Thompson's replacement theorem for elementary abelian subgroups

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This article defines a replacement theorem
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This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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Statement

For an odd prime

Suppose S is a group of prime power order for an odd prime p.

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

Suppose A \in \mathcal{A}_e(S) and B is an abelian subgroup such that A normalizes B but B does not normalize A. Then, there exists an elementary abelian subgroup A^* of AB such that:

  • |A^*| = |A|, so in particular, A^* \in \mathcal{A}_e(S).
  • A \cap B is a proper subgroup of A^* \cap B.
  • A^* normalizes A.

For the prime 2

A slight modification works for the prime 2, but we have to drop the requirement that A^* be elementary abelian and instead only have A^* abelian but of size at least that of A:

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

Suppose A \in \mathcal{A}_e(S) and B is an abelian subgroup such that A normalizes B but B does not normalize A. Then, there exists an abelian subgroup A^* of AB such that:

  • |A^*| \ge |A|.
  • A \cap B is a proper subgroup of A^* \cap B.
  • A^* normalizes A.

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