Commutators and Squares in Free Nilpotent Groups

Abstract

In a free group no nontrivial commutator is a square. And in the free group 𝐹₂=𝐹(𝑥₁,𝑥₂) freely generated by 𝑥₁,𝑥₂ the commutator [𝑥₁,𝑥₂] is never the product of two squares in 𝐹₂, although it is always the product of three squares. Let 𝐹_2,3=⟨𝑥₁,𝑥₂⟩ be a free nilpotent group of rank 2 and class 3 freely generated by 𝑥₁,𝑥₂. We prove that in 𝐹_2,3=⟨𝑥₁,𝑥₂⟩, it is possible to write certain commutators as a square. We denote by 𝑆𝑞(𝛾) the minimal number of squares which is required to write 𝛾 as a product of squares in group 𝐺. And we define 𝑆𝑞(𝐺)=sup{𝑆𝑞(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 𝑆𝑞(𝐹^′_2,3)=3.

1. Introduction

Schützenberger [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 𝑟th powers where 𝑟≥2. Lyndon and Newman [2] have shown that in the free group 𝐹₂=𝐹(𝑥₁,𝑥₂) freely generated by 𝑥₁,𝑥₂, the commutator [𝑥₁,𝑥₂] is never a product of two squares in 𝐹₂, although it is always the product of three squares. In [3] we proved that for an odd integer 𝑘,[𝑥₂,𝑥₁]^𝑘 is not a product of two squares in 𝐹₂, and it is the product of three squares. Put 𝑤=[𝑥₂,𝑥₁] and 𝑘=2𝑛+1. We presented the following expression of [𝑥₂,𝑥₁]^2𝑛+1 as a product of the minimal number of squares:

[𝑥₂,𝑥₁]^2𝑛+1=((𝑤^𝑛𝑥₂𝑥₁)^𝑤𝑛)²(𝑤^𝑛𝑥₁⁻¹)²((𝑤^−𝑛𝑥₂⁻¹)^𝑥1)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 𝑥₁,…,𝑥_2𝑛, and 𝑁 any odd integer. Then there exist elements 𝑢₁,…,𝑢_𝑚 in 𝐹 such that ([𝑥₁,𝑥₂]⋯[𝑥_2𝑛−1,𝑥_2𝑛])^𝑁=𝑢²₁…,𝑢²_𝑚 (1.3) if and only if 𝑚≥2𝑛+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 Sq(𝛾). And we define Sq(𝐺)=sup{Sq(𝛾);𝛾∈𝐺^′}.

We prove that in the free nilpotent group 𝐹_2,3=⟨𝑥₁,𝑥₂⟩ of rank 2 and class 3 freely generated by 𝑥₁,𝑥₂ 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 Sq[ℎ,𝑔] where ℎ,𝑔∈𝐹_2,3. Then we prove that Sq(𝐹^′_2,3)=3.

2. Main Results

We will prove the following theorems.

