# Finite $p$-group with cyclic abelianization is abelian

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.

• I think it's cracked: prove it by induction on $a$ when $|G| = p^a$. The case $a = 1$ is trivial. Suppose $a > 1$. Because $G$ is a $p$-group, we have that $G/Z(G)$ is a $p$-group with order $< p^a$, so we apply the induction hypothesis (it's abelianization must be a quotient of the abelianization of $G$, so it's also cyclic). This means $G/Z(G)$ is abelian, which implies $[G, G] \leq Z(G)$. So $G/Z(G)$ is cyclic, finishing the proof. – Henrique Augusto Souza Jul 27 '18 at 18:25
• Why is $G/Z(G)$ is cyclic? I don't see it. – user193319 Mar 24 at 18:24
• @user193319 it's abelianization is cyclic, but by induction this means that $G/Z(G)$ is abelian. So it is equal to it's abelianization, and thus cyclic! – Henrique Augusto Souza Mar 25 at 1:51

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.
• Thanks! I was actually using this fact to prove the exercise of section 1.4 of Serre's Galois Cohomology ($p$-completion of a group $G$ s.t. $G^{\text{ab}} = \mathbb{Z}$ is $\mathbb{Z}_p$). My first approach was trough the Frattini subgroup argument, but given that Serre only defines it three sections later, I was looking for an alternative solution. – Henrique Augusto Souza Jul 27 '18 at 19:35
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.