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

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

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

2. ## 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.

3. ## 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).

4. I think I've got it. Suppose $\displaystyle \exists x \in G | f(x) \notin A$. Let $\displaystyle u: G --> H/A, u(g) = Af(g)$. Then u is an homomorphism (easy). But $\displaystyle u(x) = Af(x) \neq A$. Hence u is onto, because |H/A| = 2. Hence $\displaystyle G/f^{-1}(A) \cong H/A$ by the homomorphism theorem because $\displaystyle Ker u = f^{-1}(A)$. Hence $\displaystyle |G:f^{-1}(A)| = 2$. (1)

Also, $\displaystyle Ker u \unlhd G$. Hence $\displaystyle f^{-1}(A) \unlhd G$. But G simple. $\displaystyle f^{-1}(A) = G$ contradicts (1). Suppose $\displaystyle f^{-1}(A) = <1>$. Then we have $\displaystyle H/A \cong G/f^{-1}(A) \cong G$ and $\displaystyle |G| = 2$, because $\displaystyle |H/A| = |H:A| = 2$ by the statement, contradicting the fact that $\displaystyle |G| > 2$. Therefore there is no such x in G such that $\displaystyle f(x) \notin A$ and $\displaystyle f(G) \leq A$ which is what we wanted to prove.

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

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

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.

6. ## 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.

7. ## 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.