Proof Involving Homomorphism and Normality

  • Thread starter rlusk35
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    Proof
In summary, we need to show that if \theta is a homomorphism from G onto H and N \triangleleft G, then \theta(N) \triangleleft H. To do this, we first observe that N is a subgroup of G and \theta(G) is a subgroup of H. Since \theta is onto, we can write any element h in H as \theta(g) for some g in G. Using the fact that \theta is a homomorphism, we can show that h\theta(N) = \theta(g)\theta(N) = \theta(gN) = \theta(N)h. Therefore, we have shown that \theta(N) is normal in H.
  • #1
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Homework Statement


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.


Homework Equations





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.
 
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  • #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.
 
  • #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.
 
  • #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.
 
  • #5
How do you know that [tex]\theta[/tex] is onto?
 
  • #6
because in the original problem you said ".. [tex]
\theta
[/tex] is a homomorphism from G onto H"
 
  • #7
I'm following you now. Thanks for the help.
 

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