POWERS OF A PRODUCT OF COMMUTATORS AS PRODUCTS OF SQUARES

IJMMS 2004:7, 373–375 PII. S0161171204304047 http://ijmms.hindawi.com

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POWERS OF A PRODUCT OF COMMUTATORS AS PRODUCTS OF SQUARES

ALIREZA ABDOLLAHI Received 1 April 2003

We prove that for any odd integerNand any integern >0, theNth power of a product of ncommutators in a nonabelian free group of countable infinite rank can be expressed as a product of squares of 2n+1 elements and, for all such oddNand integersn, there are commutators for which the number 2n+1 of squares is the minimum number such that the Nth power of its product can be written as a product of squares. This generalizes a recent result of Akhavan-Malayeri.

2000 Mathematics Subject Classification: 20F12, 20F99.

1. Introduction. Lyndon et al. [2] have shown that the product ofncommutators in a nonabelian free group can be written as a product of 2n+1 squares of elements and there are commutators for which the number 2n+1 of squares is the minimum number such that the product of these commutators can be written as a product of squares. Recently, Akhavan-Malayeri [1] proved, for an odd integern, that[x,y]ⁿ of two distinct elements of a free generating set of a nonabelian free group is not a product of two squares but it is the product of three squares. We generalize these results in the following theorem.

Theorem1.1. LetFbe a free group with a basis of distinct elementsx¹,...,x2n, and Nany odd integer. Then there exist elementsu1,...,uminF such that

x1,x2

···

x2n−1,x2nN=u²1···u²m (1.1) if and only ifm≥2n+1.

Note that the theorem for evenNis not true since the element in the left-hand side of the above equation is actually a square. The proof of this theorem is almostmutatis mu- tandisas the proof of the main result of [2]. Throughout this note,[x,y]=x⁻¹y⁻¹xy and [x,y,z]=[[x,y],z]for all elementsx,y, zof a groupG, andGdenotes the derived subgroup ofG.

2. Proof of the main result

Proof ofTheorem1.1. We show first that this equation has a solution form= 2n+1, hence trivially form≥2n+1. SinceNis odd, there is an integerksuch that N=2k+1. Thus it is enough to show that, for any elementv ofF, we can express the elementv²[x1,x2]···[x2n−1,x2n]as a product of 2n+1 squares. We argue by

374 ALIREZA ABDOLLAHI

induction onn. Ifn=1, then by the following well-known identity this case is proved:

A²[B,C]=

A²B⁻¹A⁻¹2

ABA⁻¹C⁻¹A⁻¹2

(AC)². (2.1)

Assumen >1 and suppose inductively that v²

x1,x2

···

x2n−3,x2n−2

=u²1···u²_2n−1 (2.2)

for some elementsu¹,...,u2n−1inF. Now by the identity (2.1) we can write u²_2n−1

x2n−1,x2n

=U²V²W² (2.3)

for some elementsU,V, andWinF, and so v²

x1,x2

···

x2n−1,x2n

=u²1···u²_2n−2U²V²W², (2.4) which completes the induction. This first part of the proof is essentially well known in a topological context: the nonorientable surface formed by attaching one cross-cap and n handles to a sphere (the connected sum of 1 projective plane and ntori) is homeomorphic to the surface obtained by attaching 2n+1 cross-caps (the connected sum of 2n+1 projective planes). In this context, the identity (2.1) is just the handle calculus that says cross-cap + handle=3 cross-caps.

For the converse, we suppose that the equation holds. LetGbe the group with the following presentation:

yi|yi²=

yi,yj,yk

=1∀i,j,k∈ {1,2,...,2n}

. (2.5)

The equation would also hold inGsinceGis a quotient ofF. So we have y1,y2

···

y2n−1,y2n

_N

=v1²···vm² (2.6) for some elementsv1,...,vminG. SinceNis odd,N=2t+1 for some integert. Since Gis nilpotent of class 2 andy_i²=1 for eachi, we have[yi,yj]²=1 and all the com- mutators are in the center ofG, so the latter equation can be rewritten as

y1,y2

···

y2n−1,y2n

=v1²···vm². (2.7)

Letcij=[yi,yj]. Then each elementvofGhas a unique expression v=y1^a1···y2n^a2n

i<j

c_ij^dij forai,dij∈Z2. (2.8)

Let

vk=y1^a1k···y_2n^a2nkzk, (2.9) whereaik∈Z2andzk∈Gfor alli∈ {1,...,2n}and allk∈ {1,...,m}. Sincez²_k=1 for allk, we have

v1²···vm² =

i<j

c

_m k=1a_ika_jk

ij . (2.10)

A PRODUCT OF COMMUTATORS 375 IfA is the matrixA=(aij)overZ2, andAi=(ai1,...,aim)is the ith row ofA, then from the relationv1²···vm² =[y1,y2]···[y2n−1,y2n]we conclude that, taking inner products,

Ai·Aj=





1 if{i,j} = {2h−1,2h}for 1≤h≤n,

0 otherwise. (2.11)

We conclude thatA·A^T=B, whereA^T is the transpose ofA, andBis the direct sum of nmatrices of the form_{1 0}

0 1

, and hence has rank 2n. It follows that rank(A)≥2n. But the equationAi·Ai=_m

j=1aijaij=0 for eachiimplies that the sum of the columns of Ais 0, whence rank(A)≤m−1. Thereforem−1≥2n.

References

[1] M. Akhavan-Malayeri,Powers of commutators as products of squares, Int. J. Math. Math. Sci.

31(2002), no. 10, 635–637.

[2] R. Lyndon, T. McDonough, and M. Newman,On products of powers in groups, Proc. Amer.

Math. Soc.40(1973), 419–420.

Alireza Abdollahi: Department of Mathematics, University of Isfahan, Isfahan 81746-73441, Iran

E-mail address:[email protected]

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