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

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) uses::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 $$A$$.