On a decomposition of an element of a free metabelian group as a productof primitive elements (стр. 2 из 3)

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