Cayley’s Theorem · Cayley’s Theorem Recall that when we first considered examples of groups,...

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.