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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
...
The action is transitive on
Fn − {0} and has two orbits on Fn
...
, wn with w1 = v1 as the matrices of the shape


1 ∗
...
∗



...


...


∗

Example 15
...
14
...


In this example, H is “the group” and G is “the set”
...
Since G =
Orb(gi ) =
Hgi we conclude that |G| = i |Orb(gi )| =
P
P
i |H| and therefore that |H| | |G| and that [G : H], the number of cosets, is
i |H|/|Stab(gi )| =
|G|/|H|
...

Example 15
...
15
...
Let S = {gH : g ∈ G} be the set of left cosets of H
in G
...


Here there is a unique orbit (we say G acts transitively)
...

Example 15
...
16
...
It acts on the sphere by rotations: an element θ ∈ G rotates
the sphere by angle θ around the north-south axes
...
See Figure 15
...


θ

Figure 15
...
Action on the sphere by rotation
...
0
...
Let G be the dihedral group D16
...

G = {e, x, x2 ,
...
, yx7 },
where x is the rotation clockwise by angle 2π/8 and y is the reflection through the y-axis
...

We let S be the set of colorings of the octagon ( = necklaces laid on the table) having 4 red vertices
(rubies) and 4 green vertices (sapphires)
...

For example, the coloring s0 in Figure 15
...
Therefore,
the stabilizer of s0 contains at least the set of eight elements
{e, x2 , x4 , x6 , y, yx2 , yx4 , yx6 }
...
1)

Remember that the stabilizer is a subgroup and, by Lagrange’s theorem, of order dividing 16 = |G|
...
2
...

On the other hand, Stab(s0 ) 6= G because x 6∈ Stab(s0 )
...
1)
...
0
...
By Lagrange’s theorem there are two cosets
...
The proof of Lemma 14
...
10 tells us how to find the orbit: it is the
set
{s0 , xs0 },
portrayed in Figure 15
...


Figure 15
...
The orbit of the necklace
...
Cayley’s theorem
Dummit & Foote
§4
...
0
...
Every finite group of order n is isomorphic to a subgroup of Sn
...
Let A be a finite set
...
Then, ΣA is a group
...
, sn be
the elements of A, we can identify bijective functions A −→ A with permutations of {1,
...

Lemma 16
...
19
...
The kernel of this homomorphism is ∩a∈A Stab(a)
...
An element g define a function φg : A −→ A by φg (a) = ga
...
Note that φh φg (a) = φh (ga) = hga = φhg (a) for every a and hence φh φg = φhg
...
This shows that every φg is a bijection and the map
Ψ : G −→ ΣA ,

Ψ

g 7→ φg ,

is a homomorphism
...
)
The kernel of this homomorphism is the elements g such that φg is the identity, i
...
, φg (a) = a for
all a ∈ A
...
The set of such elements g is just ∩a∈A Stab(a)
...
(of Theorem) Consider the action of G on itself by multiplication (Example 15
...
14), (g, g 0 ) 7→
gg 0
...
Thus this action gives an injective homomorphism
G −→ ΣG ∼
= Sn ,
where n = |G|
...
1
...
Let G be a group and H a subgroup of
finite index n
...
The kernel K of Ψ is
∩a∈G/H Stab(a) = ∩g∈G Stab(gH) = ∩g∈G gHg −1
...
Furthermore, since the
resulting homomorphism G/K −→ Sn is injective we get that |G/K| = [G : K] divides [G : H]! = |Sn |
...
Thus, for example, a simple infinite group has no subgroups of finite
index
...
Indeed, if K 0 C G, K 0 < H then K = gKg −1 ⊂ gHg −1 and we see that K 0 ⊆ K
...
1

GROUP THEORY

29

17
...
1
...

Theorem 17
...
1
...
Let N be the number of
orbits of G in S
...
Then
1 X
(17
...

|G|
g∈G

Remark 17
...
2
...
It means exactly that: For every
s1 , s2 ∈ S there exists g ∈ G such that g ? s1 = s2
...
We define a function

(

T : G × S −→ {0, 1},
Note that for a fixed g ∈ G we have
I(g) =

T (g, s) =
X

1
0

g?s=s

...


g∈G

Let us fix representatives s1 ,
...
Now,


Ã
!
X
X X
X X

I(g) =
T (g, s) =
T (g, s)
g∈G

g∈G

=

X

s∈S

|Stab(s)| =

s∈S

=

N
X

=

i=1

X
s∈S

X

i=1 s∈Orb(si )
N
X

s∈S

g∈G

|G|
|Orb(s)|
N

X
|G|
=
|Orb(s)|
i=1

|G|
· |Orb(si )| =
|Orb(si )|

X
s∈Orb(si )

N
X

|G|
|Orb(si )|

|G|

i=1

= N · |G|
...
1
...
Let G be a finite group acting transitively on a finite S
...