Theorem 2.1. Let 𝐹_2,3=⟨𝑥₁,𝑥₂⟩ be a free nilpotent group of rank 2 and class 3 freely generated by 𝑥₁,𝑥₂. Then Sq(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,𝑦]^𝑟[𝑥,𝑦,𝑥]^{𝑟(𝑟−1)/2},[𝑥^𝑟,𝑦^𝑠]=[𝑥,𝑦]^𝑟𝑠[𝑥,𝑦,𝑥]^{𝑟𝑠(𝑟−1)/2}[𝑥,𝑦,𝑦]^{𝑟𝑠(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⧵𝛾₃(𝐹_2,3). First we study the form of the element [ℎ,𝑔]. Since 𝛾₃(𝐹_2,3) lies in the center of 𝐹_2,3 we may express ℎ as 𝑥^𝑟1₁𝑥^𝑟2₂[𝑥₂,𝑥₁]^𝛽 and 𝑔 as 𝑥^𝑠1₁𝑥^𝑠2₂[𝑥₂,𝑥₁]^𝛼. We have shown in [5] that. [ℎ,𝑔]=[𝑥₂,𝑥₁]^𝜆[𝑥₂,𝑥₁,𝑥₂]^𝜇[𝑥₂,𝑥₁,𝑥₁]^𝜈, (3.1) where 𝜆=𝑟₂𝑠₁−𝑟₁𝑠₂,𝑠𝜇=¹𝑟₂(𝑟₂−1)2−𝑟₁𝑠₂(𝑠₂−1)2−𝑟₁𝑟₂𝑠₂+𝑟₂𝑠₁𝑠₂+𝛽𝑠₂−𝛼𝑟₂,𝑟𝜈=²𝑠₁(𝑠₁−1)2−𝑠₂𝑟₁(𝑟₁−1)2+𝛽𝑠₁−𝛼𝑟₁. (3.2)
Now we consider the equation [ℎ,𝑔]=𝑢²(⋄).The element 𝑢 has a presentation of the following form: 𝑢=𝑥^𝑟′1₁𝑥^𝑟′2₂[𝑥₂,𝑥₁]^𝛼′[𝑥₂,𝑥₁,𝑥₂]^𝛾′[𝑥₂,𝑥₁,𝑥₁]^𝛽′, (3.3) where 𝑟^′₁,𝑟^′₂,𝛼^′,𝛽^′,and𝛾^′ are unique integer elements.
Lemma 2.3 implies that 𝑢²=𝑥^2𝑟′1₁𝑥^2𝑟′2₂[𝑥₂,𝑥₁]^{2𝛼′+𝑟′1𝑟′2}[𝑥₂,𝑥₁,𝑥₂]^{2𝛾′+𝛼′𝑟′2+𝑟′1𝑟′2(𝑟′2−1)/2+𝑟′1𝑟′22}×[𝑥₂,𝑥₁,𝑥₁]^{2𝛽′+𝛼′𝑟′1+𝑟′1𝑟′2(𝑟′1−1)/2}. (3.4)
Thus equation (⋄) holds in 𝐹_2,3 if and only if 𝑟^′₁=𝑟^′₂=0,2𝛼^′=𝜆,2𝛽^′=𝜈,2𝛾^′=𝜇. (3.5)
In particular the equation (⋄) has a solution only if 𝜆,𝜇,and𝜈 are even. Put 𝑐₁=𝛼𝑟₂−𝛽𝑠₂,𝑐₂=𝛼𝑟₁−𝛽𝑠₁, then |𝛼=^{𝑐1−𝑠2𝑐2−𝑠1}||^{𝑟2−𝑠2}_{𝑟1−𝑠1}|=𝑠₁𝑐₁−𝑠₂𝑐₂2𝛼^′|,𝛽=^{𝑟2𝑐1𝑟1𝑐2}|−2𝛼^′=𝑟₁𝑐₁−𝑟₂𝑐₂2𝛼^′. (3.6) Hence we need 𝑠₁𝑐₁−𝑠₂𝑐₂ and 𝑟₁𝑐₁−𝑟₂𝑐₂ to be even. We have the following two cases.
