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 ∩f(G)). [/tex] But A and f(G) are normal in H and so is [tex]A ∩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)) ⊆A[/tex].

Case [tex]f^-1(A) = ¡1¿[/tex]: But here the only thing I can infer is that [tex]A ∩f(G) = ¡1¿[/tex]. Any ideas?

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## Well this is not true.

Well this is not true.

Let G=Z_2 simple group of order 2. Let A be any group and let H be a direct product H=GxA. Let f:G->H be an embedding f(g)=(g,e).

Note that A is a normal subgroup of H od index 2, but f(G) is not a subgroup of A.