If $G$ a finite $p$-group s.t $G/[G;G]$ cyclic then $G$ abelian. Let $G$ be a finite $p$-group and $[G,G]$ its commutator sub-group, I
need to show that if the group quotient $G/[G,G]$ is cyclic then $G$
is an abelian group.
My attempt is to let $g\in G$ s.t $g$ modulo $[G,G]$ be a generator of
the group quotient, and thus for all elements $y$ of $G$ we have $y=g^m$
modulo $[G,G]$, that is $y=g^mu$ for some $u\in [G,G]$. So for
arbitrary $x,y\in G$, I must to prove that $xy=yx$. In writing
$x=g^nv$ and  $y=g^mu$,  I get $xy=g^nvg^mu$  and $yx=g^mug^nv$
Here I find myself stuck and I don't know what can I do, also I may be stuck because I do not see where I can use  the
assumption that $G$ is $p$-group, it may be that this path is solely
based on the definitions and the universal property of the
commutator group fails, could you help me to achieve this? Thanks in advance for all  participation.

Hint: look at the Frattini subgroup $\Phi(G)=G'G^p$. $G/\Phi(G)$ is cyclic. And remember that $\Phi(G)$ are non-generators: mouse over for a proof.

If $G$ is a (nil)$p$(otent)-group, then $G'\subseteq \Phi(G)$. So of $G/G'$ is cyclic, then $G/\Phi(G) \cong (G/G')/(\Phi(G)/G')$ is cyclic, say $G/\Phi(G)=\langle \bar g\Phi(G) \rangle$. This implies that $G=\langle g \rangle\Phi(G)$, and since $\Phi(G)$ are non-generators, $G=\langle g \rangle$ is cyclic.

