|On a decomposition of an element of a free metabelian group as a productof primitive elements
Тема: On a decomposition of an element of a free metabelian group as a productof primitive elements
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Создан: 30 июня 2007 года
On a decomposition
of an element of a free metabelian group as a productof primitive elements
E.G. Smirnova, Omsk State
University, Mathematical Department
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
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
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
The action in the module M'2 is
determined as ,where is any preimage of element in M2, and
Note that for , we have
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  is inner, thus for some we have
Consider a primitive element of the
form ux, . By the definition there
exists an automorphism such that
Using elementary transformations we
can find a IA-automorphism with a first row of the form(1). Then by
mentioned above Bachmuth's theorem
In particular the elements of type
u1-xx, u1-yy, are primitive.
Предложение 2. Every element of the derived
subgroup of a free metabelian group M2 can be presented as a product of not
more then three primitive elements.
Доказательство. Every element can be written as , and can be presented as
A commutator , by well-known commutator
identities can be presented as
The last commutator in (3) can be
added to first one in (2). We get [y-1 , that is a product of three
4. A decomposition
of an element of a free metabelian group of rank 2 as a product of primitive
For further reasonings we need the
following fact: any primitive element of a group A2 is induced by a
primitive element , . It can be explained in such way.
One can go from the basis to some other basis by using a
sequence of elementary transformations, which are in accordance with elementary
transformations of the basis <x,y> of the group M2.
The similar assertions are valid for
any rank .
Предложение 3. Any element of group M2 can be
presented as a product of not more then four primitive elements.
Доказательство. At first consider the elements in
form . An element is primitive in A2 by lemma 1,
consequently there is a primitive element of type . Hence, Since, an element is primitive, it can be included
into some basis inducing the same basis of A2. After rewriting in this new
basis we have:
and so as before
Obviously, two first elements above
are primitive. Denote them as p1, p2. Finally, we have
, a product of three primitive
If , then by proposition 1 we can find
an expansion as a product of two primitive elements,
which correspond to primitive elements of M2: v1xk1yl1,v2xk2yl2,v1,v2 .
we have the expansion
The element w(v1xk1yl1) can be
presented as a product of not more then three primitive elements. We have a
product of not more then four primitive elements in the general case.
5. A decomposition of elements of a
free metabelian group of rank as a product of primitive elements
Consider a free metabelian group
Mn=<x1,...,xn> of rank .
Предложение 4. Any element can be presented as a product of not
more then four primitive elements.
Доказательсво. It is well-known , that M'n as
a module is generated by all commutators . Therefore, for any there exists a
Separate the commutators from (4)
into three groups in the next way.
1) - the commutators not including the
element x2 but including x1.
2) - the other commutators not
including the x1.
3) And the third set consists of the
Consider an automorphism of Mn,
defining by the following map:
The map determines automorphism, since the
Jacobian has a form
and hence, det Jk=1.
Since element can be included into a basis of Mn,
it is primitive. Thus any element can be presented in form
[x1-1x2-1x3-1]. =p1p2p3p4 a product
of four primitive elements.
Note that the last primitive element
p4=x1-1x2-1x3-1 can be arbitrary.
Предложение 5. Any element of a free metabelian
group Mn can be presented as a product of not more then four primitive
Доказательство. Case 1. Consider an element , so that g.c.m.(k1,...,kn)=1. An
element is primitive by lemma 1 and there
exists a primitive element ,
An element from derived subgroup can
be presented as a product of not more then four primitive elements with a fixed
one of them:
Case 2. If , then by lemma 2 , where are primitive in An. There exist
primitive elements So We have just proved that the element
wp1 can be presented as a product of not more then three primitive elements
p1'p2'p3'. Finally we have c=p1'p2'p3'p2, a product of not more then four
Bachmuth S. Automorphisms of free
metabelian groups // Trans.Amer.Math.Soc. 1965. V.118. P. 93-104.
Р., Шупп П. Комбинаторная теория групп. М.: Мир, 1980.
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