Finite $p$-group with cyclic abelianization is abelian

Question

Is the following result true? If so, how to prove it?

Let $G$ be a finite $p$-group whose abelianization $G^{\text{ab}} = G/[G,G]$ is cyclic. Then, $G$ is abelian.

I found some similar results relation normal cyclic subgroups of $G$, the commutator subgroup $[G, G]$ but nothing seems to fit properly on a proof of this proposition. I still couldn't think of a counter-example either.

My latest strategy was trying to prove that $G/Z(G)$ is cyclic, but I'm not sure how.

Answer 1

Let $\Phi(G)$ be the Frattini subgroup of $p$-group $G$, that is, the intersection of all maximal subgroups of $G$. In a $p$-group, every maximal subgroups has index $p$ and is normal. Hence $G' \subseteq \Phi(G)$ (One can prove that in fact $\Phi(G)=G^{p}G'$). If $G/G'$ is cyclic it follows that $G/\Phi(G)$ is cyclic, say $G/\Phi(G)=\langle \bar{g}\rangle$. But then $G=\langle g \rangle \Phi(G)$. A well-known fact states that $\Phi(G)$ consists of so-called non-generators (see here), whence $G=\langle g \rangle$ and thus cyclic.

Answer 2

We may also prove without using the properties of Frattini group.

Suppose that $G/G'$ is cyclic where $G$ is a $p$-group. Then $G/G'$ has a unique maximal subgroup $M/G'$. Then $M$ is a maximal subgroup of $G$. If $N$ is also a maximal subgroup of $G$ then $G/N\cong \mathbb Z_p$, and hence $G'\leq N$. As a result, $N/G'$ is a maximal subgroup of $G/G'$.

Then $M=N$, that is, $M$ is the only maximal subgroup of $G$. Pick $x\notin M$. Then we see that $\langle x\rangle=G$, concluding the proof.