Let $G$ be a group, $G'=[G,G]$ and $G''=[G',G']$ the first and second derived subgroups and assume $G''$ is cyclic. Prove that $G''\subset Z(G')$.

Question

I'm trying to prove the following, but I'm stuck and I don't see how to continue. Any help is much appreciated!

Let $G$ be a group, $G'=[G,G]$ and $G''=[G',G']$ the first and second derived subgroups and assume $G''$ is cyclic. Prove that $G''\subset Z(G')$.

My work this far: Since $G''$ is cyclic, it is abelian. Since it is the commutator subgroup of $G'$, it is also known that $G''$ is normal. Thus (using a theorem from my syllabus), there exists a homomorphism $g:G'/G''\to\mathrm{Aut}(G'')$ such that $g(aG'')=\phi_{a|G''}$ for $a\in G'$, where $\phi_{a|G'}:G''\to G'':x\mapsto axa^{-1}$, thus the conjugation map by $a$. Now it suffices to proof $\phi_{a|G'}=Id_{G'}$ for all $a\in G'$.

But how to prove this? I don't see it. The work I did this far follows a hint that was given for the question.

Answer 1

In general, if $H$ is a subgroup of $G$, then $N_G(H)/C_G(H) \hookrightarrow Aut(H)$, by conjugation. Now take $H=G''$, being cyclic. Then $Aut(G'')$ is abelian. Since $G''$ is normal in $G$ we have $N_G(G'')=G$ and so $G/C_G(G'')$ is abelian, whence $G' \subseteq C_G(G'')$. That is, $G'$ centralizes $G''$, or equivalenty $G'' \subseteq Z(G')$.