International Journal of Mathematics and Mathematical Sciences
Volume 2009 (2009), Article ID 264150, 10 pages
doi:10.1155/2009/264150
Research Article

Commutators and Squares in Free Nilpotent Groups

Mehri Akhavan-Malayeri

Department of Mathematics, Alzahra University, Vank, Tehranm 19834, Iran

Received 2 August 2009; Revised 24 November 2009; Accepted 1 December 2009

Academic Editor: Howard Bell

Copyright © 2009 Mehri Akhavan-Malayeri. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In a free group no nontrivial commutator is a square. And in the free group 𝐹 2 = 𝐹 ( 𝑥 1 , 𝑥 2 ) freely generated by 𝑥 1 , 𝑥 2 the commutator [ 𝑥 1 , 𝑥 2 ] is never the product of two squares in 𝐹 2 , although it is always the product of three squares. Let 𝐹 2 , 3 = ⟨ 𝑥 1 , 𝑥 2 ⟩ be a free nilpotent group of rank 2 and class 3 freely generated by 𝑥 1 , 𝑥 2 . We prove that in 𝐹 2 , 3 = ⟨ 𝑥 1 , 𝑥 2 ⟩ , it is possible to write certain commutators as a square. We denote by 𝑆 𝑞 ( 𝛾 ) the minimal number of squares which is required to write 𝛾 as a product of squares in group 𝐺 . And we define 𝑆 𝑞 ( 𝐺 ) = s u p { 𝑆 𝑞 ( 𝛾 ) ; 𝛾 ∈ 𝐺 } . We discuss the question of when the square length of a given commutator of 𝐹 2 , 3 is equal to 1 or 2 or 3. The precise formulas for expressing any commutator of 𝐹 2 , 3 as the minimal number of squares are given. Finally as an application of these results we prove that 𝑆 𝑞 ( 𝐹 2 , 3 ) = 3 .

1. Introduction

Sch ̈ u tzenberger [1] proved that in a free group the equation

[ ] 𝑥 , 𝑦 = 𝑧 𝑟 , 𝑟 ≥ 2 ( 1 . 1 ) implies 𝑧 = 1 ; that is, no nontrivial commutator is a proper power. It means that it is impossible to write [ 𝑥 , 𝑦 ] as an 𝑟 th powers where 𝑟 ≥ 2 . Lyndon and Newman [2] have shown that in the free group 𝐹 2 = 𝐹 ( 𝑥 1 , 𝑥 2 ) freely generated by 𝑥 1 , 𝑥 2 , the commutator [ 𝑥 1 , 𝑥 2 ] is never a product of two squares in 𝐹 2 , although it is always the product of three squares. In [3] we proved that for an odd integer 𝑘 , [ 𝑥 2 , 𝑥 1 ] 𝑘 is not a product of two squares in 𝐹 2 , and it is the product of three squares. Put 𝑤 = [ 𝑥 2 , 𝑥 1 ] and 𝑘 = 2 𝑛 + 1 . We presented the following expression of [ 𝑥 2 , 𝑥 1 ] 2 𝑛 + 1 as a product of the minimal number of squares:

𝑥 2 , 𝑥 1 2 𝑛 + 1 = 𝑤 𝑛 𝑥 2 𝑥 1 𝑤 𝑛 2 𝑤 𝑛 𝑥 1 − 1 2 𝑤 − 𝑛 𝑥 2 − 1 𝑥 1 2 . ( 1 . 2 ) Recently Abdollahi [4] generalized these results as the following theorem.

Theorem 1.1 (Abdollahi [4]). Let 𝐹 be a free group with a basis of distinct elements 𝑥 1 , … , 𝑥 2 𝑛 , and 𝑁 any odd integer. Then there exist elements 𝑢 1 , … , 𝑢 𝑚 in 𝐹 such that 𝑥 1 , 𝑥 2 ⋯ 𝑥 2 𝑛 − 1 , 𝑥 2 𝑛 𝑁 = 𝑢 2 1 … , 𝑢 2 𝑚 ( 1 . 3 ) if and only if 𝑚 ≥ 2 𝑛 + 1 .

