M.Sc Student Shpigelman Yuval Solution of the Word Problem in Certain One-Relator Free Product Department of Mathematics Mr. Arie Juhasz Abstract

According to a basic theorem in group theory, every group G is homomorphic image of a free group.

In other words, there is a free group F with set of generators X={x1, x2?}  and a set R in F such that  G=F/<<R>> , where <<R>> is the normal closure of R in F. <X|R>, is called a Presentation of G by generators X and relators R.

If X and R are finite G is called finitely presented. From now onwards all the groups will be finitely presented.

Let  X={x1, x2?xn}  and Y=X{x1-1, x2-1?xn-1} . A basic problem in combinatorial group theory is the word problem which is the following: Let G be presented by X and R Is there an algorithm which decides whether a reduce word

W=y1y2?ym  , yi in Y is equal to in G?  In other words, whether W is in <<R>>.  In general there is no solution to the word problem, but there are classes of groups in which the word problem is solvable, and this allows to say much about them.

In the case R={Um??}, m>1 and U is a word on Y, it is known that there is a solution to the word problem. Can we say the same thing when F=G1*G2 (a free product of two groups) instead of free group?

The word problem for this group was studied but not fully, For example, there is a solution to the word problem under every one of the following conditions:
1. m>3
2. m=3, and U  have no subwords of order 2.
3. m=2 , and G1  and G2 are locally indictable(A strong version of not containing elements of finite order) .

We will focus on the case  m=3 and explicitly allow U  having subwords of order 2.

We are using van Kampen diagrams. It is known that if certain geometric conditions holds for those diagrams then there is a solution to the word problem. Such condition is the locally geometric condition W(6).
In this paper we translate, in certain cases, the geometric condition "van Kampen diagrams for U3 doesn't satisfy W(6)" to word equations over the free product.

Using methods from diagrams theory on the results of this paper can give classification of all the cases that the geometric method of using the condition W(6) doesn’t work.