Case. If 𝑟₁𝑠₂=2𝑘, for some integer 𝑘, then 𝑟₂𝑠₁=2𝛼^′+2𝑘, and hence 𝑟₂𝑠₁≡₂0.And we have 𝑐₁=−𝛼^′+𝛼^′(𝑟₂+𝑠₂)−𝑘𝑟₂+(𝛼^′+𝑘)𝑠₂−2𝛾^′,𝑐₂=𝛼^′+(𝛼^′+𝑘)𝑠₁−𝑘𝑟₁−2𝛽^′. (3.7) Further, 0≡₂𝑠₁𝑐₁+𝑠₂𝑐₂≡₂𝛼^′(𝑠₁+𝑠₁𝑠₂+𝑠₂),0≡₂𝑟₁𝑐₁+𝑟₂𝑐₂≡₂𝛼^′(𝑟₁+𝑟₁𝑟₂+𝑟₂). (3.8) Now if 𝛼^′ is an odd integer, then we have 0≡₂𝑟₁+𝑟₁𝑟₂+𝑟₂≡₂𝑠₁+𝑠₁𝑠₂+𝑠₂. (3.9) It follows that 𝑟₁,𝑟₂,𝑠₁, and 𝑠₂ are all even. Hence 𝜆=𝑟₂𝑠₁−𝑟₁𝑠₂ is divisible by 4. But 𝜆=2𝛼^′ implies that 𝛼^′≡₂0, a contradiction. Hence in Case 1 we have 𝛼^′≡₂0 and 𝜆≡₄0.
Now 𝑟₁𝑠₂=2𝑘, and 𝑟₂𝑠₁=2𝛼^′+2𝑘 imply that 𝜇=𝛼^′𝑟₂−𝑘𝑟₂−𝛼^′+𝑘𝑠₂+2𝛼^′𝑠₂+𝛽𝑠₂−𝛼𝑟₂=2𝛾^′,𝜈=𝛼^′𝑠₁+𝑘𝑠₁−𝛼^′−𝑘𝑟₁+𝛽𝑠₁−𝛼𝑟₁=2𝛽^′. (3.10) Hence we have 𝜇≡₂𝑟₂(𝑘+𝛼)+𝑠₂(𝑘+𝛽),𝜈≡₂𝑟₁(𝑘+𝛼)+𝑠₁(𝑘+𝛽). (3.11) And we have the following cases. Subcase 1.1. If 𝑟₁≡₂𝑟₂≡₂𝑠₁≡₂𝑠₂≡₂0, then it is clear that for any integer numbers 𝛼 and 𝛽 we have; 𝜆≡₄0,𝜇≡₂𝜈≡₂0. (3.12) And the equation (⋄) has solution.Subcase 1.2. If 𝑟₁≡₂𝑟₂≡₂𝑠₁≡₂0 and 𝑠₂≡₂1, then 𝑟₁𝑠₂≡₄𝜆≡₄0. We have the following two cases.(1.2.1) If 𝑟₁≡₄0, then we have 𝜆≡₄0. Also from 𝑟₁𝑠₂=2𝑘, it follows that 𝑘≡₂0. Now if we choose 𝛽≡₂0, then from (3.11) it follows that 𝜇≡₂0 and 𝜈≡₂0 for any 𝛼∈ℤ. And in this case the equation (⋄) has a solution.(1.2.2) If 𝑟₁≡₄2, then 𝜆≡₄2, and the equation (⋄) has no solution.
Hence in Subcase 1.2 if 𝑟₁≡₄0,𝑟₂≡₂𝑠₁≡₂0,𝑠₂≡₂1, and 𝛽≡₂0, for any 𝛼∈ℤ the equation (⋄) has a solution.Subcase 1.3. If 𝑟₁≡₂𝑟₂≡₂𝑠₂≡₂0 and 𝑠₁≡₂1, then 𝑠₁𝑟₂≡₄𝜆≡₄0. We have two cases.(1.3.1) If 𝑟₂≡₄0, then 𝜆≡₄0. Since 𝑟₁𝑠₂=2𝑘, and 𝑟₁≡₂𝑠₂≡₂0, hence 𝑘≡₂0. Now if we identify 𝛽≡₂0, then from (3.11) it follows that 𝜇≡₂0 and 𝜈≡₂0. And the equation (⋄) has a solution.(1.3.2) If 𝑟₂≡₄2, then 𝜆≡₄2, and the equation (⋄) has no solution.
