Proof
...
We define f 0 by
f 0 (aN ) = f (a)
...
Then f 0 (a) = f (a) = f (bn) =
f (b)f (n) = f (b) = f 0 (bN )
...
Now f 0 (aN bN ) = f 0 (abN ) = f (ab) =
f (a)f (b) = f 0 (aN )f 0 (bN ), which shows f 0 is a homomorphism
...
That is, f 0 is injective and surjective onto its image
...

Finally, f 0 (πN (a)) = f 0 (aN ) = f (a) so f 0 ◦ πN = f
...

¤

GROUP THEORY

19

G

G/N

K

N

K/N

end of lecture 8
Example 9
...
4
...

We get the first homomorphism f1 be looking at the action of the symmetries on the axes {a, b}
...

Similarly, if we let A, B be the lines whose equation is a = b and a = −b, then D8 acts as permutations
on {A, B} and we get a homomorphism f2 : D8 −→ S2 such that f2 (x) = (AB), f2 (y) = (AB)
...
We
find that N1 = {1, x2 , y, x2 y} and N2 = {1, x2 , xy, x3 y}
...

Now, quite generally, if gi : G −→ Hi are group homomorphisms then g : G −→ H1 ×H2 , defined by
g(r) = (g1 (r), g2 (r)) is a group homomorphism with kernel Ker(g1 ) ∩ Ker(g2 )
...
Applying this to our situation, we get a homomorphism
f = (f1 , f2 ) : D8 −→ S2 × S2 ,
whose kernel is {1, x2 }
...
That
is,
D 8 / < x 2 >∼
= S2 × S2
...
The second isomorphism theorem
Dummit & Foote
§3
...
0
...
Let G be a group
...
Then
BN/N ∼
= B/(B ∩ N )
...
Recall from Lemma 7
...
12 that B ∩ N C B
...


We need first to show it is well defined
...
0
...
Therefore, b · B ∩ N = by · B ∩ N = b0 · B ∩ N and f is well defined
...
Note that (bn)(b1 n1 ) = (bb1 )(b−1
1 nb1 )n1 and so f (bn ·
b1 n1 ) = bb1 · B ∩ N = b · B ∩ N · b1 · B ∩ N = f (b)f (b1 ), which shows f is a homomorphism
...

The kernel of f is {bn : f (b) = e, b ∈ B, n ∈ N } = {bn : b ∈ B ∩N, b ∈ B, n ∈ N } = (B ∩N )N = N
...

Remark 10
...
6
...
Let B < G
...
We conclude that f (B) ∼
= BN/N ∼
= B/(B ∩ N )
...
Consider the homomorphism π : G −→ G/N
...
The restriction of f to B is also a group homomorphism with kernel B ∩ N
...
But, f (B) = f (BN ) and we are done
...
The third isomorphism theorem
Dummit & Foote
§3
...
0
...
Let N < K < G be groups, such that N C G, KC G
...
In particular K/N C G/N and furthermore
(G/N )/(K/N ) ∼
= G/K
...
By Proposition 8
...
1 if N < A < G then πN (A) < G/N and if B < G/N is a (normal)
−1
subgroup then πN
(B) < G is a (normal) subgroup clearly containing N
...
Namely, πN πN
Lemma 8
...
5
...
Let a ∈ A and g ∈ G
...
But gN aN (gN )−1 = gag −1 N and gag −1 = a0 ∈ A
because AC G
...


First, f is well defined: f (gnN ) = gnK = gK for n ∈ N
...
Clearly, f is surjective
...
e
...
That is, the kernel of f is just K/N
...

