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

Statement
Suppose $$p$$ is a prime number and $$G$$ is an infinite fact about::p-group that is fact about::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) $$\Phi(H) \le Z(H)$$, viz., the Frattini subgroup is contained inside the center (i.e., $$H$$ is a fact about::Frattini-in-center group).
 * 2) $$[G,H] \le Z(H)$$ (i.e., $$H$$ is a fact about::commutator-in-center subgroup of $$G$$).
 * 3) $$C_G(H)= Z(H)$$ (i.e., $$H$$ is a fact about::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) uses::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) uses::Every abelian characteristic subgroup is contained in a maximal among abelian characteristic subgroups
 * 3) uses::Characteristic implies normal
 * 4) uses::Third isomorphism theorem
 * 5) uses::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 fact about::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.