If S is a subset of G with G finite, while f:G->H is a morphism with kernel N, prove that:(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Homework Help: Question from Saunders Maclane and Birkhoff Algebra book.

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