Hence in Subcase 1.3 if 𝑟₁≡₂𝑠₂≡₂0,𝑟₂≡₄0, and 𝛽≡₂0, for any 𝛼∈ℤ the equation (⋄) has a solution.Subcase 1.4. If 𝑟₁≡₂𝑟₂≡₂0 and 𝑠₁≡₂𝑠₂≡₂1, then we have the following two cases.(1.4.1) If 𝑟₁≡₄0, then 𝜆≡₄𝑠₁𝑟₂≡₄0. Now 𝑠₁≡₂1 implies 𝑟₂≡₄2. If we choose 𝛽≡₂0, then for any 𝛼∈ℤ the equation (⋄) has a solution. Hence if 𝑟₁≡₄𝑟₂≡₄0,𝑠₁≡₂𝑠₂≡₂1, and 𝛽≡₂0, then for any 𝛼∈ℤ, the equation (⋄) has a solution.(1.4.2) If𝑟₁≡₄2. Since 𝜆≡₄𝑠₁𝑟₂−𝑟₁𝑠₂≡₄0, hence 𝑟₂≡₄2.If we identify 𝛽≡₂1, for any 𝛼∈ℤ then 𝜇≡₂𝜈≡₂0. And the equation (⋄) has a solution.Subcase 1.5. If 𝑟₁≡₂𝑠₁≡₂𝑟₂≡₂0, and 𝑟₂≡₂1, we have the following two cases.(1.5.1) If 𝑠₁≡₄0, then 𝜆≡₄0. Since 𝑟₁𝑠₂=2𝑘, hence 𝑘≡₂0. If we identify 𝛼≡₂0, for any 𝛽∈ℤ, then 𝜇≡₂𝜈≡₂0. And the equation (⋄) has a solution.(1.5.2) If 𝑠₁≡₄2, then 𝜆≡₄2. And the equation (⋄) has no solution. Hence in this case only if 𝑠₁≡₄0, the equation (⋄) has a solution.Subcase 1.6. If 𝑟₁≡₂𝑠₁≡₂0 and 𝑟₂≡₂𝑠₂≡₂1, then similar to Case 4,if 𝑟₁≡₄𝑠₁≡₄0 or 𝑟₁≡₄𝑠₁≡₄2 then 𝜆≡₄0. And for any 𝛼≡₂𝛽,𝜇≡₂𝜈≡₂0, the equation (⋄) has a solution.Subcase 1.7. If 𝑟₁≡₂𝑠₂≡₂0 and 𝑟₂≡₂𝑠₁≡₂1, then 𝜆≡₂1. Hence the equation (⋄) has no solution.Subcase 1.8. If 𝑟₁≡₂0 and 𝑟₂≡₂𝑠₂≡₂𝑠₁≡₂1, then 𝜆≡₂1. Hence the equation (⋄) has no solution.Subcase 1.9. If 𝑟₁≡₂1 and 𝑟₂≡₂𝑠₂≡₂𝑠₁≡₂0, we have two cases.