Dummit and Foote, 6.1.18
Suppose both $G'/G''$ and $G''/G'''$ are cyclic, then $G''=G'''$ (you may assume $G'''=1$).
Where $G' = [G,G], G'' = [G',G'],$ etc.
Question
The assumption that $G'''=1$ apparently follows from the third isomorphism theorem, but I don't see how.
I know $G', G''$, and $G'''$ are all characteristic (and hence normal) in $G$.
Then the third isomorphism theorem says $$ G''/G''' \triangleleft G/G''' $$ and $$ (G/G''') \; \big{/} (G'/G''') \cong G/G' $$ I don't see why that this implies $G''' = \{ e\}$ ?