The conjugacy stability problem for parabolic subgroups in artin groups

Mediterr. J. Math. (2022) 19:237 https://doi.org/10.1007/s00009-022-02153-9 1660-5446/22/050001-22 published online September 14, 2022 c The Author(s) 2022 The Conjugacy Stability Problem for Parabolic Subgroups in Artin Groups Mar´ıa Cumplido Abstract. Given an Artin group Aand a parabolic subgroup P, we study if every two elements of Pthat are conjugate in A, are also conjugate in P. We provide an algorithm to solve this decision problem if Asatisfies 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 Ahas FC-type, and we totally solve it if Ais isomorphic to a free product of Artin groups of spherical type. In particular, we show that in this latter case, every element of Ais contained in a unique minimal (by inclusion) parabolic subgroup. Mathematics Subject Classification. 20F36, 20F10. Keywords. Artin groups, conjugacy stability, conjugacy classes, algorithmic in group theory. 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 Sand at most one relation of the form stst ···=tsts ···, for every pair s, t ∈S, with the same number of letters ms,t at each side of the equality. If there is no relation associated with a pair of generators s, t ∈S, then we denote ms,t =∞. Then, the presentation of an Artin group is as follows: AS=S|sts...   ms,t elements =tst...  ms,t elements ∀s, t ∈S, s =t, ms,t =∞. These groups are algebraic generalisations of the well-known braid groups on n+ 1 strands [2]: An=σ1,...,σ n σiσj=σjσi,|i−j|>1 σiσjσi=σjσiσj,|i−j|=1. A fundamental tool for the study of braid groups is the action by isometries of Anon the curve complex of the n+1-punctured disk Dn+1. The curve complex has as vertices (isotopy classes of non-degenerated) simple closed curves 237 Page 2 of 22 M. Cumplido MJOM in Dn+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 Anand the simple closed curves in Dn+1 (see an explanation in [9, Section 2]. Astandard parabolic subgroup AXis a subgroup generated by a subset of generators X⊆S. The conjugate of any standard parabolic subgroup by an element of ASis 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 [6,11] or the complex of irreducible parabolic subgroups [9]. 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 [24] proved that a standard parabolic subgroup is again an Artin group, and we also know that they are convex in every case [7]. The structure of centralisers of parabolic subgroups and many of their properties have been well studied only certain cases by Paris [21] and Godelle [12–14], among others; and we only know if the intersection of parabolic subgroups is again a parabolic subgroup for a few families of Artin groups [9,10,20]. 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 1. A parabolic subgroup Pof an Artin group Ais conjugacy stable in Aif for every x, y ∈Psuch that g−1xg =y,g∈G, there is ˆg∈Psuch that ˆg−1xˆg=y.IfPis not conjugacy stable in Awe say that the inclusion of Pinto Amerges conjugacy classes. This problem has been solved only for some specific families of Artin groups. [16] 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. [5], 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 s2=1foreverys∈S. For large type and FC-type, a simpler question was addressed by Godelle [15]: He studied what happens if in the definition of conjugacy stable we impose gto be an element of S. At the end of Cumplido et al. [10], 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 αin an Artin group there is a unique minimal (with respect to the