# 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 $P$.

## Facts used

1. Thompson's replacement theorem for abelian subgroups

## Proof

### 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 $B and define [itex]A_n = A_{n-1}^*$. 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.