TITLE: If $G'/G''$ and $G''/G'''$ are cyclic then $G''=G'''$
QUESTION [9 upvotes]: Prove that, if $G'/G''$ and $G''/G'''$ are both cyclic then $G''=G'''$.
I was expecting this proof would be similar to the proof of

$$G/Z(G)~~~~ \text{is cyclic} \Rightarrow G~~~ \text{is abelian}$$

I tried in that way but i could reach nowhere.
Any help would be appreciated.
Thank You.

REPLY [7 votes]: 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.