# Cyclic normal subgroup of perfect group is in the center

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.

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 $$H \leq \operatorname{Z}(G)$$.