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?
If A normal in G and f:G --> H homomorphism, then f(A) normal in H. In particular f(G) is normal in H, because G is normal in G, I think.
EDIT: I'm wrong. f(A) is normal in f(G).
I think I've got it. Suppose . Let . Then u is an homomorphism (easy). But . Hence u is onto, because |H/A| = 2. Hence by the homomorphism theorem because . Hence . (1)
Also, . Hence . But G simple. contradicts (1). Suppose . Then we have and , because by the statement, contradicting the fact that . Therefore there is no such x in G such that and which is what we wanted to prove.
Yes, quite clear. It only took me some work to see that H = Af(G). However, if g<>0, H = A U Af(g) subset of A(Af(g)) = Af(g) subset of Af(G). So, H subset of Af(G) and H = Af(G). Thank you very much for your kind posts.