The conjugacy stability problem for parabolic subgroups in artin groups

Medi e . J. Ma h. (2022) 19:237 h ps://doi.o g/10.1007/s00009-022-02153-9 1660-5446/22/050001-22 published online Sep embe 14, 2022 c The Au ho (s) 2022 The Conjugacy S abili y P oblem o Pa abolic Subg oups in A in G oups Ma ´ıa Cumplido Abs ac . Gi en an A in g oup Aand a pa abolic subg oup P, we s udy i e e y wo elemen s o P ha a e conjuga e in A, a e also conjuga e in P. We p o ide an algo i hm o sol e his decision p oblem i Asa isfies h ee p ope ies ha a e conjec u ed o be ue o e e y A in g oup. This allows o sol e he p oblem o new amilies o A in g oups. We also pa ially sol e he p oblem i Ahas FC- ype, and we o ally sol e i i Ais isomo phic o a ee p oduc o A in g oups o sphe ical ype. In pa icula , we show ha in his la e case, e e y elemen o Ais con ained in a unique minimal (by inclusion) pa abolic subg oup. Ma hema ics Subjec Classi ica ion. 20F36, 20F10. Keywo ds. A in g oups, conjugacy s abili y, conjugacy classes, algo i hmic in g oup heo y. 1. In oduc ion A in (o A in–Ti s) g oups we e defined by Jacques Ti s in he 60’s. They a e g oups p esen ed by a fini e se o gene a o s Sand a mos one ela ion o he o m s s ···= s s ···, o e e y pai s, ∈S, wi h he same numbe o le e s ms, a each side o he equali y. I he e is no ela ion associa ed wi h a pai o gene a o s s, ∈S, hen we deno e ms, =∞. Then, he p esen a ion o an A in g oup is as ollows: AS=S|s s...   ms, elemen s = s ...  ms, elemen s ∀s, ∈S, s = , ms, =∞. These g oups a e algeb aic gene alisa ions o he well-known b aid g oups on n+ 1 s ands [2]: An=σ1,...,σ n σiσj=σjσi,|i−j|>1 σiσjσi=σjσiσj,|i−j|=1. A undamen al ool o he s udy o b aid g oups is he ac ion by isome ies o Anon he cu e complex o he n+1-punc u ed disk Dn+1. The cu e com- plex has as e ices (iso opy classes o non-degene a ed) simple closed cu es 237 Page 2 o 22 M. Cumplido MJOM in Dn+1. Fo A in g oups, he analogous o simple closed cu es a e i e- ducible pa abolic subg oups. In ac , he e is a bijec ion be ween he p ope i educible pa abolic subg oups o Anand he simple closed cu es in Dn+1 (see an explana ion in [9, Sec ion 2]. As anda d pa abolic subg oup AXis a subg oup gene a ed by a subse o gene a o s X⊆S. The conjuga e o any s anda d pa abolic subg oup by an elemen o ASis called a pa abolic subg oup. The s udy o pa abolic subg oups has been an impo an sou ce o esea ch in A in g oups o e he las o y yea s. These a e na u al and easy- o-define subg oups. They a e he main ing edien o complexes in which A in g oups ac , as he Deligne complex [6,11] o he complex o i educible pa abolic subg oups [9]. Howe e , as i happens o mos ques ions in A in g oups, basic p ope ies o pa abolic subg oups a e in gene al unknown. Some o he ac s we know abou a e he ollowing: In his hesis, Van de Lek [24] p o ed ha a s anda d pa abolic subg oup is again an A in g oup, and we also know ha hey a e con ex in e e y case [7]. The s uc u e o cen alise s o pa abolic subg oups and many o hei p ope ies ha e been well s udied only ce ain cases by Pa is [21] and Godelle [12–14], among o he s; and we only know i he in e sec ion o pa abolic subg oups is again a pa abolic subg oup o a ew amilies o A in g oups [9,10,20]. In his pape , we discuss in which cases embeddings o pa abolic sub- g oups in o he A in g oup me ge conjugacy classes. This is also called he conjugacy s abili y p oblem o pa abolic subg oups. De ini ion 1. A pa abolic subg oup Po an A in g oup Ais conjugacy s able in Ai o e e y x, y ∈Psuch ha g−1xg =y,g∈G, he e is ˆg∈Psuch ha ˆg−1xˆg=y.I Pis no conjugacy s able in Awe say ha he inclusion o Pin o Ame ges conjugacy classes. This p oblem has