If $K$ is a normal subgroup of a group $G$, then the center $Z(K)$ of $K$ is also a normal subgroup of $G$.
I want to prove this statement. But without using characteristic subgroups and automorphisms. Since in our course we did not learn yet. I hope there is another proof that only uses conjugation, (normal) subgroups, and center/centralizers.
Let $z \in Z(H)$ and let $g \in G$. Since $H$ is a normal subgroup of $G$ and $z \in H$, we know that $gzg^{-1} \in H$. We claim that $gzg^{-1} \in Z(H)$. Let $h \in H$. We want to show that $hgzg^{-1}=gzg^{-1}h$. Since $H$ is normal in $G$, we know that $g^{-1}hg=h'$ for some $h' \in H$.
Try to take it from here.