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, $\displaystyle f^{-1}(A) = f^{-1}(A \cap f(G)). $ But A and f(G) are normal in H and so is $\displaystyle A \cap f(G)$. So $\displaystyle f^{-1}(A) $ is normal in G. But G is simple. Therefor, either $\displaystyle f^{-1}(A) $ = <1> or $\displaystyle f^{-1}(A) $ = G.

Case $\displaystyle f^{-1}(A) = G$: then $\displaystyle f(G) = f(f^{-1}(A)) \subseteq A$.

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