been sol ed only o some specific amilies o A in g oups. [16] p o ed ha pa abolic subg oups o b aid g oups a e always con- jugacy s able. Howe e , o A in g oups his is no always he case. In Cal ez e al. [5], we gi e an explici classifica ion o sphe ical- ype (o fini e ype) A in g oups, which a e he g oups ha become fini e when adding o hei p esen a ion he ela ions s2=1 o e e ys∈S. Fo la ge ype and FC- ype, a simple ques ion was add essed by Godelle [15]: He s udied wha happen- s i in he defini ion o conjugacy s able we impose g o be an elemen o S. A he end o Cumplido e al. [10], we comple ely classi y he pa abolic subg oups o la ge A in g oups up o conjugacy s abili y, using he a o e- men ioned esul s o Pa is and Godelle. The aim o his a icle is o use hese esul s o p o e ha conjugacy s abili y p oblem can be sol ed o e e y A in g oup sa is ying h ee p ope ies ha a e conjec u ed o always hold in A in g oups. I o an elemen αin an A in g oup he e is a unique minimal (wi h espec o he inclusion) pa abolic subg oup Pαcon aining α, we say ha Pα is he pa abolic closu e o α. We will show: Theo em A. Le Abe a s anda disable (Defini ion 14)A in g oup sa is ying he ibbon p ope y (Defini ion 13)and such ha e e y elemen in Ahas a MJOM The Conjugacy S abili y P oblem Page 3 o 22 237 pa abolic closu e. Then, he e is an algo i hm ha decides whe he a pa abolic subg oup Po Ais conjugacy s able in Ao no . The exis ence o pa abolic closu es—which is a consequence o he in e sec- ion o wo pa abolic subg oups being a pa abolic subg oup—and he o h- e wo hypo heses o he heo em a e conjec u ed o be ue o all A in g oups. In pa icula , hey a e known o be ue o sphe ical- ype A in g oups [9,12]. The s anda disa ion and ibbon p ope ies a e ue o FC- ype and wo-dimensional A in g oups [13,14]. Fo FC- ype, he p oblem o he in e sec ion o pa abolic subg oups is sol ed o sphe ical- ype pa a- bolic subg oups— he conjuga es o some sphe ical- ype s anda d pa abolic subg oup—by Mo is-W igh [20]. Using hese esul s and he ac ha FC- ype A in g oups can be seen as amalgama ed ee-p oduc s o sphe ical- ype A in g oups, we will pa ially sol e he conjugacy s abili y p oblem o pa abolic subg oups o a FC- ype A in g oup A. We will o ally sol e he p oblem i Ais isomo phic o a ee-p oduc o sphe ical- ype A in g oups, by p o ing he exis ence o pa abolic closu es in his case (P oposi ion 28). This is summa ized in Theo em B. De ini ion 2. Gi en an A in g oup Aand a pa abolic subg oup Po A, we say ha Pis conjugacy quasi-s able i o e e y wo elemen s x, y ∈P con ained in (possibly diffe en ) sphe ical- ype pa abolic subg oups o Asuch ha g−1xg =ywi h g∈A, he e is z∈Psuch ha z−1xz =y. Rema k 3.No ice ha o sphe ical- ype A in g oups being conjugacy quasi- s able is equi alen o be conjugacy s able. Theo em B. Le Abe an FC- ype A in g oup. The e is an algo i hm ha decides whe he a gi en pa abolic subg oup Po Ais conjugacy quasi-s able in A.In pa icula , his algo i hm can ell whe he a sphe ical- ype pa abolic subg oup is conjugacy s able o no . Mo eo e , i Ais isomo phic o a ee p oduc o sphe ical ype A in g oups, hen e e y elemen o Ahas a pa abolic closu e and he e is an algo- i hm ha sol es he conjugacy s abili y p oblem o e e y pa abolic subg oup o A. This a icle is s uc u ed in he ollowing way: In Sec . 2we will desc ibe a esul o Pa is [21] ha gi es an algo i hm o decide when wo s anda d pa abolic subg oups a e conjuga e in any A in g oup, and we will gi e an explici o m o his algo i hm; in Sec . 3we will explain how o modi y his algo i hm o sol e he conjugacy s abili y p oblem o pa abolic subg oups o A in g oups ha sa is y he h ee hypo hesis o Theo em A; in Sec . 