Definition 1.2. Let 𝐺 be a group and 𝛾 ∈ 𝐺 . The minimal number of squares which is required to write 𝛾 as a product of squares in 𝐺 is called the square length of 𝛾 and denoted by S q ( 𝛾 ) . And we define S q ( 𝐺 ) = s u p { S q ( 𝛾 ) ; 𝛾 ∈ 𝐺 } .

We prove that in the free nilpotent group 𝐹 2 , 3 = ⟨ 𝑥 1 , 𝑥 2 ⟩ of rank 2 and class 3 freely generated by 𝑥 1 , 𝑥 2 it is possible to write certain nontrivial commutators as a proper power. We consider certain equations over free group 𝐹 2 , 3 . Using this, we find S q [ ℎ , 𝑔 ] where ℎ , 𝑔 ∈ 𝐹 2 , 3 . Then we prove that S q ( 𝐹 2 , 3 ) = 3 .

2. Main Results

We will prove the following theorems.

Theorem 2.1. Let 𝐹 2 , 3 = ⟨ 𝑥 1 , 𝑥 2 ⟩ be a free nilpotent group of rank 2 and class 3 freely generated by 𝑥 1 , 𝑥 2 . Then S q ( 𝐹 2 , 3 ) = 3 .

An application of Theorem 2.1 is displayed in the next result.

Corollary 2.2. In a free nilpotent group of rank 2 and class 3 , it is possible to find nontrivial solutions for the equation [ ] 𝑥 , 𝑦 = 𝑧 𝑟 , 𝑟 ≥ 2 . ( 2 . 1 )

We will use the following well-known identities regarding groups which are nilpotent of class 3.

Lemma 2.3. Let 𝐺 = ⟨ 𝑥 , 𝑦 ⟩ be nilpotent of class 3. Then, for all integers 𝑟 , 𝑠 the following hold: 𝑥 𝑟 = [ ] , 𝑦 𝑥 , 𝑦 𝑟 [ ] 𝑥 , 𝑦 , 𝑥 𝑟 ( 𝑟 − 1 ) / 2 , 𝑥 𝑟 , 𝑦 𝑠 = [ ] 𝑥 , 𝑦 𝑟 𝑠 [ ] 𝑥 , 𝑦 , 𝑥 𝑟 𝑠 ( 𝑟 − 1 ) / 2 [ ] 𝑥 , 𝑦 , 𝑦 𝑟 𝑠 ( 𝑠 − 1 ) / 2 . ( 2 . 2 )