¤
Example 11
...
8
...
0
...
Using the third isomorphism theorem we conclude that the graph of the subgroups of
D8 containing < x2 > is exactly that of S2 × S2 (analyzed in Example 2
...
1)
...
Here we have
D1 =< x >,
D2 =< y, x2 >,
D3 =< xy, x2 >
and
H1 = f (D1 ) = {(1, 1), ((ab), (AB))},
H2 = f (D2 ) = {(1, 1), (1, (AB))},
H3 = f (D3 ) = {(1, 1), ((ab), 1)}
...
0
...
Let F be a field and let N = {diag[f, f,
...
We proved in an assignment that
N = Z(GLn (F)) and is therefore a normal subgroup
...

Let Pn−1 (F) be the set of equivalence classes of non-zero vectors in Fn under the equivalence v ∼ w
if there is f ∈ F∗ such that f v = w; that is, the set of lines through the origin
...

Let
π : GLn (F) −→ PGLn (F)
be the canonical homomorphism
...

Consider the image of SLn (F) in PGLn (F); it is denoted PSLn (F)
...

The group PSLn (F) is equal to π(SLn (F)) = π(SLn (F)N ) and is isomorphic to SLn (F)N/N ∼
=
SLn (F)/SLn (F) ∩ N = SLn (F)/µn (F), where by µN (F) we mean the group {f ∈ F× : f n = 1} (where
we identify f with diag[f, f,
...
Therefore,
PSLn (F) ∼
= SLn (F)/µn (F)
...

×(n)
×
n
×
Let F
be the subgroup of F consisting of the elements {f : f ∈ F }
...
We conclude that
PGLn (F)/PSLn (F) ∼
= F× /F×(n)
...
The lattice of subgroups of a group
Let G be a group
...
Define an order on this set by
A ≤ B if A is a subgroup of B
...
That is,
the relation is really an order
...
Every two elements A, B have a minimum A ∩ B (that is if C ≤ A, C ≤ B
then C ≤ A ∩ B) and a maximum < A, B > - the subgroup generated by A and B (that is C ≥
A, C ≥ B then C ≥< A, B >)
...
It is a lattice in its own right
...

Here is the lattice of subgroups of D8
...

VVVV
D8 L
LL VVVV

LL
LL VVVVVVVV
LL
VVVV
L
VVV



< y, x2 >

< yx, x2 >

UUUUii
TTTT
t
TTTT
iiii UUUUUUU
i
tt
i
i
TTTT
t
UUUU
tt iiiiiii
TTTT
U
t
U
U
t
i
U
TTT
i
UUU
tiii
2
2
<
yx
>
<
y
>
< yx >
< yx3 >

eeeeedddddddddddddd
r
hhh
e
e
h
e
r
h
e
e
h
r
r
hhh
eeeee ddddddd
rrr hhhhhheheheeedededededededddddddd
r
r
e
d
h
d
e
h e d
rherhderhedhedhedhedededededdddd
{e} d

subgroups of order 4

subgroups of order 2

How to prove that these are all the subgroups? Note that every proper subgroup has order 2 or 4 by
Lagrange’s theorem
...
Else, it can only be of order 4 and every element different from e has
order 2
...

end of lecture 10

GROUP THEORY

23

Part 3
...
Basic definitions
Let G be a group and let S be a non-empty set
...

Given an action of G on S we can define the following sets
...
Define the orbit of s
Orb(s) = {g ? s : g ∈ G}
...
We also define the stabilizer of s to be
Stab(s) = {g ∈ G : g ? s = s}
...
In fact, it is a subgroup, as the next Lemma states
...
We’ll make
more precise later
...
This function, we’ll see later, is bijective
...
Basic properties
Lemma 14
...
10
...
We say that s1 is related to s2 , i
...
, s1 ∼ s2 , if there exists
g ∈ G such that
g ? s1 = s2
...
The equivalence class of s1 is its orbit Orb(s1 )
...
The set Stab(s) is a subgroup of G
...
Then
|Orb(s)| =
Proof
...

|Stab(s)|

(1) We need to show reflexive, symmetric and transitive
...
Second, if s1 ∼ s2 then for a suitable g ∈ G we

Dummit & Foote
§4
...
Therefore
g −1 ? (g ? s1 ) = g −1 ? s2
⇒

(g −1 g) ? s1 = g −1 ? s2

⇒

e ? s1 = g −1 ? s2

⇒

s1 = g −1 ? s2

⇒

g −1 ? s2 = s1

⇒

s2 ∼ s1
...
If s1 ∼ s2 and s2 ∼ s3 then for suitable g1 , g2 ∈ G we have
g1 ? s1 = s2 ,

g 2 ? s 2 = s3
...

