Proof Involving Homomorphism and Normality

[rlusk35]

Homework Statement

Prove that if \theta is a homomorphism from G onto H, and N \triangleleft G, then \theta(N) \triangleleft 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 \triangleleft G so I know that gng^{-1} \in N for all n\inN and all g\inG. I also know that N is a subgroup of G.

I know that \theta(G), the image of \theta, is a subgroup of H. Because of this, I would also think that \theta(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.

 

[VeeEight]

You are trying to show that <br /> \theta<br />(N) is normal in H, so you should start with h <br /> \theta<br />(N), for some h in H. <br /> \theta<br /> is onto, so h= <br /> \theta<br />(g) for some g in G. Now use the fact that <br /> \theta<br /> is a homomorphism and that N is normal in G.

 

[rlusk35]

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.

 

[VeeEight]

No, the whole point of the proof is to show that <br /> \theta<br />(N) is normal in H. To show that, you must show that for all h in H, h<br /> \theta<br />(N) = <br /> \theta<br />(N)h.

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

Thus you have h <br /> \theta<br />(N)= <br /> \theta<br />(g) <br /> \theta<br />(N). Now you must use the fact that <br /> \theta<br /> is a homomorphism.

 

[rlusk35]

How do you know that \theta is onto?

 

[VeeEight]

because in the original problem you said ".. <br /> \theta<br /> is a homomorphism from G onto H"

 

[rlusk35]

I'm following you now. Thanks for the help.