Thompson's replacement theorem for elementary abelian subgroups

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., $|A|\geq |B|$ 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., $|A|\geq |B|$ 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:

• $|A^{*}|\geq |A|$.
• $A\cap B$ is a proper subgroup of $A^{*}\cap B$.
• $A^{*}$ normalizes $A$.

Related facts

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