inclusion) parabolic subgroup Pαcontaining α, we say that Pα is the parabolic closure of α. We will show: Theorem A. Let Abe a standardisable (Definition 14)Artin group satisfying the ribbon property (Definition 13)and such that every element in Ahas a MJOM The Conjugacy Stability Problem Page 3 of 22 237 parabolic closure. Then, there is an algorithm that decides whether a parabolic subgroup Pof Ais conjugacy stable in Aor 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 [9,12]. The standardisation and ribbon properties are true for FC- type and two-dimensional Artin groups [13,14]. 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 [20]. Using these results and the fact that FC- type Artin groups can be seen as amalgamated free-products of sphericaltype 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 Ais isomorphic to a free-product of spherical-type Artin groups, by proving the existence of parabolic closures in this case (Proposition 28). This is summarized in Theorem B. Definition 2. Given an Artin group Aand a parabolic subgroup Pof A, we say that Pis conjugacy quasi-stable if for every two elements x, y ∈P contained in (possibly different) spherical-type parabolic subgroups of Asuch that g−1xg =ywith g∈A, there is z∈Psuch that z−1xz =y. Remark 3.Notice that for spherical-type Artin groups being conjugacy quasistable is equivalent to be conjugacy stable. Theorem B. Let Abe an FC-type Artin group. There is an algorithm that decides whether a given parabolic subgroup Pof Ais conjugacy quasi-stable in A.In particular, this algorithm can tell whether a spherical-type parabolic subgroup is conjugacy stable or not. Moreover, if Ais isomorphic to a free product of spherical type Artin groups, then every element of Ahas 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 Sect. 2we will describe a result of Paris [21] 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 Sect. 3we 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 Sect. 4we will discuss the case of FC-type Artin groups. Remark 4.After the first preprint of this paper, [3] generalised the results in Cumplido et al. [10] 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 4of Sect. 3to solve the conjugacy problem in this case. 237 Page 4 of 22 M. Cumplido MJOM Figure 1. Classification of irreducible Coxeter graphs of finite type Haettel [17] has also proved the three conjectures for Euclidean Artin groups of type ˜ Aand ˜ 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, [21] to decide when two standard parabolic subgroups are conjugate in an Artin group AS. 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, [18]. 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 5. The Coxeter graph ΓSof the Artin group ASis the graph defined by the following data: •The set of vertices of ΓSis S. •There is an edge connecting sand tif and only if ms,t >2. This edge if labeled with ms,t if ms,t >3. If ΓSis connected, we say that ASis irreducible. In Fig. 1, the reader can find the classification [8] 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 AS as An,Bn,Dn..., but normally we will say that the Artin group and Coxeter graph are of type An,Bn,Dn... We denote the generators of ASby s1,s 2,s 3,..., accordingly with the numbering of Fig. 1. Given an Artin group AS, the submonoid A+ Sof ASgenerated by S has the exactly same presentation as AS(seen as monoid) [22]. If ASis an MJOM The Conjugacy Stability Problem Page 5 of 22 237 Artin group of spherical type, it has a Garside structure. This implies that if AShas spherical