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On intersections of two non-incident subgroups of finite p-groups

  • Jiao Wang EMAIL logo
Published/Copyright: August 28, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

In this paper, we investigate finite p-groups G such that whenever A , B < G are non-incident, then A ∩ B ⊴ ⟨ A , B ⟩ . This partially solves a problem proposed by Y. Berkovich.

Keywords: finite p-group; metacyclic p-group; commutator subgroup
MSC 2010: 20D15

1 Introduction

All groups considered in this paper are finite.

Let G be a group. If H and K are normal subgroups of G , then H ∩ K ⊴ H and H ∩ K ⊴ K . It is interesting to investigate the structure of a group G by using the intersection of two subgroups of finite p-groups. For example, Janko in [1] determined completely the structure of 2-groups G in which any two distinct maximal abelian subgroups have cyclic intersection. Janko in [2] also classified non-abelian 2-groups such that any two distinct minimal non-abelian subgroups have cyclic intersection. We also remark that many authors investigate the structure of groups by using information about the intersection of the subgroups, for example [3,4].

In fact, it is easy to see that G is a Dedekind group if H ∩ K ⊴ H and H ∩ K ⊴ K for any two subgroups H and K of G . Now it is natural to ask the following question:

What can be said about the structure of a p-group G in which H ∩ K ⊴ H and H ∩ K ⊴ K for any two non-incident subgroups H and K of G ?

The question above was first posed by Berkovich (see Problem 1342 in [5] and Problem 1546 in [6]). In this paper, we hope to investigate the structure of a p-group G in which H ∩ K ⊴ H and H ∩ K ⊴ K for any two non-incident subgroups H and K of G (this means that H ≰ K and K ≰ H ). For convenience, we call this kind of p-groups Q -p-groups. In this paper, we get the following main results:

Main result 1.1

Suppose that G is a non-abelian 2-group of order 2 n and d ( G ) = 2 . Then G is a Q -2-group if and only if G is one of the following groups:

  1. G is a minimal non-abelian 2-group;

  2. 2-groups of maximal class;

  3. ⟨ a , b ∣ a 8 = b 2 n − 3 = 1 , [ a , b ] = b 2 ⟩ , n ≥ 6 ;

  4. ⟨ a , b ∣ a 8 = b 2 n − 3 = 1 , [ a , b ] = b 2 n − 4 + 2 ⟩ , n ≥ 6 ;

  5. ⟨ a , b ∣ a 4 = b 2 n − 2 = 1 , [ a , b ] = b 2 ⟩ , n ≥ 5 ;

  6. ⟨ a , b ∣ a 4 = b 2 n − 2 = 1 , [ a , b ] = b 2 n − 3 + 2 ⟩ , n ≥ 5 ;

  7. ⟨ a , b ∣ a 8 = b 2 n − 2 = 1 , a 4 = b 2 n − 3 , [ a , b ] = b 2 ⟩ , n ≥ 5 ;

  8. ⟨ a , b , c ∣ a 8 = c 2 = 1 , a 4 = b 2 , [ a , b ] = c , [ c , a ] = a 4 , [ c , b ] = 1 ⟩ ;

  9. ⟨ a , b , c ∣ a 8 = 1 , a 4 = b 2 = c 2 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ ;

  10. ⟨ a , b , c ∣ a 2 n − 2 = b 2 = 1 , c 2 = a 2 n − 3 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ .

Main result 1.2

Let G be a Q -2-group and d ( G ) ≥ 3 . Then exp ( G ′ ) ≤ 2 and c ( G ) ≤ 2 .

Main result 1.3

Let G be a Q -2-group with d ( G ) ≥ 4 and ∣ G ∣ ≥ 2 7 . Then G is Dedekind.

Main result 1.4

Let G be a Q -p-group and p > 2 . Then exp ( G ′ ) ≤ p and c ( G ) ≤ 3 .

Main result 1.5

Let G be a Q -p-group with d ( G ) ≥ 4 and p > 2 . Then G is abelian.

2 Preliminaries

First we introduce some notions and notations.

Let G be a group. If every subgroup of G is normal, then G is called a Dedekind group.

Let G be a p-group. We use D 2 n , Q 2 n and S D 2 n to denote the dihedral group, the generalized quaternion group and the semi-dihedral group of order 2 n , respectively. We also use d ( G ) and C p m to denote the minimal number of generators of G and the cyclic group of order p m . Let G n ( G ) = ⟨ [ g 1 , g 2 , … , g n ] ∣ g i ∈ G ⟩ . Then c ( G ) denotes the nilpotency class of G . If H and K are groups, then H ∗ K denotes a central product of H and K . H G = ⋂ x ∈ G x − 1 H x is the normal core of the subgroup H in G . M ⋖ G means M is a maximal subgroup of G . For other notations and terminologies the reader is referred to [7].

Now we list some known results which will be used later.

Lemma 2.1

[8, Section 36, Remark 1] Let G be a p-group. If there is a normal subgroup R < G ′ such that G / R is metacyclic, then G is also metacyclic.

Lemma 2.2

[8, Section 1, Exercise 8a] Let G be a minimal non-abelian p-group. Then G is one of the following groups:

  1. Q 8 ;

  2. M p ( n , m ) = ⟨ a , b ∣ a p n = b p m = 1 , [ a , b ] = a p n − 1 ⟩ is metacyclic with n ≥ 2 and m ≥ 1 ;

  3. M p ( n , m , 1 ) = ⟨ a , b , c ∣ a p n = b p m = c p = 1 , [ a , b ] = c , [ c , a ] = [ c , b ] = 1 ⟩ is non-metacyclic with n ≥ m ≥ 1 and m + n ≥ 3 when p = 2 .

Lemma 2.3

[8, Section Appendix 1, Theorem A.1.4] Let G be a p-group and x , y ∈ G .

  1. ( x y ) p ≡ x p y p (mod ℧ 1 ( G ′ ) G p ).

  2. [ x p , y ] ≡ [ x , y ] p (mod ℧ 1 ( N ′ ) N p ), where N = ⟨ x , [ x , y ] ⟩ .

