Abelian Frattini subgroup implies centralizer is critical

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

Note that the critical subgroup we obtain this way is of a special kind: it is a fact about::c-closed critical subgroup.

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
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 $$P$$.

Related facts

 * Centralizer of commutator subgroup has class at most two
 * Centralizer of Frattini subgroup is Frattini-in-center

Facts used

 * 1) uses::Frattini subgroup is characteristic
 * 2) uses::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, $$C_P(C) \le C$$.