Short exact sequences and group homomorphisms

[myownsavior]

Abstract algebra question. Given the short exact sequence

[tex]$ 1 \longrightarrow N \longrightarrow^{\phi} G \longrightarrow^{\psi} H \longrightarrow 1 $[/tex]

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

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

Now since I know [tex]$ H \cong G/N $[/tex], for this isomorphism [tex]$ G \cong N \times H[/tex] 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.