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

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

Case [tex]f^{-1}(A) = G[/tex]: then [tex]f(G) = f(f^{-1}(A)) \subseteq A[/tex].

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

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