On a decomposition of an element of a free metabelian group as a productof primitive elements

On a decomposition of an element of a free metabelian group as a productof primitive elements

On a decomposition of an element of a free metabelian group as a productof primitive elements

E.G. Smirnova, Omsk State University, Mathematical Department

1. Introduction

Let G=Fn/V be a free in some variety group of rank n. An element is called primitive if and only if g can be included in some basis g=g1,g2,...,gn of G. The aim of this note is to consider a presentation of elements of free groups in abelian and metabelian varieties as a product of primitive elements. A primitive length |g|pr of an element is by definition a smallest number m such that g can be presented as a product of m primitive elements. A primitive length |G|pr of a group G is defined as , so one can say about finite or infinite primitive length of given relatively free group.

Note that |g|pr is invariant under action of Aut G. Thus this notion can be useful for solving of the automorphism problem for G.

This note was written under guideness of professor V. A. Roman'kov. It was supported by RFFI grant 95-01-00513.

2. Presentation of elements of a free abelian group of rank n as a product of primitive elements

Let An be a free abelian group of rank n with a basis a1,a2,...,an. Any element can be uniquelly written in the form

.

Every such element is in one to one correspondence with a vector . Recall that a vector (k1,...,kn) is called unimodular, if g.c.m.(k1,...,kn)=1.

Лемма 1. An element of a free abelian group An is primitive if and only if the vector (k1,...,kn) is unimodular.

Доказательство. Let , then . If c is primitive, then it can be included into a basis c=c1,c2,...,cn of the group An. The group (n factors) in such case, has a basis , where means the image of ci. However, , that contradics to the well-known fact: An(d) is not allowed generating elements. Conversely, it is well-known , that every element c=a1k1,...,ankn such that g.c.m.(k1,...,kn)=1 can be included into some basis of a group An.

Note that every non unimodular vector can be presented as a sum of two unimodular vectors. One of such possibilities is given by formula (k1,...,kn)=(k1-1,1,k3,...,kn)+(1,k2-1,0,...,0).

Предложение 1. Every element , , can be presented as a product of not more then two primitive elements.

Доказательсво. Let c=a1k1...ankn for some basis a1,...an of An. If g.c.m.(k1,...,kn)=1, then c is primitive by Lemma 1. If , then we have the decomposition (k1,...,kn)=(s1,...,sn)+(t1,...,tn) of two unimodular vectors. Then c=(a1s1...ansn)(a1t1...antn) is a product of two primitive elements.

Corollary.It follows that |An|pr=2 for . ( Note that .

3. Decomposition of elements of the derived subgroup of a free metabelian group of rank 2 as a product of primitive ones

Let be a free metabelian group of rank 2. The derived subgroup M'2 is abelian normal subgroup in M2. The group is a free abelian group of rank 2. The derived subgroup M'2 can be considered as a module over the ring of Laurent polynomials

.

The action in the module M'2 is determined as ,where is any preimage of element in M2, and

.

Note that for , we have

(u,g)=ugu-1g-1=u1-g.

Any automorphism is uniquelly determined by a map

.

Since M'2 is a characteristic subgroup, induces automorphism of the group A2 such that

Consider an automorphism of the group M2, identical modM'2, which is defined by a map

,

By a Bachmuth's theorem from [1] is inner, thus for some we have

Consider a primitive element of the form ux, . By the definition there exists an automorphism such that

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