4we will discuss he case o FC- ype A in g oups. Rema k 4.A e he fi s p ep in o his pape , [3] gene alised he esul s in Cumplido e al. [10] and showed ha he in e sec ion o pa abolic sub- g oups is a pa abolic subg oup o wo-dimensional A in g oups wi h a Cox- e e g aph—see nex sec ion—in which e e y e ex is disconnec ed om a mos one o he e ex. This comple ed he se o h ee hypo heses needed in Theo em A and allowed him wo apply Algo i hm 4o Sec . 3 o sol e he conjugacy p oblem in his case. 237 Page 4 o 22 M. Cumplido MJOM Figu e 1. Classifica ion o i educible Coxe e g aphs o fini e ype Hae el [17] has also p o ed he h ee conjec u es o Euclidean A in g oups o ype ˜ Aand ˜ C, so we know ha he main heo em wo ks o hese g oups. 2. Conjuga e S anda d Pa abolic Subg oups In his sec ion, we explain in de ail he esul s in Pa is, [21] o decide when wo s anda d pa abolic subg oups a e conjuga e in an A in g oup AS. This wo k is based on he pa o Daan K amme ’s hesis ha sol es he conjugacy p oblem in Coxe e g oups, which is published in K amme , [18]. To begin, we fi s need o know how o define he Coxe e g aph o an A in g oup and he classifica ion o A in g oups o sphe ical ype. De ini ion 5. The Coxe e g aph ΓSo he A in g oup ASis he g aph de- fined by he ollowing da a: •The se o e ices o ΓSis S. •The e is an edge connec ing sand i and only i ms, >2. This edge i labeled wi h ms, i ms, >3. I ΓSis connec ed, we say ha ASis i educible. In Fig. 1, he eade can find he classifica ion [8] o he en ypes o i e- ducible A in g oups o sphe ical ype. All he o he A in g oups o sphe ical ype a e di ec p oduc s o i educible ones. When use ul, we will e e o AS as An,Bn,Dn..., bu no mally we will say ha he A in g oup and Cox- e e g aph a e o ype An,Bn,Dn... We deno e he gene a o s o ASby s1,s 2,s 3,..., acco dingly wi h he numbe ing o Fig. 1. Gi en an A in g oup AS, he submonoid A+ So ASgene a ed by S has he exac ly same p esen a ion as AS(seen as monoid) [22]. I ASis an MJOM The Conjugacy S abili y P oblem Page 5 o 22 237 A in g oup o sphe ical ype, i has a Ga side s uc u e. This implies ha i AShas sphe ical ype he e is a la ice o de defined by “abiff ∃c∈A+ S,ac=b”. The leas common mul iple o all gene a o s o Sis called he Ga side elemen o ASand is deno ed by Δ. By B iesko n and Sai o, [4] we know ha he cen e Z(AS)o ASis gene a ed ei he by Δ o by Δ2. Fo A in g oups o ype An(n≥2),D n(n≥5),E 6and I2(m)(m≥ 5andodd),Δ 2gene a es he cen e o he g oup. O he wise, Δ gene a es he cen e o AS. In he fi s case, he conjuga ion by Δ can be seen as a eflec ion au omo phism o ΓS. These conjuga ions, ha a e well-known by expe s, a e de ailed in wha ollows: •Fo An,n≥2, one has ha Δ−1siΔ=sn−i+1. •Fo Dn,wi h n≥5andnodd, he conjuga ion by Δ pe mu es s1and s2and fixes he o he gene a o . •Fo E6, he conjuga ion by Δ fixes s1and Δ−1siΔ=s8−i o i=1. •Fo I2(m),wi h m≥5andmodd, he conjuga ion by Δ pe mu es s1 and s2. We will be specially in e es ed in he A in g oups such ha he conju- ga ion by Δ can be seen as a eflec ion au omo phism o ΓS: De ini ion 6. We say ha ASis a wis able A in g oup i i is one o he ollowing A in g oups o sphe ical ype: An,n≥2; Dn,n≥5andnodd; E6;I2(m),m≥5andmodd. Thanks o [24], we also know ha a s anda d pa abolic subg oup AYis an A in g oup ha ing as Coxe e g aph ΓY⊂ΓS.I AYhas sphe ical ype, we deno e i s Ga side elemen by ΔY. Suppose ha AXis a maximal p ope s anda d pa abolic subg oup o a wis able s anda d pa abolic subg oup AYo AS, in o he wo ds, X=Y { }, ∈X.I AYis o ype An,nodd, suppose ha is no he cen al gene a o o AY.I AYis o ype E6suppose ha does no co espond o s1o s4 and i AYis o ype Dnsuppose ha co esponds o ei he s1o s2. Then Δ−1 YAXΔYis a s anda d pa abolic subg oup o AYdiffe en om AX.