Then there exists g ∈ G without fixed points
...
, g } then P
n
1
g∈G I(g) is
i=1 I(gi ) =
P
P

I(g1 ) + I(g2 ) + · · · + I(gn )
...
, sm } then the double sum is T (g1 , s1 ) + T (g1 , s2 ) + · · · + T (g1 , sm ) + T (g2 , s1 ) + T (g2 , s2 ) + · · · + T (g2 , sm ) +
· · · + T (gn , s1 ) + T (gn , s2 ) + · · · + T (gn , sm )
...
cit
...
3
...
By contradiction
...
That is, suppose that for
every g ∈ G we have
I(g) ≥ 1
...


g∈G

By Cauchy-Frobenius formula, the number of orbits N is greater than 1
...


¤
end of lecture 12

17
...
Applications to combinatorics
...
2
...
How many roulettes with 11 wedges painted 2 blue, 2 green and 7 red are there
when we allow rotations?
Let S be the set of painted roulettes
...
, 11
...

Let G be the group Z/11Z
...
The element 1 rotates a painted roulette by
angle 2π/11 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 ≤ 10
then i doesn’t fix any element of S
...
2
...
Let G be a finite group of prime order p
...
Then
hgi = G
...

Proof
...
By Lagrange’s theorem
| hgi | divides |G| = p
...

Suppose that 1 ≤ i ≤ 10 and i fixes s
...

single colored! Contradiction
...

11

Example 17
...
3
...
Let us enumerate the sectors of a roulette by the numbers
µ ¶µ ¶
12
10
1,
...
The set S is a set of
= 2970 elements (choose which 2 are blue, and then
2
2
choose out of the ten left which 2 are green)
...
It acts on S by rotations
...
The element n rotates a painted roulette by angle 2nπ/12 anti-clockwise
...
We use CFF
...
Thus I(0) = 2970
...
Indeed, suppose that i fixes a painted roulette
...
Then so must be the i + r sector (because the r-th sector goes
under the action of i to the r + i-th sector)
...
But there are
only 2 blue sectors! The only possibility is that the r + 2i sector is the same as the r sector, namely,
i = 6
...
, 12 we may write
i as the permutation
(1 7)(2 8)(3 9)(4 10)(5 11)(6 12)
...
We may choose one pair for blue, one pair for green
...
Thus there
are 30 = 6 · 5 possible choices
...

Example 17
...
4
...
We ask how many necklaces there are when we allow rotations and flipping-over
...

The group G = D16 acts on S and we are interested in the number of orbits for the group G
...
, 7

I(g)
70
0
2
6
6

We explain how the entries in the table are obtained:

µ ¶
8
The identity always fixes the whole set S
...

The element x cannot fix any coloring, because any coloring fixed by x must have all sections of
2
the same color (because x = (1 2 3 4 5 6 7 8))
...
Apply that for r = 3, 5, 7 to see that if xr fixes a coloring so does x ,
which is impossible
...
We see that if, say 1 is green so are 3, 5, 7 and
the rest must be red
...
This gives us two colorings fixed by x2
...

Consider now x4
...
In any coloring
µ ¶
4
4
fixed by x each of the cycles (1 5)(2 6)(3 7) and (4 8) must be single colored
...

It remains to deal with the elements yxi
...
There are two
kinds of reflections
...
For example y = (2 8)(3 7)(4 6) is of this form)
...

(For example yx = (1 8)(2 7)(3 6)(4 5) is of this sort)
...

One needs only apply CFF to get that there are
N=

1
(70 + 2 · 2 + 6 + 8 · 6) = 8
16

distinct necklaces
...
3
...
Sam Loyd (1841-1911) was America’s greatest puzzle expert and
invented thousands of ingenious and tremendously popular puzzles
...
, 15 and one free spot
...
For example
1
5
9
13

2
6
10
14

3
7
11
15

4
1 2
8
5 6
7→
12
9 10
13 14

3
7
11

4
8
7→
12
15

1
5
9
13

2
6
10
14

3
7
11

4
8
7→
12
15

1 2
5 6
9
13 14

3
7
10
11

4
8
7→
12
15

1
5
9
13

2
6
14

3
7
10
11

4
8
12
15

Can one pass from the original position to the position below?
1 2 3
5 6 7
9 10 11
13 15 14

4
8
12

It turns out that the answer is no
...


