# Thread: Groups G and G', M is a normal subgroup of G'

1. ## Groups G and G', M is a normal subgroup of G'

We are given:
1. h: G -> G' is a group homomorphism.
2. M is a normal subgroup of G'.
3. N is the set of x in G such that h(x) is in M.

Prove that N is a normal subgroup of G.

It's not that hard for me to prove that N is a subgroup of G--that just comes from the fact that h is a group homomorphism. But I really have no idea how to prove that N is not just a subgroup but a normal subgroup. In fact, these normal subgroup proofs are just plain difficult for me right now. Please help.

2. ## Re: Groups G and G', M is a normal subgroup of G'

You need to show that $\displaystyle yxy^{-1}\in N$ for all $\displaystyle y\in G$ and $\displaystyle x\in N$. What happens when you apply the definition of N to $\displaystyle yxy^{-1}$?

3. ## Re: Groups G and G', M is a normal subgroup of G'

Originally Posted by emakarov
You need to show that $\displaystyle yxy^{-1}\in N$ for all $\displaystyle y\in G$ and $\displaystyle x\in N$. What happens when you apply the definition of N to $\displaystyle yxy^{-1}$?
Do I just take h(yxy^-1), which clearly is an element in M, use the homomorphism property to split it up into h(y)h(x)h(y^-1), and I'm pretty much done?

EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?

4. ## Re: Groups G and G', M is a normal subgroup of G'

Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.

5. ## Re: Groups G and G', M is a normal subgroup of G'

Originally Posted by emakarov
Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.
So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right? If so, I'm done.

6. ## Re: Groups G and G', M is a normal subgroup of G'

Originally Posted by phys251
EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?
Yes, you need to use both directions of z ∈ N ⇔ h(z) ∈ M (which holds by definition of N): left-to-right for z = x and right-to-left for z = yxy^(-1).

Originally Posted by phys251
So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right?
Yes.

7. ## Re: Groups G and G', M is a normal subgroup of G'

Awesome. Thanks!