(I is one o he o bidden gene a o s, hen Δ−1 YAXΔY=AX). This is he main ing edien o Pa is’ esul , which s a es ha wo s anda d pa abolic subg oups AXand AXa e conjuga e i and only i i is possible o go om one o he o he by pe o ming hose ypes o conjuga ions o “ wis s”. Le X⊂Sand define Adj(X) as he se o e ices in ΓS ha a e adjacen oΓ X. We will conside lis s o couples (Y,c), whe e Y⊂Sis a subse o gene a o s and c∈ASis an elemen ha conjuga es he se X o he se Y. Fo a gi en X⊂S, we will ecu si ely cons uc he lis VXas ollows. S a he lis wi h he couple (X,1). Fo e e y (Y,c) in he lis and o e e y ∈Adj(Y), ake he connec ed componen ΓYo ΓY∪{ }con aining . I his componen is wis able, conjuga e Yby he Ga side elemen ΔYo he componen . I he esul Zis a subse o gene a o s ha is no con ained in some couple o he lis , add he couple (Z,cΔY). Repea his p ocess. To p o e ha he p ocess s ops a some poin , jus obse e ha he se o s anda d pa abolic subg oups o an A in g oup is fini e. 237 Page 6 o 22 M. Cumplido MJOM Theo em 7. Gi en an A in g oup ASand wo s anda d pa abolic subg oups AXand AX,AXis conjuga e o AXi and only i he e is a couple (X,c) in VX,inwhichcasecis a conjugacy elemen . P oo . This heo em is a e o mula ion o [21, Theo em 4.1]. We can see ha cis a conjugacy elemen by i s own cons uc ion.  In Algo i hm 1, we gi e o Pa is’ esul an explici algo i hmic o m. The algo i hm ells us when wo s anda d pa abolic subg oups AXand AX a e conjuga e. I hey a e no , i cons uc s he whole lis VX. Algo i hm 1: Algo i hm ha finds a conjuga ing elemen be ween wo s anda d pa abolic subg oups o ells ha i does no exis . Inpu : The Coxe e g aph ΓSo an A in g oup ASand wo subse s X,X⊂S. Ou pu : A conjuga ing elemen be ween he pa abolic subg oups AXand AXo “The e is no conjuga ing elemen ”. i |X| =|X| hen e u n “The e is no conjuga ing elemen ”; V={(X,1)}; o (Y,c)∈Vdo o ∈Adj(Y)do i he connec ed componen ΓYo ΓY∪{ }con aining is wis able hen Z=Δ −1 YYΔY; i Zis no he fi s elemen o any couple in V hen V=V∪{(Z,cΔY)}; i Z=X hen e u n cΔY; e u n “The e is no conjuga ing elemen ”; Example. Conside he sphe ical- ype A in g oup E7, as depic ed in Fig. 1. We a e going o see ha he pa abolic subg oup AXwi h X={s1,s 2,s 3,s 4,s 6} is conjuga e o AX, whe e X={s2,s 4,s 5,s 6,s 7}. Fi s , we ake s5∈ Adj(X). The se o gene a o s X∪{s5}={s1,s 2,s 3,s 4,s 5,s 6}defines a con- nec ed sphe ical- ype pa abolic subg oup isomo phic o E6, which is wis able. I we conjuga e Xby he Ga side elemen o AX∪{s5}, we ob ain he se o gene a o s Y={s1,s 2,s 4,s 5,s 6}. Now ake s7∈Adj(Y). The g oup de- fined by Y∪{s7}has he connec ed componen AZ,Z={s1,s 4,s 5,s 6,s 7}, which is a ( wis able) b aid g oup. Conjuga ing by he co esponding Ga side elemen , we finally ob ain Δ−1 ZYΔZ=X. MJOM The Conjugacy S abili y P oblem Page 7 o 22 237 3. Solu ion o he Conjugacy S abili y P oblem In his sec ion, we will explain wo o he h ee hypo heses o Theo em A, namely he ibbon p ope y and he p ope y o being s anda disable. A e ha , we will cons uc he main algo i hm o his pape o know when he embedding o a s anda d pa abolic subg oup me ges conjugacy classes. We fi s desc ibe he esul s o Godelle [12–14] conce ning he se o elemen s conjuga ing wo s anda d pa abolic subg oup o an A in g oup AS. Suppose ha AX,X⊂S T, is a s anda d pa abolic subg oup o sphe ical ype and le X=X { }, o some ∈X. Since Xhas a sphe ical ype, X also has a sphe ical ype and we can conside ΔXand ΔX. We ha e ha Δ−1 XΔXAXΔ−1 XΔX=Δ −1 XAXΔX=AY, o some subse Y⊂X. The conjuga ing elemen Δ−1 XΔXand i s in e se is wha Godelle espec i ely calls an elemen a y (X,Y)– ibbon and an ele- men a y (Y,X)– ibbon. In