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Complete decomposition of the generalized quaternion groups

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Open Mathematics
From the journal Open Mathematics Volume 21 Issue 1

Abstract

Let G be a finite nonabelian group. For any integer m ≥ 2 , let A 1 , … , A m be nonempty subsets of G . If A 1 , … , A m are mutually disjoint and if the subset product A 1 … A m = { α 1 … α m ∣ α v ∈ A v , v = 1 , 2 , … , m } coincides with G , then ( A 1 , … , A m ) is called a complete decomposition of G of order m . In this article, we let G be the generalized quaternion groups Q 2 n , which is a finite nonabelian group of order 2 n with group presentation given by ⟨ x , y ∣ x 2 n − 1 = 1 , y 2 = x 2 n − 2 , y x = x 2 n − 1 − 1 y ⟩ for positive integer n ≥ 3 . We determine the existence of complete decompositions of Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , and show that Q 2 n can be written in the product of 2 n − 1 subsets, i.e., Q 2 n = A 1 A 2 ⋯ A 2 n − 1 , where ∣ A j ∣ = 2 , for j ∈ { 1 , 2 , … , 2 n − 1 } . In addition, we construct the non-complete decomposition of Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , using the non-exhaustive subsets of Q 2 n .

Keywords: generalized quaternion groups; complete decomposition; subsets product
MSC 2010: 05A18; 20D60

1 Introduction

A lot of research has been carried out on group coverings over the years. Group factorization is one type of group covering. The first result related to group factorization was showed by Cauchy [1] in the eighteenth century and rediscovered by Davenport [2] in 1935. Over the past few decades, there have been numerous results related to group factorization. De Bruijn [3] studied the factorization of cyclic groups after Hajós discovered the existence of cyclic groups that admit a non-trivial factorization. Lucchini and Garonzi classified the finite noncyclic groups, all of whose irredundant covers are minimal in [4]. Factorizations of nonabelian groups have also been studied by some researchers recently (see, e.g., [5,6]).

The results on factoring the abelian groups into subsets have various applications in different fields, including cryptography, geometry of tiling, and coding theory. The researchers used group factorization as a trapdoor function to design some public-key cryptosystems (see, e.g., [7,8]). Rivest et al. [9] studied new public key cryptosystems using logarithmic signatures over a finite group. Szabo and Ward [10] showed that there exists an application in the geometry of tiling by factoring the abelian groups. Amin [11] and Szabo [12] showed some applications of group factorization in the construction of error-correcting codes.

Chin and Chen studied the complete decomposition of finite abelian groups, and the existence of the order of complete decomposition of Z n ( n ≥ 6 ) was determined in [13]. As an application of the complete decomposition of nonabelian group, we note that a group-based key exchange protocol was proposed by Sin and Chen [14]. In this article, we investigate the complete decomposition of a finite nonabelian group, the generalized quaternion group Q 2 n , of order 2 n with two cyclic subgroups, which are ⟨ x ⟩ and ⟨ y ⟩ . Let G be a finite nonabelian group. For any integer m ≥ 2 , let A 1 , … , A m be nonempty subsets of G . If A 1 , … , A m are mutually disjoint and if the subset product A 1 … A m = { α 1 … α m ∣ α v ∈ A v , v = 1 , 2 , … , m } coincides with G , then ( A 1 , … , A m ) is a complete decomposition of G of order m . Throughout this article, let Q 2 n = ⟨ x , y ∣ x 2 n − 1 = 1 , x 2 n − 2 = y 2 , y x = x 2 n − 1 − 1 y ⟩ be the generalized quaternion group of order 2 n , where n ≥ 3 .

Getting motivation from the works of Chin and Chen [13] and Wong et al. [6], we investigate the product of the mutually disjoint subsets of the generalized quaternion groups Q 2 n . In Section 2, we determine the existence of the complete decomposition ( A 1 , … , A k ) of generalized quaternion groups Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , where the cardinality of every subset is larger than or equal to 2. We provide some constructions of complete decompositions of Q 2 n for cases ∑ i = 1 k ∣ A i ∣ = 2 n and ∑ i = 1 k ∣ A i ∣ < 2 n , respectively. In Section 3, we show some constructions of non-complete decompositions of Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , using the non-exhaustive subsets of Q 2 n .

2 Some constructions of complete decomposition of Q 2 n

Let A = { 1 , x , x 2 y , x 3 y } and B = { x 2 , x 3 , y , x y } , where A , B ⊆ Q 8 . Then, ( A , B ) is a complete decomposition of Q 8 of order 2. By calculations, we note that there are 44 complete decompositions ( A , B ) of Q 8 of order 2 as given in Table 1. It can be shown that ( B , A ) is also the complete decomposition of Q 8 of order 2 if ( A , B ) is a complete decomposition of Q 8 .

Table 1

Complete decomposition ( A , B ) of Q 8 of order 2

∣ A ∣ ∣ B ∣ Number of complete decomposition
3 5 24
4 4 20

In general, if ( A 1 , A 2 , … , A m ) is a complete decomposition of Q 2 n of order m , we wish to clarify whether ( A j 1 , A j 2 , … , A j m ) is also a complete decomposition of Q 2 n , where { A j 1 , A j 2 , … , A j m } = { A 1 , A 2 , … , A m } . We answer this in the negative with an example. Let A 1 = { 1 , x } , A 2 = { x 2 , y } , A 3 = { x y , x 2 y } , and A 4 = { x 3 , x 3 y } be the subsets of Q 8 . We observe that ( A 1 , A 2 , A 3 , A 4 ) is a complete decomposition of Q 8 of order 4 but ( A 2 , A 1 , A 3 , A 4 ) is not a complete decomposition of Q 8 . Hence, the order of the subsets in the construction of the complete decomposition is important since Q 2 n is a non-abelian group.

