M.Sc Student | Shpigelman Yuval |
---|---|
Subject | Solution of the Word Problem in Certain One-Relator Free Product |
Department | Department of Mathematics |
Supervisor | Mr. Arie Juhasz |
Full Thesis text - in Hebrew |
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={x_{1}, x_{2}?} 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={x_{1}, x_{2}?x_{n}} and Y=X∪{x_{1}^{-1}, x_{2}^{-1}?x_{n}^{-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=y_{1}y_{2}?y_{m} , y_{i} 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={U^{m}??}, 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=G_{1}*G_{2} (a free product of two groups) instead of free group?
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 U^{3}
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.