# ZJ-subgroup contains D*-subgroup

This article gives the statement, and possibly proof, of the subgroup obtained from one subgroup-defining function, namely D*-subgroup, always being contained in the subgroup obtained from another subgroup-defining function, namely ZJ-subgroup.
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## 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 $P$.

## Definitions used

### 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).

## 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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $D$ is an abelian characteristic subgroup, hence abelian normal subgroup of $P$. Fact (1) $D$ is the D*-subgroup abelianness follows from containment in $\mathcal{D}^*(P)$, and characteristic follows from it being the unique maximal element.
2 The product $BD$ is a subgroup of $P$ that is metabelian. Step (1) [SHOW MORE]
3 $B$ is an abelian subgroup of maximum order in $BD$. $B$ is abelian of maximum order in $P$ Given-direct
4 There exists a subgroup $C$ of $BD$ that is abelian and normalized by $D$, is contained in the normal closure $B^{BD}$, and has the same order as $B$ (and hence maximum order among abelian subgroups of $BD$). In particular, $D$ normalizes $C$. Fact (2) Apply Fact (2) in the ambient group $BD$.
5 $CD$ has nilpotency class at most two. Fact (3) Steps (1), (4) Apply Fact (3) and Steps (1) and (4) inside the ambient group $CD$. Note here that $C$ is not necessarily normal, but since it is normalized by $D$, it is normal in the join $CD$, which is the group within which we apply Fact (3).
6 For any $x \in C$, the subgroup $\langle D , x \rangle$ is a subgroup of $P$ of class at most two. Fact (4) Step (5) Fact-step combination direct
7 For any $x \in C$, $x$ centralizes $D$. $D = D^*(P)$ and hence $D \in \mathcal{D}^*(P)$. Step (6) Apply Step (6) and the condition that $D \in \mathcal{D}^*(P)$ to get the result.
8 The product $CD$ is an abelian subgroup of $P$ Steps (1), (4), (7) By Steps (1), (4), both are abelian. By Step (7), they centralize each other. Hence their product is also abelian.
9 $CD = C$, so $D \le C$. Steps (4), (8) $CD$ is a subgroup of $BD$ that contains $C$. By Step (4), $C$ is abelian of maximum order in $BD$. This forces $CD = C$.
10 $D \le B^{BD} = B^D$. Steps (4), (9) $D \le C$, and by Step (4), $C \le B^{BD}$.
11 $BD = B^{BD}$, so $B$ is a contranormal subgroup of $BD$. Step (10) By the previous step, $BD \le B^{BD}$. The other direction, $B^{BD} \le BD$, is definitional.
12 $B = BD$, so $D \le B$. Fact (5) we are working with $p$-groups Step (11) Step-fact-given combination direct
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