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Cayley's theorem 
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#1
Sep109, 08:20 AM

P: 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 = λa1 what does this remaining part of the proof mean? thanks! 


#2
Sep609, 08:07 PM

Sci Advisor
HW Helper
P: 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 P_{a}:G>G which sends x to ax. P_{a} is a permutation because, as a function, its inverse is P_{a1}. In other words, P_{a} lives in S_{g}. Now consider the map F:G>S_{g} sending a to P_{a}. 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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