(1.9.1) If 𝑠₂≡₄0, then 𝜆≡₄0. Since 𝑟₁𝑠₂=2𝑘, hence 𝑘≡₂0. If we identify 𝛼≡₂0, for any 𝛽∈ℤ, then 𝜇≡₂𝜈≡₂0. And the equation (⋄) has a solution.(1.9.2) If 𝑠₂≡₄2, then 𝜆≡₄2. And the equation (⋄) has no solution.Subcase 1.10. If 𝑟₁≡₂𝑠₂≡₂1 and 𝑟₂≡₂𝑠₁≡₂0, then 𝑟₁𝑠₂≡₂1. And the equation (⋄) has no solution.Subcase 1.11. If 𝑟₁≡₂𝑠₁≡₂1 and 𝑟₂≡₂𝑠₂≡₂0, then similar to Subcase 1.6, if 𝑟₂≡₄𝑠₂≡₄0 or 𝑟₂≡₄𝑠₂≡₄2 then 𝜆≡₄0. And for any 𝛼≡₂𝛽,𝜇≡₂𝜈≡₂0, the equation (⋄) has a solution.Subcase 1.12. If 𝑟₁≡₂𝑠₁≡₂𝑠₂≡₂1 and 𝑟₂≡₂0, then 𝑟₁𝑠₂≡₂1. And the equation (⋄) has no solution.Subcase 1.13. If 𝑟₁≡₂𝑟₂≡₂1 and 𝑠₁≡₂𝑠₂≡₂0, then we have two cases.(1.13.1) If 𝑠₁≡₄0, then 𝜆≡₄0 implies 𝑠₂≡₂0. If we identify 𝛼≡₂0, for any 𝛽∈ℤ, the equation (⋄) has a solution.(1.13.2) If 𝑠₁≡₄2, then 𝑠₂≡₄2. And if 𝛼≡₂1, for any 𝛽∈ℤ, the equation (⋄) has a solution.Subcase 1.14. If 𝑟₁≡₂𝑟₂≡₂𝑠₂≡₂1 and 𝑠₁≡₂0, then 𝑟₁𝑠₂≡₂1. In this case the equation (⋄) has no solution.Subcase 1.15. If 𝑟₁≡₂𝑟₂≡₂𝑠₁≡₂1 and 𝑠₁≡₂0, then 𝑟₂𝑠₁≡₂1. In this case the equation (⋄) has no solution.Case 2. If 𝑟₁𝑠₂≡₂1. Since 𝜆=𝑠₁𝑟₂−𝑟₁𝑠₂≡₂0, hence 𝑟₁≡₂𝑟₂≡₂𝑠₁≡₂𝑠₂≡₂1. If we identify 𝛼≡₂𝛽, then 𝜇≡₂𝜈≡₂0. In this case the equation (⋄) has a solution.
Hence we show that in the following twelve cases the equation (⋄) has solution. And Sq[ℎ,𝑔]=1.
