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Question from Saunders Maclane and Birkhoff Algebra book.

  1. Feb 26, 2009 #1


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    If S is a subset of G with G finite, while f:G->H is a morphism with kernel N, prove that:
    [G:S]=[f(G):f(S)][N:SnN] where 'n' stands for intersection.(The question is on page 78 exercise 10).

    Now as far I can tell [G:S] is the number of right or left cosets of S in G, if the morphism was monomorphism it would be equal [f(G):f(S)], now [N:(SnN)] is the number of cosets m(SnN) where m is in N, I must find a way to show that in order to count the number of gS's for each f(g)f(S) I need to count the number of preimages, but f(gS)=f(g)f(S), I don't know how to procceed.

    Any hints?

    Thanks in advance.
  2. jcsd
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