The article gives an algorithm that allows solving the problem of weak power-law conjugacy of words in Artin groups, which are tree products of Artin groups with a tree structure and Artin groups of extra-large type.
Artin group, tree structure, weak power conjugacy, diagram
Let 
– be a finitely generated Artin group with co-representation
where 
– consisting of 
alternating generators 
, – element of the matrix 
: 
[1].
The extra-large type of groups
was introduced by K. Appel and P. Schupp [1]. It is typical for it that 
for all 
For we construct a graph 
so that 
correspond to the vertices of the graph , and 
– an edge with ends 
. has a tree structure if – is a tree.
which has a tree structure, is a free product with amalgamation by cyclic subgroups of Artin groups on two generators.
The woody structure for in 2003 was identified by V.N. Bezverkhny [2].
Now
tree product 
with tree structure or extra-large type, where the union 
, 
is taken over 
, 
, where 
– is the generator of , 
– is the generator of . 
will be called an Artin group with a generalized tree structure or a special class of Artin groups.
It is known that in Artin groups of extra-large type, with a tree structure, and in a special class of Artin groups, the problems of equality and conjugacy of words are solvable [1] - [3].
Let an annular map M, be given on the plane, that is, its complement - two components. We assume that K is unbounded, and H is the bounded component of M, while 
is the outer, and 
is the inner boundary of M. The cycle σ of the smallest length, including the edges , is the outer boundary cycle of M. Similarly, τ is the inner boundary cycle of M. Further, we denote a free product as F, R – symmetrized subset of F, 
– normal closure. If u and z are cyclically reduced in F, do not lie in N, are not conjugate in F, but conjugate in 
, then there exists a reduced ring R-diagram M with at least one domain, and σ, τ are the inner and outer boundary cycles of M, 
. Let u, z be two cyclically reduced words, 
, are not conjugate in F and are conjugate in G. The boundary marks of the regions D of the R-diagram M are relations from R. Consider the transformations of the R – diagram of M. If 
intersect along the edge e so that the common label is either a relation from 
, or a single word in 
, then we erase e, 
and 
are united into a single region, and in the latter case we cut out the resulting region by gluing it border. As a result, we obtain a reduced, that is, invariant under the considered transformations, ring R-diagram M' with boundary marks u and 
. In what follows, we will consider only such diagrams.
is a regular part of M if is the union of a sequence of closed edges occurring in the same order in 
and in 
. 
is simple when 
is the correct part of 
. Further 
is the number of edges of ; 
is the degree of the vertex 
; 
is the number of internal edges, 
is the syllable length w. A simple domain 
is Denov's [2] for 
, the distance 
of the Denov's domain is Denov's reduction 
.
Definition 1 [2]. A sequence 
of domains
forms a strip in M with 
, if:
-
, 
– connected path of syllable length at least 1; -
, boundaries 
, 
are crossed by e; 
;
.
Removing 
in is an 
-reduction [2].
Lemma 1 [2]. There is an algorithm that finds out for each cyclically reduced word w of the group G its R, -irreducibility.
If D is located on both sides of the edge e and 
, the glued edges e and 
intersect 
D, then we have (s – i) – domain [2].
Theorem 1 [3]. If M is the R-conjugacy diagram of words ϕ(σ), ϕ(τ) ∈ G without (s − i)-domains; moreover ϕ(σ), ϕ(τ) are cyclically R, -uncancellable, then M is one-layer.
The R-diagram M of conjugacy of cyclically R and -irreducible words 
from G is considered especially special if M has a unique D with 
. Replacing 
by 
is a special ring R-cancellation [2].
Definition 2 [2]. w is dead-end if it is cyclically R, -irreducible and cannot be specially annular R-reduced.
It follows from [1] - [3] that one can effectively establish whether w is dead-end.
Lemma2. For conjugate dead-end words 
their syllable lengths are equal and minimal in the sense that a word of shorter syllable length v is not conjugate to w.
The argument is obvious.
Definition 3 [2]. In G the weak power conjugacy problem is algorithmically solvable if, for any , 
the existence of an integer n, is effectively defined for which 
are conjugate in G.
Lemma 3. Given any dead-end w from G, we can construct a word 
conjugate to it or its square, any power of which is R, -irreducible.
The proof is similar to the proof of lemma 11 in [4]. This is possible because we introduce the concept of a strip, which is identical to Artin groups with a tree structure, which is identical to that given in Definition 1, as well as the one-layer structure of the R-diagram M of conjugacy of words considered in theorem 1.
Theorem2. For any w from G one can construct a word conjugate to it or its square, any power of which is R, -irreducible.
Proof. Let us execute R, -reductions in cyclic w, then a special ring R-reduction, if any. We get the dead-end word conjugate to w. Next, we apply lemma 2 to the resulting word.
Lemma 3. For every cyclically irreducible v from G one can construct its cyclically conjugate R, -irreducible.
Theorem 3. In G, the weak power conjugacy problem is algorithmically solvable.
Proof. Suppose that 
, v are conjugate in G. We must show that n is bounded.
Based on Theorem 2, we pass from w to
, each degree of which is R, -irreducible. Note that is conjugate to either w or its square.
Lemma 4. Let 
be conjugate. Then 
(
Proof. It follows from [1], [5] that these problems have been solved in the factors of G.
The following cases are possible:
- From w go to .
Suppose that 
, that is 
.
Consider the ring diagram of M conjugacy of 
with
.
According to theorem 1, M is a one-layer diagram. Let t denote the number of domains with access to γ and δ. The number of edges going to
does not exceed 
, the number of vertices does not exceed 
, then t, obviously, does not exceed the sum of edges and vertices, that is 
. We have 
, 
R and has maximum length. Obviously, 
. Hence (
2. From 
go to . 
. Then 
. We replace by its conjugate , any degree of which is cyclically R, -irreducible according to theorem 2. We obtain 
. Replace 
by its equal in the group G R, -irreducible 
according to lemma 3. Next, we use case 1.
The theorem is proved.
Note also that the special class of Artin groups considered in this paper belongs to the almost large Artin groups considered in [6]. Investigations of a special class of Artin groups by diagrammatic methods lead to simpler algorithms for solving the problems of equality, conjugacy, and weak power conjugacy. These methods are supposed to be used in the further study of these groups, in particular, to solve the problems of power conjugacy and intersection of cyclic subgroups.
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2. Bezverkhny V.N., Karpova O. Yu. Problems of equality and conjugacy of words in Artin groups with tree structure. Izvestiya TulSU. Series Mathematics. Mechanics. Computer science. 2006. V. 12, № 1. P. 67-82.
3. Dobrynina I. V., Ugarov A. S. On the problem of conjugacy of words in generalized tree structures of Artin groups // International conference dedicated to the 90th anniversary of the Department of Higher Algebra, Faculty of Mechanics and Mathematics, MSU. MSU, 2019. P. 28-29.
4. Bezverkhny V.N. and Karpova O. Yu., "The problem of entering a cyclic subgroup in Artin groups with tree structure," Chebyshevskii Sbornik. 2008. V.9, №1. P. 30-49.
5. Bezverkhny V.N., Kuznetsova A.N. Solvability of the problem of power conjugacy of words in Artin groups of extra-large type // Chebyshev collection. 2008. V.9, №1. P. 50-68.
6. Derek F. Holt, Sarah Rees. Biatomatic structures in systolic Artin groups. 2018. arXiv:1807.02532v1 [math. GR].
