Stable version of Thompson's replacement theorem for abelian subgroups

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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

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<?matH> and define <math>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.