Lemma 2.4

[8, Section 1, Proposition 1.23] If a p-group G is non-Dedekind, then there exists a normal subgroup K of index p in G ′ such that G / K is non-Dedekind.

Lemma 2.5

[9, Lemma 2.2] Suppose that G is a finite non-abelian p-group. Then the following conditions are equivalent:

  1. G is minimal non-abelian;

  2. d ( G ) = 2 and ∣ G ′ ∣ = p ;

  3. d ( G ) = 2 and Φ ( G ) = Z ( G ) .

Lemma 2.6

[10, Theorem 2] Let G be a metacyclic 2-group. Then G has one presentation of the following three kinds:

  1. G has a cyclic maximal subgroup.

  2. Ordinary metacyclic 2-groups G = ⟨ a , b ∣ a 2 r + s + u = 1 , b 2 r + s + t = a 2 r + s , a b = a 1 + 2 r ⟩ , where r , s , t , u are non-negative integers with r ≥ 2 and u ≤ r .

  3. Exceptional metacyclic 2-groups G = ⟨ a , b ∣ a 2 r + s + v + t ′ + u = 1 , b 2 r + s + t = a 2 r + s + v + t ′ , a b = a − 1 + 2 r + v ⟩ , where r , s , v , t , t ′ , u are non-negative integers with r ≥ 2 , t ′ ≤ r , u ≤ 1 , t t ′ = s v = t v = 0 , and if t ′ ≥ r − 1 , then u = 0 .

Groups of different types or of the same type but with different values of parameters are not isomorphic to each other.

Lemma 2.7

[11, Theorem 3] Let G be a 3-group in which ∣ ⟨ x ⟩ / ⟨ x ⟩ G ∣ ≤ 3 , for any x ∈ G . Then c ( G ) ≤ 4 if d ( G ) = 2 and c ( G ) ≤ 5 if d ( G ) > 2 .

Lemma 2.8

[11, Theorem 6] Let G be a 3-group in which ∣ ⟨ x ⟩ / ⟨ x ⟩ G ∣ ≤ 3 , for any x ∈ G . If d ( G ) = 2 , then exp ( G ′ ) ≤ 9 and exp ( G 3 ) ≤ 3 .

Lemma 2.9

[12, Corollary 2] Let G be a p-group with G ≠ 1 and p > 2 . Then d ( G ) ≤ log p ∣ Ω 1 ( G ) ∣ .

3 Some properties of Q -p-groups

In this section, we discuss the properties of Q -p-groups which will be used later.

Lemma 3.1

Let G be a Q -p-group. If H is a subgroup of G and N is a normal subgroup of G , then H and G / N are Q -p-groups.

Proof

Let A and B be subgroups of H and A ≰ B , B ≰ A . Then, by the hypotheses of the lemma, A ∩ B ⊴ ⟨ A , B ⟩ .

Set G ¯ = G / N . Let K ¯ = K / N and L ¯ = L / N be subgroups of G ¯ and K ¯ ≰ L ¯ , L ¯ ≰ K ¯ . Then K ≰ L , L ≰ K . By the hypotheses, we see K ∩ L ¯ ⊴ ⟨ K , L ⟩ ¯ . It follows that K ¯ ∩ L ¯ ⊴ ⟨ K ¯ , L ¯ ⟩ . Thus, the lemma is true.□

Lemma 3.2

Let G be a Q -p-group. Then, for any x ∈ G , ⟨ x p ⟩ ⊴ G .

Proof

Without loss of generality, we may assume x ∉ Z ( G ) and o ( x ) > p . Take y ∈ G . If x ∈ ⟨ y ⟩ or y ∈ ⟨ x ⟩ , then [ x , y ] = 1 and so y ∈ C G ( x ) ≤ C G ( x p ) . If y ∉ ⟨ x ⟩ , then ⟨ y , x p ⟩ ≰ ⟨ x ⟩ . If x ∉ ⟨ y ⟩ , then ⟨ x ⟩ ≰ ⟨ y , x p ⟩ . Otherwise, we see x p ∈ Φ ( ⟨ y , x p ⟩ ) and so ⟨ y , x p ⟩ = ⟨ y ⟩ , a contradiction. So ⟨ y , x p ⟩ and ⟨ x ⟩ are two non-incident subgroups of G . Since G is a Q -p-group, we see y ∈ N G ( x p ) . So ⟨ x p ⟩ ⊴ G .□

Lemma 3.3

Let G be a Q -p-group. Then, for any a ∈ G ⧹ Φ ( G ) , b ∈ G ⧹ ⟨ a , Φ ( G ) ⟩ , we see ⟨ a , b ⟩ ⊴ G .

Proof

Without loss of generality, we may assume d ( G ) = 3 and c ∈ G ⧹ ⟨ a , b ⟩ . Let A = ⟨ a , b ⟩ , B = ⟨ b , c ⟩ and C = ⟨ a , c ⟩ . Then b ∈ A ∩ B ⊴ G . By the hypotheses, [ b , a ] , [ b , c ] ∈ A ∩ B . Similarly, [ a , c ] ∈ A ∩ C ⊴ G . It follows that G ′ ≤ A and the lemma is true.□

4 Q -p-groups of even order

Theorem 4.1

Suppose that G is a non-abelian metacyclic 2-group of order 2 n . Then G is a Q -2-group if and only if G is one of the following groups:

  1. G is a minimal non-abelian 2-group;

  2. 2-groups of maximal class;

  3. ⟨ a , b ∣ a 8 = b 2 n − 3 = 1 , [ a , b ] = b 2 ⟩ , n ≥ 6 ;

  4. ⟨ a , b ∣ a 8 = b 2 n − 3 = 1 , [ a , b ] = b 2 n − 4 + 2 ⟩ , n ≥ 6 ;

  5. ⟨ a , b ∣ a 4 = b 2 n − 2 = 1 , [ a , b ] = b 2 ⟩ , n ≥ 5 ;

  6. ⟨ a , b ∣ a 4 = b 2 n − 2 = 1 , [ a , b ] = b 2 n − 3 + 2 ⟩ , n ≥ 5 ;

  7. ⟨ a , b ∣ a 8 = b 2 n − 2 = 1 , a 4 = b 2 n − 3 , [ a , b ] = b 2 ⟩ , n ≥ 5 .

