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..
$ 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..