Join of abelian subgroups of maximum order in Sylow subgroup is characteristic in every p-subgroup containing it

Statement
Suppose $$G$$ is a finite group, $$P$$ is a $$p$$-Sylow subgroup, and $$J(P)$$ is the subgroup of $$P$$ generated by all the abelian subgroups of maximum order in $$P$$. Then, if $$Q$$ is a $$p$$-subgroup of $$G$$ containing $$J(P)$$, $$J(Q) = Q$$. In particular, $$J(Q)$$ is a fact about::characteristic subgroup of $$Q$$.

Facts used

 * 1) uses::Sylow implies order-dominating

Proof
Given: A finite group $$G$$, a prime $$p$$. $$P$$ is a $$p$$-Sylow subgroup. $$J(P)$$ is the subgroup of $$P$$ generated by all the Abelian $$p$$-subgroups of maximum order in $$P$$. $$Q$$ is a $$p$$-subgroup of $$G$$ containing $$J(P)$$.

To prove: $$J(P) = J(Q)$$.

Proof: Let $$\mathcal{A}(P)$$ denote the set of abelian subgroups of maximum order in $$P$$.


 * 1) By fact (1), there exists a conjugate subgroup, say $$S$$, to $$P$$ in $$G$$, such that $$Q \le S$$. In particular, $$J(P) \le Q \le S$$, so $$J(P) \le S$$.
 * 2) $$J(P) = J(S)$$: Since $$P$$ and $$S$$ are conjugate, they are in particular isomorphic. Thus, the maximum possible order of an abelian subgroup in $$P$$ is the same as the maximum possible order of an abelian subgroup in $$S$$. All abelian subgroups of maximum order in $$P$$ are in $$J(P)$$, hence in $$S$$. Thus, $$\mathcal{A}(P) \subseteq \mathcal{A}(S)$$, so $$J(P) \le J(S)$$. On the other hand, since $$P \cong S$$, the orders of $$J(P)$$ and $$J(S)$$ are equal, so $$J(P) = J(S)$$.
 * 3) $$J(P) = J(S) = J(Q)$$: We thus have $$J(P) = J(S) \le Q \le S$$. Since $$Q \le S$$, the maximum possible order of an abelian subgroup of $$Q$$ is at most equal to the maximum possible order of an abelian subgroup of $$S$$. Since $$J(S) \le Q$$, there do exist subgroups of $$Q$$ whose order equals the maximum possible order of an abelian subgroup of $$S$$. Thus, the maximum possible order of an abelian subgroup of $$Q$$ equals the maximum possible order of an abelian subgroup of $$S$$. In particular, $$\mathcal{A}(Q) \subseteq \mathcal{A}(S)$$. Conversely, since $$J(S) \le Q$$, we have $$\mathcal{A}(S) \subseteq \mathcal{A}(Q)$$, so $$\mathcal{A}(Q) = \mathcal{A}(S)$$, and thus, $$J(Q) = J(S)$$.