Proof

Since G is a non-abelian metacyclic 2-group, we see G is one of the groups listed in Lemma 2.6.

If G is a group listed in (1) in Lemma 2.6, then G is of maximal class or G is minimal non-abelian. So G is as given in parts (1) and (2).

If G is a group listed in (2) in Lemma 2.6, then G = ⟨ a , b ∣ a 2 r + s + u = 1 , b 2 r + s + t = a 2 r + s , [ a , b ] = a 2 r ⟩ with r ≥ 2 and u ≤ r . Since [ a , b 2 ] ∈ ⟨ a 2 r + 1 ⟩ , by Lemma 3.2, we see a 2 r + 1 ∈ ⟨ b 2 ⟩ . Let a 1 = a b − 2 t . Then ⟨ a 1 ⟩ ∩ ⟨ b ⟩ = 1 . It is easy to see that ⟨ [ a 1 2 , b ] ⟩ = ⟨ a 2 r + 1 ⟩ . Thus, ⟨ a 2 r + 1 ⟩ ≤ ⟨ a 1 ⟩ ∩ ⟨ b ⟩ = 1 . So ∣ G ′ ∣ = 2 . By Lemma 2.5, G is a minimal non-abelian 2-group.

If G is of type (3) in Lemma 2.6, then G = ⟨ a , b ∣ a 2 r + s + v + t ′ + u = 1 , b 2 r + s + t = a 2 r + s + v + t ′ , [ a , b ] = a − 2 + 2 r + v ⟩ with r ≥ 2 and u ≤ 1 . It follows from [ a , b 2 ] ∈ ⟨ b ⟩ that s + t ′ ≤ 1 and so s + t ′ + u ≤ 2 . Let o ( a ) = 2 m . If s + t ′ + u = 2 , then s + t ′ = u = 1 . Thus, b 2 r + s + t = a 2 m − 1 and [ a , b ] = a − 2 + 2 m − 2 . Let c = b 2 r + s + t − 1 a 2 m − 2 . Then c 2 = 1 . If m > 4 or r + s + t > 2 , then ( b 2 r + s + t − 2 a 2 m − 3 ) 2 = c . If m = 4 and r + s + t = 2 , then ( b a ) 2 = c . By Lemma 3.2, ⟨ c ⟩ ⊴ G , which implies c ∈ Z ( G ) . However, it is impossible. So s + t ′ + u ≤ 1 . Thus, [ a , b ] = a − 2 or a − 2 + 2 m − 1 , b 2 r + s + t = a 2 m − 1 or 1. It is easy to see that ( a b 2 ) 2 = a 2 b 4 and so ⟨ a 2 b 4 ⟩ ⊴ G . Since [ a 2 b 4 , b ] = a − 4 , we see a 4 ∈ ⟨ a 2 b 4 ⟩ . If b 2 r + s + t = 1 , then b 4 = 1 or b 8 = 1 . So G is as given in parts (3)–(6). If b 2 r + s + t = a 2 m − 1 , then b 4 = a 2 m − 1 or b 8 = a 2 m − 1 . If b 8 = a 2 m − 1 , then o ( a 2 m − 2 b 4 ) = 2 . It follows from ( a 2 m − 3 b 2 ) 2 = a 2 m − 2 b 4 that ⟨ a 2 m − 2 b 4 ⟩ ⊴ , which implies a 2 m − 2 b 4 ∈ Z ( G ) . However, [ a 2 m − 2 b 4 , b ] ≠ 1 , a contradiction. So b 4 = a 2 m − 1 and G is given in part ( 7 ) .

Conversely, every group listed in the theorem is a Q -2-group and they are pairwise non-isomorphic.□

Corollary 4.2

Let G be a Q -2-group. Then Φ ( G ) is abelian.

Proof

For any a , b ∈ G , we may assume that H = ⟨ a 2 , b ⟩ is not abelian and o ( a ) = 2 n . By the hypotheses, we see ⟨ a 2 ⟩ ⊴ G and so H is metacyclic. It follows from Theorem 4.1 that [ a 2 , b ] = a 2 n − 1 , a − 4 or a − 4 + 2 n − 1 . Then, it is easy to see that [ a 2 , b 2 ] = 1 , which implies Φ ( G ) is abelian.□

Lemma 4.3

Let G be a Q -2-group and d ( G ) = 2 . If exp ( G ′ ) = 2 , then G is one of the following groups:

  1. G is a minimal non-abelian 2-group;

  2. ⟨ a , b , c ∣ a 8 = c 2 = 1 , a 4 = b 2 , [ a , b ] = c , [ c , a ] = a 4 , [ c , b ] = 1 ⟩ .

Proof

If ∣ G ∣ ≤ 2 4 , then the conclusion holds by checking the list of groups of order 2 3 and 2 4 . Assume ∣ G ∣ ≥ 2 5 . If ∣ G ′ ∣ = 2 , then by Lemma 2.5, G is a minimal non-abelian 2-group. So we may assume ∣ G ′ ∣ > 2 , G is non-metacyclic and G = ⟨ a , b ⟩ with o ( a ) = 2 n , o ( b ) = 2 m , [ a , b ] = c , [ c , a ] = d . Then [ a 2 , b ] = d . By Lemma 3.2, we see d ∈ ⟨ a 2 n − 1 ⟩ ≤ Z ( G ) . Similarly, [ c , b ] ∈ ⟨ b 2 m − 1 ⟩ ≤ Z ( G ) . Let A = ⟨ a , c ⟩ and B = ⟨ b , c ⟩ . Noting that G is a Q -2-group, we see G ′ ≤ A ∩ B . Since Ω 1 ( A ) = ⟨ c ⟩ × ⟨ a 2 n − 1 ⟩ , G ′ = Ω 1 ( B ) = ⟨ c ⟩ × ⟨ a 2 n − 1 ⟩ . So for any x ∈ G ⧹ Φ ( G ) , y ∈ G ⧹ ⟨ Φ ( G ) , x ⟩ , we see G ′ = Ω 1 ( ⟨ c , x ⟩ ) = Ω 1 ( ⟨ c , y ⟩ ) . It follows that ⟨ x ⟩ ∩ ⟨ y ⟩ ≠ 1 . Without loss of generality, we may assume a 2 r = b 2 t , [ c , a ] = a 2 n − 1 and [ c , b ] = 1 , where r ≥ 2 .

If r ≤ t , then, by letting a 1 = a b − 2 t − r , we see ⟨ a 1 ⟩ ∩ ⟨ b ⟩ = 1 . If r > t > 1 or r > t , r > 2 , then, by letting b 1 = a − 2 r − t b , we see ⟨ a ⟩ ∩ ⟨ b 1 ⟩ = 1 . So r = 2 , t = 1 , which implies a 4 = b 2 . If n ≥ 4 , then, by letting b 1 = a − 2 n − 2 − 1 b , we also have ⟨ a ⟩ ∩ ⟨ b 1 ⟩ = 1 . If n = 3 , then o ( a ) = 8 and G is given in part ( 2 ) .□

Lemma 4.4

Let G be a non-metacyclic Q -2-group and d ( G ) = 2 . If G ′ is cyclic with order 4, then G is one of the following groups:

  1. ⟨ a , b , c ∣ a 2 n − 2 = b 2 = 1 , c 2 = a 2 n − 3 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ ;

  2. ⟨ a , b , c ∣ a 8 = 1 , a 4 = b 2 = c 2 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ .

Proof

We may assume G = ⟨ a , b ⟩ , [ b , a ] = c , o ( a ) = 2 m , o ( b ) = 2 n , o ( c ) = 4 , [ c , a ] = 1 , [ c , b ] = 1 or c 2 . Then [ a 2 , b ] = c 2 . By Lemma 3.2, we see c 2 = a 2 m − 1 .

If ⟨ a ⟩ ∩ ⟨ b ⟩ ≠ 1 , then we may assume a 2 r = b 2 t , r ≥ 2 . If r < t , then, by letting a 1 = a b − 2 t − r , we see ⟨ a 1 ⟩ ∩ ⟨ b ⟩ = 1 . If r ≥ t > 2 or r > t = 2 or r ≥ 3 , t = 1 , then, by letting b 1 = a − 2 r − t b , we see ⟨ a ⟩ ∩ ⟨ b 1 ⟩ = 1 .

If r = 2 , t = 1 , then, by letting b 1 = a − 2 b , we see b 1 2 = c 2 = a 2 m − 1 . We may assume m − 1 = 2 and so a 4 = b 2 = c 2 . If [ c , b ] = 1 , then ( b c ) 2 = 1 . However, [ ( b c ) 2 , a ] = c 2 ≠ 1 , a contradiction. So [ c , b ] = c 2 and G is given in part ( 2 ) .

If r = t = 2 , then, by letting b 1 = a − 2 r − t b , we see b 1 4 = c 2 . We may assume a 4 = b 1 4 = c 2 . Let A = ⟨ a 2 c , b 1 2 c ⟩ , B = ⟨ b , a 2 c ⟩ . Then A ∩ B = ⟨ a 2 c ⟩ . However, [ a 2 c , b ] = c 2 ∉ ⟨ a 2 c ⟩ , a contradiction.

Now we may assume ⟨ a ⟩ ∩ ⟨ b ⟩ = 1 . If [ c , b ] = 1 , then [ a , b 2 ] = c 2 and so c 2 ∈ ⟨ b ⟩ ∩ ⟨ a ⟩ = 1 , a contradiction. So [ c , b ] = c 2 . If o ( a ) = 8 , then a 4 = c 2 . Let A = ⟨ a 2 c , b a 2 ⟩ and B = ⟨ b , a 2 ⟩ . Then A ≰ B and B ≰ A . However, A ∩ B = ⟨ b a 2 ⟩ ⋬ B , a contradiction. So o ( a ) ≥ 16 . Let A = ⟨ a 2 n − 2 c , b 2 ⟩ and B = ⟨ a 2 n − 2 c , a 2 b ⟩ . If o ( b ) > 2 , then A ≰ B and B ≰ A . However, A ∩ B ⋬ ⟨ A , B ⟩ , a contradiction. So b 2 = 1 and we get the group in part ( 1 ) .□

According to Lemma 4.4, we have the following result:

Corollary 4.5

Let G be a Q -2-group and d ( G ) = 2 . If c ( G ) = 2 , then exp ( G ′ ) = 2 .

Lemma 4.6

Let G be a non-metacyclic Q -2-group and d ( G ) = 2 . Then ∣ G ′ ∣ ≤ 4 .

Proof

Otherwise, ∣ G ′ ∣ ≥ 8 . Assume ∣ G ∣ = 2 n . Then n ≥ 6 . Let N ≤ G ′ such that ∣ N ∣ = 2 and N ⊴ G . Then, by Lemmas 2.1 and 3.1, G / N is a non-metacyclic Q -2-group. By induction, ∣ G ′ / N ∣ ≤ 4 and so ∣ G ′ ∣ = 8 . Set G ¯ = G / N and N = ⟨ x ⟩ . Then G ¯ ≅ ⟨ a ¯ , b ¯ , c ¯ ∣ a ¯ 2 n − 3 = b ¯ 2 = 1 ¯ , c ¯ 2 = a ¯ 2 n − 4 , [ b ¯ , a ¯ ] = c ¯ , [ c ¯ , b ¯ ] = c ¯ 2 , [ c ¯ , a ¯ ] = 1 ¯ ⟩ by Lemma 4.4.

If G ′ is cyclic, then c 4 = a 2 n − 3 = x and ⟨ c 2 ⟩ = ⟨ a 2 n − 4 ⟩ . It follows from [ b 2 , a ] = 1 that [ c , b ] = c − 2 and therefore [ c 2 , b ] = c 4 , which implies [ a 2 n − 4 , b ] = a 2 n − 3 . On the other hand, we see ⟨ [ b , a 2 ] ⟩ = ⟨ c 2 ⟩ and so [ b , a 4 ] = c 4 = a 2 n − 3 , which implies n = 6 . Thus, c 2 = a 4 and o ( a ) = 16 . Let A = ⟨ a 2 c , b a 2 ⟩ and B = ⟨ b , a 2 ⟩ . Then A ≰ B and B ≰ A . However, A ∩ B = ⟨ b a 2 ⟩ ⋬ B , a contradiction.

If G ′ is not cyclic, then G ′ = ⟨ c , x ⟩ . c 4 = a 2 n − 3 = 1 . It follows from [ b 2 , a ] = c 2 [ c , b ] = 1 that [ c , b ] = c 2 . Since x ∈ G ′ , we see [ c , a ] = x . Thus, [ ( b a ) 2 , a ] = [ c , a ] = x ∈ ⟨ ( b a ) 2 ⟩ . However, it is impossible.□

According to Lemmas 4.3, 4.4 and 4.6, we have the following results:

Theorem 4.7

Let G be a non-metacyclic Q -2-group, ∣ G ∣ = 2 n and d ( G ) = 2 . Then G is one of the following pairwise non-isomorphic groups:

  1. G is a minimal non-abelian 2-group;

  2. ⟨ a , b , c ∣ a 8 = c 2 = 1 , a 4 = b 2 , [ a , b ] = c , [ c , a ] = a 4 , [ c , b ] = 1 ⟩ ;

  3. ⟨ a , b , c ∣ a 8 = 1 , a 4 = b 2 = c 2 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ ;

  4. ⟨ a , b , c ∣ a 2 n − 2 = b 2 = 1 , c 2 = a 2 n − 3 , [ b , a ] = c , [ c , a ] = 1 , [ c , b ] = c 2 ⟩ .

Lemma 4.8

Let G be a Q -2-group and d ( G ) = 3 . Then G has no subgroup A such that A ≅ ⟨ a , b , x ∣ a 8 = x 2 = 1 , a 4 = b 2 , [ a , b ] = x , [ x , a ] = a 4 , [ x , b ] = 1 ⟩ .

Proof

Otherwise, we assume A = ⟨ a , b , x ∣ a 8 = x 2 = 1 , a 4 = b 2 , [ a , b ] = x , [ x , a ] = a 4 , [ x , b ] = 1 ⟩ and A ≤ G . Then [ a , x ] ≠ 1 and [ a 2 , b ] = a 4 ≠ 1 . Since a 2 and x ∈ Φ ( G ) , by Corollary 4.2, we see a ∉ Φ ( G ) and b ∉ Φ ( G ) . Since d ( G ) = 3 , we may assume that there exists c ∈ G ⧹ A such that G = ⟨ a , b , c ⟩ . Let B = ⟨ a , c ⟩ . Then A ∩ B ⊴ G . Since a ∈ A ∩ B , we see ⟨ a ⟩ G ≤ A ∩ B , which implies ⟨ a , x ⟩ ≤ A ∩ B . Noting that ∣ A ∣ = 2 5 and ∣ ⟨ a , x ⟩ ∣ = 2 4 , we see A ∩ B = ⟨ a , x ⟩ . Thus, [ c , a ] , [ c , x ] ∈ ⟨ a , x ⟩ . It follows that [ c , x ] ∈ ⟨ a 4 ⟩ . Since [ a , x ] ≠ 1 , we see B is non-abelian. If [ a , c ] = a i x , then [ a , b c ] ∈ ⟨ a ⟩ . So, without loss of generality, we may assume [ a , c ] ∈ ⟨ a ⟩ , which implies B is metacyclic. By Theorem 4.1, we see C B ( a ) ⋖ B . It follows that B = C B ( a ) ⟨ x ⟩ = ⟨ a , c 2 , x ⟩ = ⟨ a , x ⟩ . Thus, c ∈ ⟨ a , x ⟩ ≤ A , a contradiction.□

Theorem 4.9

Let G be a Q -2-group and d ( G ) ≥ 3 . Then exp ( G ′ ) ≤ 2 .

Proof

Otherwise, exp ( G ′ ) > 2 . Let N ≤ G ′ such that ∣ N ∣ = 2 and N ⊴ G . By Lemma 3.1, exp ( G ′ / N ) ≤ 2 and so exp ( G ′ ) = 4 . Then there exist a , b ∈ G such that o ( [ a , b ] ) = 4 . We may assume d ( G ) = 3 and G = ⟨ a , b , c ⟩ , [ a , b ] = x , [ b , c ] = y . By Corollary 4.5, x ∉ Z ( G ) . We may assume [ b , x ] ≠ 1 . If o ( y ) = 4 , then, by calculation, we see o ( [ b , a c ] ) ≤ 2 . So we may assume o ( y ) ≤ 2 . Let A = ⟨ a , b ⟩ and B = ⟨ b , c ⟩ . By Lemma 3.3, we see B ⊴ G and so x ∈ B , which implies that B is non-abelian. According to Lemmas 4.3 and 4.8, ⟨ b , c ⟩ is minimal non-abelian. It follows from [ b , x ] ≠ 1 that B = ⟨ b , x ⟩ . However, it is impossible.□

Theorem 4.10

Let G be a Q -2-group and d ( G ) ≥ 3 . Then c ( G ) ≤ 2 .

Proof

For any g 1 , g 2 ∈ G , by Lemmas 4.3, 4.8 and 4.9, we see ⟨ g 1 , g 2 ⟩ is abelian or minimal non-abelian. We may assume d ( G ) = 3 , G = ⟨ a , b , c ⟩ and [ a , b ] = x . Let A = ⟨ a , b ⟩ and B = ⟨ b , c ⟩ . By Lemma 3.3, B ⊴ G and so x ∈ Φ ( B ) . According to Lemma 2.5, x ∈ Z ( B ) . So x ∈ Z ( G ) and the theorem is true.□

Lemma 4.11

Let G be a non-Dedekind 2-group with d ( G ) ≥ 4 and ∣ G ′ ∣ = 2 . Then G is a Q -2-group if and only if G is one of the following pairwise non-isomorphic groups:

  1. D 8 ∗ Q 8 ;

  2. M 2 ( 2 , 1 , 1 ) ∗ Q 8 .

Proof

Since ∣ G ′ ∣ = 2 , there exist a , b ∈ G such that ⟨ a , b ⟩ is minimal non-abelian. By Lemma 2.2, ⟨ a , b ⟩ is isomorphic to Q 8 or M 2 ( n , m ) , n ≥ 2 or M 2 ( n , m , 1 ) with n ≥ m ≥ 1 and m + n ≥ 3 . For any x , y ∈ G , since ∣ G ′ ∣ = 2 , we have [ x , y 2 ] = 1 , which implies Φ ( G ) ≤ Z ( G ) . So a ∉ Φ ( G ) and b ∉ Φ ( G ) . Assume d ( G ) = t + 2 . Then there exist x 1 , x 2 , … , x t with o ( x i ) = 2 e i , where 1 ≤ i ≤ t , such that G = ⟨ a , b , x 1 , x 2 , … , x t ⟩ . Without loss of generality, we may assume e 1 ≥ e 2 ≥ ⋯ ≥ e t and [ x i , a ] = [ x i , b ] = 1 .

If ⟨ a , b ⟩ ≅ M 2 ( n , m ) , then we may assume a 2 n = b 2 m = 1 , [ a , b ] = a 2 n − 1 , where n ≥ 2 . For any x i , by Lemma 3.3, we see G ′ ≤ ⟨ b , x i ⟩ , where 1 ≤ i ≤ t . It follows that a 2 n − 1 ∈ ⟨ x i ⟩ ∩ ⟨ x j ⟩ ≠ 1 , where 1 ≤ i , j ≤ t and i ≠ j . We may assume x i 2 s = x j 2 k , where s ≥ k ≥ 1 . If s > t ≥ 1 or s = t ≥ 2 , then, by letting x ′ = x i − 2 s − t x j , we see ⟨ x ′ ⟩ ∩ ⟨ x i ⟩ = 1 . So s = k = 2 and x 1 2 = x 2 2 = ( x 1 x 2 ) 2 . Then, for any 1 ≤ i , j ≤ t and i ≠ j , we see ⟨ x i , x j ⟩ ≇ Q 8 . If follows that t = 2 . If n ≥ 3 , then ( x 1 a 2 n − 2 ) 2 = 1 , a contradiction. So n = 2 . If m > 1 , then, by letting A = ⟨ b x 1 , a ⟩ and B = ⟨ b x 1 , a x 2 ⟩ , we see A ∩ B ⋬ A . So m = 1 and G ≅ D 8 ∗ Q 8 .

If ⟨ a , b ⟩ ≅ M 2 ( n , m , 1 ) , then we may assume a 2 n = b 2 m = c 2 = 1 , [ a , b ] = c , [ c , a ] = [ c , b ] = 1 , where n ≥ m ≥ 1 and m + n ≥ 3 . For any x i , by Lemma 3.3, we see G ′ ≤ ⟨ b , x i ⟩ , where 1 ≤ i ≤ t . By the above, we see t = 2 and ⟨ x 1 , x 2 ⟩ ≇ Q 8 . Since n ≥ m ≥ 1 and m + n ≥ 3 , we see n ≥ 2 . Let A = ⟨ a 2 x 1 , a 2 x 2 ⟩ and B = ⟨ a 2 x 1 , b ⟩ . If n ≥ 3 , then A ∩ B = ⟨ a 2 x 1 ⟩ ⋬ A . So n = 2 . Let C = ⟨ a x 1 , b ⟩ and D = ⟨ a x 1 , b x 2 ⟩ . If m = 2 , then C ∩ D = ⟨ a x 1 ⟩ ⋬ C . So m = 2 and G ≅ M 2 ( 2 , 1 , 1 ) ∗ Q 8 .

If ⟨ a , b ⟩ ≅ Q 8 , then a 4 = 1 , a 2 = b 2 , [ a , b ] = a 2 . For any 1 ≤ i , j ≤ t and i ≠ j , by the above, we may assume ⟨ x i , x j ⟩ ≅ Q 8 or ⟨ x i , x j ⟩ is abelian. If ⟨ x i , x j ⟩ ≅ Q 8 , then x i 2 = x j 2 = a 2 . Let A = ⟨ a x 1 , b ⟩ and B = ⟨ a x 1 , b x 2 ⟩ . Then A ∩ B = ⟨ a x 1 ⟩ ⋬ A . So ⟨ x i , x j ⟩ is abelian and we may assume ⟨ x i ⟩ ∩ ⟨ x j ⟩ = 1 . If ⟨ a , b ⟩ ∩ ⟨ x i ⟩ = ⟨ a 2 ⟩ , then ⟨ a x i 2 e i − 2 , b x i 2 e i − 2 ⟩ ≅ M 2 ( 2 , 1 ) . Thus, we may assume ⟨ a , b ⟩ ∩ ⟨ x i ⟩ = 1 , where 1 ≤ i ≤ t . Let C = ⟨ a x 1 , b ⟩ and D = ⟨ a x 1 , x 2 ⟩ . If o ( x 1 ) > 2 , then C ∩ D = ⟨ a x 1 ⟩ ⋬ D . Thus, o ( x 1 ) = 2 and so o ( x i ) = 2 . So G is a Dedekind group, in contradiction to the hypothesis.

Conversely, every group listed in the lemma is a non-Dedekind Q -2-group and they are pairwise non-isomorphic.□

Lemma 4.12

Let G be a Q -2-group and N ≤ G ′ ∩ Z ( G ) with ∣ N ∣ = 2 . Then the quotient group G / N is not isomorphic to M 2 ( 2 , 1 , 1 ) ∗ Q 8 .