gene al, o any (no necessa ily o sphe ical ype) pa abolic subg oup ATo AS, i he e is s∈Ssuch ha he componen ΓUo ΓT∪{s} ha con ains sis o sphe ical ype, we call T,s := Δ−1 U {s}ΔUand i s in e se elemen a y ibbons—no ice ha ΓUdoes no need o be wis able—. We say ha an elemen = 1 2··· qis a (T,T)— ibbon i and only i he e is a sequence o se s o gene a o s T=T1,T 2,...,T q+1 =Tsuch ha each iis an elemen a y (Ti,T i+1)— ibbon. The se o all (T,T)— ibbons is deno ed by Ribb(T,T). When e e ing o a (T,T)— ibbon wi hou ca ing abou he specific T, we will use he e m (T,−)— ibbon. Now we will see some p ope ies abou ibbons. The ollowing lemma will allow us o wo k on some o he p oo s using posi i e elemen a y ibbons and ea he nega i e ones as an analogous case: Lemma 8. Le ASbe an A in g oup, X⊂Sand a∈S X. Suppose ha ΓYis he componen o ΓX∪{a} ha con ains a. I he e is an elemen a y ibbon X,a =Δ −1 Y {a}ΔY, hen he e a e T⊂S,s∈S Tsuch ha T,s = Δ−1 Y {s}ΔYand −1 X,a =Δ Y {s}Δ−1 Y. P oo . We ha e ha −1 X,a =Δ −1 YΔY {a}=Δ Y {s}Δ−1 Y, whe e s=Δ −1 YaΔY. To see ha he e is a posi i e elemen a y ibbon o he o m Δ−1 Y {s}ΔY,le T=(X∪{a}) {s}. Hence, ΓYis he componen o ΓT∪{s} ha con ains s and T,s =Δ −1 Y {s}ΔY. Rema k 9.In he abo e lemma, T,s =Δ −1 Y {s}ΔYand −1 X,a =Δ Y {s}Δ−1 Y a e (posi i e and nega i e) elemen a y (T,X)— ibbons. Simila ly, X,a = Δ−1 Y {a}ΔYand −1 T,s =Δ Y {a}Δ−1 Ya e (posi i e and nega i e) elemen a y (X,T)— ibbons. The nex wo lemmas help us unde s and how he conjuga ion by ib- bons ans o ms he gene a o s o s anda d pa abolic subg oups: 237 Page 8 o 22 M. Cumplido MJOM Lemma 10. Le ASbe an A in g oup and X⊂S.Le ∈S Xand Z⊂ X∪{ }be such ha ΓZis he connec ed componen o ΓX∪{ }con aining and i is o sphe ical ype. Le X⊆Xdeno e any subse defining a connec ed componen ΓXo ΓX.Then •I AXdefines a componen which is no o sphe ical ype, hen −1 X, s X, =s, o e e y s∈X. •I AXis o sphe ical ype and o ype diffe en om A, D, E6and I2(m), hen −1 X, s X, =s, o e e y s∈X. •I AXis o ype ei he E6o I2(m), hen ei he −1 X, s X, =s o e e y s∈Xo −1 X, s X, =Δ XsΔ−1 X o e e y s∈X. •I AXis o ype ei he Ao D, hen −1 X, X X, ⊂X∪{ }. P oo . I Xis no o sphe ical ype, hen Xcanno be con ained in Zand Xand Za e no adjacen , so he conjuga ion by he elemen a y ibbon X, does no modi y X. Suppose ha Xis diffe en om A,D,E6and I2(m). I Xis no con ained in Z, again Xand Za e no adjacen and he e o e Xcanno be modified by a conjuga ion by X, .I X⊂Z, hen X⊂Z { }so (X,Z)∈{(Bm1,B m2),(B3,F 4),(H3,H 4),(E7,E 8)}, o 1<m 1<m 2. In his case, bo h ΔXand ΔZa e cen al in AXand AZ, e- spec i ely. This means ha −1 X, s X, =s, o e e y s∈X.I Xis o ype E6 o I2(m), we can suppose as be o e ha X⊂Z. In his case, (X,Z)∈ {(E6,E 7),(E6,E 8)(I2(5),H 3),(I2(5),H 4)},soΔ Zis cen al in AZand ΔX is no cen al in AX.Thus, −1 X, s X, =Δ −1 ZΔXsΔ−1 XΔZ=Δ XsΔ−1 X o e e y s∈X. The las i em ollows by defini ion.  Rema k 11.By Lemma 8, he p e ious lemma wo ks analogously i we e- place he posi i e elemen a y (X,−)— ibbon X, by a nega i e elemen a y (X,−)— ibbon. Lemma 12. Le ASbe an A in g oup, X⊂S,andαbe an (X, X)— ibbon. Le X⊆Xdeno e any subse defining a connec ed componen ΓXo ΓX. Then, •I AXhas no sphe ical ype o has a sphe ical ype diffe en om A, D,E6and I2(m), hen α−1sα =s, o e e y s∈X. •I AXis o ype E6o I2(m), hen ei he α−1sα =s o e e y s∈X o α−1sα =Δ XsΔ−1 X o e e y s∈X. •I AXis o ype Ao D, hen α−1Xα=X,whe eΓX is isomo phic o ΓX. P oo . By defini ion, αis a p oduc k i=1 io elemen a y (Xi,Y i)— ibbons, i, whe e Yi=Xi+1 and X1=Yk=X. When we conjuga e AXby an elemen a y X— ibbon, we ob ain a pa abolic subg oup o he same ype. The e o e, by Lemma 10 and Rema k 11 we can dis inguish h ee cases. I AXhas non-sphe ical ype o has a sphe ical ype diffe en om A,D,E6 and I2(m), hen all he conjuga ions by he elemen a y ibbons a e i ial. I AXis o ype E6o I2(m), hen Δ2 Xis he smalles posi i e powe o ΔX ha is cen al and all conjuga ions a e as indica ed in he second i em o Lemma 10.I AXis o ype Ao D, he esul is i ial.  