I've been trying for a while to solve an exercise/prove a proposition, which at first seemed elementary, but now I even doubt if it's a true proposition. The proposition is:

Let $G$ be a perfect group and let $K$ be a cyclic, normal subgroup of $G$. Show that $K$ is contained in the center of $G$ (i.e. $Z(G)$).

Obviously it's enough to show that the generator of $K$ is in the center, i.e. it commutes with every element of the group, but I can't figure out why that's true.

I thought about using Grun's lemma and show that if the generator of $K$ wasn't in the center, then its $Z(G)$-coset would be in the center of $G/Z(G)$, but it turned out to be the same approach as the first one.

Then I thought about showing that the orbit (under the action of conjugation) of the generator is a singleton and I found that $\sqrt{|G|}$ is a lower bound for the size of the stabilizer $|Stab(x)|$, where $x$ is the generator, but that's just for a finite group $G$, and I couldn't actually find a higher bound (it'd be fine if I could show that x is actually stable under conjugation from any $g\in G$).

Thank you in advance for any help.


You need to take advantage of the fact that the normal subgroup is cyclic somehow.

So let $H$ be cyclic and normal in $G$, where $G$ is perfect. Then by the $N/C$ Theorem, $G/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$. But since $H$ is cyclic, $\operatorname{Aut}(H)$ is abelian. Thus $G/C_G(H)$ is abelian which implies that $C_G(H) \geq G'=G$. Therefore, $C_G(H)=G$ and that is equivalent to $H \leq \operatorname{Z}(G)$.

  • 1
    $\begingroup$ Thank you very much! I wasn't familiar with the N/C Theorem so I guess I was dealing with the question out of the proper context. $\endgroup$ – Nothing just that Nov 30 '20 at 3:28

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