Proof

Otherwise, we may assume N = ⟨ x ⟩ and G ¯ = G / N = ⟨ a ¯ , b ¯ , c ¯ , d ¯ ∣ a ¯ 4 = b ¯ 2 = c ¯ 4 = d ¯ 4 = 1 ¯ , c ¯ 2 = d ¯ 2 , [ c ¯ , d ¯ ] = c ¯ 2 , [ a ¯ , b ¯ ] = c ¯ 2 , [ a ¯ , c ¯ ] = [ a ¯ , d ¯ ] = [ b ¯ , c ¯ ] = [ d ¯ , d ¯ ] = 1 ¯ ⟩ . By Lemma 4.9, we see o ( c ) = o ( d ) = 4 and G ′ = ⟨ c , x ⟩ ≅ C 2 × C 2 . According to Lemma 3.3, we see G ′ ≤ ⟨ c , b ⟩ . If b 2 = 1 , then, since x ∈ ⟨ c , b ⟩ , we see [ c , b ] = x . Similarly, [ d , b ] = x and [ c d , b ] = x . However, it is impossible. So b 2 = x . Similarly, a 4 = x . By letting b 1 = b a 2 , we see b 1 2 = 1 . So we also have a contradiction.□

Theorem 4.13

Let G be a Q -2-group with d ( G ) ≥ 4 and ∣ G ∣ ≥ 2 7 . Then G is Dedekind.

Proof

Otherwise, we may assume ∣ G ∣ = 2 7 by Lemma 2.4. Let N ≤ G ′ ∩ Z ( G ) such that ∣ N ∣ = 2 . Set G ¯ = G / N . Then, by Lemma 2.4 again, we may assume G ¯ is a non-Dedekind group. Thus, G ¯ is of the type (2) with order 2 6 in Lemma 4.11. However, it is impossible by Lemma 4.12.□

5 Q -p-groups of odd order

Lemma 5.1

Let G be a Q -p-group with d ( G ) = 2 and p > 3 . If exp ( G ′ ) = p , then c ( G ) ≤ 3 .

Proof

We may assume G = ⟨ a , b ⟩ , [ a , b ] = c and [ c , a ] = d ≠ 1 . Assume A = ⟨ a , c ⟩ , B = ⟨ b , c ⟩ , C = ⟨ c , d ⟩ , D = ⟨ a , d ⟩ and o ( a ) = p n . Then, by the hypotheses, we see G ′ ≤ A ∩ B and A ′ ≤ C ∩ D . Now we prove A ′ ≤ ⟨ a p n − 1 , d ⟩ . Since [ a , d ] ∈ ⟨ [ c , d ] ⟩ , it is easy to see that there exists an element g ∈ A ⧹ ⟨ c , Φ ( A ) ⟩ such that [ g , d ] = 1 . Noting that A ′ ≤ ⟨ g , d ⟩ , we see A ′ ≤ ⟨ ℧ 1 ( g ) , d ⟩ . Since p > 3 and c ( A ) ≤ 3 , by Lemma 2.3, we see ℧ 1 ( A ) = ⟨ a p ⟩ , which implies A ′ ≤ ⟨ a p n − 1 , d ⟩ . It follows that G ′ ≤ ⟨ a p n − 1 , c , d ⟩ . If d ∈ ⟨ a ⟩ , then G 3 ≤ Z ( G ) . So we may assume d ∉ ⟨ a ⟩ . Without loss of generality, we may assume [ c , b ] ∈ ⟨ a p n − 1 ⟩ . It follows from G ′ ≤ ⟨ c , b ⟩ that d ∈ Φ ( ⟨ c , b ⟩ ) = ⟨ b p ⟩ ⊴ G . So d ∈ Z ( G ) , G 3 ≤ Z ( G ) and the lemma is proved.□

Theorem 5.2

Let G be a Q -3-group with d ( G ) = 2 . Then c ( G ) ≤ 3 .

Proof

We may assume G = ⟨ a , b ⟩ , o ( a ) = 3 n , [ a , b ] = c and [ c , a ] = d ≠ 1 . By Lemma 3.2, for any x ∈ G , we see ∣ ⟨ x ⟩ / ⟨ x ⟩ G ∣ ≤ 3 . Thus, by Lemmas 2.7 and 2.8, we see c ( G ) ≤ 4 , exp ( G ′ ) ≤ 9 and exp ( G 3 ) ≤ 3 . Then [ c , d ] ∈ [ G 2 , G 3 ] ≤ G 5 = 1 , o ( d ) = 3 and c ( ⟨ a , d ⟩ ) ≤ 2 . Let A = ⟨ c , d ⟩ , B = ⟨ a , d ⟩ . Then [ a , d ] ∈ A and so [ a , d ] ∈ ⟨ c 3 ⟩ . Noting that G ′ ≤ ⟨ a , c ⟩ , we see [ c , b ] ∈ Ω 1 ( ⟨ a , c ⟩ ) = ⟨ a 3 n − 1 , c 3 , d ⟩ . Without loss of generality, we may assume [ c , b ] ∈ ⟨ a i 3 n − 1 c 3 j ⟩ , which implies c ( ⟨ c , b ⟩ ) < 2 . It follows from G ′ ≤ ⟨ c , b ⟩ that d ∈ Φ ( ⟨ c , b ⟩ ) = ⟨ b 3 , c 3 ⟩ ⊴ G . Thus, d ∈ Z ( G ) and G 3 ≤ Z ( G ) . So c ( G ) ≤ 3 .□

Theorem 5.3

Let G be a Q -p-group and p > 2 . Then exp ( G ′ ) ≤ p .

