Let A and A' be abelian groups and B a subgroup of A. Let f: A -> A' be a group homomorphism. Let [itex] A^f = Im(f), B^f = Im(f|_B), A_f = Ker(f), B_f=Ker(f|_B) [/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Show that [itex] (A:B) = (A^f:B^f)(A_f:B_f) [/itex].

This is the work I have done: We have [itex]B \subset E=f^{-1}(f(B)) [/itex]. We also have an isomorphism from [itex] A/E [/itex] to [itex] A^f/B^f [/itex]. We have the inclusion map [itex] \iota : A_f \rightarrow E [/itex]. We can use this to define a homomorphism [itex] \iota_* : A_f/B_f \rightarrow E/B [/itex] by xB_f goes to xB. This is well defined because B_f is contained in B. It is also one to one for xB=B implies x is in B and x in A_f which gives x in B_f. The last thing I need to show is surjectivity...

This just seems hopeless to me to. But, if I can show that then we have the theorem. Note, I really haven't used the fact that the groups are abelian which is almost surely cause for thinking that I'm not on the right track.

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