# Proof Involving Homomorphism and Normality

## Homework Statement

Prove that if $$\theta$$ is a homomorphism from G onto H, and N $$\triangleleft$$ G, then $$\theta$$(N) $$\triangleleft$$ H.

## 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$$\in$$N and all g$$\in$$G. 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.

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

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.

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

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

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

How do you know that $$\theta$$ is onto?

because in the original problem you said ".. $$\theta$$ is a homomorphism from G onto H"

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