# The conjugacy stability problem
for parabolic subgroups in Artin groups

María Cumplido

(August 27, 2024)

###### Abstract

Given an Artin group $A$ and a parabolic subgroup $P$, we study if every two
elements of $P$ that are conjugate in $A$, are also conjugate in $P$. We
provide an algorithm to solve this decision problem if $A$ satisfies three
properties that are conjectured to be true for every Artin group. This allows
to solve the problem for new families of Artin groups. We also partially solve
the problem if $A$ has $FC$-type, and we totally solve it if $A$ is isomorphic
to a free product of Artin groups of spherical type. In particular, we show
that in this latter case, every element of $A$ is contained in a unique
minimal (by inclusion) parabolic subgroup.

_2020 Mathematics Subject Classification._ 20F36, 20F10.

_Key words._ Artin groups; conjugacy stability; conjugacy classes; algorithmic
in group theory.

lemmatheorem lemma

propositiontheorem proposition

corollarytheorem corollary

conjecturetheorem conjecture

claimtheorem remarktheorem notationtheorem notation

claim

remark

definitiontheorem definition

exampletheorem example

##  1 Introduction

Artin (or Artin–Tits) groups were defined by Jacques Tits in the 60’s. They
are groups presented by a finite set of generators $S$ and at most one
relation of the form $stst\cdots=tsts\cdots$, for every pair $s,t\in S$, with
the same number of letters $m_{s,t}$ at each side of the equality. If there is
no relation associated to a pair of generators $s,t\in S$, then we denote
$m_{s,t}=\infty$. Then, the presentation of an Artin group is as follows:

$A_{S}=\langle S\,|\,\underbrace{sts\dots}_{m_{s,t}\text{
elements}}=\underbrace{tst\dots}_{m_{s,t}\text{ elements}}\forall s,t\in
S,\,s\neq t,\,m_{s,t}\neq\infty\rangle.$

These groups are algebraic generalisations of the well-known braid groups on
$n+1$ strands (Artin,, 1947):

$A_{n}=\left\langle\sigma_{1},\dots,\sigma_{n}\,\begin{array}[]{|lr}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i},&|i-j|>1\\\
\sigma_{i}\sigma_{j}\sigma_{i}=\sigma_{j}\sigma_{i}\sigma_{j},&|i-j|=1\end{array}\right\rangle.$

A fundamental tool for the study of braid groups is the action by isometries
of $A_{n}$ on the curve complex of the $n+1$-punctured disk
$\mathcal{D}_{n+1}$. The curve complex has as vertices (isotopy classes of
non-degenerated) simple closed curves in $\mathcal{D}_{n+1}$. For Artin
groups, the analogous of simple closed curves are irreducible parabolic
subgroups. In fact, there is a bijection between the proper irreducible
parabolic subgroups of $A_{n}$ and the simple closed curves in
$\mathcal{D}_{n+1}$ (see an explanation in Cumplido et al.,, 2019, Section 2).

A _standard parabolic subgroup_ $A_{X}$ is a subgroup generated by a subset of
generators $X\subseteq S$. The conjugate of any standard parabolic subgroup by
an element of $A_{S}$ is called a _parabolic subgroup_. The study of parabolic
subgroups has been an important source of research in Artin groups over the
last forty years. These are natural and easy to define subgroups. They are the
main ingredient of complexes in which Artin groups act, as the Deligne complex
(Deligne,, 1972, Charney & Davis,, 1995) or the complex of irreducible
parabolic subgroups (Cumplido et al.,, 2019). However, as it happens for most
questions in Artin groups, basic properties of parabolic subgroups are in
general unknown. Some of the facts we know about are the following: In his
thesis, Van der Lek, (1983) proved that a standard parabolic subgroup is again
an Artin group, and we also know that they are convex in every case (Charney &
Paris,, 2014). The structure of centralisers of parabolic subgroups and many
of their properties have been well studied only certain cases by Paris, (1997)
and Godelle, (2003a, 2007, b), among others; and we only know if the
intersection of parabolic subgroups is again a parabolic subgroup for a few
families of Artin groups (Cumplido et al.,, 2019, Morris-Wright,, 2021,
Cumplido et al.,, 2020).

In this paper we discuss in which cases embeddings of parabolic subgroups into
the Artin group merge conjugacy classes. This is also called the _conjugacy
stability_ problem for parabolic subgroups.

######  Definition .

A parabolic subgroup $P$ of an Artin group $A$ is _conjugacy stable_ in $A$ if
for every $x,y\in P$ such that $g^{-1}xg=y$, $g\in G$, there is $\hat{g}\in P$
such that $\hat{g}^{-1}x\hat{g}=y$. If $P$ is not conjugacy stable in $A$ we
say that the inclusion of $P$ into $A$ _merges conjugacy classes_.

This problem has been solved only for some specific families of Artin groups.
González-Meneses, (2014) proved that parabolic subgroups of braid groups are
always conjugacy stable. However, for Artin groups this is not always the
case. In (Calvez et al.,, 2020), we give an explicit classification for
_spherical-type_ (or finite type) Artin groups, which are the groups that
become finite when adding to their presentation the relations $s^{2}=1$ for
every $s\in S$. For large type and FC-type, a simpler question was addressed
by Godelle, (2021): He studied what happens if in the definition of conjugacy
stable we impose $g$ to be an element of $S$. At the end of (Cumplido et al.,,
2020), we completely classify the parabolic subgroups of large Artin groups up
to conjugacy stability, using the aforementioned results of Paris and Godelle.
The aim of this article is to use these results to prove that conjugacy
stability problem can be solved for every Artin group satisfying three
properties that are conjectured to always hold in Artin groups.

If for an element $\alpha$ in an Artin group there is a unique minimal (with
respect to the inclusion) parabolic subgroup $P_{\alpha}$ containing $\alpha$,
we say that $P_{\alpha}$ is the _parabolic closure_ of $\alpha$. We will show:

Theorem A. Let $A$ be a standardisable (Section 3) Artin group satisfying the
ribbon property (Section 3) and such that every element in $A$ has a parabolic
closure. Then, there is an algorithm that decides whether a parabolic subgroup
$P$ of $A$ is conjugacy stable in $A$ or not.

The existence of parabolic closures –which is a consequence of the
intersection of two parabolic subgroups being a parabolic subgroup– and the
other two hypotheses of the theorem are conjectured to be true for all Artin
groups. In particular, they are known to be true for spherical-type Artin
groups (Godelle,, 2003a, Cumplido et al.,, 2019). The standardisation and
ribbon properties are true for FC-type and two-dimensional Artin groups
(Godelle,, 2007, 2003b). For FC-type, the problem of the intersection of
parabolic subgroups is solved for spherical-type parabolic subgroups –the
conjugates of some spherical-type standard parabolic subgroup– by Morris-
Wright, (2021). Using these results and the fact that FC-type Artin groups can
be seen as amalgamated free-products of spherical-type Artin groups, we will
partially solve the conjugacy stability problem for parabolic subgroups of a
FC-type Artin group $A$. We will totally solve the problem if $A$ is
isomorphic to a free-product of spherical-type Artin groups, by proving the
existence of parabolic closures in this case (Section 4.1). This is summarized
in Theorem B.

######  Definition .

Given an Artin group $A$ and a parabolic subgroup $P$ of $A$, we say that $P$
is _conjugacy quasi-stable_ if for every two elements $x,y\in P$ contained in
(possibly different) spherical-type parabolic subgroups of $A$ such that
$g^{-1}xg=y$ with $g\in A$, there is $z\in P$ such that $z^{-1}xz=y$.

######  Remark .

Notice that for spherical-type Artin groups being conjugacy quasi-stable is
equivalent to be conjugacy stable.

Theorem B. Let $A$ be an FC-type Artin group. There is an algorithm that
decides whether a given parabolic subgroup $P$ of $A$ is conjugacy quasi-
stable in $A$. In particular, this algorithm can tell whether a spherical-type
parabolic subgroup is conjugacy stable or not.

Moreover, if $A$ is isomorphic to a free product of spherical type Artin
groups, then every element of $A$ has a parabolic closure and there is an
algorithm that solves the conjugacy stability problem for every parabolic
subgroup of $A$.

This article is structured in the following way: In Section 2 we will describe
a result of Paris, (1997) that gives an algorithm to decide when two standard
parabolic subgroups are conjugate in any Artin group, and we will give an
explicit form for this algorithm; in Section 3 we will explain how to modify
this algorithm to solve the conjugacy stability problem for parabolic
subgroups of Artin groups that satisfy the three hypothesis of Theorem A; in
Section 4 we will discuss the case of FC-type Artin groups.

######  Remark .

After a first preprint of this paper, Blufstein, (2021) generalised the
results in (Cumplido et al.,, 2020) and showed that the intersection of
parabolic subgroups is a parabolic subgroup for two-dimensional Artin groups
with a Coxeter graph –see next section– in which every vertex is disconnected
from at most one other vertex. This completed the set of three hypotheses
needed in Theorem A and allowed him two apply Algorithm 4 of Section 3 to
solve the conjugacy problem in this case.

Haettel, (2021) has also proved the three conjectures for Euclidean Artin
groups of type $\tilde{A}$ and $\tilde{C}$, so we know that the main theorem
works for these groups.

##  2 Conjugate standard parabolic subgroups

In this section we explain in detail the results in (Paris,, 1997) to decide
when two standard parabolic subgroups are conjugate in an Artin group $A_{S}$.
This work is based on the part of Daan Krammer’s thesis that solves the
conjugacy problem in Coxeter groups, which is published in (Krammer,, 2009).
To begin, we first need to know how to define the Coxeter graph of an Artin
group and the classification of Artin groups of spherical type.

######  Definition .

The _Coxeter graph_ $\Gamma_{S}$ of the Artin group $A_{S}$ is the graph
defined by the following data:

  * •

The set of vertices of $\Gamma_{S}$ is $S$.

  * •

There is an edge connecting $s$ and $t$ if and only if $m_{s,t}>2$. This edge
if labeled with $m_{s,t}$ if $m_{s,t}>3$.

If $\Gamma_{S}$ is connected, we say that $A_{S}$ is _irreducible_.

In Figure 1, the reader can find the classification (Coxeter,, 1935) of the
ten types of irreducible Artin groups of spherical type. All the other Artin
groups of spherical type are direct products of irreducible ones. When useful,
we will refer to $A_{S}$ as $A_{n}$, $B_{n}$, $D_{n}\dots$, but normally we
will say that the Artin group and Coxeter graph are of type $A_{n}$, $B_{n}$,
$D_{n}\dots$ We denote the generators of $A_{S}$ by $s_{1},s_{2},s_{3},\dots$,
accordingly with the numbering of Figure 1.

