TITLE: Let $G$ be a finite nilpotent group and $G'$ its commutator subgroup. Show that if $G/G'$ is cyclic then $G$ is cyclic.
QUESTION [7 upvotes]: So I thought the cleanest way to do this was to simply prove $G' = 1$ since if $G$ is cyclic $G' = 1$ and then $G \cong G/G'$, but I got no where with this.
My next idea was since $G$ is nilpotent I know it's the direct product of normal Sylow subgroups which commute with one another, so it suffices to show that each Sylow subgroup is cyclic. If $P$ is a Sylow $p$ subgroup then $PG'/G'$ is cyclic so by the second isomorphism theorem so is $P/P\cap G'$. So if $P$ intersects $G'$ trivial y then $P$ is cyclic. But I don't know what to do with the case where $P\cap G' \neq 1$.

REPLY [2 votes]: Let $G$ be finite and nilpotent. Then $G' \subseteq \Phi(G)$, the Frattini subgroup. So, if $G/G'$ is cyclic, then $G/\Phi(G)$ is cyclic. Hence, there is a $g \in G$ with $G/\Phi(G)=\langle \overline{g} \rangle$. But then $G=\langle g \rangle\Phi(G)=\langle g \rangle$, since the elements of the Frattini subgroup are non-generators.