Short exact sequences and group homomorphisms

myownsavior
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Abstract algebra question. Given the short exact sequence

$ 1 \longrightarrow N \longrightarrow^{\phi} G \longrightarrow^{\psi} H \longrightarrow 1 $

I need to show that given a mapping $ j: H \longrightarrow G, and $ \psi \circ j = Id_h $ (the identity on H), then $ G \cong N \times H. (The internal direct product).

So far I have proved the following: $ \phi $ is injective, $ \psi $ is surjective, N is normal in G, and $ H \cong G/N $.

Now since I know $ H \cong G/N $, for this isomorphism $ G \cong N \times H to be true, wouldn't I need to show that G/N is normal in G?

I think the natural projection mapping is the right path (mapping g in G to its coset in G/N), but I can't get it to be an internal direct product..
 
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You have to use j at some point. Note that the image of j is isomorphic to H.

Edit: Are N, G, H supposed to be abelian? Otherwise you only get a "twisted direct product".
 
I think I got it, K: NxH -> G, taking (n,h) -> n*j(h). I forgot that internal direct products are isomorphic to the external direct products in the finite case.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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