$1$$2$${n-1}$${n}$$4$$1$$2$${n-1}$${n}$$3$${n-1}$${n}$$2$$1$$2$$3$${5}$${6}$$4$$1$$2$$3$$5$$6$$4$$1$${7}$$2$$3$$5$$6$$4$$1$${7}$${8}$$1$$2$$3$$5$$1$$2$$3$$5$$4$$1$$2$$3$$4$$4$$1$$2$$m$$A_{n}:$$B_{n}:$$D_{n}:$$H_{3}:$$H_{4}:$$E_{6}:$$E_{7}:$$E_{8}:$$F_{4}:$$I_{2}(m):$$n\geq
1$$n\geq 2$$n\geq 4$$m\geq 5$ Figure 1: Classification of irreducible Coxeter
graphs of finite type

Given an Artin group $A_{S}$, the submonoid $A_{S}^{+}$ of $A_{S}$ generated
by $S$ has the exactly same presentation as $A_{S}$ (seen as monoid) (Paris,,
2002). If $A_{S}$ is an Artin group of spherical type, it has a Garside
structure. This implies that if $A_{S}$ has spherical type there is a lattice
order $\preccurlyeq$ defined by “$a\preccurlyeq b$ iff $\exists c\in
A_{S}^{+},$ $ac=b$”. The least common multiple of all generators of $S$ is
called the Garside element of $A_{S}$ and is denoted by $\Delta$. By
(Brieskorn & Saito,, 1972) we know that the centre $Z(A_{S})$ of $A_{S}$ is
generated either by $\Delta$ or by $\Delta^{2}$.

For Artin groups of type $A_{n}\,(n\geq 2),D_{n}\,(n\geq 5),E_{6}$ and
$I_{2}(m)\,(m\geq 5\text{ and odd})$, $\Delta^{2}$ generates the centre of the
group. Otherwise, $\Delta$ generates the centre of $A_{S}$. In the first case,
the conjugation by $\Delta$ can be seen as a reflection automorphism of
$\Gamma_{S}$. These conjugations, that are well-known by experts, are detailed
in what follows:

  * •

For $A_{n},\,n\geq 2$, one has that $\Delta^{-1}s_{i}\Delta=s_{n-i+1}$.

  * •

For $D_{n},$ with $n\geq 5$ and $n\text{ odd}$, the conjugation by $\Delta$
permutes $s_{1}$ and $s_{2}$ and fixes the other generator.

  * •

For $E_{6}$, the conjugation by $\Delta$ fixes $s_{1}$ and
$\Delta^{-1}s_{i}\Delta=s_{8-i}$ for $i\neq 1$.

  * •

For $I_{2}(m),$ with $m\geq 5$ and $m\text{ odd}$, the conjugation by $\Delta$
permutes $s_{1}$ and $s_{2}$.

We will be specially interested in the Artin groups such that the conjugation
by $\Delta$ can be seen as a reflection automorphism of $\Gamma_{S}$:

######  Definition .

We say that $A_{S}$ is a _twistable_ Artin group if it is one of the following
Artin groups of spherical type:

$A_{n},\,n\geq 2;\quad D_{n},\,n\geq 5\text{ and }n\text{ odd;}\quad
E_{6};\quad I_{2}(m),\,m\geq 5\text{ and }m\text{ odd.}$

Thanks to Van der Lek, (1983), we also know that a standard parabolic subgroup
$A_{Y}$ is an Artin group having as Coxeter graph
$\Gamma_{Y}\subset\Gamma_{S}$. If $A_{Y}$ has spherical type, we denote its
Garside element by $\Delta_{Y}$.

Suppose that $A_{X}$ is a maximal proper standard parabolic subgroup of a
twistable standard parabolic subgroup $A_{Y}$ of $A_{S}$, in other words,
$X=Y\setminus\\{t\\}$, $t\in X$. If $A_{Y}$ is of type $A_{n}$, $n$ odd,
suppose that $t$ is not the central generator of $A_{Y}$. If $A_{Y}$ is of
type $E_{6}$ suppose that $t$ does not correspond to $s_{1}$ or $s_{4}$ and if
$A_{Y}$ is of type $D_{n}$ suppose that $t$ corresponds to either $s_{1}$ or
$s_{2}$. Then $\Delta_{Y}^{-1}A_{X}\Delta_{Y}$ is a standard parabolic
subgroup of $A_{Y}$ different from $A_{X}$. (If $t$ is one of the forbidden
generators, then $\Delta_{Y}^{-1}A_{X}\Delta_{Y}=A_{X}$). This is the main
ingredient of Paris’ result, which states that two standard parabolic
subgroups $A_{X}$ and $A_{X^{\prime}}$ are conjugate if and only if it is
possible to go from one to the other by performing those types of conjugations
or “twists”.

Let $X\subset S$ and define $\mathrm{Adj}(X)$ as the set of vertices in
$\Gamma_{S}$ that are adjacent to $\Gamma_{X}$. We will consider lists of
couples $(Y,c)$, where $Y\subset S$ is a subset of generators and $c\in A_{S}$
is an element that conjugates the set $X$ to the set $Y$. For a given
$X\subset S$, we will recursively construct the list $\mathcal{V}_{X}$ as
follows. Start the list with the couple $(X,1)$. For every $(Y,c)$ in the list
and for every $t\in\mathrm{Adj}(Y)$, take the connected component
$\Gamma_{Y^{\prime}}$ of $\Gamma_{Y\cup\\{t\\}}$ containing $t$. If this
component is twistable, conjugate $Y$ by the Garside element
$\Delta_{Y^{\prime}}$ of the component. If the result $Z$ is a subset of
generators that is not contained in some couple of the list, add the couple
$(Z,c\Delta_{Y^{\prime}})$. Repeat this process. To prove that the process
stops at some point, just observe that the set of standard parabolic subgroups
of an Artin group is finite.

######  Theorem 1.

Given an Artin group $A_{S}$ and two standard parabolic subgroups $A_{X}$ and
$A_{X^{\prime}}$, $A_{X}$ is conjugate to $A_{X^{\prime}}$ if and only if
there is a couple $(X^{\prime},c)$ in $\mathcal{V}_{X}$, in which case $c$ is
a conjugacy element.

###### Proof.

This theorem is a reformulation of (Paris,, 1997, Theorem 4.1). We can see
that $c$ is a conjugacy element by its own construction. ∎

In Algorithm 1, we give to Paris’ result an explicit algorithmic form. The
algorithm tells us when two standard parabolic subgroups $A_{X}$ and
$A_{X^{\prime}}$ are conjugate. If they are not, it constructs the whole list
$\mathcal{V}_{X}$.

Input : The Coxeter graph $\Gamma_{S}$ of an Artin group $A_{S}$ and two
subsets $X,{X^{\prime}}\subset S$.

Output : A conjugating element between the parabolic subgroups $A_{X}$ and
$A_{X^{\prime}}$ or “There is no conjugating element”.

if  _$|X|\neq|{X^{\prime}}|$ _ then

return _“There is no conjugating element”;_

$\mathcal{V}=\\{(X,1)\\}$;

for  _$(Y,c)\in\mathcal{V}$_ do

for  _$t\in\mathrm{Adj}(Y)$_ do

if  _the connected component $\Gamma_{Y^{\prime}}$ of $\Gamma_{Y\cup\\{t\\}}$
containing $t$ is twistable_ then

$Z=\Delta_{Y^{\prime}}^{-1}Y\Delta_{Y^{\prime}}$;

if  _ $Z$ is not the first element of any couple in $\mathcal{V}$_ then

$\mathcal{V}=\mathcal{V}\cup\\{(Z,c\Delta_{Y^{\prime}})\\}$;

if  _$Z={X^{\prime}}$_ then

return _$c\Delta_{Y^{\prime}}$ ;_

return _“There is no conjugating element”;_

Algorithm 1 Algorithm that finds a conjugating element between two standard
parabolic subgroups or tells that it does not exist.

Example. Consider the spherical-type Artin group $E_{7}$, as depicted in
Figure 1. We are going to see that the parabolic subgroup $A_{X}$ with
$X=\\{s_{1},s_{2},s_{3},s_{4},s_{6}\\}$ is conjugate to $A_{X^{\prime}}$,
where $X^{\prime}=\\{s_{2},s_{4},s_{5},s_{6},s_{7}\\}$. First, we take
$s_{5}\in\mathrm{Adj}(X)$. The set of generators
$X\cup\\{s_{5}\\}=\\{s_{1},s_{2},s_{3},s_{4},s_{5},s_{6}\\}$ defines a
connected spherical-type parabolic subgroup isomorphic to $E_{6}$, which is
twistable. If we conjugate $X$ by the Garside element of
$A_{X\cup\\{s_{5}\\}}$, we obtain the set of generators
$Y=\\{s_{1},s_{2},s_{4},s_{5},s_{6}\\}$. Now take $s_{7}\in\mathrm{Adj}(Y)$.
The group defined by $Y\cup\\{s_{7}\\}$ has the connected component $A_{Z}$,
$Z=\\{s_{1},s_{4},s_{5},s_{6},s_{7}\\}$, which is a (twistable) braid group.
Conjugating by the corresponding Garside element, we finally obtain
$\Delta_{Z}^{-1}Y\Delta_{Z}=X^{\prime}$.

##  3 Solution to the conjugacy stability problem

In this section we will explain two of the three hypothesis of Theorem A,
namely the ribbon property and the property of being standardisable. After
that, we will construct the main algorithm of this paper to know when the
embedding of a standard parabolic subgroup merges conjugacy classes.

We first describe the results of (Godelle,, 2003a, 2007, b) concerning the set
of elements conjugating two standard parabolic subgroup of an Artin group
$A_{S}$. Suppose that $A_{X}$, $X\subset S\setminus T$, is a standard
parabolic subgroup of spherical type and let $X^{\prime}=X\setminus\\{t\\}$,
for some $t\in X$. Since $X$ has spherical type, $X^{\prime}$ also has
spherical type and we can consider $\Delta_{X}$ and $\Delta_{X^{\prime}}$. We
have that

$\Delta_{X}^{-1}\Delta_{X^{\prime}}A_{X^{\prime}}\Delta_{X^{\prime}}^{-1}\Delta_{X}=\Delta_{X}^{-1}A_{X^{\prime}}\Delta_{X}=A_{Y},$

for some subset $Y\subset X$. The conjugating element
$\Delta_{X^{\prime}}^{-1}\Delta_{X}$ and its inverse is what Godelle
respectively calls an elementary $(X^{\prime},Y)$–ribbon and an elementary
$(Y,X^{\prime})$–ribbon.

