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Short exact sequences and group homomorphisms

  1. Mar 9, 2009 #1
    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..
  2. jcsd
  3. Mar 10, 2009 #2
    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".
  4. Mar 10, 2009 #3
    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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