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Cayley's theorem

  1. Sep 1, 2009 #1

    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 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?

  2. jcsd
  3. Sep 6, 2009 #2


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    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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