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 $$p$$.

Applications

 * Equivalence of definitions of special group: A special group is a group where the center, commutator subgroup and Frattini subgroup are all equal. In this case, this subgroup is elementary Abelian.
 * Frattini-in-center odd-order p-group implies p-power map is endomorphism
 * Frattini-in-center odd-order p-group implies (p plus 1)-power map is automorphism
 * Commutator subgroup lemma for Frattini-in-center cyclic-center p-group

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 $$p$$.