Abstract Algebra help please

Let G be a group and H be a normal subgroup of G. Prove Z(H) is a normal subgroup of G where Z(H) is the center of H.

To prove this i need to prove that for all g in G the ghg^-1 is in Z(H) for h in H

That isn't quite what you need to prove. To prove that Z(H) is a normal subgroup of G, you need to prove two things:

(1) Z(H) is a subgroup of G.
(2) For all g in G and z in Z(H), one has that gzg^(-1) is in Z(H).

The proof of (1) is relatively straightforward: you verify in turn that (i) Z(H) contains the identity element of G, (ii) Z(H) is closed under multiplication, and (iii) Z(H) is closed under inversion. [You may already have proved (1), if you know the general fact that whenever K is a group, Z(K) is a subgroup of K. It follows from this general fact that Z(H) is a subgroup of H, and since H is a subgroup of G, it follows that Z(H) is a subgroup of G.]

So let's focus on (2).

Fix arbitrary elements g of G and z of Z(H). We want to prove that gzg^(-1) is in Z(H). This requires proving two things:

(a) gzg^(-1) is in H.
(b) if h is any element of H, then h gzg^(-1) = gzg^(-1) h.

First let us prove (a). The set Z(H) is a subset of H by definition. Since z is in Z(H) and Z(H) is a subset of H, we deduce that z is in H. Since z is in H, g is in G, and H is a normal subgroup of G, it follows that gzg^(-1) is in H. This concludes the proof of (a).

Now let us prove (b). Fix any h in H. Since h is in H, and g^(-1) is in G, and H is normal in G, we know that g^(-1) h (g^(-1))^(-1) is in H. It follows that

h gzg^(-1) = g g^(-1) h g z g^(-1)
= g [(g^(-1) h (g^(-1))^(-1) z] g^(-1)
= g [z (g^(-1) h (g^(-1))^(-1)] g^(-1) [since g^(-1) h (g^(-1))^(-1) is in H and z is in Z(H)]
= g z g^(-1) h g g^(-1)
= g z g^(-1) h.

This concludes the proof of (b).