In general, for any (not necessarily of spherical type) parabolic subgroup
$A_{T}$ of $A_{S}$, if there is $s\in S$ such that the component $\Gamma_{U}$
of $\Gamma_{T\cup\\{s\\}}$ that contains $s$ is of spherical type, we call
$r_{T,s}:=\Delta_{U\setminus\\{s\\}}^{-1}\Delta_{U}$ and its inverse
_elementary ribbons_ –notice that $\Gamma_{U}$ does not need to be twistable–.
We say that an element $r=r_{1}r_{2}\cdots r_{q}$ is a _$(T,T^{\prime})$
–ribbon_ if and only if there is a sequence of sets of generators
${T=T_{1}},T_{2},\dots,T_{q+1}=T^{\prime}$ such that each $r_{i}$ is an
elementary $(T_{i},T_{i+1})$–ribbon. The set of all $(T,T^{\prime})$–ribbons
is denoted by $\mathrm{Ribb}(T,T^{\prime})$. When referring to a
$(T,T^{\prime})$–ribbon without caring about the specific $T^{\prime}$, we
will use the term $(T,-)$–ribbon.

Now we will see some properties about ribbons. The following lemma will allow
us to work some of the proofs using positive elementary ribbons and treat the
negative ones as an analogous case:

######  Lemma .

Let $A_{S}$ be an Artin group, $X\subset S$ and $a\in S\setminus X$. Suppose
that $\Gamma_{Y}$ is the component of $\Gamma_{X\cup\\{a\\}}$ that contains
$a$. If there is an elementary ribbon
$r_{X,a}=\Delta_{Y\setminus\\{a\\}}^{-1}\Delta_{Y}$, then there are $T\subset
S$, $s\in S\setminus T$ such that
$r_{T,s}=\Delta^{-1}_{Y\setminus\\{s\\}}\Delta_{Y}$ and
$r_{X,a}^{-1}=\Delta_{Y\setminus\\{s\\}}\Delta_{Y}^{-1}$.

###### Proof.

We have that
$r^{-1}_{X,a}=\Delta_{Y}^{-1}\Delta_{Y\setminus\\{a\\}}=\Delta_{Y\setminus\\{s\\}}\Delta_{Y}^{-1}$,
where $s=\Delta_{Y}^{-1}a\Delta_{Y}$. To see that there is a positive
elementary ribbon of the form $\Delta_{Y\setminus\\{s\\}}^{-1}\Delta_{Y}$, let
$T=(X\cup\\{a\\})\setminus\\{s\\}$. Hence, $\Gamma_{Y}$ is the component of
$\Gamma_{T\cup\\{s\\}}$ that contains $s$ and
$r_{T,s}=\Delta_{Y\setminus\\{s\\}}^{-1}\Delta_{Y}$.

∎

######  Remark .

In the above lemma, $r_{T,s}=\Delta^{-1}_{Y\setminus\\{s\\}}\Delta_{Y}$ and
$r^{-1}_{X,a}=\Delta_{Y\setminus\\{s\\}}\Delta_{Y}^{-1}$ are (positive and
negative) elementary $(T,X)$–ribbons. Similarly,
$r_{X,a}=\Delta_{Y\setminus\\{a\\}}^{-1}\Delta_{Y}$ and
$r_{T,s}^{-1}=\Delta_{Y\setminus\\{a\\}}\Delta_{Y}^{-1}$ are (positive and
negative) elementary $(X,T)$–ribbons.

The next two lemmas help us understand how the conjugation by ribbons
transforms the generators of standard parabolic subgroups:

######  Lemma .

Let $A_{S}$ be an Artin group and $X\subset S$. Let $t\in S\setminus X$ and
$Z\subset X\cup\\{t\\}$ be such that $\Gamma_{Z}$ is the connected component
of $\Gamma_{X\cup\\{t\\}}$ containing $t$ and it is of spherical type. Let
$X^{\prime}\subseteq X$ denote any subset defining a connected component
$\Gamma_{X^{\prime}}$ of $\Gamma_{X}$. Then

  * •

If $A_{X^{\prime}}$ defines a component which is not of spherical type, then
$r_{X,t}^{-1}sr_{X,t}=s$, for every $s\in X^{\prime}$.

  * •

If $A_{X^{\prime}}$ is of spherical type and of type different from
$A,D,E_{6}$ and $I_{2}(m)$, then $r_{X,t}^{-1}sr_{X,t}=s$, for every $s\in
X^{\prime}$.

  * •

If $A_{X^{\prime}}$ is of type either $E_{6}$ or $I_{2}(m)$, then either
$r_{X,t}^{-1}sr_{X,t}=s$ for every $s\in X^{\prime}$ or
$r_{X,t}^{-1}sr_{X,t}=\Delta_{X^{\prime}}s\Delta_{X^{\prime}}^{-1}$ for every
$s\in X^{\prime}$.

  * •

If $A_{X^{\prime}}$ is of type either $A$ or $D$, then
$r_{X,t}^{-1}X^{\prime}r_{X,t}\subset X\cup\\{t\\}$.

###### Proof.

If $X^{\prime}$ is not of spherical type, then $X^{\prime}$ cannot be
contained in $Z$ and $X^{\prime}$ and $Z$ are not adjacent, so the conjugation
by the elementary ribbon $r_{X,t}$ does not modify $X^{\prime}$. Suppose that
$X^{\prime}$ is different from $A$, $D$, $E_{6}$ and $I_{2}(m)$. If
$X^{\prime}$ is not contained in $Z$, again $X^{\prime}$ and $Z$ are not
adjacent and therefore $X^{\prime}$ cannot be modified by a conjugation by
$r_{X,t}$. If $X^{\prime}\subset Z$, then $X^{\prime}\subset
Z\setminus\\{t\\}$ so
$(X^{\prime},Z)\in\\{(B_{m_{1}},B_{m_{2}}),(B_{3},F_{4}),(H_{3},H_{4}),(E_{7},E_{8})\\}$,
for $1<m_{1}<m_{2}$. In this case, both $\Delta_{X^{\prime}}$ and $\Delta_{Z}$
are central in $A_{X^{\prime}}$ and $A_{Z}$ respectively. This means that
$r_{X,t}^{-1}sr_{X,t}=s$, for every $s\in X^{\prime}$. If $X^{\prime}$ is of
type $E_{6}$ or $I_{2}(m)$, we can suppose as before that $X^{\prime}\subset
Z$. In this case,
$(X^{\prime},Z)\in\\{(E_{6},E_{7}),(E_{6},E_{8})(I_{2}(5),H_{3}),(I_{2}(5),H_{4})\\}$,
so $\Delta_{Z}$ is central in $A_{Z}$ and $\Delta_{X^{\prime}}$ is not central
in $A_{X^{\prime}}$. Thus,
$r_{X,t}^{-1}sr_{X,t}=\Delta_{Z}^{-1}\Delta_{X^{\prime}}s\Delta_{X^{\prime}}^{-1}\Delta_{Z}=\Delta_{X^{\prime}}s\Delta_{X^{\prime}}^{-1}$
for every $s\in X^{\prime}$. The last item follows by definition.

∎

######  Remark .

By Section 3, the previous lemma works analogously if we replace the positive
elementary $(X,-)$–ribbon $r_{X,t}$ by a negative elementary $(X,-)$–ribbon.

######  Lemma .

Let $A_{S}$ be an Artin group, $X\subset S$, and $\alpha$ be an
$(X,X)$–ribbon. Let $X^{\prime}\subseteq X$ denote any subset defining a
connected component $\Gamma_{X^{\prime}}$ of $\Gamma_{X}$. Then,

  * •

If $A_{X^{\prime}}$ has not spherical type or has a spherical type different
from $A$, $D$, $E_{6}$ and $I_{2}(m)$, then $\alpha^{-1}s\alpha=s$, for every
$s\in X^{\prime}$.

  * •

If $A_{X^{\prime}}$ is of type $E_{6}$ or $I_{2}(m)$, then either
$\alpha^{-1}s\alpha=s$ for every $s\in X^{\prime}$ or
$\alpha^{-1}s\alpha=\Delta_{X^{\prime}}s\Delta_{X^{\prime}}^{-1}$ for every
$s\in X^{\prime}$.

  * •

If $A_{X^{\prime}}$ is of type $A$ or $D$, then
$\alpha^{-1}X^{\prime}\alpha=X^{\prime\prime}$, where
$\Gamma_{X^{\prime\prime}}$ is isomorphic to $\Gamma_{X^{\prime}}$.

###### Proof.

By definition, $\alpha$ is a product $\prod_{i=1}^{k}r_{i}$ of elementary
$(X_{i},Y_{i})$–ribbons, $r_{i}$, where $Y_{i}=X_{i+1}$ and $X_{1}=Y_{k}=X$.
When we conjugate $A_{X^{\prime}}$ by an elementary $X$–ribbon, we obtain a
parabolic subgroup of the same type. Therefore, by Section 3 and Section 3 we
can distinguish three cases. If $A_{X^{\prime}}$ has non-spherical type or has
spherical type different from $A$, $D$, $E_{6}$ and $I_{2}(m)$, then all the
conjugations by the elementary ribbons are trivial. If $A_{X^{\prime}}$ is of
type $E_{6}$ or $I_{2}(m)$, then $\Delta_{X^{\prime}}^{2}$ is the smallest
positive power of $\Delta_{X^{\prime}}$ that is central and all conjugations
are as indicated in the second item of Section 3. If $A_{X^{\prime}}$ is of
type $A$ or $D$, the result is trivial. ∎

Now we define the two main properties that used ribbons that are conjectured
to be true for every Artin group:

######  Definition .

Given an Artin group $A_{S}$ and $S^{\prime}\subseteq S$, we say that a pair
$(X,Y)$, $X,Y\subseteq S^{\prime}$, is _conjugate by ribbons_ in
$A_{S^{\prime}}$ if, for any $g\in A_{S^{\prime}}$,

$g^{-1}A_{X}g=A_{Y}\text{ if and only if }g\in
A_{X}\cdot(\mathrm{Ribb}(X,Y)\cap A_{S^{\prime}}).$

We say that $A_{S}$ satisfies the _ribbon property_ if, for any two sets of
generators $X,Y\subset S$, the pair $(X,Y)$ is conjugate by ribbons in $A_{Z}$
for every $Z\in\\{T\subseteq S\,|\,X,Y\subseteq T\\}$.

######  Definition .

Let $A_{S}$ be an Artin group and $X,Y\subset S$. We say that the pair $(X,Y)$
is _standardisable_ in $A_{S}$ if

$\forall g\in A_{S}\text{ such that }g^{-1}A_{Y}g\subseteq A_{X}\text{ there
are }h\in A_{X}\text{ and }Z\subseteq X\text{ such that
}h^{-1}g^{-1}A_{Y}gh=A_{Z}.$

In particular, if there is no $g\in A_{S}\text{ such that
}g^{-1}A_{Y}g\subseteq A_{X}$, then $(X,Y)$ is standardisable. We say that
$A_{S}$ is _standardisable_ if every pair $(X,Y)$, $X,Y\subset S$, is
standardisable.

Godelle conjectures that every Artin group is standardisable and has the
ribbon property (Godelle,, 2007, Conjecture 1, Conjecture 4.2) after the first
article by (Paris,, 1997) showing the ribbon property and other results about
normalizers for spherical-type Artin groups. Godelle proves that FC-type Artin
groups satisfy the ribbon property in (Godelle,, 2003b, Theorem 3.2) and in
(Godelle,, 2007, Proposition 4.3) he uses the ribbon property to prove that
they are also standardisable. He also shows that all two-dimensional Artin
groups are standardisable and satisfy the ribbon property, and this is what we
use in (Cumplido et al.,, 2020) to solve the conjugacy stability problem for
large Artin groups.

###  3.1 Proof of Theorem A

To prove Theorem A we will first prove the following theorem:

######  Theorem 2.

Let $A_{S}$ be an Artin group and let $X\subset S$. There is an algorithm that
decides whether $A_{X}$ is conjugacy stable if the three following properties
hold:

  * •

For any $Y\subset S$, the pair $(X,Y)$ is standardisable;

  * •

For any $X_{1},X_{2}\subseteq X$, the pair $(X_{1},X_{2})$ is conjugate by
ribbons in $A_{S}$ and in $A_{X}$;

  * •

Every element $\alpha\in A_{X}$ has a parabolic closure $P_{\alpha}$ in
$A_{S}$.

Let us see that the previous theorem implies Theorem A:

###### Proof of Theorem A.

Let $A_{S}$ be an Artin group. To give a solution to the conjugacy stability
problem for parabolic subgroups of Artin groups, we shall notice that the
property of being conjugacy stable is preserved under conjugation. Hence, it
suffices to give an algorithm that tells if a standard parabolic subgroup
$A_{X}$ is conjugacy stable for every $X\subset S$. In order to satisfy the
conditions of Theorem A, $A_{S}$ need to be standardisable and conjugate by
ribbons and every element $\alpha\in A_{S}$ has a parabolic closure
$P_{\alpha}$ in $A_{S}$. In particular, we have the three conditions of
Theorem 2 for every $X\subset S$, so we have the desired algorithm.

∎

Before showing Theorem 2, we will prove some lemmas:

######  Lemma .

Let $A_{S}$ be any Artin group, $X\subset S$ and $\alpha,\beta\in A_{S}$. The
parabolic closure $P_{\beta^{-1}\alpha\beta}$ exists if and only if
$P_{\alpha}$ exists and in this case
$P_{\beta^{-1}\alpha\beta}=\beta^{-1}P_{\alpha}\beta$.

###### Proof.

Suppose that $P_{\alpha}$ exists and let $Q$ be a parabolic subgroup
containing $\beta^{-1}\alpha\beta$. Then $\alpha\in\beta Q\beta^{-1}$, where
$\beta Q\beta^{-1}$ is a parabolic subgroup. Hence $P_{\alpha}\subset\beta
Q\beta^{-1}$ and $\beta^{-1}P_{\alpha}\beta\subset Q$. The converse is
symmetric. ∎

######  Remark .

Let $(*)$ be a property for parabolic subgroups that is preserved under
conjugation, such as being of spherical type. Notice that the previous proof
can be adapted to prove that, if $P(*)_{\alpha}$ is the unique minimal
parabolic subgroup containing $\alpha$ and satisfying $(*)$, then
$\beta^{-1}P(*)_{\alpha}\beta$ is the unique minimal parabolic subgroup
containing $\beta^{-1}\alpha\beta$ and satisfying $(*)$.

The _support_ , $\mathrm{supp}(g)$, of a positive element $g\in A_{S}$ is the
set of all generators that appear in any positive word representing $g$. For
Artin groups of spherical type, the parabolic closure of an element depends on
the element support. In particular, for positive elements we have the
following result:

######  Lemma  (Cumplido et al.,, 2019, Proposition 6.8).

Let $A_{S}$ be an Artin group of spherical type. The parabolic closure of a
positive element $g\in A_{S}$ is $A_{\mathrm{supp}(g)}$.

Our strategy to know if two elements are conjugate inside a parabolic subgroup
will be based on taking their parabolic closure and verifying if this
parabolic closure are conjugate inside the parabolic subgroup. However, there
several ways of sending a set of generators to another set of generators by
conjugacy. Due to that, standard parabolic subgroups of type $D_{k}$ will
produce special cases that will need to be treated separately. The following
three lemmas will help to deal with this cases.

######  Lemma .

Let $A_{S}$ be a spherical-type Artin group. Let $\Delta^{e}$ be a central
power of the Garside element and $\alpha$ and $\beta$ be two elements of
$A_{S}$. Then $\alpha$ and $\beta$ are conjugate if and only if
$\alpha^{-1}\Delta^{e}$ and $\beta^{-1}\Delta^{e}$ are conjugate.

In particular, using the same numbering as in Figure 1, the elements
$(s_{1}s_{3}s_{4}\cdots s_{n})^{-1}\Delta$ and $(s_{2}s_{3}s_{4}\cdots
s_{n})^{-1}\Delta$ are not conjugate in $D_{n}$ when $n$ is even.

###### Proof.

The first statement is quite straightforward as
$\gamma^{-1}\alpha\gamma=\beta$ if and only if
$\gamma^{-1}\alpha^{-1}\gamma=\beta^{-1}$. Since $\Delta^{e}$ is central, this
happens if and only if
$\gamma^{-1}\alpha^{-1}\Delta^{e}\gamma=\gamma^{-1}\alpha^{-1}\gamma\Delta^{e}=\beta^{-1}\Delta^{e}$.

For the second statement, notice that by Section 3.1 the parabolic closures of
$s_{1}s_{3}s_{4}\cdots s_{n}$ and $s_{2}s_{3}s_{4}\cdots s_{n}$ are
respectively $A_{S\setminus\\{s_{2}\\}}$ and $A_{S\setminus\\{s_{1}\\}}$. We
know by Algorithm 1 that these two parabolic subgroups are not conjugate in
$D_{n}$ if $n$ is even. Therefore, by Section 3.1 and the previous statement,
$(s_{1}s_{3}s_{4}\cdots s_{n})^{-1}\Delta$ and $(s_{2}s_{3}s_{4}\cdots
s_{n})^{-1}\Delta$ are not conjugate. ∎

######  Lemma .

Let $A_{S}$ be an Artin group and let $A_{X}$ and $A_{T}$ be standard
parabolic subgroups so that $T\subset X\subset S$ such that $(T,T)$ is
conjugate by ribbons in $A_{X}$. Suppose that there is $s\in S\setminus X$ so
that the connected component of $\Gamma_{T\cup\\{s\\}}$ containing $s$ is of
type $D_{2k+1}$, $k>1$, and $s=s_{5}$ with the numbering established for
$D_{2k+1}$ in Figure 1. If $A_{X}$ is conjugacy stable, then at least one of
the following situations applies:

  * •

There is $t\in X$ adjacent to $s_{4}$ such that the connected component of
$A_{T\cup\\{t\\}}$ containing $t$ and $X^{\prime}$ is of type
$D_{2k^{\prime}+1}$, $k^{\prime}>1$.

  * •

There are $t_{1}\in X$ adjacent to $s_{1}$ and $t_{2}\in X$ adjacent to
$s_{2}$, such that the connected component of $\Gamma_{T\cup\\{t_{1}\\}}$
containing $t_{1}$ is of type $D_{2k_{1}+1}$ and the connected component of
$\Gamma_{T\cup\\{t_{2}\\}}$ containing $t_{2}$ is of type $D_{2k_{2}+1}$,
$k_{1},k_{2}>1$.

###### Proof.

Let $\Gamma_{Y}$ be the connected component of $A_{T\cup\\{s_{5}\\}}$
containing $s$, which is of type $D_{2k+1}$, for some $k>1$. Notice that
$s=s_{5}$ can be adjacent to at most two connected components of $\Gamma_{T}$,
one being generated the subset $X^{\prime}=\\{s_{1},s_{2},s_{3},s_{4}\\}$ with
the numbering established for $D_{2k+1}$ in Figure 1. The element $\Delta_{Y}$
conjugates $a:=(s_{1}s_{3}s_{4})^{-1}\Delta_{X^{\prime}}\Delta_{T\setminus
X^{\prime}}=s_{2}s_{1}s_{3}s_{4}s_{2}s_{1}s_{3}s_{4}s_{2}\Delta_{T\setminus
X^{\prime}}$ to
$b:=(s_{2}s_{3}s_{4})^{-1}\Delta_{X^{\prime}}\Delta_{T\setminus
X^{\prime}}=s_{1}s_{2}s_{3}s_{4}s_{1}s_{2}s_{3}s_{4}s_{1}\Delta_{T\setminus
X^{\prime}}$.

Since $A_{X}$ is conjugacy stable, there is an element $x\in A_{X}$ such that
$x^{-1}ax=b$. As $a$ and $b$ are positive and
$\mathrm{supp}(a)=\mathrm{supp}(b)=T$, by Section 3.1 the parabolic closures
of $a$ and $b$ are both $A_{T}$. Then, by Section 3.1, any element that
conjugates $a$ to $b$ normalises $A_{T}$. In particular $x^{-1}A_{T}x=A_{T}$
and, since $(T,T)$ is conjugate by ribbons in $A_{X}$, we can write $x$ as
$x=x_{1}x_{2}$ where $x_{1}\in A_{T}$ and $x_{2}\in A_{X}$ is a
$(T,T)$–ribbon. Equivalently, we can write $x=x_{2}x_{3}$, where
$x_{3}=x_{2}^{-1}x_{1}x_{2}\in A_{T}$.

