Abelian Frattini subgroup implies centralizer is critical

Statement

Suppose $P$ is a group of prime power order such that the Frattini subgroup (?) $\Phi(P)$ of $P$ is an Abelian group (?). Then, the centralizer of the Frattini subgroup, i.e., the group $C_P(\Phi(P))$, is a Critical subgroup (?) of $P$.

Note that the critical subgroup we obtain this way is of a special kind: it is a C-closed critical subgroup (?).

Definitions used

Frattini subgroup

Further information: Frattini subgroup

The Frattini subgroup of a group is defined as the intersection of all its maximal subgroups. For a $p$-group, it is the unique smallest group such that the quotient is elementary Abelian.

In particular, for a $p$-group, the Frattini subgroup contains the commutator subgroup, and it also contains the Frattini subgroup of any intermediate subgroup.

Critical subgroup

Further information: Critical subgroup

A subgroup $C$ of a group $P$ is termed critical in $P$ if $C$ is a characteristic subgroup of $P$ and the following hold:

1. $\Phi(C) \le Z(C)$; In other words, $C$ is a Frattini-in-center group.
2. $[P,C] \le Z(C)$: In other words, $C$ is a commutator-in-center subgroup of $P$.
3. $C_P(C) \le C$: In other words, $C$ is a self-centralizing subgroup of $P$.

Facts used

1. Frattini subgroup is characteristic
2. Characteristicity is centralizer-closed: The centralizer of a characteristic subgroup is characteristic.

Proof

Given: A finite $p$-group $P$ with Abelian Frattini subgroup $\Phi(P)$. $C = C_P(\Phi(P))$.

To prove: $C$ is a critical subgroup of $P$.

Proof: By facts (1) and (2), $C$ is characteristic in $P$. Note also that since $\Phi(P)$ is Abelian, $\Phi(P) \le C$.

We now check each condition:

1. $\Phi(C) \le Z(C)$: First, observe that $\Phi(C) \le \Phi(P)$, because $C/\Phi(P)$ is elementary Abelian. By definition of $C$, we have $\Phi(P) \le Z(C)$. Thus, $\Phi(C) \le Z(C)$.
2. $[P,C] \le Z(C)$: Indeed, $[P,C] \le [P,P] \le \Phi(P) \le Z(C)$.
3. $C_P(C) \le C$: We have $C_P(C) \le C_P(\Phi(P))$, because $\Phi(P) \le C$. Thus, $C_P(C) \le C$.