Thompson's replacement theorem for elementary abelian subgroups

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 $$A$$.

Related facts

 * Thompson's replacement theorem for abelian subgroups
 * Glauberman's replacement theorem