MJOM The Conjugacy S abili y P oblem Page 9 o 22 237 Now we define he wo main p ope ies ha used ibbons ha a e con- jec u ed o be ue o e e y A in g oup: De ini ion 13. Gi en an A in g oup ASand S⊆S, we say ha a pai (X,Y ), X, Y ⊆S,isconjuga e by ibbons in ASi , o any g∈AS, g−1AXg=AYi and only i g∈AX·(Ribb(X,Y )∩AS). We say ha ASsa isfies he ibbon p ope y i , o any wo se s o gene a o s X,Y ⊂S, he pai (X,Y ) is conjuga e by ibbons in AZ o e e y Z∈{T⊆ S|X,Y ⊆T}. De ini ion 14. Le ASbe an A in g oup and X,Y ⊂S. We say ha he pai (X,Y )iss anda disable in ASi ∀g∈ASsuch ha g−1AYg⊆AX he e a e h∈AXand Z⊆X such ha h−1g−1AYgh =AZ. In pa icula , i he e is no g∈ASsuch ha g−1AYg⊆AX, hen (X, Y ) is s anda disable. We say ha ASis s anda disable i e e y pai (X,Y ), X,Y ⊂S, is s anda disable. Godelle conjec u es ha e e y A in g oup is s anda disable and has he ibbon p ope y [14, Conjec u e 1, Conjec u e 4.2] a e he fi s a icle by Pa is, [21] showing he ibbon p ope y and o he esul s abou no malize s o sphe ical- ype A in g oups. Godelle p o es ha FC- ype A in g oups sa is y he ibbon p ope y in [13, Theo em 3.2] and in [14, P oposi ion 4.3] he uses he ibbon p ope y o p o e ha hey a e also s anda disable. He also shows ha all wo-dimensional A in g oups a e s anda disable and sa is y he ibbon p ope y, and his is wha we use in Cumplido e al. [10] o sol e he conjugacy s abili y p oblem o la ge A in g oups. 3.1. P oo o Theo em A To p o e Theo em A we will fi s p o e he ollowing heo em: Theo em 15. Le ASbe an A in g oup and le X⊂S. The e is an algo i hm ha decides whe he AXis conjugacy s able i he h ee ollowing p ope ies hold: •Fo any Y⊂S, he pai (X,Y )is s anda disable; •Fo any X1,X 2⊆X, he pai (X1,X 2)is conjuga e by ibbons in AS and in AX; •E e y elemen α∈AXhas a pa abolic closu e Pαin AS. Le us see ha he p e ious heo em implies Theo em A: P oo o Theo em A. Le ASbe an A in g oup. To gi e a solu ion o he conjugacy s abili y p oblem o pa abolic subg oups o A in g oups, we shall no ice ha he p ope y o being conjugacy s able is p ese ed unde conju- ga ion. Hence, i suffices o gi e an algo i hm ha ells i a s anda d pa abolic subg oup AXis conjugacy s able o e e y X⊂S. To sa is y he condi ions o Theo em A, ASneed o be s anda disable and conjuga e by ibbons and e e y elemen α∈AShas a pa abolic closu e Pαin AS. In pa icula , we 237 Page 16 o 22 M. Cumplido MJOM P oo . Choose an ∞-label ms, in AXand ake he decomposi ions A≃ AS {s}∗AS {s, }AS { }and AX≃AX {s}∗AX {s, }AX { }. We know by Lem- ma 25 ha he amalgam no mal o m o αhas hei e ms in AXso i is also an amalgam no mal o m wi h espec o he s uc u e o AX. Then, we can ob ain a cyclically educed elemen x∈AX om αby conjuga ing by an elemen co AX. Also, we can w i e Qα=β−1AYβ, whe e AYis a sphe ical ype s anda d pa abolic subg oup o AS. Then, βαβ−1∈AYandwecan ob ain a cyclically educed elemen y∈AY om βαβ−1by conjuga ing by an elemen o AY. We will show ou lemma by induc ion on he numbe o ∞-labels o AX. Suppose ha he e is only one ∞-label ms, in AX. We fi s p o e ha xis con ained in a sphe ical- ype s anda d pa abolic subg oup AX. In his case AX {s}and AX { }ha e a sphe ical ype, so i xis con ained in any o hem we a e done. Suppose hen ha xis no con ained in any o hese wo subg oups. As xand ya e conjuga e and cyclically educed, by P oposi ion 24 xis ob ained om yby conjuga ing by an elemen in