3. Proofs of the Main Result

Proof of Theorem 2.1. Let ℎ , 𝑔 be any two elements of 𝐹 2 , 3 ⧵ 𝛾 3 ( 𝐹 2 , 3 ) . First we study the form of the element [ ℎ , 𝑔 ] . Since 𝛾 3 ( 𝐹 2 , 3 ) lies in the center of 𝐹 2 , 3 we may express ℎ as 𝑥 𝑟 1 1 𝑥 𝑟 2 2 [ 𝑥 2 , 𝑥 1 ] 𝛽 and 𝑔 as 𝑥 𝑠 1 1 𝑥 𝑠 2 2 [ 𝑥 2 , 𝑥 1 ] 𝛼 . We have shown in [5] that. [ ] = 𝑥 ℎ , 𝑔 2 , 𝑥 1 𝜆 𝑥 2 , 𝑥 1 , 𝑥 2 𝜇 𝑥 2 , 𝑥 1 , 𝑥 1 𝜈 , ( 3 . 1 ) where 𝜆 = 𝑟 2 𝑠 1 − 𝑟 1 𝑠 2 , 𝑠 𝜇 = 1 𝑟 2 𝑟 2 − 1 2 − 𝑟 1 𝑠 2 𝑠 2 − 1 2 − 𝑟 1 𝑟 2 𝑠 2 + 𝑟 2 𝑠 1 𝑠 2 + 𝛽 𝑠 2 − 𝛼 𝑟 2 , 𝑟 𝜈 = 2 𝑠 1 𝑠 1 − 1 2 − 𝑠 2 𝑟 1 𝑟 1 − 1 2 + 𝛽 𝑠 1 − 𝛼 𝑟 1 . ( 3 . 2 )
Now we consider the equation [ ℎ , 𝑔 ] = 𝑢 2 ( ⋄ ) . The element 𝑢 has a presentation of the following form: 𝑢 = 𝑥 𝑟 ′ 1 1 𝑥 𝑟 ′ 2 2 𝑥 2 , 𝑥 1 𝛼 ′ 𝑥 2 , 𝑥 1 , 𝑥 2 𝛾 ′ 𝑥 2 , 𝑥 1 , 𝑥 1 𝛽 ′ , ( 3 . 3 ) where 𝑟 1 , 𝑟 2 , 𝛼 , 𝛽 , a n d 𝛾 are unique integer elements.
Lemma 2.3 implies that 𝑢 2 = 𝑥 2 𝑟 ′ 1 1 𝑥 2 𝑟 ′ 2 2 𝑥 2 , 𝑥 1 2 𝛼 ′ + 𝑟 ′ 1 𝑟 ′ 2 𝑥 2 , 𝑥 1 , 𝑥 2 2 𝛾 ′ + 𝛼 ′ 𝑟 ′ 2 + 𝑟 ′ 1 𝑟 ′ 2 ( 𝑟 ′ 2 − 1 ) / 2 + 𝑟 ′ 1 𝑟 ′ 2 2 × 𝑥 2 , 𝑥 1 , 𝑥 1 2 𝛽 ′ + 𝛼 ′ 𝑟 ′ 1 + 𝑟 ′ 1 𝑟 ′ 2 ( 𝑟 ′ 1 − 1 ) / 2 . ( 3 . 4 )
Thus equation ( ⋄ ) holds in 𝐹 2 , 3 if and only if 𝑟 1 = 𝑟 2 = 0 , 2 𝛼 = 𝜆 , 2 𝛽 = 𝜈 , 2 𝛾 = 𝜇 . ( 3 . 5 )
In particular the equation ( ⋄ ) has a solution only if 𝜆 , 𝜇 , a n d 𝜈 are even. Put 𝑐 1 = 𝛼 𝑟 2 − 𝛽 𝑠 2 , 𝑐 2 = 𝛼 𝑟 1 − 𝛽 𝑠 1 , then | | 𝛼 = 𝑐 1 − 𝑠 2 𝑐 2 − 𝑠 1 | | | | 𝑟 2 − 𝑠 2 𝑟 1 − 𝑠 1 | | = 𝑠 1 𝑐 1 − 𝑠 2 𝑐 2 2 𝛼 | | , 𝛽 = 𝑟 2 𝑐 1 𝑟 1 𝑐 2 | | − 2 𝛼 = 𝑟 1 𝑐 1 − 𝑟 2 𝑐 2 2 𝛼 . ( 3 . 6 ) Hence we need 𝑠 1 𝑐 1 − 𝑠 2 𝑐 2 and 𝑟 1 𝑐 1 − 𝑟 2 𝑐 2 to be even. We have the following two cases.
