1. Not finding help here? Sign up for a free 30min 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!

Abstract Algebra-Isomorphism

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Let A,B be normal sub-groups of a group G.
    G=AB.

    Prove that:
    G/AnB is isomorphic to G/A*G/B

    Have no idea how to start...Maybe the second isom. theorem can help us...

    TNX!



    2. Relevant equations
    3. The attempt at a solution
     
  2. jcsd
  3. Dec 18, 2009 #2
    Use the internal characterization of direct products of groups: if [tex]G[/tex] has two normal subgroups [tex]H, K[/tex] such that [tex]HK = G[/tex] and [tex]H \cap K = 1[/tex], then [tex]G \cong H \times K[/tex].

    Also, the third isomorphism theorem may help you (if [tex]K \subset H[/tex] are both normal subgroups of [tex]G[/tex], then [tex]G/H \cong (G/K)/(H/K)[/tex]).
     
  4. Dec 18, 2009 #3
    Sry but I rly can't figure out the Latex text (I see it in black, and it's realy not clear)...
    If I understand what you're saying, then we don't have the right conditions to use "internal characterization of direct products of groups"...
    A,B are normal sub-groups of G and AB=G but who said AnB={1}? The isomorphism you've put afterwards is relevant only when G=A*B and it isn't the case///

    Am I wrong?

    TNx
     
  5. Dec 18, 2009 #4
    I've managed to prove it...TNX a lot anyway...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Abstract Algebra-Isomorphism
Loading...