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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.

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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.

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  • $\begingroup$ We have $g^{-1}hg=h'$. Multiply both sides by $g^{-1}$ on the right. $\endgroup$
    – cgb5436
    Commented Apr 28, 2023 at 5:20
  • $\begingroup$ 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 $H$. $\endgroup$
    – lego_compiler
    Commented May 1, 2023 at 1:00
  • $\begingroup$ I guess, I used unintentionally the concept of automorphism. Am I right? I have not covered that topic yet. $\endgroup$
    – lego_compiler
    Commented May 1, 2023 at 1:02
  • $\begingroup$ Did you get your question answered? I didn't see your question until now. $\endgroup$
    – cgb5436
    Commented Nov 27, 2023 at 4:16

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