ZJ-subgroup contains D*-subgroup

Statement
Suppose $$p$$ is a prime number and $$P$$ is a finite p-group. Then, the ZJ-subgroup of $$P$$ contains the D*-subgroup of $$P$$.

D*-subgroup
Denote by $$\mathcal{D}^*(P)$$ the set:

$$\mathcal{D}^*(P) = \{ A \le P \mid A \mbox{ is abelian and }\operatorname{class}(\langle A,x \rangle ) \le 2 \implies x \in C_P(A) \ \forall \ x \in P \}$$

ZJ-subgroup
Denote by $$\mathcal{A}(P)$$ the set of abelian subgroups of maximum order in $$P$$. Then, the ZJ-subgroup $$ZJ(P)$$ is defined as the intersection of all members of $$\mathcal{A}(P)$$. Equivalently, it is the center of the join of all these subgroups (see join of abelian subgroups of maximum order).

Facts used

 * 1) uses::Characteristic implies normal
 * 2) uses::Stable version of Thompson's replacement theorem for abelian subgroups
 * 3) uses::Group generated by finitely many abelian normal subgroups is nilpotent of class at most equal to the number of subgroups
 * 4) uses::Nilpotency of fixed class is subgroup-closed
 * 5) uses::Nilpotent implies no proper contranormal subgroup and uses::prime power order implies nilpotent

Proof
It suffices to show that $$D$$ is contained in every abelian subgroup of maximum order in $$P$$, because $$ZJ(P)$$ is the intersection of all these.

Given: A finite $$p$$-group $$P$$, $$D = D^*(P)$$, $$B$$ is an abelian subgroup of maximum order in $$P$$.

To prove: $$D \le B$$

Proof: