Research Article | Open Access

# Commutators and Squares in Free Nilpotent Groups

**Academic Editor:**Howard Bell

#### Abstract

In a free group no nontrivial commutator is a square. And in the free group freely generated by the commutator is never the product of two squares in , although it is always the product of three squares. Let be a free nilpotent group of rank 2 and class 3 freely generated by . We prove that in , it is possible to write certain commutators as a square. We denote by the minimal number of squares which is required to write as a product of squares in group . And we define . We discuss the question of when the square length of a given commutator of is equal to 1 or 2 or 3. The precise formulas for expressing any commutator of as the minimal number of squares are given. Finally as an application of these results we prove that .

#### 1. Introduction

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

implies ; that is, no nontrivial commutator is a proper power. It means that it is impossible to write as an th powers where . Lyndon and Newman [2] have shown that in the free group freely generated by the commutator is never a product of two squares in although it is always the product of three squares. In [3] we proved that for an odd integer is not a product of two squares in and it is the product of three squares. Put and . We presented the following expression of as a product of the minimal number of squares:

Recently Abdollahi [4] generalized these results as the following theorem.

Theorem 1.1 (Abdollahi [4]). *Let be a free group with a basis of distinct elements and any odd integer. Then there exist elements in such that
**
if and only if .*

*Definition 1.2. *Let be a group and The minimal number of squares which is required to write as a product of squares in is called *the square length of * and denoted by And we define

We prove that in the free nilpotent group of rank and class freely generated by it is possible to write certain nontrivial commutators as a proper power. We consider certain equations over free group Using this, we find where Then we prove that

#### 2. Main Results

We will prove the following theorems.

Theorem 2.1. *Let be a free nilpotent group of rank and class freely generated by Then *

An application of Theorem 2.1 is displayed in the next result.

Corollary 2.2. *In a free nilpotent group of rank and class it is possible to find nontrivial solutions for the equation
*

We will use the following well-known identities regarding groups which are nilpotent of class 3.

Lemma 2.3. *Let be nilpotent of class 3. Then, for all integers the following hold:
*

#### 3. Proofs of the Main Result

*Proof of Theorem 2.1. *Let be any two elements of First we study the form of the element Since lies in the center of we may express as and as We have shown in [5] that.
where

Now we consider the equation The element has a presentation of the following form:
where are unique integer elements.

Lemma 2.3 implies that

Thus equation holds in if and only if

In particular the equation has a solution only if are even. Put , then
Hence we need and to be even. We have the following two cases.*Case. *If for some integer then , and hence And we have
Further,
Now if is an odd integer, then we have
It follows that and are all even. Hence is divisible by 4. But implies that a contradiction. Hence in Case 1 we have and

Now and imply that
Hence we have
And we have the following cases. *Subcase 1.1. *If then it is clear that for any integer numbers and we have;
And the equation has solution.*Subcase 1.2. *If and then We have the following two cases.(1.2.1) If , then we have Also from it follows that Now if we choose then from (3.11) it follows that and for any And in this case the equation has a solution.(1.2.2) If , then , and the equation has no solution.

Hence in Subcase 1.2 if , and for any the equation has a solution.*Subcase 1.3. *If and then We have two cases.(1.3.1) If , then Since and hence Now if we identify , then from (3.11) it follows that and And the equation has a solution.(1.3.2) If , then and the equation has no solution.

Hence in Subcase 1.3 if , and for any the equation has a solution.*Subcase 1.4. *If and , then we have the following two cases.(1.4.1) If , then Now implies If we choose then for any the equation has a solution. Hence if and then for any the equation has a solution.(1.4.2) If Since hence If we identify for any then And the equation has a solution.*Subcase 1.5. *If and , we have the following two cases.(1.5.1) If , then Since hence If we identify for any then And the equation has a solution.(1.5.2) If , then And the equation has no solution. Hence in this case only if , the equation has a solution.*Subcase 1.6. *If and , then similar to Case ,if or then And for any , the equation has a solution.*Subcase 1.7. *If and , then Hence the equation has no solution.*Subcase 1.8. *If and , then Hence the equation has no solution.*Subcase 1.9. *If and , we have two cases.(1.9.1) If , then Since hence If we identify for any then And the equation has a solution.(1.9.2) If , then And the equation has no solution.*Subcase 1.10. *If and , then And the equation has no solution.*Subcase 1.11. *If and , then similar to Subcase 1.6, if or then And for any the equation has a solution.*Subcase 1.12. *If and , then And the equation has no solution.*Subcase 1.13. *If and , then we have two cases.(1.13.1) If , then implies If we identify for any the equation has a solution.(1.13.2) If , then And if for any the equation has a solution.*Subcase 1.14. *If and , then In this case the equation has no solution.*Subcase 1.15. *If and , then In this case the equation has no solution.*Case 2. *If Since hence If we identify then In this case the equation has a solution.

Hence we show that in the following twelve cases the equation has solution. And

(1) for all (2) for all (3)(4)(5)(6)(7)(8)(9)(10)(11)(12)

And more precisely we have

Now in the following ten cases the equation has no solution.(13)(14)(15)(16)(17)(18)(19)(20)(21)(22)

We consider the equation Suppose that the equation has a nontrivial solution The elements and have a representation of the following forms:
where are unique integer numbers. By applying Lemma 2.3 one obtains
Hence
where for

Hence equation holds if
Note that second equation gives ; hence equation has nontrivial solution only if In particular in the cases from (17) to (22), since is odd, the equation has no solution and

Finally it remains to consider the cases from (13) to (16). In these cases we have And we prove that if then It is clear that implies Hence In particular and Now we have

Now in the cases from (13) and (15), we have Hence And if we identify:

then for the elements

we have It covers the cases from (13) and (15).

Now we consider the cases from (14) and (16). Since in these cases hence If we identify

then for the elements

one obtains And the equation satisfies.

In particular in the cases from (13) to (16), we have =2. This completes the proof.

As an immediate consequence of Theorem 2.1, we obtain the exact value of the

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

Theorem 3.1 (Rhemtulla-Akhavan[5]). *Let be a free nilpotent group of rank and class freely generated by Then any element of can be expressed as a product of at most two commutators.*

We will also use the fact that if , and are any elements of a group , then

*Proof of Corollary 2.2. *Let be any element of We may write
where , and are suitable integer numbers. Since lies in the center of and is abelian, we may express as
There are two cases:(1)(2)*Case 1. *By (), we may write as a product of three squares.*Case 2. *We may write

Since is even, () yields In Theorem 2.1 we produce elements of square length equal to three. This shows that and completes the proof.

*Note. *Let be a free nilpotent group of rank and class freely generated by Now is a quotient of Since the equations and do not hold in the cases from (17) to (22) in these equations should not hold in . And similarly since the equation does not hold in the cases from (13) to (16) in , hence these equations will not hold in

#### 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.

#### References

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#### Copyright

Copyright © 2009 Mehri Akhavan-Malayeri. This 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.