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 $\mathrm{\Phi }(P)$ is contained in $Z(P)$ (i.e., $P$ is a Frattini-in-center group). Then, the commutator subgroup $P^{\prime }=[P,P]$ is an elementary Abelian group: it is an Abelian group of exponent $p$.

Related facts

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: $P^{\prime }$ is elementary Abelian.

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

Note that since $P^{\prime }\le Z(P)$, $P$ 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 $P^{\prime }$ is generated by elements of order $p$. Since $P^{\prime }$ is Abelian, this tells us that $P^{\prime }$ has exponent $p$.