Cayley’s Theorem

Recall that when we first considered examples of groups, we noted that there was arelationship between the symmetric groups

†

Sn and the dihedral groups

†

Dn . In particular,we noted that each element of

†

D3 could be identified with a unique element of

†

S3 andvice versa. Using what we have learned since then, we can say that

†

D3 @ S3 . When weconsidered the same correspondence for

†

D4 and

†

S4 , there was an important difference. Inthis case, each element of

†

D4 corresponded to a unique element of

†

S4 , but there wereelements of

†

S4 which did not correspond to anything from

†

D4 . In this case, ourcorrespondence defined an injective homomorphism which was not surjective. Thesetwo examples illustrate an important relationship between finite groups and thesymmetric groups

†

Sn . We will show that every finite group is isomorphic to a subgroupof

†

Sn . First, some preliminaries:

Lemma1: (a) Let

†

j :G Æ H be a homomorphism of groups. Define

†

Imj ={j(g)| g ΠG}. Then

†

Imj is a subgroup of H.(b) If

†

j is injective, then

†

G @ Imj .

Example: Recall from the last handout that we can define an injective homomorphism

†

f :C Æ GL(2,R) by

†

f (a + bi)=a -bb a

È

Î Í ˘

˚ ˙ , where C is the group of non-zero complex

numbers under multiplication. So

†

C @ Imf .

Example: Recall that

†

D4 is generated by a and r where r is the transformation defined by

rotating

†

p2

units about the z-axis, let a is rotation

†

p units about the line y=x in the x-y

plane. We can define a map

†

j :D4 Æ S4 by setting

†

j(r)= (1 4 3 2) and

†

j(a)= (4 2) .Then the definition extends to all of

†

D4 by setting

†

j(rka)= (1 4 3 2)k (4 2) . We cancheck that (1 4 3 2) and (4 2) satisfy the same relationship as a and r(namely

†

(4 2)(1 4 3 2)k = (1 4 3 2)-k (4 2) ) and consequently,

†

j is a homomorphism.Check that

†

j is injective. Thus

†

D4 is isomorphic to a subgroup of

†

S4 . (The image of

†

jis the subgroup of

†

S4 generated by (1 4 3 2) and (4 2).)

Now, we defined the symmetric group

†

Sn as the set of permutations of n objects orequivalently as the set of bijections from the set {1, 2, 3, … , n} to itself. This seconddescription has an advantage in that we can generalize it to infinite sets.

Definition: Let S be a set and define A(S) to be the set of all bijective functions from S toitself. A(S) is called the permutation group on the set S.

It is easy to see from properties of bijective functions that A(S) is a group undercomposition. Note that if S is a set of n elements, then

†

A(S)@ Sn .

Lemma 2: Let G be a group and

†

a ΠG . Define

†

ja :G Æ G by

†

ja(g)= ag . Then

†

ja ΠA(G).

Lemma 3: Let G be a group,

†

a ΠG , and

†

ja :G Æ G as above. Define

†

f :G Æ A(G) by

†

f (a)= ja for each

†

a ΠG . Then f is an injective homomorphism.

Putting Lemmas 1-3 together, we have proved Cayley’s Theorem.

Theorem (Cayley’s Theorem): Every group G is isomorphic to a subgroup of A(S).

Corollary: Every finite group G of order n is isomorphic to a subgroup of the symmetricgroup

†

Sn .

An homomorphism from a group G to a group of permutations is called a representationof G. The homomorphism

†

ja is called the left regular representation of G. We will callthis realization of G as a group of permutations the permutation representation of G.

Example: Find the permutation representation of a cyclic group of order n.