The non-trivial elementary $(T,-)$–ribbons in $A_{X}$ are the ones written as
$r_{t}:=\Delta_{Z\setminus\\{t\\}}^{-1}\Delta_{Z}$ or
$\Delta_{Z\setminus\\{t\\}}\Delta_{Z}^{-1}$ (Section 3), where $t\in X$,
$\Gamma_{Z}$ is the connected component of $\Gamma_{{T}\cup\\{t\\}}$ that
contains $t$ and $\Gamma_{Z\cup\\{t\\}}$ has spherical type. If $\Gamma_{Z}$
does not contain $X^{\prime}$, since $\Delta_{T\setminus X^{\prime}}$ is
central in $A_{T\setminus X^{\prime}}$, the conjugation by all other
elementary $(T,-)$–ribbons will commute with $X^{\prime}$ and will fix $a$. If
$\Gamma_{Z}$ contains $X^{\prime}$, as $X^{\prime}$ has type $D_{4}$, we have
that $\Gamma_{Z\cup\\{t\\}}$ has type $D_{m}$. In the case of type $D_{m}$,
$m$ even, the conjugation by $r_{t}$ centralises $A_{Z}$ and, in particular,
it fixes $a$. In the case of type $D_{m}$ with $m$ odd, $t$ is adjacent to
either $s_{1}$, or $s_{2}$, or $s_{4}$. If for that case we suppose that none
of the items of the lemma are satisfied, then all $t\in X$ are all adjacent to
$s_{1}$ or all adjacent to $s_{2}$. So suppose without loss of generality that
all $t\in X$ are adjacent to $s_{1}$. Then, the conjugation by $r_{t}$
normalises $A_{Z}$ and permutes the elements in $Z$ (it switches $s_{2}$ and
$s_{4}$). Hence, $r_{t}$ conjugates $a$ to
$c:=(s_{1}s_{3}s_{2})^{-1}\Delta_{X^{\prime}}\Delta_{T\setminus X^{\prime}}$
and it conjugates $c$ to $a$. It follows that, as $x_{2}$ is a product of
elementary ribbons, each one preserving $T$, one has
$x_{2}^{-1}ax_{2}\in\\{a,c\\}$. Then, since
$x_{3}^{-1}(x_{2}^{-1}ax_{2})x_{3}=b$ either $a$ or $c$ are conjugate to $b$
in $A_{T}$ (and so in $A_{X^{\prime}}$). However, we know by Section 3.1 that
$a$ is not conjugate to $b$ in $A_{X^{\prime}}$, and that $c$ is not conjugate
to $b$ in $A_{X^{\prime}}$. A contradiction. Hence some of the items of the
statement must be satisfied. ∎

######  Lemma .

Let $A_{S}$ be an Artin group and $A_{X}$ and $A_{T}$, $T\subset X\subset S$,
be standard parabolic subgroups such that $(T,T)$ is conjugate by ribbons in
$A_{X}$. Suppose that there is $s\in S\setminus X$ so that the connected
component of $\Gamma_{T\cup\\{s\\}}$ containing $s$ is of type $D_{2k+1}$,
$k>2$, and $s=s_{2i+1}$, $i>2$, with the numbering established for $D_{2k+1}$
in Figure 1. If $A_{X}$ is conjugacy stable in $A_{S}$, then there is $t\in X$
adjacent to $s_{2i}$ such that the connected component of
$\Gamma_{T\cup\\{t\\}}$ containing $t$ is of type $D_{2k^{\prime}+1}$,
$k^{\prime}>2$.

###### Proof.

The proof is analogous to the proof of the previous lemma. Notice that
$s=s_{2i+1}$ can be adjacent to at most two connected components of
$\Gamma_{T}$, one being generated the subset
$X^{\prime}=\\{s_{1},s_{2},s_{3},s_{4},\cdots s_{2i}\\}$ with the numbering
established for $D_{2k+1}$ in Figure 1. Suppose also that there is no $t\in X$
adjacent to $s_{2i}$ such that the connected component of
$\Gamma_{T\cup\\{t\\}}$ containing $X^{\prime}$ is of type $D_{2k^{\prime}+1}$
for some $k^{\prime}>2$. The element $\Delta_{Y}$ conjugates
$a:=(s_{1}s_{3}\cdots s_{2i})^{-1}\Delta_{X^{\prime}}\Delta_{T\setminus
X^{\prime}}$ to $b:=(s_{2}s_{3}\cdots
s_{2i})^{-1}\Delta_{X^{\prime}}\Delta_{T\setminus X^{\prime}}$. Suppose that
$A_{X}$ is conjugacy stable. Then, there is $x\in A_{X}$ such that
$x^{-1}ax=b$.

It is well-known by experts that $\Delta_{X^{\prime}}=(s_{2}s_{3}\cdots
s_{2i}s_{1})^{2i-1}=(s_{1}s_{3}\cdots s_{2i}s_{2})^{2i-1}$ (see Brieskorn &
Saito,, 1972 and Paris,, 2004). So, by Section 3.1, the parabolic closure of
both $a$ and $b$ is $A_{T}$. Then, by Section 3.1, $x$ normalises $A_{T}$. As
in the proof of the previous lemma, $(T,T)$ being conjugate by ribbons in
$A_{X}$ means that we can write $x=x_{2}x_{3}$, where $x_{2}\in A_{X}$ is an
$(T,T)$–ribbon and $x_{3}\in A_{T}$. The non-trivial elementary
$(T,-)$–ribbons belonging to $A_{X}$ are
$r_{t}:=\Delta_{Z\setminus\\{t\\}}^{-1}\Delta_{Z}$ or
$\Delta_{Z\setminus\\{t\\}}\Delta_{Z}^{-1}$ (Section 3), where $t\in X$,
$\Gamma_{Z}$ is the connected component of $\Gamma_{{T}\cup\\{t\\}}$ that
contains $t$ and $\Gamma_{Z\cup\\{t\\}}$ has spherical type. If $\Gamma_{Z}$
does not contain $X^{\prime}$, since $\Delta_{T\setminus X^{\prime}}$ is
central in $A_{T\setminus X^{\prime}}$, the conjugation by all other
elementary $(T,-)$–ribbons will commute with $X^{\prime}$ and will fix $a$.
Otherwise, in $\Gamma_{Z}$ has type $E_{7}$ or $D_{2k^{\prime}+1}$ for some
$k^{\prime}>i$. In the $E_{7}$ case, the conjugation by $r_{t}$ centralises
$A_{Z}$ and it fixes $a$. We have supposed that the $D_{2k^{\prime}+1}$ case
does not happens, so $x_{2}^{-1}ax_{2}=a$. Hence $x_{3}^{-1}ax_{3}=b$ with
$x_{3}\in A_{X^{\prime}}$, which by Section 3.1 is a contradiction. ∎

The following lemma will help to deal with fact that an element in the
stabilizer of a standard parabolic subgroup can permute its connected
components.

######  Lemma .

Let $A_{S}$ be an Artin group and $A_{X}$, $X\subset S$, a standard parabolic
subgroup of $A_{S}$ which is conjugacy stable. Suppose that for any
$X_{1},X_{2}\subseteq X$, the pair $(X_{1},X_{2})$ is conjugate by ribbons in
$A_{X}$. Let $X_{1},X_{2}\subset X$ be such that $g^{-1}{X_{1}}g={X_{2}}$, for
some $g\in A_{S}$. Then there is $g^{\prime}\in A_{X}$ such that
$g^{-1}{Y}g=g^{\prime-1}{Y}g^{\prime}$ for every connected component
$\Gamma_{Y}$ of $\Gamma_{X_{1}}$.

###### Proof.

Denote by $\Gamma_{Y_{1}},\Gamma_{Y_{2}},\dots,\Gamma_{Y_{l}}$, $Y_{i}\subset
X_{1}$, the connected components of $\Gamma_{X_{1}}$. For each $i$, denote by
$s_{i,j}$, $1\leq j\leq|Y_{i}|$ the elements in $Y_{i}$. Now define
$y_{i}:=\left(\prod_{j=1}^{|Y_{i}|}s_{i,j}\right)^{k_{i}}$, where the
$k_{i}$’s are chosen to satisfy that the number of letters in
$\left(\prod_{j=1}^{|Y_{i}|}s_{i,j}\right)^{k_{i}}$ is different from the
number of letters in
$\left(\prod_{j=1}^{|Y_{i^{\prime}}|}s_{i^{\prime},j}\right)^{k_{i^{\prime}}}$
for $i^{\prime}\neq i$.

By Section 3.1, we know that the element $a:=y_{1}y_{2}\dots y_{l}$ has
parabolic closure $A_{X_{1}}$ and that $g^{-1}ag$ has parabolic closure
$A_{X_{2}}$. Since $A_{X}$ is conjugacy stable in $A_{S}$, there must be $h\in
A_{X}$ such that $h^{-1}ah=g^{-1}ag$. Then, by Section 3.1,
$h^{-1}A_{X_{1}}h=A_{X_{2}}$. The ribbon hypothesis tells us that
$h=h_{1}h_{2}$ with $h_{1}\in A_{X_{1}}$ and
$h_{2}\in\mathrm{Ribb}(X_{1},X_{2})\cap A_{X}$. So $h_{2}$ is an element in
$A_{X}$ conjugating $X_{1}$ to $X_{2}$. Also, since $h_{1}\in A_{X_{1}}$, the
conjugation $h_{1}$ preserves each $Y_{i}$ and one has that
$h^{-1}A_{Y_{i}}h=h_{2}^{-1}A_{Y_{i}}h_{2}$ for $1\leq i\leq l$ –observe that
conjugations by $g$ and $h_{2}$ send letters to letters, but $h$ and $h_{1}$
do not have to–.

Suppose that $g^{-1}{Y_{i}}g\neq h_{2}^{-1}{Y_{i}}h_{2}$ for some $1\leq i\leq
l$. By the previous discussion, one should have that $hg^{-1}(y_{1}y_{2}\dots
y_{l})gh^{-1}=y_{1}y_{2}\dots y_{l}$, but let us see that this is impossible:
The conjugation by $gh^{-1}$ permutes non trivially the components of
$A_{X_{1}}$ and every $y_{i}$ is contained in a different component, so there
is some $y_{i}$ such that $hg^{-1}y_{i}gh^{-1}=y_{i^{\prime}}$, for
$i^{\prime}\neq i$. However, since the relations in an Artin group are
homogeneous, two positive words representing two conjugate positive elements
need to have the same number of letters. This is a contradiction and therefore
we can set ${g^{\prime}=h_{2}}$. ∎

###### Proof of Theorem 2.

We are going to prove that a standard parabolic subgroup $A_{X}$ is conjugacy
stable in $A_{S}$ if and only if the following conditions are fulfilled:

  1. 1.

For every $X^{\prime}\subset X$ such that $g^{-1}{X^{\prime}}g\subset{X}$, for
some $g\in A_{S}$, there is $h\in A_{X}$ such that
$g^{-1}{X^{\prime\prime}}g=h^{-1}{X^{\prime\prime}}h$ for every connected
component $\Gamma_{X^{\prime\prime}}$ of $\Gamma_{X^{\prime}}$.