AY∪(S {s, }). Then, x∈AX:= AY∪(S {s, }). Since AYhas sphe ical ype, Ycanno simul a- neously con ain sand . By Van de Lek, [24], he in e sec ion o s anda d pa abolic subg oups is ( he expec ed) s anda d pa abolic subg oup, meaning ha AX∩AX=AX∩X.Soxis con ained in AX∩X, which has a sphe - ical ype because i lies in AXand canno con ain simul aneously sand . Conjuga ing by c−1, we ha e ha αis in he sphe ical- ype pa abolic sub- g oup cAX∩Xc−1<A X, which mus con ain Qαbecause he sphe ical- ype pa abolic closu e is unique. This finishes he p oo o he base case o ou induc ion. To p o e he s ep case suppose ha , i αis con ained in a s anda d pa abolic subg oup wi h less han k∞-labels, hen Qαis con ained in ha pa abolic subg oup. Le AXha e klabels. I xbelongs o AX {s}o AX { }, hen xbelongs o he s anda d pa abolic subg oup con aining less han k ∞-labels. O he wise, applying he same easoning as in he ini ial case, x∈ AX∩AY∪(S {s, }), which also has less han k∞-labels. Thus, by hypo hesis, he sphe ical- ype pa abolic closu e Qxo xis in AX. The e o e, αis in he sphe ical- ype pa abolic subg oup cQxc−1<A X, which mus con ain Qα. In he pa icula case in which ASis a ee p oduc o sphe ical- ype A in g oups, we can p o e he exis ence o a pa abolic closu e, hence all he hypo heses o Theo em A will be ulfilled. Lemma 27. Suppose ha ASis an A in g oup o FC- ype such ha AS≃ AX1∗AX1∗···∗AXk, whe e e e y AXiis a sphe ical- ype A in g oup. Le α∈AS. Then, any minimal pa abolic subg oup con aining αm o any m∈Z con ains also α. P oo . I αin con ained in a single ac o AXi, hisisp o enin[9, The- o em 8.2]. Suppose o he wise and le Pbe a minimal pa abolic subg oup con aining αm.The eisanelemen βsuch ha β−1Pβ =AXis s anda d. No ice ha β−1αmβ=(β−1αβ)m. This means ha he amalgam no mal MJOM The Conjugacy S abili y P oblem Page 17 o 22 237 o m o (β−1αβ)mcan be w i en using only le e s in X(Lemma 25). By hypo hesis, he leng h o he amalgam o m o β−1αβ is bigge han 1, hence all he le e s in he amalgam no mal o m o β−1αβ a e le e s ha appea in he amalgam no mal o m o (β−1αβ)m. The e o e AXcon ains β−1αβ. Conjuga ing by β−1,weha e ha Pcon ains α. P oposi ion 28. I ASis an A in g oup o FC- ype such ha AS≃AX1∗ AX1∗···∗AXk, whe e e e y AXiis a sphe ical- ype A in g oup, hen e e y elemen αhas a pa abolic closu e Pα. P oo . We will p o e he p oposi ion by induc ion on k.I k=1,AShas sphe ical ype and he esul is ue by [9, P oposi ion 7.2]. Now suppose ha he esul is ue o k−1 and conside he ee p oduc s uc u e AX1∗B whe e B=AX2∗AX3∗···∗AXk. Suppose he e a e wo minimal pa abolic subg oups P1=β−1AYβ,P2=γ−1AZγcon aining α.By[20, Theo em 3.1], i P1and P2ha e sphe ical ype, hen αis con ained in P1∩P2,soby minimali y P1=P2. Suppose hen ha P1has non-sphe ical ype. Then, AYis a minimal pa abolic subg oup con aining α:= βαβ−1and AZis a minimal pa abolic subg oup con aining α := γαγ−1. Applying Algo i hm 1Algo i hm implies ha i AYand AZa e diffe en , hey canno be conjuga e. Le α1α2α3···α be he amalgam no mal o m o αwi h espec o AX1∗B. We also know ha αand α a e conjuga e and ha all αi’s a e con ained in AY(Lemma 25). I = 1, hen by P oposi ion 24 we ha e ha αand α belong o he same ac o Fo he ee p oduc and a e conjuga e by an elemen in ha ac o . By induc i e hypo hesis, AYis he pa abolic closu e o αin Fand AZis he pa abolic closu e o α in F,sobyLemma16 has o conjuga e AY o AZ, which is only possible i AY=AZ. Since α =γβ−1αβγ−1, we can apply again Lemma 16 o ob ain γβ−1AYβγ−1=AY,soP1=P2.I ≥2, hen α is ob ained om α by cyclically pe mu ing he αi’s. This means ha α,α  belong AY∩AZ, which by Van de Lek,[24] is a pa abolic subg oup con ained in AYand AZ. As AYand AZa e minimal, we