Re: Subgroup Index - an equality

Do you mean
[G:H] = [im f : im g] * [ker f : ker g] ?

You are right - there should be: "ker f: ker g".

(2) G is a finite group,

The claim should be true even without this.
But if you do have this condition, note that
everything is finite and you may
use [X:Y] = #X / #Y, which trivializes the task

Well... It still isn't so trivial for me... :(
.

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