1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Short exact sequences and group homomorphisms
  1. Short Exact Sequences (Replies: 2)

  2. Group Homomorphism (Replies: 1)

  3. Homomorphism of groups (Replies: 2)

  4. Group Homomorphism? (Replies: 3)

Loading...