GROUP THEORY

33

Figure 17
...
Loyd’s 14 − 15 puzzle
...
4
...
12 In the case of the Rubik cube there is a group G acting on the cube
...
It is called the
cube group
...
Namely, reverting it to its original position
...
It is natural do deal with the set of generators a±1 , b±1 ,
...
A common question is what is the maximal number of basic operations
that may be required to return a cube to its original position
...
2, thus at least 19, moves
...
2
...


12Also known as the Hungarian cube
...
The Symmetric Group
Dummit & Foote
§4
...
Conjugacy classes
Let σ ∈ Sn
...

Since disjoint cycles commute, the order does not matter and we may assume that the length of the
cycles is non-decreasing
...
it )| = t (we shall call it the length of the cycle; it
is equal to its order as an element of Sn ), then
|σ1 | ≤ |σ2 | ≤ · · · ≤ |σr |
...

We therefore get a partition p(σ) of the number n, that is, a set of non-decreasing positive integers
1 ≤ a1 ≤ a2 ≤ · · · ≤ ar such that n = a1 + a2 + · · · + ar
...

Lemma 18
...
1
...

Proof
...
This implies that
for every cycle (i1 i2
...
it )τ −1 = (τ (i1 ) τ (i2 )
...

In particular, since τ στ −1 = (τ σ1 τ −1 )(τ σ2 τ −1 ) · · · (τ σr τ −1 ), a product of disjoint cycles, we get that
p(σ) = p(τ στ −1 )
...
Say
σ = σ1 σ2
...
i1t(1) )(i21
...
(ir1
...
ρr
1
2
r
= (j11
...
jt(2)
)
...
jt(r)
)
...


¤

Corollary 18
...
2
...
13 There are p(n) conjugacy classes
in Sn
...
we get p(1) = 1, p(2) = 2, p(3) = 3, p(4) = 5, p(5) = 7, p(6) = 11,
...
Note that if σ ∈ An then since An C Sn also τ στ −1 ∈ An
...
However, we would like to consider
the An -conjugacy classes of elements of An
...
0
...
The Sn -conjugacy class of an element σ ∈ An is a disjoint union of [Sn : An CSn (σ)]
An -conjugacy classes
...

In the latter case, the Sn -conjugacy class of σ is the disjoint union of the An -conjugacy class of σ and
the An -conjugacy class of τ στ −1 , where τ can be chosen to be any odd permutation
...
Let A be the Sn -conjugacy class of σ
...
We first note that Sn acts on the set B = {Aα : α ∈ J}
...
This is well
defined: if σα0 is another representative for the An -conjugacy class of σα then σα0 = τ σα τ −1 for some
τ ∈ An
...

The action of Sn is transitive on B
...

The stabilizer of A0 is just An CSn (σ)
...
Namely, if and only if ρσρ−1 = τ στ −1 for some τ ∈ An , equivalently,
(τ −1 ρ)σ = σ(τ −1 ρ), that is (τ −1 ρ) ∈ CSn (σ) which is to say that ρ ∈ An CSn (σ)
...

Since [Sn : An ] = 2, we get that [Sn : An CSn (σ)] = 1 or 2, with the latter happening if and only
if An ⊇ CSn (σ)
...
Moreover,
the orbit consists of the An -conjugacy classes of the elements gσ, g running over a complete set of
representatives for the cosets of An CSn (σ) in Sn
...
The simplicity of An
In this section we prove that An is a simple group for n 6= 4
...
We shall focus on the case n ≥ 5 and
prove the theorem inductively
...

We make the following general observation:
Lemma 19
...
4
...

Proof
...
If N is normal
and n ∈ N then its conjugacy class {gng −1 : g ∈ G} is contained in N
...


Dummit & Foote
§§4
...
6

GROUP THEORY

36

Conjugacy classes in S5
cycle type
5
1+4
1+1+3
1+ 2+ 2
1+1+1+2
1 + 1+ 1+ 1+ 1
2+ 3

representative
(12345)
(1234)
(123)
(12)(34)
(12)
1
(12)(345)

size of conjugacy class
24
30
20
15
10
1
20

order
5
4
3
2
2
1
6

even?
X
×
X
X
×
X
×

Let τ be a permutation commuting with (12345)
...
In particular, τ = (12345)n−1 and so is an
even permutation
...

One checks that (123) commutes with the odd permutation (45)
...
Similarly, the permutation (12)(34) commutes with the odd
permutation (12)
...
We
get the following table for conjugacy classes in A5
...
One
checks that this is impossible unless N = A5
...
0
...
The group A5 is simple
...
0
...
The group An is simple for n ≥ 5
...
The proof is by induction on n
...
Let N be a normal subgroup of
An and assume N 6= {1}
...

