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

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

    Case f^{-1}(A) = <1>: But here the only thing I can infer is that A \cap f(G) = <1>. Any ideas?
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  2. #2
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    Re: G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) <

    Hi,
    Most of what you say is true, but what makes you think f(G) is normal in H? It needn't be.

    G simple, f homomorphism from G to H. If A normal in H and  |H:A|=2 then  f(G) &lt; A.-mhfgroups14.png
    Thanks from ENRIQUESTEFANINI
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  3. #3
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    Re: G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) <

    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).
    Last edited by ENRIQUESTEFANINI; June 5th 2013 at 07:51 AM.
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    I think I've got it. Suppose \exists x \in G | f(x) \notin A. Let u: G --> H/A, u(g) = Af(g). Then u is an homomorphism (easy). But u(x) = Af(x) \neq A. Hence u is onto, because |H/A| = 2. Hence G/f^{-1}(A) \cong H/A by the homomorphism theorem because Ker u = f^{-1}(A). Hence |G:f^{-1}(A)| = 2    . (1)

    Also, Ker  u \unlhd G. Hence f^{-1}(A) \unlhd G. But G simple. f^{-1}(A) = G contradicts (1). Suppose f^{-1}(A) = <1>. Then we have H/A \cong G/f^{-1}(A) \cong G and |G| = 2, because |H/A| = |H:A| = 2 by the statement, contradicting the fact that |G| > 2. Therefore there is no such x in G such that f(x) \notin A and  f(G) \leq A which is what we wanted to prove.
    Last edited by ENRIQUESTEFANINI; June 5th 2013 at 11:57 AM.
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    Re: G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) <

    Quote Originally Posted by johng View Post
    Hi,
    Most of what you say is true, but what makes you think f(G) is normal in H? It needn't be.

    Click image for larger version. 

Name:	MHFgroups14.png 
Views:	7 
Size:	10.3 KB 
ID:	28532
    I understand your proof except for a detail. The problem does not say G or H are
    finite. So, I cannot say |f(G)| = 2. I think we need to prove that H/A isomorphic to G. Then, yes, H/A is finite and |G| = 2.
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    Re: G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) <

    Hi,
    I've attached further explanation. I hope the proof is now clear.

    G simple, f homomorphism from G to H. If A normal in H and  |H:A|=2 then  f(G) &lt; A.-mhfgroups14a.png
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  7. #7
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    Re: G simple, f homomorphism from G to H. If A normal in H and |H:A|=2 then f(G) <

    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.
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