type there is a lattice order defined by “abiff ∃c∈A+ S,ac=b”. The least common multiple of all generators of Sis called the Garside element of ASand is denoted by Δ. By Brieskorn and Saito, [4] we know that the centre Z(AS)ofASis generated either by Δ or by Δ2. For Artin groups of type An(n≥2),D n(n≥5),E 6and I2(m)(m≥ 5andodd),Δ 2generates the centre of the group. Otherwise, Δ generates the centre of AS. In the first case, the conjugation by Δ can be seen as a reflection automorphism of ΓS. These conjugations, that are well-known by experts, are detailed in what follows: •For An,n≥2, one has that Δ−1siΔ=sn−i+1. •For Dn,with n≥5andnodd, the conjugation by Δ permutes s1and s2and fixes the other generator. •For E6, the conjugation by Δ fixes s1and Δ−1siΔ=s8−ifor i=1. •For I2(m),with m≥5andmodd, the conjugation by Δ permutes s1 and s2. We will be specially interested in the Artin groups such that the conjugation by Δ can be seen as a reflection automorphism of ΓS: Definition 6. We say that ASis a twistable Artin group if it is one of the following Artin groups of spherical type: An,n≥2; Dn,n≥5andnodd; E6;I2(m),m≥5andmodd. Thanks to [24], we also know that a standard parabolic subgroup AYis an Artin group having as Coxeter graph ΓY⊂ΓS.IfAYhas spherical type, we denote its Garside element by ΔY. Suppose that AXis a maximal proper standard parabolic subgroup of a twistable standard parabolic subgroup AYof AS, in other words, X=Y\{t}, t∈X.IfAYis of type An,nodd, suppose that tis not the central generator of AY.IfAYis of type E6suppose that tdoes not correspond to s1or s4 and if AYis of type Dnsuppose that tcorresponds to either s1or s2. Then Δ−1 YAXΔYis a standard parabolic subgroup of AYdifferent from AX.(If tis one of the forbidden generators, then Δ−1 YAXΔY=AX). This is the main ingredient of Paris’ result, which states that two standard parabolic subgroups AXand AX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⊂Sand define Adj(X) as the set of vertices in ΓSthat are adjacenttoΓ X. We will consider lists of couples (Y,c), where Y⊂Sis a subset of generators and c∈ASis an element that conjugates the set Xto the set Y. For a given X⊂S, we will recursively construct the list VXas follows. Start the list with the couple (X,1). For every (Y,c) in the list and for every t∈Adj(Y), take the connected component ΓYof ΓY∪{t}containing t. If this component is twistable, conjugate Yby the Garside element ΔYof the component. If the result Zis a subset of generators that is not contained in some couple of the list, add the couple (Z,cΔY). 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. 237 Page 6 of 22 M. Cumplido MJOM Theorem 7. Given an Artin group ASand two standard parabolic subgroups AXand AX,AXis conjugate to AXif and only if there is a couple (X,c) in VX,inwhichcasecis a conjugacy element. Proof. This theorem is a reformulation of [21, Theorem 4.1]. We can see that cis 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 AXand AX are conjugate. If they are not, it constructs the whole list VX. Algorithm 1: Algorithm that finds a conjugating element between two standard parabolic subgroups or tells that it does not exist. Input : The Coxeter graph ΓSof an Artin group ASand two subsets X,X⊂S. Output: A conjugating element between the parabolic subgroups AXand AXor “There is no conjugating element”. if |X| =|X|then return “There is no conjugating element”; V={(X,1)}; for (Y,c)∈Vdo for t∈Adj(Y)do if the connected component ΓYof ΓY∪{t}containing tis twistable then Z=Δ −1 YYΔY; if Zis not the first