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 sub>group 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 sub>group M'2 is
abelian normal sub>group in M2. The group
is
a free abelian group of rank 2. The derived sub>group 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 sub>group,
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
|
|
(1) |
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 sub>group 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
.
Thus,
|
|
(2) |
A commutator
,
by well-known commutator identities can be presented as
|
|
(3) |
The last commutator in (3) can be added to first
one in (2). We get
[y-1
,
that is a product of three primitive elements.
4. A decomposition of an element of a free metabelian group of rank 2 as a product of primitive elements
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 elements.
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
.
Further 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 [2], that M'n as a module is generated by all
commutators
.
Therefore, for any
there
exists a presentation
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 commutator
.
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
x3x2x1]
[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 elements.
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 sub>group can be presented as a product of not more then four primitive elements with a fixed one of them:
Then
.
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 primitive elements.
Список литературы
Bachmuth S. Automorphisms of free metabelian groups // Trans.Amer.Math.Soc. 1965. V.118. P. 93-104.
Линдон Р., Шупп П. Комбинаторная теория групп. М.: Мир, 1980.
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