# Commutators and Squares in Free Nilpotent Groups

Authors: Mehri Akhavan-Malayeri
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

---

## Abstract

In a free group no nontrivial commutator is a square. And in the
free groupF2=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 F2,3=〈x1,x2〉 be a free nilpotent group of rank 2 and class
3 freely generated by x1,x2. We prove that in F2,3=〈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
F2,3 is equal to 1 or 2 or 3. The precise formulas for expressing any commutator of F2,3 as the minimal number of squares are given. Finally as an application of these results we prove that Sq(F′2,3)=3.


---

## Body

## 1. Introduction

Schützenberger [1] proved that in a free group the equation(1.1)[x,y]=zr,r≥2
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:(1.2)[x2,x1]2n+1=((wnx2x1)wn)2(wnx1-1)2((w-nx2-1)x1)2.
Recently Abdollahi [4] generalized these results as the following theorem.Theorem 1.1 (Abdollahi [4]).
LetF be a free group with a basis of distinct elements x1,…,x2n, and N any odd integer. Then there exist elements u1,…,um in F such that
(1.3)([x1,x2]⋯[x2n-1,x2n])N=u12…,um2
if and only if m≥2n+1.Definition 1.2.
LetG 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 ofγ and denoted by Sq(γ). And we define Sq(G)=sup{Sq(γ);γ∈G′}.We prove that in the free nilpotent groupF2,3=〈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 F2,3. Using this, we find Sq[h,g] where h,g∈F2,3. Then we prove that Sq(F2,3′)=3.


## 2. Main Results

We will prove the following theorems.Theorem 2.1.
LetF2,3=〈x1,x2〉 be a free nilpotent group of rank 2 and class 3 freely generated by x1,x2. Then Sq(F2,3′)=3.An application of Theorem2.1 is displayed in the next result.Corollary 2.2.
In a free nilpotent group of rank2 and class 3, it is possible to find nontrivial solutions for the equation
(2.1)[x,y]=zr,r≥2.We will use the following well-known identities regarding groups which are nilpotent of class 3.Lemma 2.3.
LetG=〈x,y〉 be nilpotent of class  3. Then, for all integers r,s the following hold:
(2.2)[xr,y]=[x,y]r[x,y,x]r(r-1)/2,[xr,ys]=[x,y]rs[x,y,x]rs(r-1)/2[x,y,y]rs(s-1)/2.


