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, But A and f(G) are normal in H and so is . So is normal in G. But G is simple. Therefor, either = <1> or = G.
Case : then .
Case : But here the only thing I can infer is that . Any ideas?