Stable version of Thompson's replacement theorem for abelian subgroups

Statement
Suppose $$P$$ is a group of prime power order (we'll call the prime $$p$$).

Let $$\mathcal{A}(P)$$ denote the set of all abelian subgroups of maximum order in $$P$$ (i.e., $$|A| \ge |B|$$ for all abelian subgroups $$B$$ of $$P$$).

Suppose $$A \in \mathcal{A}(P)$$ and $$B$$ is an abelian subgroup of $$P$$ such that $$A$$ normalizes $$B$$. Then, there exists an abelian subgroup $$A_\infty$$ of $$P$$ such that:


 * 1) $$A_\infty \in \mathcal{A}(P)$$.
 * 2) $$A_\infty \le AB$$ and $$A_\infty$$ and $$B$$ normalize each other.
 * 3) $$A \cap B$$ is contained in $$A_1 \cap B$$.
 * 4) $$A_\infty$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$, i.e., it is contained in the normal closure of $$A$$ in $$AB$$. In particular, it is contained in the normal closure $$\langle A^P \rangle$$ of $$A$$ in $$P$$.

Related facts

 * Thompson's replacement theorem for abelian subgroups
 * Gillam's abelian-to-normal replacement theorem for abelian subgroups
 * Any abelian normal subgroup normalizes an abelian subgroup of maximum order

Facts used

 * 1) uses::Thompson's replacement theorem for abelian subgroups

Proof idea
The idea behind the proof is to apply Fact (1) and iterate the replacement operation until we get to a group that is normalized by $$B$$. That the iteration must terminate follows from the fact that the intersection with $$B$$ keeps getting bigger.

Proof details
Given: $$A \in \mathcal{A}(P)$$ and $$B$$ is an abelian subgroup of $$P$$ such that $$A$$ normalizes $$B$$.

To prove: There exists an abelian normal subgroup $$A_\infty$$ of $$P$$ such that:


 * 1) $$A_\infty \in \mathcal{A}(P)$$.
 * 2) $$A_\infty \le AB$$ and $$A_\infty$$ and $$B$$ normalize each other.
 * 3) $$A \cap B$$ is contained in $$A_\infty \cap B$$.
 * 4) $$A_\infty$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$, i.e., it is contained in the normal closure of $$A$$ in $$AB$$. In particular, it is contained in the normal closure $$\langle A^P \rangle$$ of $$A$$ in $$P$$.

Proof: Define $$A_1 = A$$ and define $$A_n$$ as follows for $$n > 1$$: if $$A_{n-1}$$ is normal in $$AB$$, then $$A_n = A_{n-1}$$. Otherwise, $$A_{n+1} = A_n^*$$ using the $${}^*$$ operation defined by Fact (1). In order to show that this iteration is feasible, we need to use induction.

Inductive operation of proof
Statement being proved by induction: For all $$n \in \mathbb{N}$$, we have the following:


 * 1) $$A_n \in \mathcal{A}(P)$$.
 * 2) $$A_n \le AB$$ and in particular $$A_n$$ normalizes $$B$$.
 * 3) For $$n > 1$$, either $$A_n = A_{n-1}$$ or $$A_n \cap B$$ properly contains $$A_{n-1} \cap B$$.
 * 4) $$A_n$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$, i.e., it is contained in the normal closure of $$A$$ in $$AB$$. In particular, it is contained in the normal closure $$\langle A^P \rangle$$ of $$A$$ in $$P$$.

Proof by induction:

Base case: For the case $$n = 1$$, conditions (1) and (2) follow from the given data. Condition (3) does not apply and condition (4) is direct.

Inductive step: Suppose the result is true for $$n - 1$$. We want to show that it is true for $$n$$. We first need to establish that $$A_n$$ is well defined and satisfies the conditions.

In the case that $$B$$ normalizes $$A_{n-1}$$, we define $$A_n = A_{n-1}$$, so all the conditions follow immediately.

In the case that $$B$$ does not normalize $$A_{n-1}$$, we have, by conditions (1) and (2) on $$A_{n-1}$$, that $$A_{n-1} \in \mathcal{A}(P)$$ with $$A_{n-1}$$ normalizing $$B$$ but $$B$$ not normalizing $$A_{n-1}$$. We can thus apply Fact (1) to $$A_{n-1}$$ and B<?matH> and define $$A_n = A_{n-1}^*$$. We now establish the condition:


 * 1) $$A_n \in \mathcal{A}(P)$$: Direct from Fact (1).
 * 2) $$A_n \le AB$$ and in particular $$A_n$$ normalizes $$B$$: By Fact (1), $$A_n \le A_{n-1}B$$. Inductively, by condition (2) on $$A_{n-1}$$, $$A_{n-1} \le AB$$, so $$A_n \le AB$$. Since $$A$$ normalizes $$B$$, so does $$AB$$, hence so does $$A_n$$.
 * 3) $$A_n \cap B$$ properly contains $$A_{n-1} \cap B$$: This follows directly from Fact (1).
 * 4) $$A_n$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$: By Fact (1), $$A_n \le \langle A_{n-1}^B \rangle$$. In turn, $$A_{n-1} \le \langle A^B \rangle$$. Combining, we get that $$A_n$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$.

Proving the condition on the stable end
First, we note that the sequence $$A_1, A_2, \dots, $$ is a sequence of elements of $$\mathcal{A}(P)$$ such that the intersections $$A_i \cap B$$ keep growing as long as the sequence does not stabilize. By the finiteness of the groups involved, the sequence must thus stabilize after finitely many steps. Let $$A_\infty$$ denote the stable member.

Conditions (1), (2) and (4) for $$A_\infty$$ follow immediately from the corresponding induction conditions. Condition (3) follows from the induction condition and the observation that a chain of inclusions gives an inclusion.