Center of a normal subgroup is also a normal subgroup

Question

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.

You can start from the definition of "center" and definition of "normal subgroup". — Nobody, Apr 28 '23 at 04:44
Thanks @Nobody, but I know that primary stuff. I know Z(G), is a normal subgroup of G. — lego_compiler, Apr 28 '23 at 04:47
Does this answer your question? Prove that the center of a group is a normal subgroup — Nobody, Apr 28 '23 at 04:50
The following might be considered as an abstract duplicate: If $N$ is a normal subgroup of a group $G$, and $M$ is a characteristic subgroup of $N$, then $M$ is a normal subgroup of G. — Anne Bauval, Apr 28 '23 at 05:02
I don't know characteristic subgroups, we didn't learn that in the course. I need a more basic explanation, a different approach. — lego_compiler, Apr 28 '23 at 05:09
Well it is a good opportunity to learn this convenient notion.Next time please do not just ask. Show some work. https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question — Anne Bauval, Apr 28 '23 at 05:09
Oh, ok. You are right. I am new here, sorry. Also, thanks for the answer. I will do some research — lego_compiler, Apr 28 '23 at 05:10
The approach I gave in my answer is much simpler than what is in the linked post. — cgb5436, Apr 28 '23 at 05:13
Prove that: the center of any group is characteristic subgroup . — Anne Bauval, Apr 28 '23 at 05:15

score 2 · Answer 1 · answered Apr 28 '23 at 05:00

2

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.

answered Apr 28 '23 at 05:00

cgb5436

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We have $g^{-1}hg=h'$. Multiply both sides by $g^{-1}$ on the right. – cgb5436 Apr 28 '23 at 05:20
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 $H$. – lego_compiler May 01 '23 at 01:00
I guess, I used unintentionally the concept of automorphism. Am I right? I have not covered that topic yet. – lego_compiler May 01 '23 at 01:02
Did you get your question answered? I didn't see your question until now. – cgb5436 Nov 27 '23 at 04:16

Center of a normal subgroup is also a normal subgroup

1 Answers1