Case. If 𝑟 1 𝑠 2 = 2 𝑘 , for some integer 𝑘 , then 𝑟 2 𝑠 1 = 2 𝛼 + 2 𝑘 , and hence 𝑟 2 𝑠 1 ≡ 2 0 . And we have 𝑐 1 = − 𝛼 + 𝛼 𝑟 2 + 𝑠 2 − 𝑘 𝑟 2 + 𝛼 𝑠 + 𝑘 2 − 2 𝛾 , 𝑐 2 = 𝛼 + 𝛼 𝑠 + 𝑘 1 − 𝑘 𝑟 1 − 2 𝛽 . ( 3 . 7 ) Further, 0 ≡ 2 𝑠 1 𝑐 1 + 𝑠 2 𝑐 2 ≡ 2 𝛼 𝑠 1 + 𝑠 1 𝑠 2 + 𝑠 2 , 0 ≡ 2 𝑟 1 𝑐 1 + 𝑟 2 𝑐 2 ≡ 2 𝛼 𝑟 1 + 𝑟 1 𝑟 2 + 𝑟 2 . ( 3 . 8 ) Now if 𝛼 is an odd integer, then we have 0 ≡ 2 𝑟 1 + 𝑟 1 𝑟 2 + 𝑟 2 ≡ 2 𝑠 1 + 𝑠 1 𝑠 2 + 𝑠 2 . ( 3 . 9 ) It follows that 𝑟 1 , 𝑟 2 , 𝑠 1 , and 𝑠 2 are all even. Hence 𝜆 = 𝑟 2 𝑠 1 − 𝑟 1 𝑠 2 is divisible by 4. But 𝜆 = 2 𝛼 implies that 𝛼 ≡ 2 0 , a contradiction. Hence in Case 1 we have 𝛼 ≡ 2 0 and 𝜆 ≡ 4 0 .
Now 𝑟 1 𝑠 2 = 2 𝑘 , and 𝑟 2 𝑠 1 = 2 𝛼 + 2 𝑘 imply that 𝜇 = 𝛼 𝑟 2 − 𝑘 𝑟 2 − 𝛼 + 𝑘 𝑠 2 + 2 𝛼 𝑠 2 + 𝛽 𝑠 2 − 𝛼 𝑟 2 = 2 𝛾 , 𝜈 = 𝛼 𝑠 1 + 𝑘 𝑠 1 − 𝛼 − 𝑘 𝑟 1 + 𝛽 𝑠 1 − 𝛼 𝑟 1 = 2 𝛽 . ( 3 . 1 0 ) Hence we have 𝜇 ≡ 2 𝑟 2 ( 𝑘 + 𝛼 ) + 𝑠 2 ( 𝑘 + 𝛽 ) , 𝜈 ≡ 2 𝑟 1 ( 𝑘 + 𝛼 ) + 𝑠 1 ( 𝑘 + 𝛽 ) . ( 3 . 1 1 ) And we have the following cases. Subcase 1.1. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 1 ≡ 2 𝑠 2 ≡ 2 0 , then it is clear that for any integer numbers 𝛼 and 𝛽 we have; 𝜆 ≡ 4 0 , 𝜇 ≡ 2 𝜈 ≡ 2 0 . ( 3 . 1 2 ) And the equation ( ⋄ ) has solution.Subcase 1.2. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 and 𝑠 2 ≡ 2 1 , then 𝑟 1 𝑠 2 ≡ 4 𝜆 ≡ 4 0 . We have the following two cases.(1.2.1) If 𝑟 1 ≡ 4 0 , then we have 𝜆 ≡ 4 0 . Also from 𝑟 1 𝑠 2 = 2 𝑘 , it follows that 𝑘 ≡ 2 0 . Now if we choose 𝛽 ≡ 2 0 , then from (3.11) it follows that 𝜇 ≡ 2 0 and 𝜈 ≡ 2 0 for any 𝛼 ∈ ℤ . And in this case the equation ( ⋄ ) has a solution.(1.2.2) If 𝑟 1 ≡ 4 2 , then 𝜆 ≡ 4 2 , and the equation ( ⋄ ) has no solution.
Hence in Subcase 1.2 if 𝑟 1 ≡ 4 0 , 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 , 𝑠 2 ≡ 2 1 , and 𝛽 ≡ 2 0 , for any 𝛼 ∈ ℤ the equation ( ⋄ ) has a solution.Subcase 1.3. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 2 ≡ 2 0 and 𝑠 1 ≡ 2 1 , then 𝑠 1 𝑟 2 ≡ 4 𝜆 ≡ 4 0 . We have two cases.(1.3.1) If 𝑟 2 ≡ 4 0 , then 𝜆 ≡ 4 0 . Since 𝑟 1 𝑠 2 = 2 𝑘 , and 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , hence 𝑘 ≡ 2 0 . Now if we identify 𝛽 ≡ 2 0 , then from (3.11) it follows that 𝜇 ≡ 2 0 and 𝜈 ≡ 2 0 . And the equation ( ⋄ ) has a solution.(1.3.2) If 𝑟 2 ≡ 4 2 , then 𝜆 ≡ 4 2 , and the equation ( ⋄ ) has no solution.
Hence in Subcase 1.3 if 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑟 2 ≡ 4 0 , and 𝛽 ≡ 2 0 , for any 𝛼 ∈ ℤ the equation ( ⋄ ) has a solution.Subcase 1.4. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 0 and 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , then we have the following two cases.(1.4.1) If 𝑟 1 ≡ 4 0 , then 𝜆 ≡ 4 𝑠 1 𝑟 2 ≡ 4 0 . Now 𝑠 1 ≡ 2 1 implies 𝑟 2 ≡ 4 2 . If we choose 𝛽 ≡ 2 0 , then for any 𝛼 ∈ ℤ the equation ( ⋄ ) has a solution. Hence if 𝑟 1 ≡ 4 𝑟 2 ≡ 4 0 , 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , and 𝛽 ≡ 2 0 , then for any 𝛼 ∈ ℤ , the equation ( ⋄ ) has a solution.(1.4.2) If 𝑟 1 ≡ 4 2 . Since 𝜆 ≡ 4 𝑠 1 𝑟 2 − 𝑟 1 𝑠 2 ≡ 4 0 , hence 𝑟 2 ≡ 4 2 . If we identify 𝛽 ≡ 2 1 , for any 𝛼 ∈ ℤ then 𝜇 ≡ 2 𝜈 ≡ 2 0 . And the equation ( ⋄ ) has a solution.Subcase 1.5. If 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑟 2 ≡ 2 0 , and 𝑟 2 ≡ 2 1 , we have the following two cases.(1.5.1) If 𝑠 1 ≡ 4 0 , then 𝜆 ≡ 4 0 . Since 𝑟 1 𝑠 2 = 2 𝑘 , hence 𝑘 ≡ 2 0 . If we identify 𝛼 ≡ 2 0 , for any 𝛽 ∈ ℤ , then 𝜇 ≡ 2 𝜈 ≡ 2 0 . And the equation ( ⋄ ) has a solution.(1.5.2) If 𝑠 1 ≡ 4 2 , then 𝜆 ≡ 4 2 . And the equation ( ⋄ ) has no solution. Hence in this case only if 𝑠 1 ≡ 4 0 , the equation ( ⋄ ) has a solution.Subcase 1.6. If 𝑟 1 ≡ 2 𝑠 1 ≡ 2 0 and 𝑟 2 ≡ 2 𝑠 2 ≡ 2 1 , then similar to Case 4 ,if 𝑟 1 ≡ 4 𝑠 1 ≡ 4 0 or 𝑟 1 ≡ 4 𝑠 1 ≡ 4 2 then 𝜆 ≡ 4 0 . And for any 𝛼 ≡ 2 𝛽 , 𝜇 ≡ 2 𝜈 ≡ 2 0 , the equation ( ⋄ ) has a solution.Subcase 1.7. If 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 and 𝑟 2 ≡ 2 𝑠 1 ≡ 2 1 , then 𝜆 ≡ 2 1 . Hence the equation ( ⋄ ) has no solution.Subcase 1.8. If 𝑟 1 ≡ 2 0 and 𝑟 2 ≡ 2 𝑠 2 ≡ 2 𝑠 1 ≡ 2 1 , then 𝜆 ≡ 2 1 . Hence the equation ( ⋄ ) has no solution.Subcase 1.9. If 𝑟 1 ≡ 2 1 and 𝑟 2 ≡ 2 𝑠 2 ≡ 2 𝑠 1 ≡ 2 0 , we have two cases.(1.9.1) If 𝑠 2 ≡ 4 0 , then 𝜆 ≡ 4 0 . Since 𝑟 1 𝑠 2 = 2 𝑘 , hence 𝑘 ≡ 2 0 . If we identify 𝛼 ≡ 2 0 , for any 𝛽 ∈ ℤ , then 𝜇 ≡ 2 𝜈 ≡ 2 0 . And the equation ( ⋄ ) has a solution.(1.9.2) If 𝑠 2 ≡ 4 2 , then 𝜆 ≡ 4 2 . And the equation ( ⋄ ) has no solution.Subcase 1.10. If 𝑟 1 ≡ 2 𝑠 2 ≡ 2 1 and 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 , then 𝑟 1 𝑠 2 ≡ 2 1 . And the equation ( ⋄ ) has no solution.Subcase 1.11. If 𝑟 1 ≡ 2 𝑠 1 ≡ 2 1 and 𝑟 2 ≡ 2 𝑠 2 ≡ 2 0 , then similar to Subcase 1.6, if 𝑟 2 ≡ 4 𝑠 2 ≡ 4 0 or 𝑟 2 ≡ 4 𝑠 2 ≡ 4 2 then 𝜆 ≡ 4 0 . And for any 𝛼 ≡ 2 𝛽 , 𝜇 ≡ 2 𝜈 ≡ 2 0 , the equation ( ⋄ ) has a solution.Subcase 1.12. If 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 and 𝑟 2 ≡ 2 0 , then 𝑟 1 𝑠 2 ≡ 2 1 . And the equation ( ⋄ ) has no solution.Subcase 1.13. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 1 and 𝑠 1 ≡ 2 𝑠 2 ≡ 2 0 , then we have two cases.(1.13.1) If 𝑠 1 ≡ 4 0 , then 𝜆 ≡ 4 0 implies 𝑠 2 ≡ 2 0 . If we identify 𝛼 ≡ 2 0 , for any 𝛽 ∈ ℤ , the equation ( ⋄ ) has a solution.(1.13.2) If 𝑠 1 ≡ 4 2 , then 𝑠 2 ≡ 4 2 . And if 𝛼 ≡ 2 1 , for any 𝛽 ∈ ℤ , the equation ( ⋄ ) has a solution.Subcase 1.14. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 2 ≡ 2 1 and 𝑠 1 ≡ 2 0 , then 𝑟 1 𝑠 2 ≡ 2 1 . In this case the equation ( ⋄ ) has no solution.Subcase 1.15. If 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 1 ≡ 2 1 and 𝑠 1 ≡ 2 0 , then 𝑟 2 𝑠 1 ≡ 2 1 . In this case the equation ( ⋄ ) has no solution.Case 2. If 𝑟 1 𝑠 2 ≡ 2 1 . Since 𝜆 = 𝑠 1 𝑟 2 − 𝑟 1 𝑠 2 ≡ 2 0 , hence 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 . If we identify 𝛼 ≡ 2 𝛽 , then 𝜇 ≡ 2 𝜈 ≡ 2 0 . In this case the equation ( ⋄ ) has a solution.
Hence we show that in the following twelve cases the equation ( ⋄ ) has solution. And S q [ ℎ , 𝑔 ] = 1 .
