I am trying to understand the proof of the following question:
Show that if $G'/G^{''}$ and $G^{''}/G^{'''}$ are both cyclic then $G^{''} = G^{'''}.$[you may assume $G^{'''} = 1.$ Then $G/G^{''}$ acts by conjigation on the cyclic group $G^{''}.$
I have found the following solution here If $G'/G''$ and $G''/G'''$ are cyclic then $G''=G'''$:
By the third isomorphism theorem, we may as well assume that $G''' = 1$. So we are given that $G'/G''$ and $G''/G''' = G''$ are cyclic, and we want to show that $G'' = 1$, or rather that $G'$ is abelian.
We will use the fact that for a subgroup $H$ of $G$, the quotient $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$.
Since $G''$ is normal, we have $G = N_G(G'')$. Then by the fact we just mentioned, $G/C_G(G'')$ is isomorphic to a subgroup of $\operatorname{Aut}(G'')$. But $G''$ is cyclic, so $\operatorname{Aut}(G'')$ is an abelian group. Thus $G/C_G(G'')$ is abelian, and so $G' \subset C_G(G'')$. But this means that $G'' \subset Z(G')$.
From here the proof follows from the fact
$$ N \subset Z(G), ~ G/N ~~ \text{cyclic} \Rightarrow G~~ \text{is abelian} ,$$
the proof of which is basically the same as the proof of the fact about $G/Z(G)$ you mentioned.
My question is:
I do not understand from where the statement "this means that $G'' \subset Z(G')$" in the solution, is there a theorem saying so?
$C_G(G'')$ is the centralizer of $G''$: it means the set of all elements that commute with every element in $G''$. So $G'\subseteq C_G(G'')$ means that every element in $G'$ commutes with every element in $G''$, so $G''$ is in the center of $G'$.