# Cayley's theorem

1. ### alexandrabel

1
Hello!

Can someone explain to me how Isomorphism is linked to cayley's theorem?

Using cayley's theorem, it is stated that ' every group is isomorphic to a group of permutations'

Proof:

Step 1: Let G be a given group and set G' of permutations form a grp isomorphic to G. Let Sg be the grp of all permutations of G. For a in G, let Pa be the mapping of G into G given by xPa = xa for x in G.

We then proceed by proving that Pa is one- to - one and onto.

May I know why there is a need to prove that Pa is one to one and onto?

Step 2: Claiming that G' is a subgroup of Sg, we then show that it is closed under permutation mulitplication, has identity permutation and an inverse.

This shows that G' is a subgroup of G but is this needed to prove the theorem?

Step 3: lastly, defining a mapping Ø: G -> G' and show that Ø is an isomorphism of G with G'.

define Ø: G -> G' by aØ = Pa for a in G

aØ = bØ
then Pa and Pb must be in the same permutations of G.
ePa = ePb
so a = b. thus Ø is one to one.

why do we have to prove that Ø is one to one when we have earlier proved that Pa is one to one?

my notes then continue to state that :

for the proof of the theorem, we consider the permutations xλa = xa for x in G
these permutations would have formed a subgroup G'' of Sg, again isomorphic to G but under the map ψ: G -> G'' defined by
aψ = λa-1

what does this remaining part of the proof mean?

thanks!

2. ### morphism

2,020
I am confused by your definition of G'. Could you clarify it a bit?

The gist of the proof is simple: each element a in G gives rise to a permutation Pa:G->G which sends x to ax. Pa is a permutation because, as a function, its inverse is Pa-1. In other words, Pa lives in Sg. Now consider the map F:G->Sg sending a to Pa. This map is an injective homomorphism. So G is isomorphic to F(G), and F(G) is a group of permutation. QED.