Indeed, let σ ∈ N be a non-trivial permutation and write it as a product of disjoint non-trivial
cycles, σ = σ1 σ2
...
Suppose that σ1 is (i1 i2
...
Then
conjugating by the transposition τ = (i1 i2 )(i5 i6 ), we get that τ στ −1 σ ∈ N , τ στ −1 σ(i1 ) = i1 and
if r > 3 τ στ −1 σ(i2 ) = i4 6= i2
...
Take τ = (i1 i2 )(i3 i4 ) then
τ στ −1 σ(i1 ) = i1 and τ στ −1 σ(i2 ) = τ σ(i4 ) ∈ {i3 , i5 }
...


GROUP THEORY

37

It still remains to consider the case where each σi is a transposition
...

Let τ = (i1 i2 )(i3 i5 ) then τ στ −1 σ = (i2 i1 )(i5 i4 )(i3 i6 )
...
and so
is a permutation of the sort we were seeking
...

Consider the subgroups Gi = {σ ∈ An : σ(i) = i}
...
By the preceding step, for some i we have that N ∩ Gi is a non-trivial normal
subgroup of Gi , hence equal to Gi
...
It follows that N ⊇< G1 , G2 ,
...
Now, every element in Sn is a product of
(usually not disjoint) transpositions and so every element σ in An is a product of an even number of
transpositions, σ = λ1 µ1
...
Since n > 4 every product λi µi belongs to
some Gj and we conclude that < G1 , G2 ,
...

¤

GROUP THEORY

38

Part 5
...
The class equation
Let G be a finite group
...
The class equation is the
partition of G to orbits obtained this way
...
Note
that the stabilizer of h ∈ G is CG (h) and so its orbit has length [G : CG (H)]
...
We get
X
|G|
(20
...

|CG (x)|

Dummit & Foote
§4
...
x6∈Z(G)

Remark 20
...
7
...
(One uses the following fact: Given n > 0 and a rational number q there
are only finitely many n-tuples (c1 ,
...
)
For example, the only group with one conjugacy class is the trivial group {1}; the only group with
two conjugacy classes is Z/2Z; the only groups with 3 conjugacy classes are Z/3Z and S3
...
p-groups
Let p be a prime
...


Dummit & Foote
§§4
...
1

Lemma 21
...
8
...
Then the center of G is not trivial
...
We use the class equation 20
...
Note that if x 6∈ Z(G) then CG (x) 6= G and so the integer
|G|
|CG (x)|

is divisible by p
...
x6∈Z(G)

|G|
= |Z(G)|
|CG (x)|

is divisible by p, hence so is the right hand side
...


¤

Theorem 21
...
9
...

(1) For every normal subgroup HC G there is a subgroup KC G such that H < K < G and
[K : H] = p
...

Proof
...
Take an
r−1

element e 6= x ∈ Z(G/H); its order is pr for some r
...

Let K 0 =< y >
...
Let K = πH
(K 0 )
...

(2) The proof just given shows that every p group has a normal subgroup of p elements
...

¤

GROUP THEORY

39

21
...
Examples of p groups
...
1
...
Groups of order p
...

21
...
2
...
We shall prove in the assignments that every such group is commutative
...

21
...
3
...
First, there are the abelian groups Z/p3 Z, Z/p2 Z × Z/pZ and (Z/pZ)3
...
It follows
that Z(G) ∼
= Z/pZ and G/Z(G) ∼
= (Z/pZ)2
...
Note that if p ≥ 3 then every element in this group is of order p (use (I + N )p =
I + N p ), yet the group is non-abelian
...
) More generally the upper unipotent matrices in GLn (Fp ) are
a group of order pn(n−1)/2 in which every element has order p if p ≥ n
...

Getting back to the issue of non-abelian groups of order p3 , one can prove that there is precisely
one additional non-abelian group of order p3
...
(This group is a semi-direct product (Z/p2 Z) o Z/pZ
...
, xd be formal symbols
...
, xd is the set of expressions (called “words”)
y1
...


We remark that the empty word is allowed
...

One checks that this is well defined on equivalence classes, that it is an associative operation, that
the (equivalence class of the) empty word is the identity, and that every element has an inverse:
(y1
...
y1−1
...
It
has the following properties:
(1) given a group G, and d elements s1 ,
...
wr are words in F (d), let N be the minimal normal subgroup containing all the wi
(such exists!)
...
, xd |w1 ,
...
xd and relations w1 ,
...
For example, one can prove
−1
2
3
2
∼
that Z/nZ ∼
=< x1 |xn1 >, Z2 ∼
=< x1 , x2 |x1 x2 x−1
1 x2 > and S3 =< x1 , x2 |x1 , x2 , (x1 x2 ) >
...