element of any couple in Vthen V=V∪{(Z,cΔY)}; if Z=Xthen return cΔY; return “There is no conjugating element”; Example. Consider the spherical-type Artin group E7, as depicted in Fig. 1. We are going to see that the parabolic subgroup AXwith X={s1,s 2,s 3,s 4,s 6} is conjugate to AX, where X={s2,s 4,s 5,s 6,s 7}. First, we take s5∈ Adj(X). The set of generators X∪{s5}={s1,s 2,s 3,s 4,s 5,s 6}defines a connected spherical-type parabolic subgroup isomorphic to E6, which is twistable. If we conjugate Xby the Garside element of AX∪{s5}, we obtain the set of generators Y={s1,s 2,s 4,s 5,s 6}. Now take s7∈Adj(Y). The group defined by Y∪{s7}has the connected component AZ,Z={s1,s 4,s 5,s 6,s 7}, which is a (twistable) braid group. Conjugating by the corresponding Garside element, we finally obtain Δ−1 ZYΔZ=X. MJOM The Conjugacy Stability Problem Page 7 of 22 237 3. Solution to the Conjugacy Stability Problem In this section, we will explain two of the three hypotheses 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 [12–14] concerning the set of elements conjugating two standard parabolic subgroup of an Artin group AS. Suppose that AX,X⊂S\T, is a standard parabolic subgroup of spherical type and let X=X\{t}, for some t∈X. Since Xhas a spherical type, X also has a spherical type and we can consider ΔXand ΔX. We have that Δ−1 XΔXAXΔ−1 XΔX=Δ −1 XAXΔX=AY, for some subset Y⊂X. The conjugating element Δ−1 XΔXand its inverse is what Godelle respectively calls an elementary (X,Y)–ribbon and an elementary (Y,X)–ribbon. In general, for any (not necessarily of spherical type) parabolic subgroup ATof AS, if there is s∈Ssuch that the component ΓUof ΓT∪{s}that contains sis of spherical type, we call rT,s := Δ−1 U\{s}ΔUand its inverse elementary ribbons—notice that ΓUdoes not need to be twistable—. We say that an element r=r1r2···rqis a (T,T)—ribbon if and only if there is a sequence of sets of generators T=T1,T 2,...,T q+1 =Tsuch that each riis an elementary (Ti,T i+1)—ribbon. The set of all (T,T)—ribbons is denoted by Ribb(T,T). When referring to a (T,T)—ribbon without caring about the specific T, we will use the term (T,−)—ribbon. Now we will see some properties about ribbons. The following lemma will allow us to work on some of the proofs using positive elementary ribbons and treat the negative ones as an analogous case: Lemma 8. Let ASbe an Artin group, X⊂Sand a∈S\X. Suppose that ΓYis the component of ΓX∪{a}that contains a. If there is an elementary ribbon rX,a =Δ −1 Y\{a}ΔY, then there are T⊂S,s∈S\Tsuch that rT,s = Δ−1 Y\{s}ΔYand r−1 X,a =Δ Y\{s}Δ−1 Y. Proof. We have that r−1 X,a =Δ −1 YΔY\{a}=Δ Y\{s}Δ−1 Y, where s=Δ −1 YaΔY. To see that there is a positive elementary ribbon of the form Δ−1 Y\{s}ΔY,let T=(X∪{a})\{s}. Hence, ΓYis the component of ΓT∪{s}that contains s and rT,s =Δ −1 Y\{s}ΔY. Remark 9.In the above lemma, rT,s =Δ −1 Y\{s}ΔYand r−1 X,a =Δ Y\{s}Δ−1 Y are (positive and negative) elementary (T,X)—ribbons. Similarly, rX,a = Δ−1 Y\{a}ΔYand r−1 T,s =Δ Y\{a}Δ−1 Yare (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: 237 Page 8 of 22 M. Cumplido MJOM Lemma 10. Let ASbe an Artin group and X⊂S.Lett∈S\Xand Z⊂ X∪{t}be such that ΓZis the connected component of ΓX∪{t}containing t and it is of spherical type. Let X⊆Xdenote any subset defining a connected component ΓXof ΓX.Then •If AXdefines a component which is not of spherical type, then r−1 X,tsrX,t =s, for every s∈X. •If AXis of spherical type and of type different from A, D, E6and I2(m), then r−1 X,tsrX,t =s, for every s∈X. •If AXis of type either E6or I2(m), then either r−1 X,tsrX,t =sfor every s∈Xor r−1 X,tsrX,t =Δ XsΔ−1 Xfor every s∈X. •If AXis of type either Aor D, then r−1 X,tXrX,t ⊂X∪{t}. Proof. If Xis not of spherical type, then Xcannot be contained in Zand Xand Zare not adjacent, so the conjugation by the elementary ribbon rX,t does not modify X. Suppose that Xis different from A,D,E6and I2(m). If Xis not contained in Z, again Xand Zare not adjacent and therefore Xcannot be modified by a conjugation by rX,t.IfX⊂Z, then X⊂Z\{t}so (X,Z)∈{(Bm1,B m2),(B3,F 4),(H3,H 4),(E7,E 8)},for 1<m 1<m 2. In this case, both ΔXand ΔZare central in AXand AZ,respectively. This means that r−1 X,tsrX,t =s, for every s∈X.IfXis of type E6 or I2(m), we can suppose as before that X⊂Z. In this case, (X,Z)∈ {(E6,E 7),(E6,E 8)(I2(5),H 3),(I2(5),H 4)},soΔ Zis central in AZand ΔX is not central in AX.Thus,r−1 X,tsrX,t =Δ −1 ZΔXsΔ−1 XΔZ=Δ XsΔ−1 Xfor every s∈X. The last item follows by definition.  Remark 11.By Lemma 8, the previous lemma works analogously if we replace the positive elementary (X,−)—ribbon rX,t by a negative elementary (X,−)—ribbon. Lemma 12. Let ASbe an Artin group, X⊂S,andαbe an (X, X)—ribbon. Let X⊆Xdenote any subset defining a connected component ΓXof ΓX. Then, •If AXhas not spherical type or has a spherical type different from A, D,E6and I2(m), then α−1sα =s, for every s∈X. •If AXis of type E6or I2(m), then either α−1sα =sfor every s∈X or α−1sα =Δ XsΔ−1 Xfor every s∈X. •If AXis of type Aor D, then α−1Xα=X,whereΓX is isomorphic to ΓX. Proof. By definition, αis a product k i=1 riof elementary (Xi,Y i)—ribbons, ri, where Yi=Xi+1 and X1=Yk=X. When we conjugate AXby an elementary X—ribbon, we obtain a parabolic subgroup of the same type. Therefore, by Lemma 10 and Remark 11 we can distinguish three cases. If AXhas non-spherical type or has a spherical type different from A,D,E6 and I2(m), then all the conjugations by the elementary ribbons are trivial. If AXis of type E6or I2(m), then Δ2 Xis the smallest positive power of ΔX that is central and all conjugations are as indicated in the second item of Lemma 10.IfAXis of type Aor D, the result is trivial.  MJOM The Conjugacy Stability Problem Page 9 of 22 237 Now we define the two main properties that used ribbons that are conjectured to be true for every Artin group: Definition 13. Given an Artin group ASand S⊆S, we say that a pair (X,Y ), X, Y ⊆S,isconjugate by ribbons in ASif, for any g∈AS, g−1AXg=AYif and only if g∈AX·(Ribb(X,Y )∩AS). We say that ASsatisfies the ribbon property if, for any two sets of generators X,Y ⊂S, the pair (X,Y ) is conjugate by ribbons in AZfor every Z∈{T⊆ S|X,Y ⊆T}. Definition 14. Let ASbe an Artin group and X,Y ⊂S. We say that the pair (X,Y )isstandardisable in ASif ∀g∈ASsuch that g−1AYg⊆AXthere are h∈AXand Z⊆X such that h−1g−1AYgh =AZ. In particular, if there is no g∈ASsuch that g−1AYg⊆AX, then (X, Y ) is standardisable. We say that ASis standardisable if every pair (X,Y ), X,Y ⊂S, is standardisable. Godelle conjectures that every Artin group is standardisable and has the ribbon property [14, Conjecture 1, Conjecture 4.2] after the first article by Paris, [21] 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 [13, Theorem 3.2] and in [14, 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. [10] 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 15. Let ASbe an Artin group and let X⊂S. There is an algorithm that decides whether AXis conjugacy stable if the three following properties hold: •For any Y⊂S, the pair (X,Y )is standardisable; •For any X1,X 2⊆X, the