  2. 2.

Let $A_{T}$, $T\subset X$, be a parabolic subgroup. Let $s\in S\setminus X$ be
such that the connected component of $\Gamma_{T\cup\\{s\\}}$ containing $s$ is
of type $D_{2k+1}$, $k>2$ with $s=s_{2i+1}$ following the numbering of Figure
1, then there is $s^{\prime}\in X$ adjacent to $s_{2i}$ such that the
connected component of $\Gamma_{T\cup\\{s^{\prime}\\}}$ containing
$s^{\prime}$ is of type $D_{2k^{\prime}+1}$, $k^{\prime}>2$. This condition is
checked by Algorithm 2.

  3. 3.

Let $A_{T}$, $T\subset X$, be a parabolic subgroup. Let $s\in S\setminus X$ be
such that the connected component of $\Gamma_{T\cup\\{s\\}}$ containing $s$ is
of type $D_{2k+1}$, $k>1$ and $s=s_{5}$ with the numbering of Figure 1. Then
either there is $s^{\prime}\in X$ adjacent to $s_{4}$ such that the connected
component of $\Gamma_{T\cup\\{s^{\prime}\\}}$ containing $s^{\prime}$ is of
type $D_{2k^{\prime}+1}$, $k>1$, or there are $t_{1}\in X$ adjacent to $s_{1}$
and $t_{2}\in X$ adjacent to $s_{2}$ such that the connected component of
$\Gamma_{T\cup\\{t_{1}\\}}$ containing $t_{1}$ is of type $D_{2k_{1}+1}$,
$k_{1}>1$ and the connected component of $\Gamma_{T\cup\\{t_{2}\\}}$
containing $t_{2}$ is of type $D_{2k_{2}+1}$, $k_{2}>1$. This condition is
checked by Algorithm 3.

Once this is proven, we will be able to construct an algorithm to solve the
conjugacy stability problem, explained in Algorithm 4. This is a refinement of
Algorithm 1 that considers permutations of components and the $D_{2k}$
exceptions.

By Section 3.1, Section 3.1 and Section 3.1, we know that if some of the items
is not satisfied, then $A_{X}$ cannot be conjugacy stable in $A_{S}$. So we
need to prove that if all the items are satisfied, then $A_{X}$ is conjugacy
stable.

Let $\alpha,\beta\in A_{X}$ be such that there is $g\in A_{S}$ satisfying
$g^{-1}\alpha g=\beta$. Let $P_{\alpha},P_{\beta}\subset A_{X}$ be the minimal
parabolic subgroups containing $\alpha$ and $\beta$ respectively. Suppose that
all of the items of the theorem are fulfilled. Thanks to the standardisation
condition, we know that $P_{\alpha}=g_{1}^{-1}A_{Y}g_{1}$ and
$P_{\beta}=g_{2}^{-1}A_{Z}g_{2}$, with $g_{1},g_{2}\in A_{X}$ and
$Y,Z\subseteq X$. Also, Section 3.1 implies that
$P_{\alpha}=g^{-1}P_{\beta}g$. So, up to conjugacy by elements of $A_{X}$, we
can suppose that $A_{Y}$ and $A_{Z}$ are the parabolic closures of $\alpha$
and $\beta$ and are conjugate by $g$.

$(Y,Z)$ being conjugate by ribbons in $A_{S}$ tells us that $g=a_{1}\cdot
a_{2}$ where $a_{1}\in A_{Y}$ and $a_{2}$ is a $(Y,Z)$–ribbon in $A_{S}$.
Since the first item is satisfied and $a_{1}^{-1}Ya_{1}=Z$, we know that there
is $h\in A_{X}$ such that $g^{-1}{Y_{i}}g=h^{-1}{Y_{i}}h$ for every connected
component $\Gamma_{Y_{i}}$ of $\Gamma_{Y}$. As $(Y,Z)$ is conjugate by ribbons
in $A_{X}$, $h=b_{1}\cdot b_{2}$ where $b_{1}\in A_{Y}$ and $b_{2}$ is a
$(Y,Z)$–ribbon in $A_{S}$. Also, note that $b_{1}$ cannot permute the
connected components of $Y$: $b_{1}=y_{1}\cdots y_{n}$ where $y_{i}\in Y_{i}$,
so every $y_{i}$ commutes with $Y_{j}$, $j\neq i$. Hence
$a_{2}^{-1}{Y_{i}}a_{2}=b_{2}^{-1}{Y_{i}}b_{2}$ for every connected component
$\Gamma_{Y_{i}}$ of $\Gamma_{Y}$.

Notice that $a_{2}b_{2}^{-1}$ is a $(Y,Y)$–ribbon. Since the connected
components of $Y$ are preserved, the conjugation by $a_{2}b_{2}^{-1}$ induces
an isomorphism of the subgraph $\Gamma_{Y_{i}}$. If $A_{Y_{i}}$ has not
spherical type or it has spherical type and it is non-twistable, we know by
Section 3 that $b_{2}a_{2}^{-1}sa_{2}b_{2}^{-1}=s$, for every $s\in Y_{i}$. If
$A_{Y_{i}}$ is of type $A$ or $D_{2k+1}$, the only isomorphisms of graphs that
we can have are the trivial one and a reflection switching the vertices
corresponding to the two first generators. Then we have either
$b_{2}a_{2}^{-1}sa_{2}b_{2}^{-1}=s$ or
$\Delta_{Y_{i}}^{-1}b_{2}a_{2}^{-1}sa_{2}b_{2}^{-1}\Delta_{Y_{i}}=s$. This is
also valid if $A_{Y_{i}}$ has type $E_{6}$ or $I_{2}(m)$ by Section 3. If
$A_{Y_{i}}$ is of type $D_{2k}$, the only non trivial isomorphism of
$\Gamma_{Y_{i}}$ is a switch of the two first vertices. But, since
$\Delta_{Y_{i}}$ is central, the conjugation by $\Delta_{Y_{i}}$ cannot
perform this isomorphism. Then, if the conjugation by $a_{2}b_{2}^{-1}$
switches the vertices, there must be a $t\in S$ adjacent to $\Gamma_{Y_{i}}$
such that the connected component of $\Gamma_{Y\cup\\{t\\}}$ containing
$Y_{i}$ is of type $D_{2k^{\prime}+1}$ (notice that the conjugation by
$\Delta_{2k^{\prime}+1}$ does the desired switching). Then, by conditions 2
and 3 above, either there is $t\in X$ adjacent to $Y_{i}$ such that
$\Delta_{Y^{\prime}}^{-1}b_{2}a_{2}^{-1}sa_{2}b_{2}^{-1}\Delta_{Y^{\prime}}=s$,
where $\Gamma_{Y^{\prime}}$ is the connected component of
$\Gamma_{Y\cup\\{t\\}}$ containing $t$; or there are $t_{1},t_{2}\in X$
adjacent to $Y_{i}$ such that
$\Delta_{Y^{\prime\prime}}^{-1}\Delta_{Y^{\prime}}^{-1}b_{2}a_{2}^{1}sa_{2}b_{2}^{-1}\Delta_{Y^{\prime}}\Delta_{Y^{\prime\prime}}=s$,
where $\Gamma_{Y^{\prime}}$ is the connected component of
$\Gamma_{Y\cup\\{t_{1}\\}}$ containing $\Gamma_{Y^{\prime}}$ is the connected
component of $\Gamma_{Y\cup\\{t_{2}\\}}$ containing $\Gamma_{Y^{\prime}}$.

The previous paragraph means that we can suppose that
$a_{2}^{-1}sa_{2}=b_{2}^{-1}sb_{2}$ up to conjugations by elements of the form
$\Delta_{X^{\prime}}$, $X^{\prime}\subset X$. Then, up these conjugations,
$g^{-1}\alpha g=a_{1}^{-1}b_{2}^{-1}\alpha b_{2}a_{1}$. Since $b_{2},a_{1}\in
A_{X}$, we have proven that under the three items, $A_{X}$ is conjugacy stable
in $A_{S}$. ∎

##  4 FC-type Artin groups

If every standard parabolic subgroup of an Artin group $A_{S}$ that does not
contain infinite relations has spherical type, then $A_{S}$ is said to be of
FC-type. The aim of this section is to prove Theorem B, that is, to discuss
whether we can apply the algorithm to solve the conjugacy stability problem in
Artin groups of FC-type. From now on, suppose that $A_{S}$ has FC-type.

It is shown in (Altobelli,, 1998, Proposition 2) that, if $s,t\in S$ are such
that $m_{s,t}=\infty$, then $A_{S}$ is isomorphic to the amalgamated free
product of $A_{S\setminus\\{s\\}}$ and $A_{S\setminus\\{t\\}}$ over
$A_{S\setminus\\{s,t\\}}$, denoted by
$A_{S\setminus\\{s\\}}*_{A_{S\setminus\\{s,t\\}}}A_{S\setminus\\{t\\}}$. Then,
if we give an order to the $\infty$-labels of $A_{S}$, we will obtain a
specific amalgamated product structure for $A_{S}$. The next two propositions
about canonical forms in amalgamated free products can be found in (Magnus et
al.,, 1966, Section 4) and sometimes will be used without explicit reference:

######  Proposition  (Canonical form for amalgamated free products).

Given the amalgamated free product $G=G_{1}*_{H}G_{2}$ of the groups $G_{1}$
and $G_{2}$ over $H$, we respectively denote by $C_{1}$ and $C_{2}$ the
transversals of $G_{1}/H$ and $G_{2}/H$ that contain $1$. Then, every $x\in G$
can be uniquely expressed as a product $x=x_{1}x_{2}\cdots x_{r}h$, where
$h\in H$, $x_{i}\in C_{1}\cup C_{2}$ is not trivial for $i=1,\dots,r$, and
$x_{i}$ and $x_{i+1}$ belong to different transversals if $r>1$. This
expression is called the _amalgam normal form_ of $x$, and we denote it by
$\rho(x)$.

######  Proposition  (Conjugacy in amalgamated free products).

Given the previous amalgamated free product $G=G_{1}*_{H}G_{2}$, any element
$g\in G$ is conjugate to an element $x$ with amalgam normal form
$x_{1}x_{2}\cdots x_{r}h$, in which $x_{1}$ and $x_{r}$ belong to different
transversals. We say that such an element $x$ is _cyclically reduced_.
Moreover,

  * •

if $x$ is conjugate to a element written $y=p_{1}p_{2}p_{3}\cdots p_{k}$,
$k\geq 2$, where $p_{i}$,$p_{i+1}$ as well as $p_{1}$,$p_{k}$ belong to
different transversals, then $x$ is obtained from $y$ by cyclically permuting
$p_{1}p_{2}p_{3}\cdots p_{k}$ and then conjugating by an element of $H$;

  * •

if $H=\\{1\\}$ and $x$ is conjugate to an element $y$ in one of the factors
($G_{1}$ or $G_{2}$), then $x$ and $y$ belong to the same factor and are
conjugate in that factor.