Proof

Suppose the result is not true and G is a counterexample of minimal order. Take x , y ∈ G ′ with o ( x ) ≤ p and o ( y ) ≤ p . Let K = ⟨ x , y ⟩ . Then, by Lemma 2.3, ( x y ) p = x p y p z , where z ∈ ℧ 1 ( K ′ ) K p . If p > 3 , then, by Lemma 5.1, ℧ 1 ( K ′ ) K p = 1 . If p = 3 , then, by Lemma 2.7, we see c ( G ) ≤ 5 . Thus, ℧ 1 ( K ′ ) K 3 = 1 . It follows that ( x y ) p = x p y p = 1 . So we may assume d ( G ) = 2 , G = ⟨ a , b ⟩ , [ a , b ] = c , L = ⟨ a , c ⟩ and o ( c ) ≥ p 2 . Then, according to Lemma 2.3, we see [ a p , b ] = c p w , where w ∈ ℧ 1 ( L ′ ) L p . Since L < G , by Lemma 5.1 and Theorem 5.2, we see ℧ 1 ( L ′ ) L p = 1 . Thus, w = 1 and so c p ∈ ⟨ a ⟩ . Similarly, c p ∈ ⟨ b ⟩ . By induction, o ( c p ) = p and so o ( c ) = p 2 .

Without loss of generality, we may assume ⟨ a ⟩ ∩ ⟨ b ⟩ = ⟨ a p s ⟩ = ⟨ b p t ⟩ , a p s = b p t and s ≥ t ≥ 2 . If s > t , then, by letting b 1 = a − p s − t b , we see [ a , b 1 p ] = c p and c p ∉ ⟨ b 1 p ⟩ . So s = t . Let b 2 = a − 1 b . Then, by Lemma 2.3, we see b 2 p = a p b − p g , where g ∈ ℧ 1 ( G ′ ) G p . Then o ( g ) ≤ p and o ( b 2 ) = p s . Noting that [ a , b 2 p ] = c p , we see c p ∈ ⟨ b 2 p ⟩ . If s = 2 , then b 2 p ∈ Z ( G ) , a contradiction. If s > 2 , then ⟨ c p ⟩ = ⟨ b 2 p s − 1 ⟩ = ⟨ a p s − 1 b p s − 1 ⟩ . It follows that ⟨ a ⟩ ∩ ⟨ b ⟩ = ⟨ a p s − 1 ⟩ , another contradiction.□

According to Lemma 5.1 and Theorems 5.2 and 5.3, we have the following results:

Corollary 5.4

Let G be a Q -p-group with d ( G ) = 2 and p > 2 . Then c ( G ) ≤ 3 .

Theorem 5.5

Let G be a Q -p-group and p > 2 . Then c ( G ) ≤ 3 .

Proof

We may assume d ( G ) = 3 , G = ⟨ a , b , c ⟩ , [ a , b ] = x and [ a , x ] = y ≠ 1 . We only need to prove that [ c , y ] = 1 . Let A = ⟨ a , y ⟩ , B = ⟨ c , y ⟩ , C = ⟨ x , y ⟩ . Then [ c , y ] ∈ A ∩ C . It follows that [ c , y ] = 1 , which implies c ( G ) ≤ 3 .□

Theorem 5.6

Let G be a Q -p-group with d ( G ) ≥ 4 and p > 2 . Then G is abelian.

Proof

Otherwise, we may assume d ( G ) = 4 and G ′ ≠ 1 . Let N ≤ G ′ such that ∣ N ∣ = p and N ⊴ G . Then G / N is a Q -p-group by Lemma 3.1. By induction, G / N is abelian and so ∣ G ′ ∣ = p . Assume A < G and A is minimal non-abelian with A = ⟨ a , b ⟩ , ⟨ a ⟩ ∩ ⟨ b ⟩ = 1 and [ a , b ] ∉ ⟨ b ⟩ . Then C G ( b ) ⋖ G . Assume that there exists an element g ∈ Ω 1 ( C G ( b ) ) ⧹ A . Let A = ⟨ a , b ⟩ , B = ⟨ b , g ⟩ . Then, we see A ∩ B = ⟨ b ⟩ and ⟨ b ⟩ ⋬ A , a contradiction. So, Ω 1 ( C G ( b ) ) ≤ A . Since ∣ Ω 1 ( A ) ∣ ≤ p 3 , by Lemma 2.9, we see Ω 1 ( C G ( b ) ) = Ω 1 ( A ) = p 3 . Then A = ⟨ a , b , c ∣ a p n = b p m = c p = 1 , [ a , b ] = c , [ c , a ] = [ c , b ] = 1 ⟩ is non-metacyclic with n ≥ 2 . By Lemma 2.9, we see ∣ Ω 1 ( G ) ∣ ≥ p 4 . Then there exists an element h ∈ Ω 1 ( G ) ⧹ C G ( b ) . Let A = ⟨ a , b ⟩ , B = ⟨ b , a h ⟩ . Then, we see A ∩ B = ⟨ b , a p n − 1 ⟩ ⋬ A , another contradiction. Thus, the theorem is true.□

Acknowledgments

The author would like to thank the referee for his/her valuable suggestions and comments that contributed to the final version of this paper.

  1. Funding information: This work was supported by “The Science & Technology Development Fund of Tianjin Education Commission for Higher Education (No. 2019KJ141).”

  2. Conflict of interest: The author states no conflict of interests.

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Received: 2020-10-10
Revised: 2021-05-03
Accepted: 2021-07-05
Published Online: 2021-08-28

© 2021 Jiao Wang, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  132. Long time decay for 3D Navier-Stokes equations in Fourier-Lei-Lin spaces
  133. On the extinction problem for a p-Laplacian equation with a nonlinear gradient source
  134. General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
  135. On hyponormality on a weighted annulus
  136. Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term
  137. Convergence results on Picard-Krasnoselskii hybrid iterative process in CAT(0) spaces
  138. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part I)
  139. Marangoni convection in layers of water-based nanofluids under the effect of rotation
  140. A transient analysis to the M(τ)/M(τ)/k queue with time-dependent parameters
  141. Existence of random attractors and the upper semicontinuity for small random perturbations of 2D Navier-Stokes equations with linear damping
  142. Degenerate binomial and Poisson random variables associated with degenerate Lah-Bell polynomials
  143. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  144. On the mixed fractional quantum and Hadamard derivatives for impulsive boundary value problems
  145. The Lp dual Minkowski problem about 0 < p < 1 and q > 0
https://doi.org/10.1515/math-2021-0077
Keywords for this article
finite p-group; metacyclic p-group; commutator subgroup
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