IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation26415010.1155/2009/264150264150Research ArticleCommutators and Squares in Free Nilpotent GroupsAkhavan-MalayeriMehri1BellHowardDepartment of MathematicsAlzahra UniversityVankTehranm 19834Iranalzahra.ac.ir200922201020090208200924112009011220092009Copyright © 2009This 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.

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);γG}. 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(F2,3)=3.

1. Introduction

Schützenberger  proved that in a free group the equation

[x,y]=zr,r2 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 r2. Lyndon and Newman  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  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  generalized these results as the following theorem.

Theorem 1.1 (Abdollahi [<xref ref-type="bibr" rid="B1">4</xref>]).

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 m2n+1.

Definition 1.2.

Let G be a group and γG. 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);γG}.

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,gF2,3. 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,r2.

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 F2,3γ3(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  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=x1r1x2r2[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=x12r1x22r2[x2,x1]2α+r1r2[x2,x1,x2]2γ+αr2+r1r2(r2-1)/2+r1r22×[x2,x1,x1]2β+αr1+r1r2(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 r2s120.And we have c1=-α+α(r2+s2)-kr2+(α+k)s2-2γ,c2=α+(α+k)s1-kr1-2β. Further, 02s1c1+s2c22α(s1+s1s2+s2),02r1c1+r2c22α(r1+r1r2+r2). Now if α is an odd integer, then we have 02r1+r1r2+r22s1+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 α20, a contradiction. Hence in Case 1 we have α20 and λ40.

Now r1s2=2k, and r2s1=2α+2k imply that μ=αr2-kr2-α+ks2+2αs2+βs2-αr2=2γ,ν=αs1+ks1-α-kr1+βs1-αr1=2β. Hence we have μ2r2(k+α)+s2(k+β),ν2r1(k+α)+s1(k+β).

And we have the following cases. Subcase 1.1.

If r12r22s12s220, then it is clear that for any integer numbers α and β we have; λ40,μ2ν20. And the equation () has solution.

Subcase 1.2.

If r12r22s120 and s221, then r1s24λ40. We have the following two cases.

If r140, then we have λ40. Also from r1s2=2k, it follows that k20. Now if we choose β20, then from (3.11) it follows that μ20 and ν20 for any α. And in this case the equation () has a solution.

If r142, then λ42, and the equation () has no solution.

Hence in Subcase 1.2 if r140,r22s120,s221, and β20, for any α the equation () has a solution.

Subcase 1.3.

If r12r22s220 and s121, then s1r24λ40. We have two cases.

If r240, then λ40. Since r1s2=2k, and r12s220, hence k20. Now if we identify β20, then from (3.11) it follows that μ20 and ν20. And the equation () has a solution.

If r242, then λ42, and the equation () has no solution.

Hence in Subcase 1.3 if r12s220,r240, and β20, for any α the equation () has a solution.

Subcase 1.4.

If r12r220 and s12s221, then we have the following two cases.

If r140, then λ4s1r240. Now s121 implies r242. If we choose β20, then for any α the equation () has a solution. Hence if r14r240,s12s221, and β20, then for any α, the equation () has a solution.

Ifr142. Since λ4s1r2-r1s240, hence r242.If we identify β21, for any α then μ2ν20. And the equation () has a solution.

Subcase 1.5.

If r12s12r220, and r221, we have the following two cases.

If s140, then λ40. Since r1s2=2k, hence k20. If we identify α20, for any β, then μ2ν20. And the equation () has a solution.

If s142, then λ42. And the equation () has no solution. Hence in this case only if s140, the equation () has a solution.

Subcase 1.6.

If r12s120 and r22s221, then similar to Case 4,if r14s140 or r14s142 then λ40. And for any α2β,μ2ν20, the equation () has a solution.

Subcase 1.7.

If r12s220 and r22s121, then λ21. Hence the equation () has no solution.

Subcase 1.8.

If r120 and r22s22s121, then λ21. Hence the equation () has no solution.

Subcase 1.9.

If r121 and r22s22s120, we have two cases.

If s240, then λ40. Since r1s2=2k, hence k20. If we identify α20, for any β, then μ2ν20. And the equation () has a solution.

If s242, then λ42. And the equation () has no solution.

Subcase 1.10.

If r12s221 and r22s120, then r1s221. And the equation () has no solution.

Subcase 1.11.

If r12s121 and r22s220, then similar to Subcase 1.6, if r24s240 or r24s242 then λ40. And for any α2β,μ2ν20, the equation () has a solution.

Subcase 1.12.

If r12s12s221 and r220, then r1s221. And the equation () has no solution.

Subcase 1.13.

If r12r221 and s12s220, then we have two cases.

If s140, then λ40 implies s220. If we identify α20, for any β, the equation () has a solution.

If s142, then s242. And if α21, for any β, the equation () has a solution.

Subcase 1.14.

If r12r22s221 and s120, then r1s221. In this case the equation () has no solution.

Subcase 1.15.

If r12r22s121 and s120, then r2s121. In this case the equation () has no solution.

Case 2.

If r1s221. Since λ=s1r2-r1s220, hence r12r22s12s221. If we identify α2β, then μ2ν20. 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.

r12s12r22s220,  for all α,β.

s12r220,s221,r140, for all α,β20.

r12s220,s121,r240,for  all  α,β20.

s12s221,r14r220,for  all  α,β20.

s12s221,r14r242,for  all  α,β20.

r12s220,r221,s140,α20,for  all  β.

r12s121,r22s220,α2β.

r12s120,r22s221,α2β.

r121,r22s120,s240,α20,for  all  β.

r12r221,s14s240,α20,for  all  β.

r12r221,s14s242,α21,for  all  β.

r12s12r22s221,α2β.

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.

r22s120,s221,r142.

r12s220,s121,r242.

r12s220,r221,s142.

r22s120,r121,s242.

r12s220,r22s121.

r12s221,r22s120.

r12s12s221,r220.

r12r22s221,s120.

r12s12r221,s220.

r22s12s221,r120.

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,and  γi 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,

ki=2αi+r1ir2i,mi=2βi+αir1i+r1ir2i(r1i-12),ni=2γi+αir2i+r1ir2i(r2i-12)+r1ir2i2.

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 λ20; hence equation () has nontrivial solution only if λ20. 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 λ42. And we prove that if ν21, then μ20. It is clear that ν21 implies m1+m221. Hence r11(α1+α2+r21)21. In particular r1121 and α1+α2+r2121. Now we have μ2n1+n22α1r21+r11r21(r21-12)+r11r212+α2r22+r12r22(r22-12)+r12r2222(1+r12)r1220.

Now in the cases from (13) and (15), we have ν21. Hence μ20. 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 μ21, hence ν20. If we identify

r11=r12=0,r21=1,r22=-1,α1=β2=γ1=0,α2=λ2,β1=ν2,γ1=μ+α22.

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. one obtains [h,g]=u12u22. 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(F2,3).

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

Theorem 3.1 (Rhemtulla-Akhavan[<xref ref-type="bibr" rid="B3">5</xref>]).

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+μx1ν+ν]. There are two cases:

λ+λ20,

λ+λ21.

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+μx1ν+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 c3 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.

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