The words P and P 0 contain the history of our computation, because Pi and Pi0 correspond to S-rules applied during the computation. The words T1 and Tm are labels of the top and the bottom paths of the trapezium. Let us choose disjoint alphabets in different copies of the trapezium. Pi0 Pi? T1 - Pi? Notice that if P and P 0 are copies of each other then we can also glue the right side of one trapezium to the left side of another without taking mirror images.

Of course the conjugacy relations involving letter k should be added into the set of group relations Z. Suppose that Tm is equal to q, i.

- The Conjugacy Problem and Higman Embeddings.
- Department of Mathematical Sciences.
- The Conjugacy Problem and Higman Embeddings!
- Mathematics Library - New Books Search!
- Biographies;

The words T10 , We add the hub to the list of defining relations Z. Again, if Z is chosen carefully then the converse is also true: Condition 2. The diagram on the previous Figure is called a disk. The trapezia forming this disk are numbered in the natural order from 1 to N. We assume that the copy of the alphabet A used in the first subtrapezium of a disk coincides with A itself the generating set of G.

## 3. ‘г ђ” Љw

Let us denote the group given by the presentation Z we have got so far by G M. There are several ways to use the group G M to embed G into a finitely presented group. We use another method, described in [OlSa2] and used also in [OlSa3]. Number these trapezia by 20 , Let us add the relations used in the new trapezia and the conjugation relations involving ki to Z. The inner boundary of it is labelled by the hub. If we add the hub to the diagram, we get a van Kampen diagram which also looks like a disk but has N-1 sectors.

Let us call the group given by the set of relations Z that we have constructed by H. We have proved that the identity map on A can be extended to a homomorphism from G to the subgroup hAi of the group H given by the set of relations Z.

It is possible to show that this homomorphism is injective, so G is embedded into H. Suppose that G has solvable conjugacy problem.

Is it true that H has solvable conjugacy problem? Let us give two examples. Example 1. Consider two pairs of words u1 , u2 and v1 , v2 over the alphabet A we can view these words as elements of G. Let q be the first state letter in K qu. Thus if the pair u1 , u2 is conjugated to the pair v1 , v2 in G then the words W1 and W2 are conjugated in G. One can also prove that the converse statement is also true. Therefore if the conjugacy problem is solvable in G then the conjugacy of t-tuples of elements in G is solvable.

It is known [Col] that there exists a finitely generated group G with solvable conjugacy problem and unsolvable problem of conjugacy for sequences of elements. For such a group G the group H has undecidable conjugacy problem. Example 2. This problem easily reduces exercise! It is quite possible that this problem can be undecidable even if the conjugacy problem is decidable. If we achieve that then we would have a recursive bound on the length of the word W in Example 1 and the words P, Q in Example 2, so we would be able to find these words by a simple search.

This lead us to the S-machine S considered in this paper see Sections 2.

This machine has four sets of state letters: K, L, P, R-letters. While P is moving, L, R stay next to K-letters. It is important to mention that we cannot allow all parts of the admissible words to be arbitrary group words. In order to keep these subwords positive we use a trick that goes back to Novikov, Boone and Higman see Rotman [Rot].

- Critical Incident Stress Management in Aviation.
- Research and Inequality.
- Building partnerships in the Americas : a guide for global health workers!
- Cities and Consumption (Routledge Critical Introductions to Urbanism and the City)?
- Research Publications to .
- Jake Bakes Cakes.
- Practical English Usage!
- The Conjugacy Problem and Higman Embeddings by - ounsahouvere.cf;
- No customer reviews.
- The Years Best Science Fiction, Seventeenth Annual Collection;
- Herb Gardening from the Ground Up: Everything You Need to Know about Growing Your Favorite Herbs.

We use a similar idea. So we had to use many different letters x, replace 2 by 4 in the BaumslagSolitar relations, and make some other technical modifications. Notice that Examples 1 and 2 are only the main obstacles that we had to overcome in this paper. Other technical difficulties lead to further fine tuning of our S-machine.

Sections 2. Now let us briefly describe the strategy of the proof that the conjugacy problem in H is Turing reducible to the conjugacy problem in G. In terms of annular Schupp diagrams, our task is the following: given two words u and v in generators of H, find out if there exists an annular diagram over the presentation of H with boundary labels u and v.

Different methods are used to treat different cases. Roughly speaking, the study of rings amounts to study the lengths of computations of our S-machines, the study of rolls amounts to the study of the space complexity how much space is needed by the machines during a computation , and the study of spirals amounts to the study of computations with periodic history.

## News & Events Archive - Department of Mathematics - University at Buffalo

The x-letters and the Baumslag-Solitar relations allow us to treat the case of rings, but they cause the main technical difficulties in the cases of rolls and spirals. The precise definition of the S-machines used in this paper will be given later. Let T1 , Let A be a set of admissible words of the form q1 w1 q We do not distinguish words over Ai which are equal in the group Ti.

So we can view wi as elements in Ti. Every S-rule is a partial transformation on the set of admissible word. Thus we can view an S-machines as a semigroup of partial transformations of the set of admissible words. The collection of groups Ti and the set of admissible words form the hardware of an S-machine. So we shall assume that G has solvable conjugacy problem. Letters from K are also called basic letters.

The projection of y In addition, each ui resp. This important property will hold for all other rules of S. This transition rule prepares the machine for step 2. The meaning of this rule is that the machine can start step 3 when Lj and Pj meet that is when the inner parts of Lj z-sectors are empty. This rule can be applied when the state letters Lj , Pj , Rj meet together; it inserts r to the left of the state letters Lj , and prepares the machine for step 4. A graphical description of the work of the S-machine S is presented on Figure 5 below.

Remark 2. Hence the inner part must be empty, and W is not reduced, a contradiction. Lemma 2. Without loss of generality we can assume that w0 is obtained from w by inserting r. Then a rule from j next to Pj S 23 will change the coordinates to r, 3. Then a rule from S 34 will insert r resp. The proof is similar to the proof of Lemma 2.

Then there exist two sequences Lemma 2. W is in the domain of h; 3.

### Product details

Now we need to explain how to simulate the S-machines S and S 2. Let U be any word in A. In general, the projection of a word U onto an alphabet Y will be denoted by UY.