In this section, we determine the existence of a complete decomposition of the generalized quaternion groups Q 2 n of order k , where k ∈ { 2 , 3 , … 2 n − 1 } .

2.1 Complete decomposition of Q 2 n of order 2

We begin by showing some preliminary results.

Proposition 1

Let x i , y , x 2 n − 1 − i ∈ Q 2 n , for integers n ≥ 3 and 1 ≤ i ≤ 2 n − 1 − 1 . Then, the following properties hold:

  1. y x i = x 2 n − 1 − i y ;

  2. ⟨ x ⟩ y x i = ⟨ x ⟩ y .

Proof

  1. By using the group relation of Q 2 n , we have y x = x 2 n − 1 − 1 y , and hence y x i = x 2 n − 1 − 1 y x i − 1 = x 2 ( 2 n − 1 − 1 ) y x i − 2 = ⋯ = x ( i − 1 ) ( 2 n − 1 − 1 ) y x = x i ( 2 n − 1 − 1 ) y . Since x c 2 n − 1 = x 2 n − 1 for positive integer c , it follows that y x i = x 2 n − 1 − i y .

  2. Clearly, ⟨ x ⟩ y x i = ⟨ x ⟩ x 2 n − 1 − i y = { x 2 n − 1 − i y , x 2 n − 1 − i + 1 y , … , x 2 n − 1 − i y } = ⟨ x ⟩ y as it consists of 2 n − 1 consecutive terms of ⟨ x ⟩ y .□

Proposition 2

Let A = { 1 , x , x 2 , … , x 2 n − 1 − 3 , x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } and B = { x 2 n − 1 − 2 , x 2 n − 1 − 1 , y , x y , … , x 2 n − 1 − 3 y } , A , B ⊆ Q 2 n for n ≥ 3 . Then, ( A , B ) is a complete decomposition of Q 2 n of order 2.

Proof

Let L 1 = { 1 , x , x 2 , … , x 2 n − 1 − 3 } ⊆ A and M 1 = { x 2 n − 1 − 2 , x 2 n − 1 − 1 } ⊆ B . Note that L 1 M 1 = { 1 , x , … , x 2 n − 1 − 4 , x 2 n − 1 − 2 , x 2 n − 1 − 1 } and ⟨ x ⟩ \ L 1 M 1 = { x 2 n − 1 − 3 } . Let L 2 = { x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } ⊆ A and M 2 = { y , x y , … , x 2 n − 1 − 3 y } ⊆ B . Given that L 2 M 2 = { 1 , x , … , x 2 n − 2 − 1 , x 2 n − 2 + 1 , x 2 n − 2 + 2 , … , x 2 n − 1 − 1 } and ⟨ x ⟩ \ L 2 M 2 = { x 2 n − 2 } , we see that L 1 M 1 ∪ L 2 M 2 = ⟨ x ⟩ since x 2 n − 1 − 3 ≠ x 2 n − 2 . By selecting { 1 , x , x 2 } ⊆ A and M 2 ⊆ B , it is straightforward to check that { 1 , x , x 2 } M 2 = ⟨ x ⟩ y . Therefore, we conclude that ( A , B ) is a complete decomposition of Q 2 n of order 2.□

Proposition 2 gives a complete decomposition of Q 2 n of order 2, where A , B ⊂ Q 2 n , ∣ A ∣ = ∣ B ∣ = 2 n − 1 , A ∪ B = Q 2 n , and the subsets A and B are pairwise disjoint. As a consequence of Proposition 2, we have the following:

Corollary 1

Let A = ( ⟨ x ⟩ \ { x k , x k + 1 } ) ∪ { x k y , x k + 1 y } and B = ( ⟨ x ⟩ y \ { x k y , x k + 1 y } ) ∪ { x k , x k + 1 } , where A , B ⊆ Q 2 n , ∣ A ∣ = ∣ B ∣ = 2 n − 1 , and k ∈ { 3 , 4 , … , 2 n − 1 − 2 } for n ≥ 4 . Then, ( A , B ) is a complete decomposition of Q 2 n of order 2.

Let A i , B i ⊆ Q 2 n for i ∈ { 1 , 2 , 3 , 4 , 5 , 6 } and n ≥ 4 . We give other constructions of the complete decomposition ( A i , B i ) of Q 2 n of order 2, where A ∩ B = ϕ and ∣ A ∣ + ∣ B ∣ = 2 n :

  1. A 1 = { 1 , x , y } , B 1 = { x 2 , x 3 , … , x 2 n − 1 − 1 , x y , x 2 y , … , x 2 n − 1 − 1 y } ;

  2. A 2 = { 1 , y , x y } , B 2 = { x , x 2 , … , x 2 n − 1 − 1 , x 2 y , x 3 y , … , x 2 n − 1 − 1 y } ;

  3. A 3 = { 1 , x , … , x 2 n − 1 − 3 , y } , B 3 = { x 2 n − 1 − 2 , x 2 n − 1 − 1 , x y , x 2 y , … , x 2 n − 1 − 1 y } ;

  4. A 4 = { 1 , x , … , x 2 n − 1 − 3 , y , x y } , B 4 = { x 2 n − 1 − 2 , x 2 n − 1 − 1 , x 2 y , x 3 y , … , x 2 n − 1 − 1 y } ;

  5. A 5 = { 1 , x , … , x 2 n − 1 − 2 , y } , B 5 = { x 2 n − 1 − 1 , x y , x 2 y , … , x 2 n − 1 − 1 y } ;

  6. A 6 = { 1 , x , … , x 2 n − 2 − 1 , x 2 n − 2 + 1 , x 2 n − 2 + 2 , … , x 2 n − 1 − 1 } , B 6 = { x 2 n − 2 , x y , x 2 y , … , x 2 n − 1 − 1 y } .

