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 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.

1. Introduction

Schützenberger [1] proved that in a free group the equation

[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:

[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.

Let F 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
([x1,x2]⋯[x2n-1,x2n])N=u12…,um2
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 F2,3=〈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.

Let F2,3=〈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 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]=zr,r≥2.

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:
[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 ResultProof of Theorem <xref ref-type="statement" rid="thm2.1">2.1</xref>.

Let h,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. [h,g]=[x2,x1]λ[x2,x1,x2]μ[x2,x1,x1]ν,
where
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:
u=x1r1′x2r2′[x2,x1]α′[x2,x1,x2]γ′[x2,x1,x1]β′,
where r1′,r2′,α′,β′,andγ′ are unique integer elements.

Lemma 2.3 implies that
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
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
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.

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′.
Further,
0≡2s1c1+s2c2≡2α′(s1+s1s2+s2),0≡2r1c1+r2c2≡2α′(r1+r1r2+r2).
Now if α′ is an odd integer, then we have
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.

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′.
Hence we have
μ≡2r2(k+α)+s2(k+β),ν≡2r1(k+α)+s1(k+β).

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.
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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

If s1≡40, then λ≡40 implies s2≡20. If we identify α≡20, for any β∈ℤ, the equation (⋄) has a solution.

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.

r1≡2s1≡2r2≡2s2≡20, for all α,β.

s1≡2r2≡20,s2≡21,r1≡40, for all α,β≡20.

r1≡2s2≡20,s1≡21,r2≡40,forallα,β≡20.

s1≡2s2≡21,r1≡4r2≡20,forallα,β≡20.

s1≡2s2≡21,r1≡4r2≡42,forallα,β≡20.

r1≡2s2≡20,r2≡21,s1≡40,α≡20,forallβ.

r1≡2s1≡21,r2≡2s2≡20,α≡2β.

r1≡2s1≡20,r2≡2s2≡21,α≡2β.

r1≡21,r2≡2s1≡20,s2≡40,α≡20,forallβ.

r1≡2r2≡21,s1≡4s2≡40,α≡20,forallβ.

r1≡2r2≡21,s1≡4s2≡42,α≡21,forallβ.

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.

Now in the following ten cases the equation (⋄) has no solution.

r2≡2s1≡20,s2≡21,r1≡42.

r1≡2s2≡20,s1≡21,r2≡42.

r1≡2s2≡20,r2≡21,s1≡42.

r2≡2s1≡20,r1≡21,s2≡42.

r1≡2s2≡20,r2≡2s1≡21.

r1≡2s2≡21,r2≡2s1≡20.

r1≡2s1≡2s2≡21,r2≡20.

r1≡2r2≡2s2≡21,s1≡20.

r1≡2s1≡2r2≡21,s2≡20.

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:
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
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
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,

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).
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)+r11r212+α2r22+r12r22(r22-12)+r12r222≡2(1+r12)r12≡20.

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=-r21=0,α1=β2=γ2=0,α2=λ2,β1=ν+r11α22,γ1=μ2.

then for the elements
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 cases μ≡21, hence ν≡20. If we identify

Let F2,3=〈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 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 <xref ref-type="statement" rid="coro2.2">2.2</xref>.

Let ζ=[x,y][w,z] be any element of F2,3′. We may write
[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
ζ=[x2,x1]λ+λ′[x2,x1,x2μ+μ′x1ν+ν′].
There are two cases:

λ+λ′≡20,

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]x2μ+μ′x1ν+ν′.

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(F2,3′)=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 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.

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.

SchützenbergerM. P.Sur l'equation a2+n=b2+mc1+p dans un groupe libreLyndonR. C.NewmanM.Commutators as products of squaresAkhavan-MalayeriM.Powers of commutators as products of squaresAbdollahiA.Powers of a product of commutators as products of squaresAkhavan-MalayeriM.RhemtullaA.Commutator length of abelian-by-nilpotent groups