## 3. Proofs of the Main Result

Proof of Theorem2.1.
Leth,g be any two elements of F2,3∖γ3(F2,3). First we study the form of the element [h,g]. Since γ3(F2,3) lies in the center of F2,3 we may express h as x1r1x2r2[x2,x1]β and g as x1s1x2s2[x2,x1]α. We have shown in [5] that. 							(3.1)[h,g]=[x2,x1]λ[x2,x1,x2]μ[x2,x1,x1]ν,
where
(3.2)λ=r2s1-r1s2,μ=s1r2(r2-1)2-r1s2(s2-1)2-r1r2s2+r2s1s2+βs2-αr2,ν=r2s1(s1-1)2-s2r1(r1-1)2+βs1-αr1.
Now we consider the equation[h,g]=u2(⋄).The element u has a presentation of the following form:
(3.3)u=x1r1′x2r2′[x2,x1]α′[x2,x1,x2]γ′[x2,x1,x1]β′,
where r1′,r2′,α′,β′,andγ′ are unique integer elements.
Lemma2.3 implies that
(3.4)u2=x12r1′x22r2′[x2,x1]2α′+r1′r2′[x2,x1,x2]2γ′+α′r2′+r1′r2′(r2′-1)/2+r1′r2′2×[x2,x1,x1]2β′+α′r1′+r1′r2′(r1′-1)/2.
Thus equation(⋄) holds in F2,3 if and only if
(3.5)r1′=r2′=0,2α′=λ,2β′=ν,2γ′=μ.
In particular the equation(⋄) has a solution only if λ,μ,andν are even. Put c1=αr2-βs2,c2=αr1-βs1, then
(3.6)α=|c1-s2c2-s1||r2-s2r1-s1|=s1c1-s2c22α′,β=|r2c1r1c2|-2α′=r1c1-r2c22α′.
Hence we need s1c1-s2c2 and r1c1-r2c2 to be even. We have the following two cases.
Case.
Ifr1s2=2k, for some integer k, then r2s1=2α′+2k, and hence r2s1≡20.And we have
(3.7)c1=-α′+α′(r2+s2)-kr2+(α′+k)s2-2γ′,c2=α′+(α′+k)s1-kr1-2β′.
Further,
(3.8)0≡2s1c1+s2c2≡2α′(s1+s1s2+s2),0≡2r1c1+r2c2≡2α′(r1+r1r2+r2).
Now if α′ is an odd integer, then we have
(3.9)0≡2r1+r1r2+r2≡2s1+s1s2+s2.
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.
Nowr1s2=2k, and r2s1=2α′+2k imply that
(3.10)μ=α′r2-kr2-α′+ks2+2α′s2+βs2-αr2=2γ′,ν=α′s1+ks1-α′-kr1+βs1-αr1=2β′.
Hence we have
(3.11)μ≡2r2(k+α)+s2(k+β),ν≡2r1(k+α)+s1(k+β).  And we have the following cases.  Subcase 1.1.
Ifr1≡2r2≡2s1≡2s2≡20, then it is clear that for any integer numbers α and β we have;
(3.12)λ≡40,μ≡2ν≡20.
And the equation (⋄) has solution.Subcase 1.2.
Ifr1≡2r2≡2s1≡20 and s2≡21, then r1s2≡4λ≡40. We have the following two cases.(1.2.1)
Ifr1≡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)
Ifr1≡42, then λ≡42, and the equation (⋄) has no solution.
Hence in Subcase 1.2 ifr1≡40,r2≡2s1≡20,s2≡21, and β≡20, for any α∈ℤ the equation (⋄) has a solution.Subcase 1.3.
Ifr1≡2r2≡2s2≡20 and s1≡21, then s1r2≡4λ≡40. We have two cases.(1.3.1)
Ifr2≡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)
Ifr2≡42, then λ≡42, and the equation (⋄) has no solution.
Hence in Subcase 1.3 ifr1≡2s2≡20,r2≡40, and β≡20, for any α∈ℤ the equation (⋄) has a solution.Subcase 1.4.
Ifr1≡2r2≡20 and s1≡2s2≡21, then we have the following two cases.(1.4.1)
Ifr1≡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.
Ifr1≡2s1≡2r2≡20, and r2≡21, we have the following two cases.(1.5.1)
Ifs1≡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)
Ifs1≡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.
Ifr1≡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.
Ifr1≡2s2≡20 and r2≡2s1≡21, then λ≡21. Hence the equation (⋄) has no solution.Subcase 1.8.
Ifr1≡20 and r2≡2s2≡2s1≡21, then λ≡21. Hence the equation (⋄) has no solution.Subcase 1.9.
Ifr1≡21 and r2≡2s2≡2s1≡20, we have two cases.(1.9.1)
Ifs2≡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)
Ifs2≡42, then λ≡42. And the equation (⋄) has no solution.Subcase 1.10.
Ifr1≡2s2≡21 and r2≡2s1≡20, then r1s2≡21. And the equation (⋄) has no solution.Subcase 1.11.
Ifr1≡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.
Ifr1≡2s1≡2s2≡21 and r2≡20, then r1s2≡21. And the equation (⋄) has no solution.Subcase 1.13.
Ifr1≡2r2≡21 and s1≡2s2≡20, then we have two cases.(1.13.1)
Ifs1≡40, then λ≡40 implies s2≡20. If we identify α≡20, for any β∈ℤ, the equation (⋄) has a solution.(1.13.2)
Ifs1≡42, then s2≡42. And if α≡21, for any β∈ℤ, the equation (⋄) has a solution.Subcase 1.14.
Ifr1≡2r2≡2s2≡21 and s1≡20, then r1s2≡21. In this case the equation (⋄) has no solution.Subcase 1.15.
Ifr1≡2r2≡2s1≡21 and s1≡20, then r2s1≡21. In this case the equation (⋄) has no solution.Case 2.
Ifr1s2≡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(3.13)[h,g]=([x2,x1]λ/2[x2,x1,x2]μ/2[x2,x1,x1]ν/2)2.
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]=u12u22(◊). Suppose that the equation (◊) has a nontrivial solution(u1,u2). The elements u1 and u2 have a representation of the following forms:
(3.14)u1=x1r11x2r21[x2,x1]α1[x2,x1,x1]β1[x2,x1,x2]γ1,u2=x1r12x2r22[x2,x1]α2[x2,x1,x1]β2[x2,x1,x2]γ2,
where rij,αi,βi,andγi are unique integer numbers. By applying Lemma 2.3 one obtains
(3.15)ui2=x12r1ix22r2i[x2,x1]2αi+r1ir2i×[x2,x1,x1]2βi+αir1i+r1ir2i((r1i-1)/2)×[x2,x1,x2]2γi+αir2i+r1ir2i(r2i-12)+r1ir2i2.
Hence
(3.16)u12u22=x12(r11+r12)x22(r21+r122)[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),
where for i=1,2,
(3.17)ki=2αi+r1ir2i,mi=2βi+αir1i+r1ir2i(r1i-12),ni=2γi+αir2i+r1ir2i(r2i-12)+r1ir2i2.
Hence equation(◊) holds if
(3.18)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).
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 haveλ≡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
(3.19)μ≡2n1+n2≡2α1r21+r11r21(r21-12)+r11r212+α2r22+r12r22(r22-12)+r12r222≡2(1+r12)r12≡20.
Now in the cases from (13) and (15), we haveν≡21. Hence μ≡20. And if we identify:
(3.20)r11=-s1+1,r12=s1-1,r22=-r21=0,α1=β2=γ2=0,α2=λ2,β1=ν+r11α22,γ1=μ2.
then for the elements(3.21)u1=x1-s+1[x2,x1,x1](ν+r11(λ/2))/2)[x2,x1,x2]μ/2,u2=x1s1-1[x2,x1]λ/2.
we have[h,g]=u12u22. It covers the cases from (13) and (15).
Now we consider the cases from (14) and (16). Since in these casesμ≡21, hence ν≡20. If we identify
(3.22)r11=r12=0,r21=1,r22=-1,α1=β2=γ1=0,α2=λ2,β1=ν2,γ1=μ+α22.
then for the elements
(3.23)u1=x2[x2,x1,x1]ν/2,u2=x2-1[x2,x1]λ/2[x2,x1,x2](μ+λ/2)/2.
one obtains [h,g]=u12u22. And the equation (◊) satisfies.
In particular in the cases from (13) to (16), we haveSq[h,g]=2. This completes the proof.As an immediate consequence of Theorem2.1, we obtain the exact value of the Sq(F2,3′).The proof of Corollary2.2 is based on our previous result [5] which we summarize here.Theorem 3.1 (Rhemtulla-Akhavan[5]).
LetF2,3=〈x1,x2〉 be a free nilpotent group of rank 2 and class 3 freely generated by x1,x2. Then any element of F2,3′ can be expressed as a product of at most two commutators.We will also use the fact that ifa,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 Corollary2.2.
Letζ=[x,y][w,z] be any element of F2,3′. We may write
(3.24)[x,y]=[x2,x1]λ[x2,x1,x2]μ[x2,x1,x1]ν,[z,w]=[x2,x1]λ′[x2,x1,x2]μ′[x2,x1,x1]ν′,
where λ,λ′,μ,μ′,ν, and ν′ are suitable integer numbers. Since γ3(F2,3) lies in the center of F2,3 and F2,3′ is abelian, we may express ζ as
(3.25)ζ=[x2,x1]λ+λ′[x2,x1,x2μ+μ′x1ν+ν′].
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(3.26)ζ=[x2,x1]λ+λ′-1[x1,x2]x2μ+μ′x1ν+ν′.
Sinceλ+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(F2,3′)=3 and completes the proof.Note.
LetG=〈x1,x2〉 be a free nilpotent group of rank 2 and class c≥3 freely generated by x1,x2. Now F2,3 is a quotient of G. Since the equations (⋄) and (◊) do not hold in the cases from (17) to (22) in F2,3, these equations should not hold in G. And similarly since the equation (⋄) does not hold in the cases from (13) to (16) in F2,3, hence these equations will not hold in G.



---

*Source: 264150-2010-02-03.xml*