(1)𝑟₁≡₂𝑠₁≡₂𝑟₂≡₂𝑠₂≡₂0,  for all 𝛼,𝛽.(2)𝑠₁≡₂𝑟₂≡₂0,𝑠₂≡₂1,𝑟₁≡₄0, for all 𝛼,𝛽≡₂0.(3)𝑟₁≡₂𝑠₂≡₂0,𝑠₁≡₂1,𝑟₂≡₄0,forall𝛼,𝛽≡₂0.(4)𝑠₁≡₂𝑠₂≡₂1,𝑟₁≡₄𝑟₂≡₂0,forall𝛼,𝛽≡₂0.(5)𝑠₁≡₂𝑠₂≡₂1,𝑟₁≡₄𝑟₂≡₄2,forall𝛼,𝛽≡₂0.(6)𝑟₁≡₂𝑠₂≡₂0,𝑟₂≡₂1,𝑠₁≡₄0,𝛼≡₂0,forall𝛽.(7)𝑟₁≡₂𝑠₁≡₂1,𝑟₂≡₂𝑠₂≡₂0,𝛼≡₂𝛽.(8)𝑟₁≡₂𝑠₁≡₂0,𝑟₂≡₂𝑠₂≡₂1,𝛼≡₂𝛽.(9)𝑟₁≡₂1,𝑟₂≡₂𝑠₁≡₂0,𝑠₂≡₄0,𝛼≡₂0,forall𝛽.(10)𝑟₁≡₂𝑟₂≡₂1,𝑠₁≡₄𝑠₂≡₄0,𝛼≡₂0,forall𝛽.(11)𝑟₁≡₂𝑟₂≡₂1,𝑠₁≡₄𝑠₂≡₄2,𝛼≡₂1,forall𝛽.(12)𝑟₁≡₂𝑠₁≡₂𝑟₂≡₂𝑠₂≡₂1,𝛼≡₂𝛽.
And more precisely we have [ℎ,𝑔]=([𝑥₂,𝑥₁]^𝜆/2[𝑥₂,𝑥₁,𝑥₂]^𝜇/2[𝑥₂,𝑥₁,𝑥₁]^𝜈/2)². (3.13)
Now in the following ten cases the equation (⋄) has no solution.(13)𝑟₂≡₂𝑠₁≡₂0,𝑠₂≡₂1,𝑟₁≡₄2.(14)𝑟₁≡₂𝑠₂≡₂0,𝑠₁≡₂1,𝑟₂≡₄2.(15)𝑟₁≡₂𝑠₂≡₂0,𝑟₂≡₂1,𝑠₁≡₄2.(16)𝑟₂≡₂𝑠₁≡₂0,𝑟₁≡₂1,𝑠₂≡₄2.(17)𝑟₁≡₂𝑠₂≡₂0,𝑟₂≡₂𝑠₁≡₂1.(18)𝑟₁≡₂𝑠₂≡₂1,𝑟₂≡₂𝑠₁≡₂0.(19)𝑟₁≡₂𝑠₁≡₂𝑠₂≡₂1,𝑟₂≡₂0.(20)𝑟₁≡₂𝑟₂≡₂𝑠₂≡₂1,𝑠₁≡₂0.(21)𝑟₁≡₂𝑠₁≡₂𝑟₂≡₂1,𝑠₂≡₂0.(22)𝑟₂≡₂𝑠₁≡₂𝑠₂≡₂1,𝑟₁≡₂0.
We consider the equation [ℎ,𝑔]=𝑢²₁𝑢²₂(◊). Suppose that the equation (◊) has a nontrivial solution(𝑢₁,𝑢₂). The elements 𝑢₁ and 𝑢₂ have a representation of the following forms: 𝑢₁=𝑥^𝑟11₁𝑥^𝑟21₂[𝑥₂,𝑥₁]^𝛼1[𝑥₂,𝑥₁,𝑥₁]^𝛽1[𝑥₂,𝑥₁,𝑥₂]^𝛾1,𝑢₂=𝑥^𝑟12₁𝑥^𝑟22₂[𝑥₂,𝑥₁]^𝛼2[𝑥₂,𝑥₁,𝑥₁]^𝛽2[𝑥₂,𝑥₁,𝑥₂]^𝛾2, (3.14) where 𝑟_𝑖𝑗,𝛼_𝑖,𝛽_𝑖,and𝛾_𝑖 are unique integer numbers. By applying Lemma 2.3 one obtains 𝑢²_𝑖=𝑥^2𝑟1𝑖₁𝑥^2𝑟2𝑖₂[𝑥₂,𝑥₁]^{2𝛼𝑖+𝑟1𝑖𝑟2𝑖}×[𝑥₂,𝑥₁,𝑥₁]^{2𝛽𝑖+𝛼𝑖𝑟1𝑖+𝑟1𝑖𝑟2𝑖((𝑟1𝑖−1)/2)}×[𝑥₂,𝑥₁,𝑥₂]^{2𝛾𝑖+𝛼𝑖𝑟2𝑖+𝑟1𝑖𝑟2𝑖(}𝑟_2𝑖−12^{)+𝑟1𝑖𝑟22𝑖}. (3.15) Hence 𝑢²₁𝑢²₂=𝑥^{2(𝑟11+𝑟12)}₁𝑥^{2(𝑟21+𝑟122)}₂[𝑥₂,𝑥₁]^{2(𝛼1+𝛼2)+𝑟11𝑟21+𝑟12𝑟22+4𝑟21𝑟12}×[𝑥₂,𝑥₁,𝑥₂]^{𝑛1+𝑛2+2𝑘1𝑟22+4𝑟21𝑟12((2𝑟21−1)/2)+8𝑟21𝑟12𝑟22}×[𝑥₂,𝑥₁,𝑥₁]^{𝑚1+𝑚2+2𝑘1𝑟12+4𝑟21𝑟12((2𝑟12−1)/2)}, (3.16) where for 𝑖=1,2,
𝑘_𝑖=2𝛼_𝑖+𝑟_1𝑖𝑟_2𝑖,𝑚_𝑖=2𝛽_𝑖+𝛼_𝑖𝑟_1𝑖+𝑟_1𝑖𝑟_2𝑖(𝑟_1𝑖−12𝑛),_𝑖=2𝛾_𝑖+𝛼_𝑖𝑟_2𝑖+𝑟_1𝑖𝑟_2𝑖(𝑟_2𝑖−12)+𝑟_1𝑖𝑟²_2𝑖. (3.17)
