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?
 A: 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.
A: 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 $Z(H)$.
A: A subgroup $M$ of $G$ is called characteristic if it is invariant under Aut$(G)$, and one writes $M$ char $G$. The center and the commutator subgroup are examples of characteristic subgroups. A characteristic subgroup must be normal since it is invariant under the inner automorphisms in particular.The above is a special case of the following:
$$M \text { char } N \unlhd G \Rightarrow M \unlhd G$$
