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.
 A: 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.

A: This comes because of important (characterizing) property of $p$-groups (nilpotent groups): in a $p$-group, every maximal subgroup is of prime index and normal. 
Suppose $G$ is non-abelian. We show that $G/[G,G]$ is not cyclic. Let $M$ be a maximal subgroup of $G$. It can not be unique (otherwise, choosing $x\in G\setminus M$, the subgroup $\langle x\rangle$ will be certainly in some maximal subgroup, but by uniqueness of $M$, we will have $\langle x\rangle=G$, but $G$ is non-cyclic).
Let $M'$ be another maximal subgroup. Now, both $M,M'$ are normal with $G/M$ and $G/M'$ isomorphic to cyclic group of order $p$, hence $[G,G]$ is contained in $M$ as well as $M'$. 
Thus, in $G/[G,G]$, there are at least two maximal subgroups - $M/[G,G]$ and $M'/[G,G]$. This means $G/[G,G]$ can not be cyclic.