pair (X1,X 2)is conjugate by ribbons in AS and in AX; •Every element α∈AXhas a parabolic closure Pαin AS. Let us see that the previous theorem implies Theorem A: Proof of Theorem A. Let ASbe 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 AXis conjugacy stable for every X⊂S. To satisfy the conditions of Theorem A, ASneed to be standardisable and conjugate by ribbons and every element α∈AShas a parabolic closure Pαin AS. In particular, we 237 Page 16 of 22 M. Cumplido MJOM Proof. Choose an ∞-label ms,t in AXand take the decompositions A≃ AS\{s}∗AS\{s,t}AS\{t}and AX≃AX\{s}∗AX\{s,t}AX\{t}. We know by Lemma 25 that the amalgam normal form of αhas their terms in AXso it is also an amalgam normal form with respect to the structure of AX. Then, we can obtain a cyclically reduced element x∈AXfrom αby conjugating by an element cof AX. Also, we can write Qα=β−1AYβ, where AYis a spherical type standard parabolic subgroup of AS. Then, βαβ−1∈AYandwecan obtain a cyclically reduced element y∈AYfrom βαβ−1by conjugating by an element of AY. We will show our lemma by induction on the number of ∞-labels of AX. Suppose that there is only one ∞-label ms,t in AX. We first prove that xis contained in a spherical-type standard parabolic subgroup AX. In this case AX\{s}and AX\{t}have a spherical type, so if xis contained in any of them we are done. Suppose then that xis not contained in any of these two subgroups. As xand yare conjugate and cyclically reduced, by Proposition 24 xis obtained from yby conjugating by an element in AY∪(S\{s,t}). Then, x∈AX:= AY∪(S\{s,t}). Since AYhas spherical type, Ycannot simultaneously contain sand t. By Van der Lek, [24], the intersection of standard parabolic subgroups is (the expected) standard parabolic subgroup, meaning that AX∩AX=AX∩X.Soxis contained in AX∩X, which has a spherical type because it lies in AXand cannot contain simultaneously sand t. Conjugating by c−1, we have that αis in the spherical-type parabolic subgroup cAX∩Xc−1<A X, which must contain Qαbecause 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 αis contained in a standard parabolic subgroup with less than k∞-labels, then Qαis contained in that parabolic subgroup. Let AXhave klabels. If xbelongs to AX\{s}or AX\{t}, then xbelongs to the standard parabolic subgroup containing less than k ∞-labels. Otherwise, applying the same reasoning as in the initial case, x∈ AX∩AY∪(S\{s,t}), which also has less than k∞-labels. Thus, by hypothesis, the spherical-type parabolic closure Qxof xis in AX. Therefore, αis in the spherical-type parabolic subgroup cQxc−1<A X, which must contain Qα. In the particular case in which ASis a free product of spherical-type Artin groups, we can prove the existence of a parabolic closure, hence all the hypotheses of Theorem A will be fulfilled. Lemma 27. Suppose that ASis an Artin group of FC-type such that AS≃ AX1∗AX1∗···∗AXk, where every AXiis a spherical-type Artin group. Let α∈AS. Then, any minimal parabolic subgroup containing αmfor any m∈Z contains also α. Proof. If αin contained in a single factor AXi,thisisprovenin[9, Theorem 8.2]. Suppose otherwise and let Pbe a minimal parabolic subgroup containing αm.Thereisanelementβsuch that β−1Pβ =AXis standard. Notice that β−1αmβ=(β−1αβ)m. This means that the amalgam normal MJOM The Conjugacy Stability Problem Page 17 of 22 237 form of (β−1αβ)mcan be written using only letters in X(Lemma 25). By hypothesis, the length of the amalgam form of β−1αβ is bigger than 1, hence all the letters in the amalgam normal form of β−1αβ are letters that