In the following, we provide a construction of complete decomposition of Q 2 n of order 2, where A and B are pairwise disjoint subsets of Q 2 n and ∣ A ∣ + ∣ B ∣ < 2 n = ∣ Q 2 n ∣ .

Proposition 3

Let A = { 1 , x , … , x 2 n − 1 − 3 } ∪ { y , x y } and B r = ( { x 2 y , x 3 y , … , x 2 n − 1 − 1 y } ∪ { x 2 n − 1 − 2 , x 2 n − 1 − 1 } ) \ { x 2 y , x 3 y , … , x r y } be the subsets of Q 2 n for r ∈ { 2 , 3 , … , 2 n − 2 + 3 } and n ≥ 5 . Then, ( A , B r ) is a complete decomposition of Q 2 n of order 2.

Proof

Note that B 2 n − 2 + 3 ⊆ B 2 n − 2 + 2 ⊆ ⋯ ⊆ B 2 . It suffices to consider the the case ( A , B 2 n − 2 + 3 ) is a complete decomposition of Q 2 n of order 2. Let L 1 = { 1 , x , … , x 2 n − 1 − 3 } , L 2 = { y , x y } , M 1 = { x 2 n − 1 − 2 , x 2 n − 1 − 1 } , and M 2 = { x 2 n − 2 + 4 y , x 2 n − 2 + 5 y , … , x 2 n − 1 − 1 y } . Given that A = L 1 ∪ L 2 , B 2 n − 2 + 3 = M 1 ∪ M 2 , and we see that L 1 M 2 = ⟨ x ⟩ y . It remains to show that ⟨ x ⟩ ⊆ A B 2 n − 2 + 3 . We compute L 1 M 1 = { 1 , x , … , x 2 n − 1 − 4 , x 2 n − 1 − 2 , x 2 n − 1 − 1 } , L 2 M 2 = { x 2 n − 2 + 1 , x 2 n − 2 + 2 , … , x 2 n − 1 − 3 } , then we have L 2 M 2 ∪ L 1 M 1 = ⟨ x ⟩ . Hence, we conclude that ( A , B r ) is a complete decomposition of Q 2 n of order 2 for r ∈ { 2 , 3 , … , 2 n − 2 + 3 } .□

In Proposition 3, we note that ( A , B ) is a complete decomposition of Q 2 n of order 2, where ∣ A ∣ + ∣ B ∣ ∈ { 2 n − 2 + 2 n − 1 − 2 , 2 n − 2 + 2 n − 1 − 1 , … , 2 n − 1 } .

Proposition 4

Let A = { 1 , x , … , x 2 n − 2 − 2 , y } and B = { x 2 n − 2 − 1 , x 2 n − 1 − 1 , x 2 y , x 2 n − 2 + 2 y } be the subsets of Q 2 n , where ∣ A ∣ = 2 n − 2 and ∣ B ∣ = 4 for n ≥ 4 . Then, ( A , B ) is a complete decomposition of Q 2 n of order 2.

Proof

We first note that { 1 , x , … , x 2 n − 2 − 2 } { x 2 n − 2 − 1 , x 2 n − 1 − 1 } = { 1 , x , … , x 2 n − 2 − 3 } ∪ { x 2 n − 2 − 1 , x 2 n − 2 , … , x 2 n − 1 − 3 } ∪ { x 2 n − 1 − 1 } and y { x 2 y , x 2 n − 2 + 2 y } = { x 2 n − 2 − 2 , x 2 n − 1 − 2 } . Consequently, ⟨ x ⟩ ⊆ A B . In order to show that ⟨ x ⟩ y ⊆ A B , we compute

{ 1 , x , … , x 2 n − 2 − 2 } { x 2 y , x 2 n − 2 + 2 y } = { y , x 2 y , x 3 y , … , x 2 n − 2 y , x 2 n − 2 + 2 y , x 2 n − 2 + 3 y , … , x 2 n − 1 − 1 y }

and y { x 2 n − 2 − 1 , x 2 n − 1 − 1 } = { x 2 n − 2 + 1 y , x y } ⊆ A B . Hence, the result holds.□

Note that ∣ A ∣ ∣ B ∣ = 2 n − 2 ⋅ 4 = 2 n = ∣ Q 2 n ∣ . The construction of the complete decomposition of Q 2 n of order 2 given in Proposition 4 has the smallest ∣ A ∣ + ∣ B ∣ = 2 n − 2 + 4 .

Corollary 2

There exist E , B ⊆ Q 2 n such that ( E , B ) is a complete decomposition of Q 2 n of order 2, where ∣ E ∣ + ∣ B ∣ ∈ { 2 n − 2 + 4 , 2 n − 2 + 5 , … , 2 n } for n ≥ 4 .

Proof

Let E , A , and B be the subsets of Q 2 n , where A = { 1 , x , … , x 2 n − 2 − 2 , y } and B = { x 2 n − 2 − 1 , x 2 n − 1 − 1 , x 2 y , x 2 n − 2 + 2 y } are the subsets as defined in Proposition 4, E ⊇ A and E ∩ B = ∅ . Then, ( E , B ) is a complete decomposition of Q 2 n of order 2, where ∣ E ∣ + ∣ B ∣ ∈ { 2 n − 2 + 4 , 2 n − 2 + 5 , … , 2 n } .□

2.2 Complete decomposition of Q 2 n of order k for k ∈ { 3 , 4 , … , 2 n ‒ 1 }

We begin by showing the existence of complete decompositions of Q 2 n of orders 3, 4, and 5.

Proposition 5

Let A 1 = { x , x y } , A 2 = { 1 , x 2 n − 1 − 1 y } , and A 3 = { x 2 , x 3 , … , x 2 n − 1 − 1 , y , x 2 y , x 3 y , … , x 2 n − 1 − 2 y } , where A 1 , A 2 , A 3 ⊆ Q 2 n for n ≥ 4 . Then, ( A 1 , A 2 , A 3 ) is a complete decomposition of Q 2 n of order 3.

Proof

Note that A 1 A 2 = { x , x 2 n − 2 + 2 , y , x y } , and the products of the subsets A 1 A 2 A 3 = Q 2 n . Therefore, ( A 1 , A 2 , A 3 ) is a complete decomposition of Q 2 n of order 3.□

The complete decomposition of Q 2 n of order 3 given in Proposition 5 is not unique. One can easily verify that ( B 1 , B 2 , B 3 ) is another construction of the complete decomposition of Q 2 n of order 3 for n ≥ 4 :

B 1 = { 1 , y } , B 2 = { x , x 2 , … , x 2 n − 1 − 3 , x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } , and B 3 = { x y , x 2 y , … , x 2 n − 1 − 3 y , x 2 n − 1 − 2 , x 2 n − 1 − 1 } .

Proposition 6

There exists a complete decomposition of Q 2 n of orders 4 and 5 for n ≥ 4 .

Proof

Let A 1 = { x , x y } , A 2 = { 1 , x 2 n − 1 − 1 y } , A 3 = { x 2 , y } , A 4 = { x 3 , x 4 , … , x 2 n − 1 − 1 , x 2 y , x 3 y , … , x 2 n − 1 − 2 y } , where A 1 , A 2 , A 3 , A 4 ⊆ Q 2 n for n ≥ 4 . Then, ( A 1 , A 2 , A 3 , A 4 ) is a complete decomposition of Q 2 n of order 4.

To construct a complete decomposition of Q 2 n of order 5, we let A 1 = { x , x y } , A 2 = { 1 , x 2 n − 1 − 1 y } , A 3 = { x 2 , y } , A 4 = { x 3 , x 3 y } , and A 5 = { x 4 , x 5 , … , x 2 n − 1 − 1 , x 2 y , x 4 y , x 5 y , … , x 2 n − 1 − 2 y } , then A 1 , A 2 , … , A 5   ⊆ Q 2 n , ∣ A 1 ∣ = ∣ A 2 ∣ = ⋯ = ∣ A 4 ∣ = 2 , ∣ A 5 ∣ = 2 n − 8 , and A 1 , A 2 , … , A 5 are mutually disjoint. Note that A 1 A 2 = { x , x 2 n − 2 + 2 , y , x y } , A 1 A 2 A 3 = { x 3 , x 2 n − 2 , x 2 n − 2 + 1 , x 2 n − 2 + 4 , x y , x 2 n − 2 + 2 y , x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } , and

A 1 A 2 A 3 A 4 = { x 6 , x 2 n − 2 − 2 , x 2 n − 2 + 3 , x 2 n − 2 + 4 , x 2 n − 2 + 7 , x 2 n − 1 − 5 , x 2 n − 1 − 4 , x 2 n − 1 − 2 , x 2 n − 1 − 1 , x 6 y , x 2 n − 2 − 5 y , x 2 n − 2 − 1 y , x 2 n − 2 + 3 y , x 2 n − 2 + 4 y , x 2 n − 2 + 7 y , x 2 n − 1 − 4 y , x 2 n − 1 − 2 y } .

Hence, we see that A 1 A 2 A 3 A 4 A 5 = Q 2 n . Therefore, ( A 1 , A 2 , A 3 , A 4 , A 5 ) is a complete decomposition of Q 2 n of order 5.□

The following lemma is required to prove the existence of a complete decomposition of Q 2 n of order k , for k ∈ { 6 , 7 , … , 2 n − 1 } .

Lemma 1

Let A i = { x i − 1 , x i − 1 y } ⊆ Q 2 n and α i ∈ A i for i ∈ { 4 , 5 , … , k − 1 } , where n ≥ 4 . If

α i = x i − 1 , i f i ≠ j , j + 1 , x i − 1 y , e l s e ,

then α 4 α 5 … α k − 1 = x ( k − 2 ) ( k − 1 ) 2 + 2 n − 2 − 2 j − 3 for k ∈ { 6 , 7 , … , 2 n − 1 } and j ∈ { 4 , 5 , … , k − 2 } .

Proof

We separate the product α 4 α 5 … α k − 1 into three parts as follows:

(1) α 4 α 5 … α j − 1 = x 3 x 4 … x j − 2 = x ( j − 2 ) ( j − 1 ) 2 − 3 ;

(2) α j α j + 1 = x j − 1 y x j y = x 2 n − 1 + 2 n − 2 − 1 = x 2 n − 2 − 1 ;

(3) α j + 2 α j + 3 … α k − 1 = x j + 1 x j + 2 … x k − 2 = x ( k − 2 ) ( k − 1 ) 2 − ( j − 2 ) ( j − 1 ) 2 − ( j − 1 ) − j .

By summing up the powers of x in equations (1)–(3), we have

( j − 2 ) ( j − 1 ) 2 − 3 + 2 n − 2 − 1 + ( k − 2 ) ( k − 1 ) 2 − ( j − 2 ) ( j − 1 ) 2 − 2 j + 1 = ( k − 2 ) ( k − 1 ) 2 + 2 n − 2 − 2 j − 3

as required. Therefore, α 4 α 5 … α k − 1 = x ( k − 2 ) ( k − 1 ) 2 + 2 n − 2 − 2 j − 3 for j ∈ { 4 , 5 , … , k − 2 } .□

Theorem 1

Let A 1 = { x , x y } , A 2 = { 1 , x 2 n − 1 − 1 y } , A 3 = { x 2 , y } , A r = { x r − 1 , x r − 1 y } (for r ∈ { 4 , 5 , … , k − 1 } ), and A k = { x k − 1 , x k , … , x 2 n − 1 − 1 , x 2 y , x k − 1 y , x k y , … , x 2 n − 1 − 2 y } be the mutually disjoint subsets of Q 2 n for n ≥ 5 . Then, ( A 1 , A 2 , … , A k ) is a complete decomposition of Q 2 n of order k for k ∈ { 6 , 7 , … , 2 n − 1 − 1 } , where ∑ i = 1 k ∣ A i ∣ = 2 n .

Proof

Note that

A 1 A 2 A 3 = { x 3 , x 2 n − 2 , x 2 n − 2 + 1 , x 2 n − 2 + 4 , x y , x 2 n − 2 + 2 y , x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } = H ∪ { x y , x 2 n − 2 + 2 y , x 2 n − 1 − 2 y , x 2 n − 1 − 1 y } ,

where H = { x 3 , x 2 n − 2 , x 2 n − 2 + 1 , x 2 n − 2 + 4 } . By Lemma 1, we have

(4) α 4 α 5 … α k − 1 ∈ { x c + 2 n − 2 − 2 k + 1 , x c + 2 n − 2 − 2 k + 3 , … , x c + 2 n − 2 − 11 } = L ,

where c = ( k − 2 ) ( k − 1 ) 2 , L ⊆ A 4 A 5 … A k − 1 , and ∣ L ∣ = k − 5 . Now, we compute h L for each element h ∈ H as follows:

x 3 L = x 3 { x 2 n − 2 + c − 2 k + 1 , x 2 n − 2 + c − 2 k + 3 , … , x 2 n − 2 + c − 11 } = { x 2 n − 2 + c − 2 k + 4 , x 2 n − 2 + c − 2 k + 6 , … , x 2 n − 2 + c − 8 } = V 1 ,

(5) x 2 n − 2 L = x 2 n − 2 { x 2 n − 2 + c − 2 k + 1 , x 2 n − 2 + c − 2 k + 3 , … , x 2 n − 2 + c − 11 } = { x 2 n − 1 + c − 2 k + 1 , x 2 n − 1 + c − 2 k + 3 , … , x 2 n − 1 + c − 11 } = V 2 ,

(6) x 2 n − 2 + 1 L = x 2 n − 2 + 1 { x 2 n − 2 + c − 2 k + 1 , x 2 n − 2 + c − 2 k + 3 , … , x 2 n − 2 + c − 11 } = { x 2 n − 1 + c − 2 k + 2 , x 2 n − 1 + c − 2 k + 4 , … , x 2 n − 1 + c − 10 } = V 3 ,

x 2 n − 2 + 4 L = x 2 n − 2 + 4 { x 2 n − 2 + c − 2 k + 1 , x 2 n − 2 + c − 2 k + 3 , … , x 2 n − 2 + c − 11 } = { x 2 n − 1 + c − 2 k + 5 , x 2 n − 1 + c − 2 k + 7 , … , x 2 n − 1 + c − 7 } = V 4 ,

where V 1 ∪ V 2 ∪ V 3 ∪ V 4 = H L ⊆ A 1 A 2 … A k − 1 . Note that

A k = { x k − 1 , x k , … , x 2 n − 1 − 2 , x k − 1 y , x k y , … , x 2 n − 1 − 2 y } ∪ { x 2 n − 1 − 1 , x 2 y } = P 1 ∪ P 2 ,

where P 1 = { x k − 1 , x k , … , x 2 n − 1 − 2 , x k − 1 y , x k y , … , x 2 n − 1 − 2 y } and P 2 = { x 2 n − 1 − 1 , x 2 y } . Next, we compute V 1 P 1 , V 2 P 1 , V 3 P 1 , V 4 P 1 ⊆ A 1 A 2 … A k as follows:

V 1 P 1 = { x c + 2 n − 2 − k + 3 , x c + 2 n − 2 − k + 4 , … , x c + 2 n − 2 + 2 n − 1 − 10 } ∪ { x c + 2 n − 2 − k + 3 y , x c + 2 n − 2 − k + 4 y , … , x c + 2 n − 2 + 2 n − 1 − 10 y } ; V 2 P 1 = { x c + 2 n − 1 − k , x c + 2 n − 1 − k + 1 , … , x c + 2 n − 13 } ∪ { x c + 2 n − 1 − k y , x c + 2 n − 1 − k + 1 y , … , x c + 2 n − 13 y } ; V 3 P 1 = { x c + 2 n − 1 − k + 1 , x c + 2 n − 1 − k + 2 , … , x c + 2 n − 12 } ∪ { x c + 2 n − 1 − k + 1 y , x c + 2 n − 1 − k + 2 y , … , x c + 2 n − 12 y } ; V 4 P 1 = { x c + 2 n − 1 − k + 4 , x c + 2 n − 1 − k + 5 , … , x c + 2 n − 9 } ∪ { x c + 2 n − 1 − k + 4 y , x c + 2 n − 1 − k + 5 y , … , x c + 2 n − 9 y } .

Given that V 1 P 1 ∪ V 2 P 1 ∪ V 3 P 1 ∪ V 4 P 1 = { x c + 2 n − 2 − k + 3 , x c + 2 n − 2 − k + 4 , … , x c + 2 n − 9 } ∪ { x c + 2 n − 2 − k + 3 y , x c + 2 n − 2 − k + 4 y , … , x c + 2 n − 9 y } and ∣ { x c + 2 n − 2 − k + 3 , x c + 2 n − 2 − k + 4 , … , x c + 2 n − 9 } ∣ = ∣ { x c + 2 n − 2 − k + 3 y , x c + 2 n − 2 − k + 4 y , … , x c + 2 n − 9 y } ∣ = 2 n − 2 + 2 n − 1 + k − 11 ≥ 2 n − 1 for n ≥ 5 , we see that ( A 1 , A 2 , … , A k ) is a complete decomposition of Q 2 n of order k for k ∈ { 6 , 7 , … , 2 n − 1 } .□

An immediate consequence of Theorem 1 is as follows:

Corollary 3

Let A 1 , A 2 , … , A k be the mutually disjoint subsets of Q 2 n for n ≥ 5 . Then, there exists a complete decomposition of Q 2 n of order k for k ∈ { 6 , 7 , … , 2 n − 1 − 1 } , where ∑ i = 1 k ∣ A i ∣ < 2 n .

Proof

Let A 1 , A 2 , … , A k − 1 be the subsets as defined in Theorem 1 and A k = { x k − 1 , x k , … , x 2 n − 1 − 2 , x k − 1 y , x k y , … , x 2 n − 1 − 2 y } . By using a similar proof as in Theorem 1, we see that ( A 1 , A 2 , … , A k ) is a complete decomposition of Q 2 n of order k , for k ∈ { 6 , 7 , … , 2 n − 1 − 1 } , where ⋃ i = 1 k A i = Q 2 n \ { x 2 n − 1 − 1 , x 2 y } and ∑ i = 1 k ∣ A i ∣ = 2 n − 2 .□

Theorem 2

There exists a complete decomposition of Q 2 n of order 2 n − 1 , where the cardinality of every subset is equal to 2 for n ≥ 5 .

Proof

Let A 1 = { x , x y } , A 2 = { 1 , x 2 n − 1 − 1 y } , A 3 = { x 2 , y } , A r = { x r − 1 , x r − 1 y } (for r ∈ { 4 , 5 , … , 2 n − 1 − 1 } ), and A 2 n − 1 = { x 2 n − 1 − 1 , x 2 y } be the mutually disjoint subsets of Q 2 n for n ≥ 5 . From equations (5) and (6), it is clear that V 2 ∪ V 3 = { x c + 2 n − 1 − 2 n + 1 , x c + 2 n − 1 − 2 n + 2 , … , x c + 2 n − 1 − 10 } = R ⊆ A 1 A 2 … A 2 n − 1 − 1 and ∣ R ∣ = 2 n − 10 ≥ 2 n − 1 since n ≥ 5 . So, R = ⟨ x ⟩ , and we see that R A 2 n − 1 = Q 2 n . Thus, ( A 1 , A 2 , … , A 2 n − 1 ) is a complete decomposition of Q 2 n of order 2 n − 1 .□

We remark that Theorem 2 gives the smallest size for all the subsets A i with ∣ A i ∣ = 2 for i ∈ { 1 , 2 , … , 2 n − 1 } . All the elements in Q 2 n can be distributed into 2 n − 1 subsets, and finally, we see that Q 2 n can be written in the product of subsets, i.e., Q 2 n = A 1 A 2 ⋯ A 2 n − 1 .

3 Non-complete decomposition of Q 2 n

Let A , B ⊆ Q 2 n for n ≥ 3 . If A B ≠ Q 2 n , then ( A , B ) is not a complete decomposition of order 2. In order to show the existence of the non-complete decomposition of Q 8 of order 2, we let A = { 1 , x } and B = { x 2 , x 3 , y , x y , x 2 y , x 3 y } . Since A B = Q 8 \ { x } , ( A , B ) is a non-complete decomposition of Q 8 . By calculating the product of the subsets in Q 8 , we have 75 non-complete decompositions of Q 8 of order 2 with various sizes of the subsets A and B as given in Table 2.

Table 2

Non-complete decomposition ( A , B ) of Q 8 of order 2

∣ A ∣ ∣ B ∣ Number of non-complete decomposition
2 6 28
3 5 32
4 4 15

Let T be a non-empty subset of a finite group G . Let T m be the subset product given { t 1 ⋯ t m ∣ t 1 , … , t m ∈ T } . If there exists a positive integer n such that T n = G , then T is exhaustive. If T n ≠ G for any positive integer n , then T is said to be non-exhaustive.

Proposition 7

Let T 1 , … , T 5 be subsets of Q 2 n for n ≥ 3 as follows:

  1. T 1 = { y , x y , x 2 y , … , x 2 n − 1 − 1 y } ;

  2. T 2 = { 1 , x 2 , x 4 , … , x 2 k , … , x 2 n − 1 − 2 , y , x 2 y , x 4 y , … , x 2 k y , … , x 2 n − 1 − 2 y } ;

  3. T 3 = { 1 , x 2 , x 4 , … , x 2 k , … , x 2 n − 1 − 2 , x y , x 3 y , … , x 2 k + 1 y , … , x 2 n − 1 − 1 y } ;

  4. T 4 = { x , x 3 , … , x 2 k + 1 , … , x 2 n − 1 − 1 , x y , x 3 y , … , x 2 k + 1 y , … , x 2 n − 1 − 1 y } ;

  5. T 5 = { x , x 3 , … , x 2 k + 1 , … , x 2 n − 1 − 1 , y , x 2 y , x 4 y , … , x 2 k y , … , x 2 n − 1 − 2 y } .

Then, ∣ T i ∣ = 2 n − 1 and T r is non-exhaustive for r = 1 , … , 5 .

Proof

(i) Note that T 1 2 = { 1 , x , x 2 , … , x 2 n − 1 − 1 } is a subgroup of Q 2 n . Therefore, T 1 2 k ≠ Q 2 n for any integer k , which implies that T 1 is non-exhaustive.

(ii), (iii) Since x 2 i + 1 ∉ T 2 k , T 3 k for all i , k ∈ N , it follows that T 2 and T 3 are non-exhaustive.

(iv), (v) We observe that T 4 2 k , T 5 2 k ⊉ { x , x 3 , … , x 2 n − 1 − 1 } ∪ { x y , x 3 y , … , x 2 n − 1 − 1 y } and T 4 2 k + 1 , T 5 2 k + 1 ⊉ { 1 , x 2 , … , x 2 n − 1 − 2 } ∪ { y , x 2 y , … , x 2 n − 1 − 2 y } for k ∈ N . Hence, T 4 and T 5 are non-exhaustive.□

We show some constructions of non-complete decompositions ( A , B ) of Q 2 n of order 2 using the non-exhaustive subsets T r listed in Proposition 7.

Proposition 8

Let A i , B i ⊆ Q 2 n and ∣ A i ∣ = ∣ B i ∣ = 2 n − 1 as follows:

  1. A 1 = { 1 , x , … , x 2 n − 1 − 1 } , B 1 = { y , x y , … , x 2 n − 1 − 1 y } ;

  2. A 2 = { 1 , x 2 , … , x 2 n − 1 − 2 , y , x 2 y , … , x 2 n − 1 − 2 y } , B 2 = { x , x 3 , … , x 2 n − 1 − 1 , x y , x 3 y , … , x 2 n − 1 − 1 y } ;

  3. A 3 = { 1 , x 2 , … , x 2 n − 1 − 2 , x y , x 3 y , … , x 2 n − 1 − 1 y } , B 3 = { x , x 3 , … , x 2 n − 1 − 1 , y , x 2 y , … , x 2 n − 1 − 2 y } .

Then, ( A i , B i ) is a non-complete decomposition of Q 2 n of order 2 for n ≥ 3 and i ∈ { 1 , 2 , 3 } .

Proof

  1. Note that A 1 = ⟨ x ⟩ and B 1 = ⟨ x ⟩ y . One can easily verify that A 1 B 1 = ⟨ x ⟩ y , and hence the result holds.

  2. We see that A 2 B 2 = { x , x 3 , … , x 2 n − 1 − 1 } ∪ { x y , x 3 y , … , x 2 n − 1 − 1 y } . Since Q 2 n \ A 2 B 2 = { 1 , x 2 , … , x 2 n − 1 − 2 , y , x 2 y , … , x 2 n − 1 − 2 y } , so ( A 2 , B 2 ) is a non-complete decomposition of Q 2 n of order 2.

  3. We see that A 3 B 3 = { x , x 3 , … , x 2 n − 1 − 1 } ∪ { y , x 2 y , … , x 2 n − 1 − 2 y } and Q 2 n \ A 3 B 3 = { 1 , x 2 , … , x 2 n − 1 − 2 , x y , x 3 y , … , x 2 n − 1 − 1 y } . It follows that ( A 3 , B 3 ) is a non-complete decomposition of Q 2 n .□

Corollary 4

There exist A , B ⊂ Q 2 n , where ( A , B ) is a non-complete decomposition of order 2 such that ∣ A ∣ + ∣ B ∣ < 2 n for n ≥ 4 .

Proof

The result is clear from Proposition 8 since A ⊂ A i and B ⊂ B i for i ∈ { 1 , 2 , 3 } .□

Proposition 9

There exists a non-complete decomposition of Q 2 n of order 2 n − 1 for n ≥ 4 .

Proof

Let A i = { x 2 ( i − 1 ) , x 2 i − 1 } and B i = { x 2 ( i − 1 ) y , x 2 i − 1 y } for i ∈ { 1 , 2 , … , 2 n − 2 } . Then, ( A 1 , A 2 , … , A 2 n − 2 , B 1 , B 2 , … , B 2 n − 2 ) is a non-complete decomposition of order 2 n − 1 .□

By Proposition 9, we readily have the following corollary.

Corollary 5

There exists a non-complete decomposition of Q 2 n of order k for k ∈ { 3 , 4 , … , 2 n − 1 − 1 } , n ≥ 4 .

