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?

The goal is to show that $Z(H)$ is normal in $G$. Take any element $z \in Z(H)$ and any $g \in G$. We need to show that $gzg^{-1} \in Z(H)$. First, note that $gzg^{-1} \in H$, since $z \in H$ and $H$ is normal in $G$. Now, given any element $h \in H$, we need to show that $h(gzg^{-1}) = (gzg^{-1})h$.
Since $H$ is normal in $G$, we have $hg = gk$ for some $k \in H$. Therefore:
$$h(gzg^{-1}) = (hg)zg^{-1} = (gk)zg^{-1} = g(kz)g^{-1} = g(zk)g^{-1} = gz(kg^{-1})$$
where we have used the fact that $kz = zk$ because $k \in H$ and $z \in Z(H)$. Now recall that $hg = gk$. Multiplying this equation on the left and right by $g^{-1}$ gives us $g^{-1}h = kg^{-1}$. Applying this to the rightmost expression in the chain of equalities above, we get $gz(kg^{-1}) = gz(g^{-1}h) = (gzg^{-1})h$, which is exactly what we want.