Hence equation (◊) holds if 𝑟₁₁=−𝑟₁₂,𝑟₂₁=−𝑟₂₁,𝜆=2(𝛼₁+𝛼₂)−2𝑟₁₁𝑟₂₁,𝜇=2(𝛾₁+𝛾₂)+𝑟₂₁(𝛼₁−𝛼₂)−2𝑘₁𝑟₂₁+𝑟₁₁𝑟₂₁(−4𝑟₂₁+1),𝜈=2(𝛽₁+𝛽₂)+𝑟₁₁(𝛼₁−𝛼₂)−2𝑘₁𝑟₁₁+𝑟₁₁𝑟₂₁(4𝑟₁₁+1). (3.18) Note that second equation gives 𝜆≡₂0; hence equation (◊) has nontrivial solution only if 𝜆≡₂0. In particular in the cases from (17) to (22), since 𝜆 is odd, the equation has no solution and Sq[ℎ,𝑔]=3.
Finally it remains to consider the cases from (13) to (16). In these cases we have 𝜆≡₄2. And we prove that if 𝜈≡₂1, then 𝜇≡₂0. It is clear that 𝜈≡₂1 implies 𝑚₁+𝑚₂≡₂1. Hence 𝑟₁₁(𝛼₁+𝛼₂+𝑟₂₁)≡₂1. In particular 𝑟₁₁≡₂1 and 𝛼₁+𝛼₂+𝑟₂₁≡₂1. Now we have 𝜇≡₂𝑛₁+𝑛₂≡₂𝛼₁𝑟₂₁+𝑟₁₁𝑟₂₁(𝑟₂₁−12)+𝑟₁₁𝑟²₂₁+𝛼₂𝑟₂₂+𝑟₁₂𝑟₂₂(𝑟₂₂−12)+𝑟₁₂𝑟²₂₂≡₂(1+𝑟₁₂)𝑟₁₂≡₂0. (3.19)
Now in the cases from (13) and (15), we have 𝜈≡₂1. Hence 𝜇≡₂0. And if we identify: 𝑟₁₁=−𝑠₁+1,𝑟₁₂=𝑠₁−1,𝑟₂₂=−𝑟₂₁𝛼=0,₁=𝛽₂=𝛾₂=0,𝛼₂=𝜆2,𝛽₁=𝜈+𝑟₁₁𝛼₂2,𝛾₁=𝜇2. (3.20)
then for the elements 𝑢₁=𝑥₁^−𝑠+1[𝑥₂,𝑥₁,𝑥₁]^{(𝜈+𝑟11(𝜆/2))/2)}[𝑥₂,𝑥₁,𝑥₂]^𝜇/2,𝑢₂=𝑥^𝑠1₁⁻¹[𝑥₂,𝑥₁]^𝜆/2. (3.21)
we have [ℎ,𝑔]=𝑢²₁𝑢²₂. It covers the cases from (13) and (15).
Now we consider the cases from (14) and (16). Since in these cases 𝜇≡₂1, hence 𝜈≡₂0. If we identify
𝑟₁₁=𝑟₁₂=0,𝑟₂₁=1,𝑟₂₂𝛼=−1,₁=𝛽₂=𝛾₁=0,𝛼₂=𝜆2,𝛽₁=𝜈2,𝛾₁=𝜇+𝛼₂2. (3.22)
then for the elements
𝑢₁=𝑥₂[𝑥₂,𝑥₁,𝑥₁]^𝜈/2,𝑢₂=𝑥₂⁻¹[𝑥₂,𝑥₁]^𝜆/2[𝑥₂,𝑥₁,𝑥₂]^{(𝜇+𝜆/2)/2}. (3.23) one obtains [ℎ,𝑔]=𝑢²₁𝑢²₂. And the equation (◊) satisfies.
In particular in the cases from (13) to (16), we have Sq[ℎ,𝑔]=2. This completes the proof.

