# 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.

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