Frattini-in-center p-group implies derived subgroup is elementary abelian

Statement

Suppose $P$ is a group of prime power order with the property that $P/Z(P)$ is elementary Abelian, or equivalently, that $\Phi(P)$ is contained in $Z(P)$ (i.e., $P$ is a Frattini-in-center group). Then, the commutator subgroup $P' = [P,P]$ is an elementary Abelian group: it is an Abelian group of exponent $p$.

Facts used

1. Class two implies commutator map is endomorphism

Proof

Given: A finite $p$-group $P$ such that $P/Z(P)$ is elementary Abelian.

To prove: $P'$ is elementary Abelian.

Proof: First, since $P/Z(P)$ is elementary Abelian, $P' \le Z(P)$, so $P'$ is in the center. In particular, $P'$ is an Abelian group. Thus, it suffices to show that $P'$ has exponent $p$.

Note that since $P' \le Z(P)$, $P$ has nilpotence class two. Thus, by fact (1), we have that for any $x \in P$, the map $y \mapsto [x,y]$ is an endomorphism. In particular, we have that for any $x,y \in P$:

$[x,y^p] = [x,y]^p$.

Now, since $P/Z(P)$ is elementary Abelian, $y^p \in Z(P)$, so the left side is the identity element. Thus, $[x,y]^p$ is the identity element for any $x,y \in P$, and so $P'$ is generated by elements of order $p$. Since $P'$ is Abelian, this tells us that $P'$ has exponent $p$.