# Stable version of Thompson's replacement theorem for abelian subgroups

## Contents

## Statement

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

Let denote the set of all abelian subgroups of maximum order in (i.e., for all abelian subgroups of ).

Suppose and is an abelian subgroup of such that normalizes . Then, there exists an abelian subgroup of such that:

- .
- and and normalize each other.
- is contained in .
- is contained in , i.e., it is contained in the normal closure of in . In particular, it is contained in the normal closure of in .

## 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

## 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 . That the iteration must terminate follows from the fact that the intersection with keeps getting bigger.

### Proof details

**Given**: and is an abelian subgroup of such that normalizes .

**To prove**: There exists an abelian normal subgroup of such that:

- .
- and and normalize each other.
- is contained in .
- is contained in , i.e., it is contained in the normal closure of in . In particular, it is contained in the normal closure of in .

**Proof**: Define and define as follows for : if is normal in , then . Otherwise, 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 , we have the following:

- .
- and in particular normalizes .
- For , either or properly contains .
- is contained in , i.e., it is contained in the normal closure of in . In particular, it is contained in the normal closure of in .

**Proof by induction**:

**Base case**: For the case , 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 . We want to show that it is true for . We first need to establish that is well defined and satisfies the conditions.

In the case that normalizes , we define , so all the conditions follow immediately.

In the case that does not normalize , we have, by conditions (1) and (2) on , that with normalizing but not normalizing . We can thus apply Fact (1) to and . We now establish the condition:

- : Direct from Fact (1).
- and in particular normalizes : By Fact (1), . Inductively, by condition (2) on , , so . Since normalizes , so does , hence so does .
- properly contains : This follows directly from Fact (1).
- is contained in : By Fact (1), . In turn, . Combining, we get that is contained in .

### Proving the condition on the stable end

First, we note that the sequence is a sequence of elements of such that the intersections 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 denote the stable member.

Conditions (1), (2) and (4) for 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.