The amalgamated free product structure of a standard parabolic subgroup
$A_{X}$ of $A_{S}$ heavily relies on the structure of $A_{S}$. The next result
is a consequence of (Altobelli,, 1998, Theorem 2).

######  Lemma  (Godelle,, 2003b, Corollary 1.12).

Let $A_{S}\simeq
A_{S\setminus\\{s\\}}*_{A_{S\setminus\\{s,t\\}}}A_{S\setminus\\{t\\}}$ be a
$FC$-type Artin group. Let $X\subset S$. If $w\in A_{X}$, then the amalgam
normal form of $w$ has its terms in $A_{X}$.

###  4.1 Proof of Theorem B

In (Morris-Wright,, 2021, Corollary 3.2), it is proven that if an element
$\alpha$ of an FC-type Artin group is contained in a spherical-type parabolic
subgroup, then there is a unique minimal (by inclusion) spherical-type
parabolic subgroup $Q_{\alpha}$ containing $\alpha$. We call $Q_{\alpha}$ the
_spherical-type parabolic closure_ of $\alpha$. Since all parabolic subgroups
of a spherical-type Artin group have spherical type, we can use Section 3.1 to
easily adjust the proof of Theorem 2 and be able to apply Algorithm 4 to
spherical-type parabolic subgroups.

The following lemma will allow us to complete the proof of the first part of
Theorem B.

######  Lemma .

Let $A_{S}$ be an Artin group of FC type and $A_{X}$, $X\subset S$, a standard
parabolic subgroup of non-spherical type. If $\alpha\in A_{X}$ belongs to a
spherical parabolic subgroup of $A_{S}$, then the spherical-type parabolic
closure $Q_{\alpha}$ of $\alpha$ is contained in $A_{X}$.

###### Proof.

Choose an $\infty$-label $m_{s,t}$ in $A_{X}$ and take the decompositions
$A\simeq
A_{S\setminus\\{s\\}}*_{A_{S\setminus\\{s,t\\}}}A_{S\setminus\\{t\\}}$ and
$A_{X}\simeq
A_{X\setminus\\{s\\}}*_{A_{X\setminus\\{s,t\\}}}A_{X\setminus\\{t\\}}$. We
know by Section 4 that the amalgam normal form of $\alpha$ has their terms in
$A_{X}$ so it is also an amalgam normal form with respect to the structure of
$A_{X}$. Then, we can obtain a cyclically reduced element $x\in A_{X}$ from
$\alpha$ by conjugating by an element $c$ of $A_{X}$. Also, we can write
$Q_{\alpha}=\beta^{-1}A_{Y}\beta$, where $A_{Y}$ is a spherical type standard
parabolic subgroup of $A_{S}$. Then, $\beta\alpha\beta^{-1}\in A_{Y}$ and we
can obtain a cyclically reduced element $y\in A_{Y}$ from
$\beta\alpha\beta^{-1}$ by conjugating by an element of $A_{Y}$. We will show
our lemma by induction on the number of $\infty$-labels of $A_{X}$.

Suppose that there is only one $\infty$-label $m_{s,t}$ in $A_{X}$. We first
prove that $x$ is contained in a spherical-type standard parabolic subgroup
$A_{X^{\prime}}$. In this case $A_{X\setminus\\{s\\}}$ and
$A_{X\setminus\\{t\\}}$ have spherical type, so if $x$ is contained in any of
them we are done. Suppose then that $x$ is not contained in any of these two
subgroups. As $x$ and $y$ are conjugate and cyclically reduced, by Section 4
$x$ is obtained from $y$ by conjugating by an element in
$A_{Y\cup(S\setminus\\{s,t\\})}$. Then, $x\in
A_{X^{\prime}}:=A_{Y\cup(S\setminus\\{s,t\\})}$. Since $A_{Y}$ has spherical
type, $Y$ cannot simultaneously contain $s$ and $t$. By (Van der Lek,, 1983),
the intersection of standard parabolic subgroups is (the expected) standard
parabolic subgroup, meaning that $A_{X}\cap A_{X^{\prime}}=A_{X\cap
X^{\prime}}$. So $x$ is contained in $A_{X\cap X^{\prime}}$, which has
spherical type because it lies in $A_{X}$ and cannot contain simultaneously
$s$ and $t$. Conjugating by $c^{-1}$, we have that $\alpha$ is in the
spherical-type parabolic subgroup $cA_{X\cap X^{\prime}}c^{-1}<A_{X}$, which
must contain $Q_{\alpha}$ because the spherical-type parabolic closure is
unique. This finishes the proof of the base case of our induction.

To prove the step case suppose that, if $\alpha$ is contained in a standard
parabolic subgroup with less than $k$ $\infty$-labels, then $Q_{\alpha}$ is
contained in that parabolic subgroup. Let $A_{X}$ have $k$ labels. If $x$
belongs to $A_{X\setminus\\{s\\}}$ or $A_{X\setminus\\{t\\}}$, then $x$
belongs to the standard parabolic subgroup containing less than $k$
$\infty$-labels. Otherwise, applying the same reasoning as in the initial
case, $x\in A_{X}\cap A_{Y\cup(S\setminus\\{s,t\\})}$, which also has less
than $k$ $\infty$-labels. Thus, by hypothesis, the spherical-type parabolic
closure $Q_{x}$ of $x$ is in $A_{X}$. Therefore, $\alpha$ is in the spherical-
type parabolic subgroup $cQ_{x}c^{-1}<A_{X}$, which must contain $Q_{\alpha}$.
∎

In the particular case in which $A_{S}$ is a free product of spherical-type
Artin groups, we can prove the existence of a parabolic closure, hence all the
hypothesis of Theorem A will be fulfilled.

######  Lemma .

Suppose that $A_{S}$ is an Artin group of FC-type such that $A_{S}\simeq
A_{X_{1}}*A_{X_{1}}*\dots*A_{X_{k}}$, where every $A_{X_{i}}$ is a spherical-
type Artin group. Let $\alpha\in A_{S}$. Then, any minimal parabolic subgroup
containing $\alpha^{m}$ for any $m\in\mathbb{Z}$ contains also $\alpha$.

###### Proof.

If $\alpha$ in contained in a single factor $A_{X_{i}}$, this is proven in
(Cumplido et al.,, 2019, Theorem 8.2). Suppose otherwise and let $P$ be a
minimal parabolic subgroup containing $\alpha^{m}$. There is an element
$\beta$ such that $\beta^{-1}P\beta=A_{X}$ is standard. Notice that
$\beta^{-1}\alpha^{m}\beta=(\beta^{-1}\alpha\beta)^{m}$. This means that the
amalgam normal form of $(\beta^{-1}\alpha\beta)^{m}$ can be written using only
letters in $X$ (Section 4). By hypothesis, the length of the amalgam form of
$\beta^{-1}\alpha\beta$ is bigger than 1, hence all the letters in the amalgam
normal form of $\beta^{-1}\alpha\beta$ are letters that appear in the amalgam
normal form of $(\beta^{-1}\alpha\beta)^{m}$. Therefore $A_{X}$ contains
$\beta^{-1}\alpha\beta$. Conjugating by $\beta^{-1}$, we have that $P$
contains $\alpha$. ∎

######  Proposition .

If $A_{S}$ is an Artin group of FC-type such that $A_{S}\simeq
A_{X_{1}}*A_{X_{1}}*\dots*A_{X_{k}}$, where every $A_{X_{i}}$ is a spherical-
type Artin group, then every element $\alpha$ has a parabolic closure
$P_{\alpha}$.

###### Proof.

We will prove the proposition by induction on $k$. If $k=1$, $A_{S}$ has
spherical type and the result is true by (Cumplido et al.,, 2019, Proposition
7.2). Now suppose that the result is true for $k-1$ and consider the free
product structure $A_{X_{1}}*B$ where $B=A_{X_{2}}*A_{X_{3}}*\dots*A_{X_{k}}$.
Suppose there are two minimal parabolic subgroups
$P_{1}=\beta^{-1}A_{Y}\beta$, $P_{2}=\gamma^{-1}A_{Z}\gamma$ containing
$\alpha$. By (Morris-Wright,, 2021, Theorem 3.1), if $P_{1}$ and $P_{2}$ have
spherical type, then $\alpha$ is contained in $P_{1}\cap P_{2}$, so by
minimality $P_{1}=P_{2}$. Suppose then that $P_{1}$ has non-spherical type.
Then, $A_{Y}$ is a minimal parabolic subgroup containing
$\alpha^{\prime}:=\beta\alpha\beta^{-1}$ and $A_{Z}$ is a minimal parabolic
subgroup containing $\alpha^{\prime\prime}:=\gamma\alpha\gamma^{-1}$. Applying
Algorithm 1 implies that if $A_{Y}$ and $A_{Z}$ are different, they cannot be
conjugate.

