- **Linear & Abstract Algebra**
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- - **Cayley's theorem**
(*http://www.physicsforums.com/showthread.php?t=333993*)

Cayley's theoremHello!
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! |

Re: Cayley's theoremI 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_{a-1}. 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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