appear in the amalgam normal form of (β−1αβ)m. Therefore AXcontains β−1αβ. Conjugating by β−1,wehavethatPcontains α. Proposition 28. If ASis an Artin group of FC-type such that AS≃AX1∗ AX1∗···∗AXk, where every AXiis a spherical-type Artin group, then every element αhas a parabolic closure Pα. Proof. We will prove the proposition by induction on k.Ifk=1,AShas spherical type and the result is true by [9, Proposition 7.2]. Now suppose that the result is true for k−1 and consider the free product structure AX1∗B where B=AX2∗AX3∗···∗AXk. Suppose there are two minimal parabolic subgroups P1=β−1AYβ,P2=γ−1AZγcontaining α.By[20, Theorem 3.1], if P1and P2have spherical type, then αis contained in P1∩P2,soby minimality P1=P2. Suppose then that P1has non-spherical type. Then, AYis a minimal parabolic subgroup containing α:= βαβ−1and AZis a minimal parabolic subgroup containing α := γαγ−1. Applying Algorithm 1Algorithm implies that if AYand AZare different, they cannot be conjugate. Let α1α2α3···αrbe the amalgam normal form of αwith respect to AX1∗B. We also know that αand α are conjugate and that all αi’s are contained in AY(Lemma 25). If r= 1, then by Proposition 24 we have that αand α belong to the same factor Fof the free product and are conjugate by an element fin that factor. By inductive hypothesis, AYis the parabolic closure of αin Fand AZis the parabolic closure of α in F,sobyLemma16 fhas to conjugate AYto AZ, which is only possible if AY=AZ. Since α =γβ−1αβγ−1, we can apply again Lemma 16 to obtain γβ−1AYβγ−1=AY,soP1=P2.Ifr≥2, then α is obtained from α by cyclically permuting the αi’s. This means that α,α  belong AY∩AZ, which by Van der Lek,[24] is a parabolic subgroup contained in AYand AZ. As AYand AZare minimal, we have that AY=AY∩AZ=AZ. It remains to show that this implies P1=P2. Notice that P2can be obtained from P1 by using conjugation by an element that centralizes α, namely c:= βgγ−1, where gis the element that conjugates αto α.Now,by[19, Corollary 4.1.6], either cand αare in the same factor (this would be the case r=1)orα and care a power of the same element h. By Lemma 27,h∈P1, hence P2=c−1P1c=P1. ProofofTheoremB.Thanks to [13, Theorem 3.2] and [14, Proposition 4.3], we know that Ais standardisable and has the ribbon property. If Ahas a free product structure, then every element has a parabolic closure (Proposition 28) and we can apply Algorithm 4Algorithm. Now suppose that A is any FC-type Artin group and that AXis standard parabolic subgroup of A. We need to prove that there is an algorithm that takes as input A and AXand decides whether for every two elements x, y ∈AX, with x, y contained in (possibly different) spherical-type parabolic subgroups, and such that g−1xg =ywith g∈A, there is g∈Hsuch that g−1xg=y. Lemma 26 237 Page 18 of 22 M. Cumplido MJOM proves that the spherical-type parabolic closures Qxand Qy, are contained in AX. This last condition and the existence of a spherical-type parabolic closure suffice to reproduce the proof of Theorem 15—just replacing parabolic closures by spherical-type parabolic closures—and show that running Algorithm 4Algorithm will do the job—notice that the only distinct irreducible standard parabolic subgroups that can be conjugate are the spherical-type ones—.  