# G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) < A.

Hi: This is a rather elementary problem. Anyway, it is this: Let G be a simple group, |G| > 2, and f homomorphism from G to H, H some group. If H contains a normal subgroup A of index 2, then f(G) is a subgroup of A.

Notation: '<' means subgroup of.

I tried the following proof, but am stuck at some point. Ker f is normal in G. But G simple. Hence either ker f = G or ker f = <1>. If ker f = G then f(G) = <1> < A. Let ker f = <1>. Obviously, $$f^{-1}(A) = f^{-1}(A \cap f(G)).$$ But A and f(G) are normal in H and so is $$A \cap f(G)$$. So $$f^{-1}(A)$$ is normal in G. But G is simple. Therefor, either $$f^{-1}(A)$$ = <1> or $$f^{-1}(A)$$ = G.
Case $$f^{-1}(A) = G$$: then $$f(G) = f(f^{-1}(A)) \subseteq A$$.

Case $$f^{-1}(A) = <1>$$: But here the only thing I can infer is that $$A \cap f(G) = <1>$$. Any ideas?