# Center of a normal subgroup is also a normal subgroup

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 $$h' \in H$$.
• We have $g^{-1}hg=h'$. Multiply both sides by $g^{-1}$ on the right. Commented Apr 28, 2023 at 5:20
• I guess, I found the answer thanks to you. Showing $hgzg^{−1}=gzg^{−1}h$ is a good simplification. I also used the lemma $gzg^{-1}ghg^{-1} = ghg^{-1}gzg^{-1}$. Which is more critical, I guess. It indicates, after a transformation action on both $h$ and $z$, $z'$ still commutes. Then showing your idea is simple because $ghg^{-1}$ is just another element of $H$. Commented May 1, 2023 at 1:00