Moreover, by the very definition the equivalence class of an element s1 of S is all the
elements of the form g ? s1 for some g ∈ G, namely, Orb(s1 )
...
We have to show that: (i) e ∈ H; (2) If g1 , g2 ∈ H then g1 g2 ∈ H; (iii) If
g ∈ H then g −1 ∈ H
...

Therefore e ∈ H
...
e
...


Then
(g1 g2 ) ? s = g1 ? (g2 ? s) = g1 ? s = s
...
Finally, if g ∈ H then g ? s = s and so
g −1 ? (g ? s) = g −1 ? s
⇒

(g −1 g) ? s = g −1 ? s

⇒

e ? s = g −1 ? s

⇒

s = g −1 ? s,

and therefore g −1 ∈ H
...
If we
show that, then by Lagrange’s theorem,
|Orb(s)| = no
...

Define a function
φ

{left cosets of H} −→ Orb(s),
by
φ(gH) = g ? s
...
First

GROUP THEORY

25

Well-defined: Suppose that g1 H = g2 H
...

Note that

g1−1 g2

∈ H, i
...
,

(g1−1 g2 )

? s = s
...


φ is surjective: Let t ∈ Orb(s) then t = g ? s for some g ∈ G
...

φ is injective: Suppose that φ(g1 H) = φ(g2 H)
...
Indeed,
φ(g1 H) = φ(g2 H)
⇒

g1 ? s = g2 ? s

⇒

g2−1 ? (g1 ? s) = g2−1 ? (g2 ? s)

⇒

(g2−1 g1 ) ? s = (g2−1 g2 ) ? s

⇒

(g2−1 g1 ) ? s = e ? s

⇒

(g2−1 g1 ) ? s = s

⇒

g2−1 g1 ∈ Stab(s) = H

⇒

g1 H = g2 H
...
0
...
The set S is a disjoint union of orbits
...
The orbits are the equivalence classes of the equivalence relation ∼ defined in Lemma 14
...
10
...

¤
We have in fact seen that every orbit is in bijection with the cosets of some group
...
We saw that if s ∈ S then there is a natural bijection G/Stab(s) ↔ Orb(s)
...
1
...


GROUP THEORY

26

15
...
0
...
The group Sn acts on the set {1, 2,
...
The action is transitive, i
...
, there is
only one orbit
...
,i−1,i+1,
...

Example 15
...
13
...
We use the following lemma:
Lemma 17
...
2
...
Let g 6= e be an element of G
...

That is, the group G is cyclic (hence commutative), generated by any non-trivial element
...
The subgroup hgi has more than one element because e, g ∈ hgi
...

Thus, | hgi | = p and therefore hgi = G
...
Then so
But any coloring fixed under rotation by 1 must be
Applying CFF we get
10
1 X
N=
I(n) =
11 n=0

¤
does hii = Z/11Z (the stabilizer is a subgroup)
...

1
· 1980 = 180
...
2
...
How many roulettes with 12 wedges painted 2 blue, 2 green and 8 red are there
when we allow rotations?
Let S be the set of painted roulettes
...
, 12
...


GROUP THEORY

31

Let G be the group Z/12Z
...
The element 1 rotates a painted roulette by
angle 2π/12 anti-clockwise
...

We are interested in N – the number of orbits for this action
...

The identity element always fixes the whole set
...
We claim that if 1 ≤ i ≤ 11
and i 6= 6 then i doesn’t fix any element of S
...
Say in
that roulette the r-th sector is blue
...
Therefore so must be the r + 2i sector
...