Algorithm 2: Algorithm to check the D2k,k>2, exceptions described in the proof of Theorem 15 Input : The Coxeter graph ΓSof an Artin group ASand three subgraphs ΓX⊂ΓS,Γ Y⊂ΓY⊂ΓXsuch that AXand AS satisfies the hypotheses of Theorem 15 and Γ Yis a connected component of ΓYof type D2k. Output: 1 (if we know that AXis not conjugacy stable) or 0. Label the elements s1,s 2,...,s 2kof Yas in Fig. 1. for t∈Adj({s2k})∩(S\X)do if the connected component of ΓY∪{t}containing Y(and t)isof type D2m+1,forsomemthen for t∈Adj({s2k})∩Xdo if the connected component of ΓY∪{t}containing Y (and t)isoftypeD2m+1,forsomemthen return 0; return 1; return 0 MJOM The Conjugacy Stability Problem Page 19 of 22 237 Algorithm 3: Algorithm to check the D4exceptions described in the proof of Theorem 15 Input : The Coxeter graph ΓSof an Artin group ASand two subgraphs ΓX⊂ΓS,Γ Y⊂ΓY⊂ΓXsuch that AXand AS satisfy the hypotheses of Theorem 15 and ΓYis a connected component of ΓYof type D4. Output: 1 (if we know that AXis not conjugacy stable) or 0. Label the elements s1,s 2,s 3,s 4of Yas in Fig. 1. Z={s1,s 2,s 3}; for s∈Zdo for t∈Adj({s})∩(S\X)do p=0;q=0; if the connected component of ΓY∪{t}containing Y(and t) is of type D2m,forsomemthen p=1;q=1; for t∈Adj({s})∩Xdo if the connected component of ΓY∪{t}containing Y (and t)isoftypeD2m+1,forsomemthen p= 0; break loop; if p=1then for t1∈Adj(Z\{s})∩Xdo if the connected component of ΓY∪{t1}containing Y(and t1)isoftypeD2m1+1,forsomem1then for t2∈Adj(Z\{s, t1})∩Xdo if the connected component of ΓY∪{t2} containing Y(and t2)isoftypeD2m2+1, for some m1then p= 0; break loop; if p=0then break loop; if p=1then return 1; if q=1then break loop; return 0 237 Page 20 of 22 M. Cumplido MJOM Algorithm 4: Algorithm that tell us if a parabolic subgroup is conjugacy stable or not. Input : The Coxeter graph ΓSof an Artin group ASand a ΓX⊂AS such that AXand ASsatisfy the hypotheses of Theorem 15. Output:“AXis conjugacy stable” or “AXis not conjugacy stable”. for (X1,X 2)⊂(X,X)such that |X1|=|X2|do if ΓX1is of type D2kthen if k>2then run algorithm 2; if algorithm 2returns 1then return “AXis not conjugacy stable”; if k=2then run algorithm 3; if algorithm 3returns 1then return “AXis not conjugacy stable”; ΓX 1,ΓX 2,...,ΓX m:= components of ΓX1; C:= {(X 1,X 2,...,X m)}; if X1=X2then D:= {(X 1,X 2,...,X m)}; else D:= {∅}; for (Y1,Y 2,...,Y m)∈Cdo Y:= Y1∪Y2∪···∪Ym; for t∈X∩Adj(Y)do if the connected component ΓYof ΓY∪{t}containing tis twistable then Z=Δ −1 YYΔY; T=(Δ −1 YY1ΔY,Δ−1 YY2ΔY,...,Δ−1 YYmΔY); if T/∈Cthen C=C∪{T}; if Z=X2and T∈ Dthen D=D∪{T}; for (Y1,Y 2,...,Y m)∈Cdo Y:= Y1∪Y2∪···∪Ym; for t∈Adj(Y)do if the connected component ΓYof ΓY∪{t}containing tis twistable then Z=Δ −1 YYΔY T=(Δ −1 YY1ΔY,Δ−1 YY2ΔY,...,Δ−1 YYmΔY) if T/∈Cthen C=C∪{T}; if Z=X2and T∈ Dthen return “AXis not conjugacy stable”; return “AXis conjugacy stable”; MJOM The Conjugacy Stability Problem Page 21 of 22 237 Acknowledgements 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 Antol´ın for useful discussions about basics on amalgamated free products. Thanks to Juan Gonz´alez-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. Funding Open Access funding provided thanks to the CRUE-CSIC agreement with Springer Nature. Funding was provided by Andalusian Ministry of Economy and Knowledge and the Operational Program FEDER 2014-2020 (Grant no. US-1263032). Ministerio de Ciencia e Innovaci´on of Spain (Grant no. PID2020-117971GB-C21). Open Access. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. 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Ph.D. thesis, Nijmegen (1983) Mar´ıa Cumplido Instituto de Matem´aticas de la Universidad de Sevilla (IMUS), Departamento de ´ Algebra Universidad de Sevilla Seville Spain e-mail: [email protected] Received: September 7, 2021. Revised: January 29, 2022. Accepted: August 6, 2022.