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alexandrabel Sep1-09 08:20 AM

Cayley's theorem

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'


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 : G -> G' and show that is an isomorphism of G with G'.

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

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

why do we have to prove that 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?


morphism Sep6-09 08:07 PM

Re: Cayley's theorem
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.

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