As an immediate consequence of Theorem 2.1, we obtain the exact value of the Sq(𝐹^′_2,3).

The proof of Corollary 2.2 is based on our previous result [5] which we summarize here.

Theorem 3.1 (Rhemtulla-Akhavan[5]). Let 𝐹_2,3=⟨𝑥₁,𝑥₂⟩ be a free nilpotent group of rank 2 and class 3 freely generated by 𝑥₁,𝑥₂. Then any element of 𝐹^′_2,3 can be expressed as a product of at most two commutators.

We will also use the fact that if 𝑎,𝑏, and 𝑐 are any elements of a group 𝐺, then 𝑎²[𝑏,𝑐]=(𝑎²𝑏⁻¹𝑐⁻¹)²(𝑎𝑏𝑎⁻¹𝑐⁻¹𝑎⁻¹)²(𝑎𝑐)². (†)

Proof of Corollary 2.2. Let 𝜁=[𝑥,𝑦][𝑤,𝑧] be any element of 𝐹^′_2,3. We may write =[𝑥₂,𝑥₁]^𝜆[𝑥₂,𝑥₁,𝑥₂]^𝜇[𝑥₂,𝑥₁,𝑥₁]^𝜈,[𝑧,𝑤]=[𝑥₂,𝑥₁]^𝜆′[𝑥₂,𝑥₁,𝑥₂]^𝜇′[𝑥₂,𝑥₁,𝑥₁]^𝜈′, (3.24) where 𝜆,𝜆^′,𝜇,𝜇^′,𝜈, and 𝜈^′ are suitable integer numbers. Since 𝛾₃(𝐹_2,3) lies in the center of 𝐹_2,3 and 𝐹^′_2,3 is abelian, we may express 𝜁 as 𝜁=[𝑥₂,𝑥₁]^𝜆+𝜆′[𝑥₂,𝑥₁,𝑥^𝜇+𝜇′₂𝑥^𝜈+𝜈′₁]. (3.25) There are two cases:(1)𝜆+𝜆^′≡₂0,(2)𝜆+𝜆^′≡₂1.Case 1. By (†), we may write 𝜁 as a product of three squares.Case 2. We may write 𝜁=[𝑥₂,𝑥₁]^{𝜆+𝜆′−1}[𝑥₁,𝑥₂]^{𝑥′2𝜇+𝜇𝑥′1𝜈+𝜈}. (3.26)
Since 𝜆+𝜆^′−1 is even, (†) yields Sq(𝜁)≤3. In Theorem 2.1 we produce elements of square length equal to three. This shows that Sq(𝐹^′_2,3)=3 and completes the proof.

Note. Let 𝐺=⟨𝑥₁,𝑥₂⟩ be a free nilpotent group of rank 2 and class 𝑐≥3 freely generated by 𝑥₁,𝑥₂. Now 𝐹_2,3 is a quotient of 𝐺. Since the equations (⋄) and (◊) do not hold in the cases from (17) to (22) in 𝐹_2,3, these equations should not hold in 𝐺. And similarly since the equation (⋄) does not hold in the cases from (13) to (16) in 𝐹_2,3, hence these equations will not hold in 𝐺.

Acknowledgments

The author would like to thank professor Howard E. Bell and the referee who have patiently read and verified this note and also suggested valuable comments. The author also would like to acknowledge the support of the Alzahra University.