ZJ-subgroup contains D*-subgroup

From Groupprops
Jump to: navigation, search
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.
View other such results

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).

Facts used

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

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
This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format