Другое : 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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