Short exact sequences and group homomorphisms

[myownsavior]

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..

 

[yyat]

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".

 

[myownsavior]

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.