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,

.

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 subgroup 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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