Groups of Prime Power Order. Volume 6

Yakov G. Berkovich, Zvonimir Janko

• Publisher: De Gruyter
• Year: 2018
• ISBN:
  9783110531008
• Edition: 1
• Topic: Mathematics
• Subtopic: Group Theory
• Index: Mathematics

This is the sixth volume of a comprehensive and elementary treatment of finite group theory. This volume contains many hundreds of original exercises (including solutions for the more difficult ones) and an extended list of about 1000 open problems. The current book is based on Volumes 1–5 and it is suitable for researchers and graduate students working in group theory.

§ 263Nonabelian 2-groups G with CG(x) ≤ H for all H ∈ Γ1 and x ∈ H − Z(G)

In this section we solve Problem 3702 for p = 2 and prove the following result.

Theorem 263.1. Let G be a nonabelian 2-group with CG(x) ≤ H for all H ∈ Γ1 and x ∈ H − Z(G). Then we have Φ(G) = Z(G) and each maximal abelian subgroup of G has order 2|Z(G)|.

Conversely, all such nonabelian 2-groups satisfy the assumptions of our theorem.

Proof. Let G be a nonabelian 2-group with CG(x) ≤ H for all H ∈ Γ1 and x ∈ H − Z(G).

Let H be a fixed maximal subgroup in G. Then Z(G) < H so that Z(G) ≤ Φ(G). For each g ∈ G − H, g2 ∈ Z(G) (indeed, CG(g2) ≰ H) and so by a result of Burnside applied to G/Z(G), we have that H/Z(G) is abelian and g inverts each element of H/Z(G) (indeed, all elements in the set G/Z(G) − H/Z(G) have order 2). Hence each maximal subgroup of G/Z(G) is abelian and so we have either G' ≤ Z(G) (and then cl(G) = 2) or G/Z(G) is minimal nonabelian. Suppose that we have the second case. Since minimal nonabelian (two generator) group G/Z(G) is generated with involutions, we get G/Z(G) ≅ D8 and since g inverts each element of H/Z(G), it follows that H/Z(G) ≅ C4 so that H is abelian and G is not of class 2. But H was an arbitrary maximal subgroup in G so that each maximal subgroup of G is abelian. This implies that |G'| = 2 and so G' ≤ Z(G) and G is of class 2, a contradiction.

We have proved that G is of class 2. Since G/Z(G) is abelian and is generated by involutions, G/Z(G) is elementary abelian and so Φ(G) = ℧1(G) ≤ Z(G). But we have also Z(G) ≤ Φ(G) and so Φ(G) = Z(G).

Suppose that G possesses an abelian subgroup A > Z(G) such that A/Z(G) ≅ E4. Then A < G. Let K be a maximal subgroup of G which does not contain A. Then (A ∩ K) > Z(G) and if x ∈ (A ∩ K) − Z(G), then CG(x) ≰ K since A ≤ CG(a), a contradiction. We have proved that each maximal abelian subgroup in G is of order 2|Z(G)|. The theorem is proved since the converse is clear. $\square$

§ 264Nonabelian 2-groups of exponent ≥ 16 all of whose minimal nonabelian subgroups, except one, have order 8

In § 90 we have determined the nonabelian 2-groups all of whose minimal nonabelian subgroups are of order 8, i.e., they are isomorphic to D8 or Q8. According to an idea of the first author, we study here such 2-groups G...