If i is equal to 6 and we enumerate the sectors of a roulette by the numbers 1,
...

In any coloring fixed by i = 6 the colors of the pairs (1 7), (2 8), (3 9), (4 10), (5 11) and (6 12) must
be the same
...
The rest would be red
...
We summarize:
element g
0
i 6= 6
i=6

I(g)
2970
0
30

Applying CFF we get that there are
N=

1
(2970 + 30) = 250
12

different roulettes
...
2
...
In this example S is the set of necklaces made of four rubies and four sapphires
laid on the table
...

We may talk of S as the colorings of a regular octagon, four vertices are green and four are red
...

The results are the following
element g
e
x, x3 , x5 , x7
x2 , x6
x4
yxi for i = 0,
...
The number of elements in S is
= 70 (chossing
4
which 4 would be green)
...
If xr fixes a coloring s0 so does (xr )r = x(r ) because

GROUP THEORY

32

the stabilizer is a subgroup
...
11
Now, x2 written as a permutation is (1 3 5 7)(2 4 6 8)
...
That is, all the freedom we have is to choose whether the cycle (1 3 5 7) is green
or red
...
The same rational applies to x6 = (8 6 4 2)(7 5 3 1)
...
It may written in permutation notation as (1 5)(2 6)(3 7)(4 8)
...
There are thus
=6
2
possibilities (Choosing which 2 out of the four cycles would be green)
...
We recall that these are all reflections
...
One may be written using permutation notation as
(i1 i2 )(i3 i4 )(i5 i6 )
(with the other two vertices being fixed
...
The other
kind is of the form
(i1 i2 )(i3 i4 )(i5 i6 )(i7 i8 )
...
Whatever is the case, one uses similar reasoning
to deduce that there are 6 colorings preserved by a reflection
...


17
...
The game of 16 squares
...

In this game, we are given a 4 × 4 box with 15 squares numbered 1, 2,
...
At
every step one is allowed to move an adjacent square into the vacant spot
...
Can you prove it? Apparently, the puzzle was originally marketed
with the tiles in the impossible position with the challenge to rearrange them into the initial position!
11x(32 ) = x9 = x because x8 = e, etc
...
1
...

17
...
Rubik’s cube
...
The
group G is generated by 6 basic moves a, b, c, d, e, f (each is a rotation of a certain “third of the cube”)
and could be thought of as a subgroup of the symmetric group on 54 = 9 × 6 letters
...
The order of the Cube Group is 227 · 314 · 53 · 72 · 11 = 43, 252, 003, 274, 489, 856, 000
One is usually interested in solving the cube
...
Since
the current position was gotten by applying an element τ of G, in group theoretic terms we attempt
to find an algorithm of writing every G in terms of the generators a, b, c, d, e, f since then also τ −1
will have such an expression, which is nothing else than a series of moves that return the cube to its
original position
...
, f ±1 (why do 3 times
a when you can do a−1 ??)
...
Otherwise said, what is the diameter
of the Cayley graph? But more than that, is there a simple algorithm of finding for every element of
G an expression in terms of the generators?
Now, since the Cayley graph of G has 12 edges emanating from each vertex (and is connected
by definition of the cube group) it follows that to reach all positions one is forced to allow at least
log12 |G| ∼ 18
...


Figure 17
...
The Rubik Cube
...


GROUP THEORY

34

Part 4
...
3

18
...
We write σ as a product of disjoint cycles:
σ = σ1 σ2 · · · σr
...
Namely, if we let |(i1 i2
...

We may also allow cycles of length 1 (they simple stand for the identity permutation) and then we
find that
n = |σ1 | + |σ2 | + · · · + |σr |
...
Note that every partition is obtained from
a suitable σ
...
0
...
Two permutations, σ and ρ, are conjugate (namely there is a τ such that τ στ −

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