Let $\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{r}$ be the amalgam normal
form of $\alpha^{\prime}$ with respect to $A_{X_{1}}*B$. We also know that
$\alpha^{\prime}$ and $\alpha^{\prime\prime}$ are conjugate and that all
$\alpha_{i}$’s are contained in $A_{Y}$ (Section 4). If $r=1$, then by Section
4 we have that $\alpha^{\prime}$ and $\alpha^{\prime\prime}$ belong to the
same factor $F$ of the free product and are conjugate by an element $f$ in
that factor. By inductive hypothesis, $A_{Y}$ is the parabolic closure of
$\alpha^{\prime}$ in $F$ and $A_{Z}$ is the parabolic closure of
$\alpha^{\prime\prime}$ in $F$, so by Section 3.1 $f$ has to conjugate $A_{Y}$
to $A_{Z}$, which is only possible if $A_{Y}=A_{Z}$. Since
$\alpha^{\prime\prime}=\gamma\beta^{-1}\alpha^{\prime}\beta\gamma^{-1}$, we
can apply again Section 3.1 to obtain
$\gamma\beta^{-1}A_{Y}\beta\gamma^{-1}=A_{Y}$, so $P_{1}=P_{2}$. If $r\geq 2$,
then $\alpha^{\prime\prime}$ is obtained from $\alpha^{\prime}$ by cyclically
permuting the $\alpha_{i}$’s. This means that
$\alpha^{\prime},\alpha^{\prime\prime}$ belong $A_{Y}\cap A_{Z}$, which by
(Van der Lek,, 1983) is a parabolic subgroup contained in $A_{Y}$ and $A_{Z}$.
As $A_{Y}$ and $A_{Z}$ are minimal, we have that $A_{Y}=A_{Y}\cap
A_{Z}=A_{Z}$. It remains to show that this implies $P_{1}=P_{2}$. Notice that
$P_{2}$ can be obtained from $P_{1}$ by using conjugation by an element that
centralizes $\alpha$, namely $c:=\beta g\gamma^{-1}$, where $g$ is the element
that conjugates $\alpha^{\prime}$ to $\alpha^{\prime\prime}$. Now, by (Magnus
et al.,, 1966, Corollary 4.1.6), either $c$ and $\alpha$ are in the same
factor (this would be the case $r=1$) or $\alpha$ and $c$ are a power of the
same element $h$. By Section 4.1, $h\in P_{1}$, hence
$P_{2}=c^{-1}P_{1}c=P_{1}$. ∎

###### Proof of Theorem B.

Thanks to (Godelle,, 2003b, Theorem 3.2) and (Godelle,, 2007, Proposition
4.3), we know that $A$ is standardisable and has the ribbon property. If $A$
has a free product structure, then every element has a parabolic closure
(Section 4.1) and we can apply Algorithm 4. Now suppose that $A$ is any FC-
type Artin group and that $A_{X}$ is standard parabolic subgroup of $A$. We
need to prove that there is an algorithm that takes as input $A$ and $A_{X}$
and decides whether for every two elements $x,y\in A_{X}$, with $x,y$
contained in (possibly different) spherical-type parabolic subgroups, and such
that $g^{-1}xg=y$ with $g\in A$, there is $g^{\prime}\in H$ such that
$g^{\prime-1}xg^{\prime}=y$. Section 4.1 proves that the spherical-type
parabolic closures $Q_{x}$ and $Q_{y}$, are contained in $A_{X}$. This last
condition and the existence of a spherical-type parabolic closure suffice to
reproduce the proof of Theorem 2 –just replacing parabolic closures by
spherical-type parabolic closures– and show that running Algorithm 4 will do
the job –notice that the only distinct irreducible standard parabolic
subgroups that can be conjugate are the spherical-type ones–.

∎

Acknowledgments

The idea of writing this paper came to me while doing a collaboration with
Alexandre Martin, to whom I am very grateful for the year I spent in Edinburgh
working under his supervision. Thanks to Yago Antolín for useful discussions
about basics on amalgamated free products. Thanks to Juan González-Meneses for
reading this paper, his suggestions and numerous helpful conversations. I also
very much appreciate the comments and remarks made by the referee of this
article.

Input : The Coxeter graph $\Gamma_{S}$ of an Artin group $A_{S}$ and three
subgraphs $\Gamma_{X}\subset\Gamma_{S}$,
$\Gamma_{Y^{\prime}}\subset\Gamma_{Y}\subset\Gamma_{X}$ such that $A_{X}$ and
$A_{S}$ satisfies the hypotheses of Theorem 2 and $\Gamma_{Y}^{\prime}$ is a
connected component of $\Gamma_{Y}$ of type $D_{2k}$.

Output : $1$ (if we know that $A_{X}$ is not conjugacy stable) or $0$.

Label the elements $s_{1},s_{2},\dots,s_{2k}$ of $Y$ as in Figure 1.

for  _$t\in\mathrm{Adj}(\\{s_{2k}\\})\cap(S\setminus X)$_ do

if  _the connected component of $\Gamma_{Y\cup\\{t\\}}$ containing
$Y^{\prime}$ (and $t$) is of type $D_{2m+1}$, for some $m$_ then

for  _$t^{\prime}\in\mathrm{Adj}(\\{s_{2k}\\})\cap X$_ do

if  _the connected component of $\Gamma_{Y\cup\\{t^{\prime}\\}}$ containing
$Y^{\prime}$ (and $t^{\prime}$) is of type $D_{2m^{\prime}+1}$, for some
$m^{\prime}$_ then

return _$0$_ ;

return _$1$_ ;

return _$0$_

Algorithm 2 Algorithm to check the $D_{2k}$, $k>2$, exceptions described in
the proof of Theorem 2

Input : The Coxeter graph $\Gamma_{S}$ of an Artin group $A_{S}$ and two
subgraphs $\Gamma_{X}\subset\Gamma_{S}$,
$\Gamma_{Y^{\prime}}\subset\Gamma_{Y}\subset\Gamma_{X}$ such that $A_{X}$ and
$A_{S}$ satisfy the hypotheses of Theorem 2 and $\Gamma_{Y^{\prime}}$ is a
connected component of $\Gamma_{Y}$ of type $D_{4}$.

Output : $1$ (if we know that $A_{X}$ is not conjugacy stable) or $0$.

Label the elements $s_{1},s_{2},s_{3},s_{4}$ of $Y$ as in Figure 1.

$Z=\\{s_{1},s_{2},s_{3}\\}$;

for  _$s\in Z$_ do

for  _$t\in\mathrm{Adj}(\\{s\\})\cap(S\setminus X)$_ do

$p=0$; $q=0$;

if  _the connected component of $\Gamma_{Y\cup\\{t\\}}$ containing
$Y^{\prime}$ (and $t$) is of type $D_{2m}$, for some $m$_ then

$p=1$; $q=1$;

for  _$t^{\prime}\in\mathrm{Adj}(\\{s\\})\cap X$_ do

if  _the connected component of $\Gamma_{Y\cup\\{t^{\prime}\\}}$ containing
$Y^{\prime}$ (and $t^{\prime}$) is of type $D_{2m^{\prime}+1}$, for some
$m^{\prime}$_ then

$p=0;$ break loop;

if  _$p=1$_ then

for  _$t_{1}\in\mathrm{Adj}(Z\setminus\\{s\\})\cap X$_ do

if  _the connected component of $\Gamma_{Y\cup\\{t_{1}\\}}$ containing
$Y^{\prime}$ (and $t_{1}$) is of type $D_{2m_{1}+1}$, for some $m_{1}$_ then

for  _$t_{2}\in\mathrm{Adj}(Z\setminus\\{s,t_{1}\\})\cap X$_ do

if  _the connected component of $\Gamma_{Y\cup\\{t_{2}\\}}$ containing
$Y^{\prime}$ (and $t_{2}$) is of type $D_{2m_{2}+1}$, for some $m_{1}$_ then

$p=0$; break loop;

if  _$p=0$_ then

break loop;

if  _$p=1$_ then

return _$1$_ ;

if  _$q=1$_ then

break loop;

return _$0$_

Algorithm 3 Algorithm to check the $D_{4}$ exceptions described in the proof
of Theorem 2

Input : The Coxeter graph $\Gamma_{S}$ of an Artin group $A_{S}$ and a
$\Gamma_{X}\subset A_{S}$ such that $A_{X}$ and $A_{S}$ satisfy the hypotheses
of Theorem 2.

Output : “$A_{X}$ is conjugacy stable” or “$A_{X}$ is not conjugacy stable”.

for  _$(X_{1},X_{2})\subset(X,X)$ such that $|X_{1}|=|X_{2}|$_ do

if  _$\Gamma_{X_{1}}$ is of type $D_{2k}$_ then

if  _$k >2$_ then

run algorithm 2;

if  _ algorithm 2 returns $1$_ then

return “$A_{X}$ is not conjugacy stable”;

if  _$k=2$_ then

run algorithm 3;

if  _ algorithm 3 returns $1$_ then

return “$A_{X}$ is not conjugacy stable”;

$\Gamma_{X^{\prime}_{1}},\Gamma_{X^{\prime}_{2}},\dots,\Gamma_{X^{\prime}_{m}}:=$
components of $\Gamma_{X_{1}}$;

$C:=\\{(X^{\prime}_{1},X^{\prime}_{2},\dots,X^{\prime}_{m})\\}$;

if  _$X_{1}=X_{2}$_ then

$D:=\\{(X^{\prime}_{1},X^{\prime}_{2},\dots,X^{\prime}_{m})\\}$;

else

$D:=\\{\emptyset\\}$;

for  _$(Y_{1},Y_{2},\dots,Y_{m})\in C$_ do

$Y:=Y_{1}\cup Y_{2}\cup\cdots\cup Y_{m}$;

for  _$t\in X\cap\mathrm{Adj}(Y)$_ do

if  _the connected component $\Gamma_{Y^{\prime}}$ of $\Gamma_{Y\cup\\{t\\}}$
containing $t$ is twistable_ then

$Z=\Delta_{Y^{\prime}}^{-1}Y\Delta_{Y^{\prime}}$;

$T=(\Delta_{Y^{\prime}}^{-1}Y_{1}\Delta_{Y^{\prime}},\Delta_{Y^{\prime}}^{-1}Y_{2}\Delta_{Y^{\prime}},\dots,\Delta_{Y^{\prime}}^{-1}Y_{m}\Delta_{Y^{\prime}})$;

if  _$T\notin C$_ then

$C=C\cup\\{T\\}$;

if  _$Z=X_{2}$ and $T\not\in D$_ then

$D=D\cup\\{T\\}$;

for  _$(Y_{1},Y_{2},\dots,Y_{m})\in C$_ do

$Y:=Y_{1}\cup Y_{2}\cup\cdots\cup Y_{m}$;

for  _$t\in\mathrm{Adj}(Y)$_ do

if  _the connected component $\Gamma_{Y^{\prime}}$ of $\Gamma_{Y\cup\\{t\\}}$
containing $t$ is twistable_ then

$Z=\Delta_{Y^{\prime}}^{-1}Y\Delta_{Y^{\prime}}$

$T=(\Delta_{Y^{\prime}}^{-1}Y_{1}\Delta_{Y^{\prime}},\Delta_{Y^{\prime}}^{-1}Y_{2}\Delta_{Y^{\prime}},\dots,\Delta_{Y^{\prime}}^{-1}Y_{m}\Delta_{Y^{\prime}})$

if  _$T\notin C$_ then

$C=C\cup\\{T\\}$;

if  _$Z=X_{2}$ and $T\not\in D$_ then

return _“ $A_{X}$ is not conjugacy stable”;_

return _“ $A_{X}$ is conjugacy stable”;_

Algorithm 4 Algorithm that tell us if a parabolic subgroup is conjugacy stable
or not.

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María Cumplido
Instituto de Matemáticas de la Universidad de Sevilla (IMUS)
Departamento de Álgebra, Universidad de Sevilla, Spain.

E-mail address<EMAIL_ADDRESS>