Proof

Let A i = { x 2 ( i − 1 ) , x 2 i − 1 } and B i = { x 2 ( i − 1 ) y , x 2 i − 1 y } for i ∈ { 1 , 2 , … , p } and p ∈ { 1 , 2 , … , 2 n − 2 − 1 } . When k = 2 p is even, we see that ( A 1 , … , A p , B 1 , … , B p ) is a non-complete decomposition of Q 2 n . When k = 2 p + 1 is odd, ( A 1 , … , A p , B 1 , … , B p , B p + 1 ) is a non-complete decomposition of Q 2 n , where B p + 1 = ⟨ x ⟩ y \ ⋃ j = 1 p B j .□

4 Conclusion

Group factorization is one type of group covering. Results on group factorization into subsets have found applications in various fields. In this article, we study the complete decomposition of the generalized quaternion group, which is obtained by factoring the generalized quaternion group into mutually disjoint subsets. We determine the existence of a complete decomposition of Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , where ∑ i = 1 k ∣ A i ∣ = 2 n and ∑ i = 1 k ∣ A i ∣ < 2 n . We show that Q 2 n can be written in the product of 2 n − 1 subsets, i.e., Q 2 n = A 1 A 2 ⋯ A 2 n − 1 where ∣ A j ∣ = 2 for j ∈ { 1 , 2 , … , 2 n − 1 } . In addition, we provide some constructions of non-complete decompositions of Q 2 n of order k , for k ∈ { 2 , 3 , … , 2 n − 1 } , using the non-exhaustive subsets of Q 2 n .

Acknowledgement

The authors are sincerely grateful to the anonymous referees for their careful comments.

  1. Conflict of interest: The authors state that there is no conflict of interest.

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Received: 2022-11-15
Revised: 2023-08-10
Accepted: 2023-08-13
Published Online: 2023-09-04

© 2023 the author(s), published by De Gruyter

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

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https://doi.org/10.1515/math-2023-0111
Keywords for this article
generalized quaternion groups; complete decomposition; subsets product
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Articles in the same Issue

  1. Special Issue on Future Directions of Further Developments in Mathematics
  2. What will the mathematics of tomorrow look like?
  3. On H 2-solutions for a Camassa-Holm type equation
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  90. Mean ergodic theorems for a sequence of nonexpansive mappings in complete CAT(0) spaces and its applications
  91. On some spaces via topological ideals
  92. Linear maps preserving equivalence or asymptotic equivalence on Banach space
  93. Well-posedness and stability analysis for Timoshenko beam system with Coleman-Gurtin's and Gurtin-Pipkin's thermal laws
  94. On a class of stochastic differential equations driven by the generalized stochastic mixed variational inequalities
  95. Entire solutions of two certain Fermat-type ordinary differential equations
  96. Generalized Lie n-derivations on arbitrary triangular algebras
  97. Markov decision processes approximation with coupled dynamics via Markov deterministic control systems
  98. Notes on pseudodifferential operators commutators and Lipschitz functions
  99. On Graham partitions twisted by the Legendre symbol
  100. Strong limit of processes constructed from a renewal process
  101. Construction of analytical solutions to systems of two stochastic differential equations
  102. Two-distance vertex-distinguishing index of sparse graphs
  103. Regularity and abundance on semigroups of partial transformations with invariant set
  104. Liouville theorems for Kirchhoff-type parabolic equations and system on the Heisenberg group
  105. Spin(8,C)-Higgs pairs over a compact Riemann surface
  106. Properties of locally semi-compact Ir-topological groups
  107. Transcendental entire solutions of several complex product-type nonlinear partial differential equations in ℂ2
  108. Ordering stability of Nash equilibria for a class of differential games
  109. A new reverse half-discrete Hilbert-type inequality with one partial sum involving one derivative function of higher order
  110. About a dubious proof of a correct result about closed Newton Cotes error formulas
  111. Ricci ϕ-invariance on almost cosymplectic three-manifolds
  112. Schur-power convexity of integral mean for convex functions on the coordinates
  113. A characterization of a ∼ admissible congruence on a weakly type B semigroup
  114. On Bohr's inequality for special subclasses of stable starlike harmonic mappings
  115. Properties of meromorphic solutions of first-order differential-difference equations
  116. A double-phase eigenvalue problem with large exponents
  117. On the number of perfect matchings in random polygonal chains
  118. Evolutoids and pedaloids of frontals on timelike surfaces
  119. A series expansion of a logarithmic expression and a decreasing property of the ratio of two logarithmic expressions containing cosine
  120. The 𝔪-WG° inverse in the Minkowski space
  121. Stability result for Lord Shulman swelling porous thermo-elastic soils with distributed delay term
  122. Approximate solvability method for nonlocal impulsive evolution equation
  123. Construction of a functional by a given second-order Ito stochastic equation
  124. Global well-posedness of initial-boundary value problem of fifth-order KdV equation posed on finite interval
  125. On pomonoid of partial transformations of a poset
  126. New fractional integral inequalities via Euler's beta function
  127. An efficient Legendre-Galerkin approximation for the fourth-order equation with singular potential and SSP boundary condition
  128. Eigenfunctions in Finsler Gaussian solitons
  129. On a blow-up criterion for solution of 3D fractional Navier-Stokes-Coriolis equations in Lei-Lin-Gevrey spaces
  130. Some estimates for commutators of sharp maximal function on the p-adic Lebesgue spaces
  131. A preconditioned iterative method for coupled fractional partial differential equation in European option pricing
  132. A digital Jordan surface theorem with respect to a graph connectedness
  133. A quasi-boundary value regularization method for the spherically symmetric backward heat conduction problem
  134. The structure fault tolerance of burnt pancake networks
  135. Average value of the divisor class numbers of real cubic function fields
  136. Uniqueness of exponential polynomials
  137. An application of Hayashi's inequality in numerical integration
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