Groups of Prime Power Order. Volume 5

Yakov G. Berkovich, Zvonimir Janko

• Verlag: De Gruyter
• Jahr: 2016
• ISBN:
  9783110389043
• Auflage: 1
• Thema: Mathématiques
• Thema: Théorie des groupes

This is the fifth volume of a comprehensive and elementary treatment of finite p -group theory. Topics covered in this volume include theory of linear algebras and Lie algebras.

The book contains many dozens of original exercises (with difficult exercises being solved) and a list of about 900 research problems and themes.

§ 225 Nonabelianp-groups in which anys(a fixed s∈{3,...,p +1}) pairwise noncommuting elements generate a group of maximal class

The nonabelian 2-groups G all of whose two-generator subgroups are of maximal class, have classified. Indeed, if M ≤ G is minimal nonabelian, then it is of maximal class so of order 8. Now the classification of such 2-groups follows from Theorem 90.1. For an independent proof, see Appendix 96. For p > 2, see [GMS1].

Any nonabelian p-group G contains p. 1 ≥ 3 pairwise noncommuting elements. Indeed, let M ≤ G be minimal nonabelian and let L1,..., Lp+1 be all maximal subgroups of M. Take, for each i, xi ∈ Li − Φ (G). Then the elements x1,..., xp+1 are pairwise noncommuting since 〈xi, xj〉= M is nonabelian for i≠ j.

Let s ∈ {3,..., p. 1} be fixed. In this section we classify the nonabelian p-groups in which any s pairwise noncommuting elements generate a p-group of maximal class.

Remark 1. Let G be a nonabelian p-group. Assume that {x1, x2,..., xs} be a set of pairwise noncommuting elements of G and 1 < s < p. 1. Then

by Lemma 116.3 (a). Take

Then {y, x1, x2,..., xs} is the set of s+ 1 pairwise noncommuting elements. It follows that any set of s < p. 1 of pairwise non-commuting elements of a (nonabelian) p-group G is a subset of a set of p. 1pairwise noncommuting elements of G.

Theorem 225.1. Let s be a fixed member of the set {3,..., p. 1} and p > 2. If any s pairwise noncommuting elements of a nonabelian p-group G generate a p-group of maximal class, then G is also of maximal class with abelian subgroup of index p.

Proof. (a) We claim that G is of maximal class. Let A ≤ G be a minimal nonabelian subgroup of the least order. If A. G, thenany s pairwise noncommuting elements of G generate G (see the second paragraph of the section) so G is of maximal class. Next we assume that A < G. As d(A) = 2 and s > 2, there are pairwise noncommuting a1,..., as−1 ∈ A that generate A. Let A < B ≤ G, where | B: A| = p. Then d(B) ≤ d(A) + 1 = 3. By Remark 1, there is as ∈ B− A such that elements a1,..., as−1, as are pairwise noncommuting and these elements generate B. By hypothesis, B is of maximal class. Thus, all subgroups of G containing A as a subgroup of index p, areof maximal class. In that case, by Exercise 10.10, G is of maximal class, as required.

(b) It remains to prove that G has an abelian subgroup of index p. One may assume that |G| > p4. Assume that G1, the fundamental subgroup of G (see Theorem 9.6), is nonabelian. By definition, G1 is not of maximal class. Then, by (a), the subgroup G1 contains s pairwise noncommuting elements that do not generate the subgroup of ma...