Analogue of critical subgroup theorem for infinite abelian-by-nilpotent p-groups

Statement

Suppose $p$ is a prime number and $G$ is an infinite P-group (?) that is Abelian-by-nilpotent group (?): $G$ has an abelian normal subgroup $N$ such that the quotient $G/N$ is a nilpotent group. Then, $G$ has a characteristic subgroup $H$ satisfying the following properties:

1. $\Phi (H)\leq Z(H)$, viz., the Frattini subgroup is contained inside the center (i.e., $H$ is a Frattini-in-center group (?)).
2. $[G,H]\leq Z(H)$ (i.e., $H$ is a Commutator-in-center subgroup (?) of $G$).
3. $C_{G}(H)=Z(H)$ (i.e., $H$ is a Self-centralizing subgroup (?) of $G$).
4. $H$ is coprime automorphism-faithful in $G$: If $\sigma$ is a non-identity automorphism of $G$ such that the order of $\sigma$ is relatively prime to $p$, then the restriction of $\sigma$ to $H$ is a non-identity automorphism of $H$.

This a the generalization of a critical subgroup to a possibly infinite p-group.

Related facts

• Thompson's critical subgroup theorem: Thompson's critical subgroup theorem is the version of this result for finite groups.

Facts used

1. Equivalence of definitions of abelian-by-nilpotent group: This states that $G$ is abelian-by-nilpotent if and only if some member of its lower central series is abelian; in particular, that member of the lower central series is an abelian characteristic subgroup with a nilpotent quotient group.
2. Every abelian characteristic subgroup is contained in a maximal among abelian characteristic subgroups
3. Characteristic implies normal
4. Third isomorphism theorem
5. Nilpotence is quotient-closed

Proof

The proof of the result is exactly the same as for finite groups. The main tricky first step is to show that there exists a subgroup that is Maximal among abelian characteristic subgroups (?), and such that the quotient is a nilpotent group. We do this first step in a separate subsection.

There exists a subgroup maximal among abelian characteristic subgroups with a nilpotent quotient group

Given: A group $G$ with an abelian normal subgroup $N$ such that $G/N$ is nilpotent.

To prove: There exists a subgroup $K$ of $G$ such that $G/K$ is nilpotent and $K$ is maximal among abelian characteristic subgroups.

Proof:

1. There exists an abelian characteristic subgroup $L$ of $G$ such that $G/L$ is nilpotent: This is a consequence of fact (1).
2. $L$ is contained in a subgroup $K$ that is maximal among abelian characteristic subgroups: This follows from fact (2).
3. $K$ is abelian characteristic and $G/K$ is abelian: By fact (3), both $K$ and $L$ are normal in $G$, and by fact (4), we have $(G/L)/(K/L)\cong G/K$. Thus, $G/K$ is isomorphic to a quotient of a nilpotent group. By fact (5), $G/K$ is nilpotent.

Construction of the critical subgroup

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