If $H$ is a normal subgroup of $G$ then $Z(H)$ is normal in $G$ 
Let $G$ be a group and $H$ a normal subgroup of $G$. Let $Z(H)$ be the center of $H$. Show that $Z(H)$ is normal in $G$.

So I know that if $h' \in Z(H)$ then $h'h=hh'$ for all $h \in H$ and I tried proving that $gZ(H)=Z(H)g$. Let $h' \in Z(H)$, so $gh'=hg$ for some $h \in H$ since $H$ is normal and $Z(H)$ is a subgroup of $H$. In particular, $z(H)$ is normal in $H$, so $gh'g^{-1} \in Z(H)$. Therefore $h$ has to be in $Z(H)$. That proves that $gZ(H) \subseteq Z(H)g$. Similarly, we can prove the reverse inclusion. Did I do it correctly?

When trying to prove that a subgroup is normal, it is always instructive to consider of which homomorphism it is the kernel.
Consider this sequence of group homomorphisms:
$$
H \to G \to Inn(G) \to Aut(H)
$$
where the first one is inclusion, the second one takes $g$ to conjugation by $g$, and the last one is restriction to $H$ (which makes sense because $H$ is normal).
The kernel of this composition is the set of $h \in H$ such that conjugation by $h$ is the identity on $H$. This set is exactly $Z(H)$.