(1) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑟 2 ≡ 2 𝑠 2 ≡ 2 0 ,   for all 𝛼 , 𝛽 . (2) 𝑠 1 ≡ 2 𝑟 2 ≡ 2 0 , 𝑠 2 ≡ 2 1 , 𝑟 1 ≡ 4 0 ,  for all 𝛼 , 𝛽 ≡ 2 0 . (3) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑠 1 ≡ 2 1 , 𝑟 2 ≡ 4 0 , f o r a l l 𝛼 , 𝛽 ≡ 2 0 . (4) 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , 𝑟 1 ≡ 4 𝑟 2 ≡ 2 0 , f o r a l l 𝛼 , 𝛽 ≡ 2 0 . (5) 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , 𝑟 1 ≡ 4 𝑟 2 ≡ 4 2 , f o r a l l 𝛼 , 𝛽 ≡ 2 0 . (6) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑟 2 ≡ 2 1 , 𝑠 1 ≡ 4 0 , 𝛼 ≡ 2 0 , f o r a l l 𝛽 . (7) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 1 , 𝑟 2 ≡ 2 𝑠 2 ≡ 2 0 , 𝛼 ≡ 2 𝛽 . (8) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 0 , 𝑟 2 ≡ 2 𝑠 2 ≡ 2 1 , 𝛼 ≡ 2 𝛽 . (9) 𝑟 1 ≡ 2 1 , 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 , 𝑠 2 ≡ 4 0 , 𝛼 ≡ 2 0 , f o r a l l 𝛽 . (10) 𝑟 1 ≡ 2 𝑟 2 ≡ 2 1 , 𝑠 1 ≡ 4 𝑠 2 ≡ 4 0 , 𝛼 ≡ 2 0 , f o r a l l 𝛽 . (11) 𝑟 1 ≡ 2 𝑟 2 ≡ 2 1 , 𝑠 1 ≡ 4 𝑠 2 ≡ 4 2 , 𝛼 ≡ 2 1 , f o r a l l 𝛽 . (12) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑟 2 ≡ 2 𝑠 2 ≡ 2 1 , 𝛼 ≡ 2 𝛽 .
And more precisely we have [ ] = 𝑥 ℎ , 𝑔 2 , 𝑥 1 𝜆 / 2 𝑥 2 , 𝑥 1 , 𝑥 2 𝜇 / 2 𝑥 2 , 𝑥 1 , 𝑥 1 𝜈 / 2 2 . ( 3 . 1 3 )
Now in the following ten cases the equation ( ⋄ ) has no solution.(13) 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 , 𝑠 2 ≡ 2 1 , 𝑟 1 ≡ 4 2 . (14) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑠 1 ≡ 2 1 , 𝑟 2 ≡ 4 2 . (15) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑟 2 ≡ 2 1 , 𝑠 1 ≡ 4 2 . (16) 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 , 𝑟 1 ≡ 2 1 , 𝑠 2 ≡ 4 2 . (17) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 0 , 𝑟 2 ≡ 2 𝑠 1 ≡ 2 1 . (18) 𝑟 1 ≡ 2 𝑠 2 ≡ 2 1 , 𝑟 2 ≡ 2 𝑠 1 ≡ 2 0 . (19) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , 𝑟 2 ≡ 2 0 . (20) 𝑟 1 ≡ 2 𝑟 2 ≡ 2 𝑠 2 ≡ 2 1 , 𝑠 1 ≡ 2 0 . (21) 𝑟 1 ≡ 2 𝑠 1 ≡ 2 𝑟 2 ≡ 2 1 , 𝑠 2 ≡ 2 0 . (22) 𝑟 2 ≡ 2 𝑠 1 ≡ 2 𝑠 2 ≡ 2 1 , 𝑟 1 ≡ 2 0 .
We consider the equation [ ℎ , 𝑔 ] = 𝑢 2 1 𝑢 2 2 ( ◊ ) . Suppose that the equation ( ◊ ) has a nontrivial solution ( 𝑢 1 , 𝑢 2 ) . The elements 𝑢 1 and 𝑢 2 have a representation of the following forms: 𝑢 1 = 𝑥 𝑟 1 1 1 𝑥 𝑟 2 1 2 𝑥 2 , 𝑥 1 𝛼 1 𝑥 2 , 𝑥 1 , 𝑥 1 𝛽 1 𝑥 2 , 𝑥 1 , 𝑥 2 𝛾 1 , 𝑢 2 = 𝑥 𝑟 1 2 1 𝑥 𝑟 2 2 2 𝑥 2 , 𝑥 1 𝛼 2 𝑥 2 , 𝑥 1 , 𝑥 1 𝛽 2 𝑥 2 , 𝑥 1 , 𝑥 2 𝛾 2 , ( 3 . 1 4 ) where 𝑟 𝑖 𝑗 , 𝛼 𝑖 , 𝛽 𝑖 , a n d 𝛾 𝑖 are unique integer numbers. By applying Lemma 2.3 one obtains 𝑢 2 𝑖 = 𝑥 2 𝑟 1 𝑖 1 𝑥 2 𝑟 2 𝑖 2 𝑥 2 , 𝑥 1 2 𝛼 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 × 𝑥 2 , 𝑥 1 , 𝑥 1 2 𝛽 𝑖 + 𝛼 𝑖 𝑟 1 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 ( ( 𝑟 1 𝑖 − 1 ) / 2 ) × 𝑥 2 , 𝑥 1 , 𝑥 2 2 𝛾 𝑖 + 𝛼 𝑖 𝑟 2 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 ( 𝑟 2 𝑖 − 1 2 ) + 𝑟 1 𝑖 𝑟 2 2 𝑖 . ( 3 . 1 5 ) Hence 𝑢 2 1 𝑢 2 2 = 𝑥 2 ( 𝑟 1 1 + 𝑟 1 2 ) 1 𝑥 2 ( 𝑟 2 1 + 𝑟 1 2 2 ) 2 𝑥 2 , 𝑥 1 2 ( 𝛼 1 + 𝛼 2 ) + 𝑟 1 1 𝑟 2 1 + 𝑟 1 2 𝑟 2 2 + 4 𝑟 2 1 𝑟 1 2 × 𝑥 2 , 𝑥 1 , 𝑥 2 𝑛 1 + 𝑛 2 + 2 𝑘 1 𝑟 2 2 + 4 𝑟 2 1 𝑟 1 2 ( ( 2 𝑟 2 1 − 1 ) / 2 ) + 8 𝑟 2 1 𝑟 1 2 𝑟 2 2 × 𝑥 2 , 𝑥 1 , 𝑥 1 𝑚 1 + 𝑚 2 + 2 𝑘 1 𝑟 1 2 + 4 𝑟 2 1 𝑟 1 2 ( ( 2 𝑟 1 2 − 1 ) / 2 ) , ( 3 . 1 6 ) where for 𝑖 = 1 , 2 ,
𝑘 𝑖 = 2 𝛼 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 , 𝑚 𝑖 = 2 𝛽 𝑖 + 𝛼 𝑖 𝑟 1 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 𝑟 1 𝑖 − 1 2 , 𝑛 𝑖 = 2 𝛾 𝑖 + 𝛼 𝑖 𝑟 2 𝑖 + 𝑟 1 𝑖 𝑟 2 𝑖 𝑟 2 𝑖 − 1 2 + 𝑟 1 𝑖 𝑟 2 2 𝑖 . ( 3 . 1 7 )
Hence equation ( ◊ ) holds if 𝑟 1 1 = − 𝑟 1 2 , 𝑟 2 1 = − 𝑟 2 1 , 𝛼 𝜆 = 2 1 + 𝛼 2 − 2 𝑟 1 1 𝑟 2 1 , 𝛾 𝜇 = 2 1 + 𝛾 2 + 𝑟 2 1 𝛼 1 − 𝛼 2 − 2 𝑘 1 𝑟 2 1 + 𝑟 1 1 𝑟 2 1 − 4 𝑟 2 1 , 𝛽 + 1 𝜈 = 2 1 + 𝛽 2 + 𝑟 1 1 𝛼 1 − 𝛼 2 − 2 𝑘 1 𝑟 1 1 + 𝑟 1 1 𝑟 2 1 4 𝑟 1 1 . + 1 ( 3 . 1 8 ) Note that second equation gives 𝜆 ≡ 2 0 ; hence equation ( ◊ ) has nontrivial solution only if