Any class two normal subgroup whose derived subgroup is in the ZJ-subgroup normalizes an abelian subgroup of maximum order

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

Suppose $p$ is an odd prime, $P$ is a finite $p$-group, and $B$ is a class two normal subgroup of $P$ such that its commutator subgroup $[B,B]$ is contained in the ZJ-subgroup $Z(J(P))$. Then, there exists an abelian subgroup of maximum order $A$ of $P$ such that $B$ normalizes $A$.

Related facts

• Glauberman's replacement theorem
• Thompson's replacement theorem
• Any abelian normal subgroup normalizes an abelian subgroup of maximum order: A corollary to Thompson's replacement theorem in much the same way as this result is a corollary to Glauberman's replacement theorem.

For more replacement theorems, refer Category:Replacement theorems.

Applications

• Glauberman's theorem on intersection with the ZJ-subgroup

Facts used

1. Glauberman's replacement theorem

Proof

Given: An odd prime $p$, a finite $p$-group $P$, a class two normal subgroup $B$ of $P$ such that $[B,B] \le Z(J(P))$.

To prove: There exists an abelian subgroup $A$ of maximum order in $P$ such that $B$ normalizes $A$.

Proof: Let $\mathcal{A}(P)$ be the set of abelian subgroups of maximum order in $P$. Let $A$ be a member of $\mathcal{A}(P)$ such that $A \cap B$ has maximum order.

If $B$ normalizes $A$, we are done. Otherwise, fact (1) guarantees that there exists $A^* \in \mathcal{A}(P)$ such that $A \cap B$ is a proper subgroup of $A^* \cap B$. This contradicts the choice of $A$ as the subgroup for which $A \cap B$ has maximal order, so $B$ must normalize $A$.

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 278, Theorem 2.9, Chapter 8 (p-constrained and p-stable groups), Section 2 (Glauberman's theorem), ^{More info}