If $G$ nilpotent and $G/G'$ is cyclic then $G$ is cyclic?
It is very easy to see that this is true when $G$ is finite: If $G$ is finite nilpotent then all maximal subgroups are normal, so $G/N$ has to be cyclic of other $p$ for some prime $p$. So all maximal subgroups contain $G'=[G,G]$. And so $G'$ is contained in the frattini subgroup $\text{Frat}(G)$. It follows that $G=\langle x,\text{Frat}(G) \rangle=\langle x\rangle$. In the last step we also use that $G$ is finite.
But what happens if $G$ is infinite. Is the statement still true?
