# Frattini subgroup contained in center implies derived subgroup is elementary abelian

## Statement

Suppose $P$ is a group of prime power order: in other words, $P$ has order $p^n$ for some prime $p$ and integer $n$. Further, suppose that the Frattini subgroup of $P$ is contained in the center of $P$; in symbols: $\Phi(P) \le Z(P)$

Then the commutator subgroup (or derived subgroup) of $P$ is elementary Abelian.

## Facts used

1. For a group of prime power order, any subgroup containing the Frattini subgroup, has a quotient which is an elementary Abelian group
2. For a group of prime power order, the derived subgroup is contained in the Frattini subgroup. For full proof, refer: Nilpotent implies derived in Frattini
3. In a group of nilpotence class two, the map $y \mapsto [x,y]$, for fixed $x$, is an endomorphism. For full proof, refer: Class two implies commutator map is endomorphism

## Proof $P' \le \Phi(P)$, and since $\Phi(P) \le Z(P)$, and $Z(P)$ is Abelian, $P'$ is Abelian. Hence, to show that it is elementary Abelian, it suffices to show that it is generated by elements of order $p$.

We know that $P'$ is generated by commutators, i.e. elements of the form $[x,y]$ where $x,y \in P$, so it suffices to show that any commutator has order $P$.

Let's do this. Since $P' \le Z(P)$, the group $P$ has nilpotence class two. By fact (2), the map $y \mapsto [x,y]$ is, for fixed $x$, an endomorphism of $P$. Thus, we have: $[x,y^p] = [x,y]^p$ $Z(P)$ contains $\Phi(P)$, so by fact (1), the quotient $P/Z(P)$ is elementary Abelian. Equivalently, for any $y \in P$, the element $y^p \in Z(P)$. Thus, the left side in the above equation is the identity element, showing that forany $x,y \in G$, $[x,y]^p = e$, completing the proof.