Dummit & Foote
pp
...
The Burnside problem asks if a group generated by d elements in
which every element x satisfies xn = 1 is finite
...
, xd moded out by the minimal normal subgroup
containing the expressions f n where f is an element of F (d)
...
There are some instances where it is finite:
d ≥ 2, n = 2, 3, 4, 6
...
Zelmanov,
building on the work of many others, proved that the answer is yes
...


22
...

Theorem 22
...
1
...
Then
G has an element of order p
...
Let S be the set consisting of p-tuples (g1 ,
...
Thus if T is the set of p-tuples (g1 ,
...
One may therefore apply CFF and get
|S| =

np − n
+ n
...

Now define an action of G on S
...
, gp ) ∈ S we define
g(g1 ,
...
, ggp )
...

Since the order of G is n, since n 6 ||S|, and since S is a disjoint union of orbits of G, there must be
an orbit Orb(s) whose size is not n
...
, gp ) in S with a non-trivial stabilizer
...
, ggp ) is equal to (g1 ,
...

This means that for some i we have
(gg1 ,
...
, gp , g1 , g2 ,
...

Therefore, gg1 = gi+1 , g 2 g1 = ggi+1 = g2i+1 ,
...
That is, there exists g 6= e with
g p = e
...
Sylow’s Theorems
Let G be a finite group and let p be a prime dividing its order
...

By a p-subgroup of G we mean a subgroup whose order is a power of p
...


Dummit & Foote
§4
...
0
...
Every maximal p-subgroup of G has order pr (such a subgroup is called a Sylow
p-subgroup) and such a subgroup exists
...
The
number np of Sylow p-subgroups satisfies: (i) np |m; (ii) np ≡ 1 (mod p)
...
0
...
To say that P is conjugate to Q means that there is a g ∈ G such that gP g −1 = Q
...
This implies that P and Q are isomorphic
as groups
...
This is often used this way: given a finite group G the first check in ascertaining
whether it is simple or not is to check whether the p-Sylow subgroup is unique for some p dividing
the order of G
...

We first prove a lemma that is an easy case of Cauchy’s Theorem 22
...
1:
Lemma 23
...
4
...
Then A
has an element of order p
...
We prove the result by induction on |A|
...

If p divides the order of N we are done by induction
...
By
maximality the subgroup BN is equal to A
...
Thus, p
divides the order of B
...
¤
Proposition 23
...
5
...

Proof
...
Assume first that p divides the order of
Z(G)
...
The order of
G/N is pr−1 m and by induction it has a p-subgroup H 0 of order pr−1
...

It is a subgroup of G such that H/N ∼
= H 0 and thus H has order |H 0 | · |N | = pr
...
Consider the class equation
|G| = |Z(G)| +

X
reps
...

|CG (x)|

We see that for some x 6∈ Z(G) we have that p does not divide

|G|
|CG (x)|
...
The

subgroup CG (x) is a proper subgroup of G because x 6∈ Z(G)
...

¤
Lemma 23
...
6
...

Proof
...
Let H = Q ∩ NG (P )
...
Its order is |H| · |P |/|H ∩ P | and so a power of p
...

¤

Dummit & Foote
§3
...
(Of Theorem) Let P be a Sylow subgroup of G
...
0
...
Let
S = {P1 ,
...
That is, the subgroups gP g −1 one gets by letting g vary over G
...
Thus, every Pi is
a Sylow p-subgroup
...

Let Q be any p-subgroup of G
...
The size of Orb(Pi ) is
|Q|/|StabQ (Pi )|
...
0
...
Thus, the orbit consists
of one element if Q ⊂ Pi and is a proper power of p otherwise
...
Then, the orbit of P1 has size 1
...
It follows, using that
S is a disjoint union of orbits, that a = 1 + tp for some t
...

We now show that all maximal p-subgroups are conjugate
...
Thus, for all i, Q 6= Pi and so Q ∩ Pi is a proper
subgroup of Q
...

Thus, p|a
...

¤
23
...
Examples and applications
...
1
...
p-groups
...

23
...
2
...
In every abelian group the p-Sylow subgroups are normal and unique
...

23
...
3
...
Consider the symmetric group S3
...

Note that indeed 3|3 = 3!/2 and 3 ≡ 1 (mod 2)
...

This is expected since n3 |2 = 3!/3 and n3 ≡ 1 (mod 3) implies n3 = 1
...
1
...
S4
...
Their number n2 |3 = 24/8

---

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