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Proof Involving Homomorphism and Normality

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that if [tex]\theta[/tex] is a homomorphism from G onto H, and N [tex]\triangleleft[/tex] G, then [tex]\theta[/tex](N) [tex]\triangleleft[/tex] H.


    2. Relevant equations



    3. The attempt at a solution
    I think I have a good idea of what is going on, but I'm struggling to tie it all together.

    It's given that N [tex]\triangleleft[/tex] G so I know that gng[tex]^{-1}[/tex] [tex]\in[/tex] N for all n[tex]\in[/tex]N and all g[tex]\in[/tex]G. I also know that N is a subgroup of G.

    I know that [tex]\theta[/tex](G), the image of [tex]\theta[/tex], is a subgroup of H. Because of this, I would also think that [tex]\theta[/tex](N) is also a subgroup of H because of the homomorphism.

    From here I need someone to lead me in the right direction. I've been trying to solve this problem for four days so any help is greatly appreciated.
     
  2. jcsd
  3. Mar 7, 2010 #2
    You are trying to show that [tex]
    \theta
    [/tex](N) is normal in H, so you should start with h [tex]
    \theta
    [/tex](N), for some h in H. [tex]
    \theta
    [/tex] is onto, so h= [tex]
    \theta
    [/tex](g) for some g in G. Now use the fact that [tex]
    \theta
    [/tex] is a homomorphism and that N is normal in G.
     
  4. Mar 7, 2010 #3
    I don't think I'm following entirely. Are you saying that since I know N is in G and because the homomorphism sends elements in G to H, I can say that N is in H? If this is true, I don't understand how to show that N is normal in H.
     
  5. Mar 7, 2010 #4
    No, the whole point of the proof is to show that [tex]
    \theta
    [/tex](N) is normal in H. To show that, you must show that for all h in H, h[tex]
    \theta
    [/tex](N) = [tex]
    \theta
    [/tex](N)h.

    Starting off with some h in H, you can observe that h = [tex]
    \theta
    [/tex](g) for some g in G. This is because [tex]
    \theta
    [/tex] is onto.

    Thus you have h [tex]
    \theta
    [/tex](N)= [tex]
    \theta
    [/tex](g) [tex]
    \theta
    [/tex](N). Now you must use the fact that [tex]
    \theta
    [/tex] is a homomorphism.
     
  6. Mar 7, 2010 #5
    How do you know that [tex]\theta[/tex] is onto?
     
  7. Mar 7, 2010 #6
    because in the original problem you said ".. [tex]
    \theta
    [/tex] is a homomorphism from G onto H"
     
  8. Mar 7, 2010 #7
    I'm following you now. Thanks for the help.
     
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