ha e ha AY=AY∩AZ=AZ. I emains o show ha his implies P1=P2. No ice ha P2can be ob ained om P1 by using conjuga ion by an elemen ha cen alizes α, namely c:= βgγ−1, whe e gis he elemen ha conjuga es α o α.Now,by[19, Co olla y 4.1.6], ei he cand αa e in he same ac o ( his would be he case =1)o α and ca e a powe o he same elemen h. By Lemma 27,h∈P1, hence P2=c−1P1c=P1. P oo o Theo emB.Thanks o [13, Theo em 3.2] and [14, P oposi ion 4.3], we know ha Ais s anda disable and has he ibbon p ope y. I Ahas a ee p oduc s uc u e, hen e e y elemen has a pa abolic closu e (P opo- si ion 28) and we can apply Algo i hm 4Algo i hm. Now suppose ha A is any FC- ype A in g oup and ha AXis s anda d pa abolic subg oup o A. We need o p o e ha he e is an algo i hm ha akes as inpu A and AXand decides whe he o e e y wo elemen s x, y ∈AX, wi h x, y con ained in (possibly diffe en ) sphe ical- ype pa abolic subg oups, and such ha g−1xg =ywi h g∈A, he e is g∈Hsuch ha g−1xg=y. Lemma 26 237 Page 18 o 22 M. Cumplido MJOM p o es ha he sphe ical- ype pa abolic closu es Qxand Qy, a e con ained in AX. This las condi ion and he exis ence o a sphe ical- ype pa abolic clo- su e suffice o ep oduce he p oo o Theo em 15—jus eplacing pa abolic closu es by sphe ical- ype pa abolic closu es—and show ha unning Algo- i hm 4Algo i hm will do he job—no ice ha he only dis inc i educible s anda d pa abolic subg oups ha can be conjuga e a e he sphe ical- ype ones—.  Algo i hm 2: Algo i hm o check he D2k,k>2, excep ions desc ibed in he p oo o Theo em 15 Inpu : The Coxe e g aph ΓSo an A in g oup ASand h ee subg aphs ΓX⊂ΓS,Γ Y⊂ΓY⊂ΓXsuch ha AXand AS sa isfies he hypo heses o Theo em 15 and Γ Yis a connec ed componen o ΓYo ype D2k. Ou pu : 1 (i we know ha AXis no conjugacy s able) o 0. Label he elemen s s1,s 2,...,s 2ko Yas in Fig. 1. o ∈Adj({s2k})∩(S X)do i he connec ed componen o ΓY∪{ }con aining Y(and )iso ype D2m+1, o somem hen o ∈Adj({s2k})∩Xdo i he connec ed componen o ΓY∪{ }con aining Y (and )iso ypeD2m+1, o somem hen e u n 0; e u n 1; e u n 0 MJOM The Conjugacy S abili y P oblem Page 19 o 22 237 Algo i hm 3: Algo i hm o check he D4excep ions desc ibed in he p oo o Theo em 15 Inpu : The Coxe e g aph ΓSo an A in g oup ASand wo subg aphs ΓX⊂ΓS,Γ Y⊂ΓY⊂ΓXsuch ha AXand AS sa is y he hypo heses o Theo em 15 and ΓYis a connec ed componen o ΓYo ype D4. Ou pu : 1 (i we know ha AXis no conjugacy s able) o 0. Label he elemen s s1,s 2,s 3,s 4o Yas in Fig. 1. Z={s1,s 2,s 3}; o s∈Zdo o ∈Adj({s})∩(S X)do p=0;q=0; i he connec ed componen o ΓY∪{ }con aining Y(and ) is o ype D2m, o somem hen p=1;q=1; o ∈Adj({s})∩Xdo i he connec ed componen o ΓY∪{ }con aining Y (and )iso ypeD2m+1, o somem hen p= 0; b eak loop; i p=1 hen o 1∈Adj(Z {s})∩Xdo i he connec ed componen o ΓY∪{ 1}con aining Y(and 1)iso ypeD2m1+1, o somem1 hen o 2∈Adj(Z {s, 1})∩Xdo i he connec ed componen o ΓY∪{ 2} con aining Y(and 2)iso ypeD2m2+1, o some m1 hen p= 0; b eak loop; i p=0 hen b eak loop; i p=1 hen e u n 1; i q=1 hen b eak loop; e u n 0 237 Page 20 o 22 M. Cumplido MJOM Algo i hm 4: Algo i hm ha ell us i a pa abolic subg oup is conju- gacy s able o no . Inpu : The Coxe e g aph ΓSo an A in g oup ASand a ΓX⊂AS such ha AXand ASsa is y he hypo heses o Theo em 15. Ou pu :“AXis conjugacy s able” o “AXis no conjugacy s able”. o (X1,X 2)⊂(X,X)such ha |X1|=|X2|do i ΓX1is o ype D2k hen i k>2 hen un algo i hm 2; i algo i hm 2 e u ns 1 hen e u n “AXis no conjugacy s able”; i k=2 hen un algo i hm 3; i algo i hm 3 e u ns 1 hen e u n “AXis no conjugacy s able”; ΓX 1,ΓX 2,...,ΓX m:= componen s o ΓX1; C:= {(X 1,X 2,...,X m)}; i X1=X2 hen D:= {(X 1,X 2,...,X m)}; else D:= {∅}; o (Y1,Y 2,...,Y m)∈Cdo Y:= Y1∪Y2∪···∪Ym; o ∈X∩Adj(Y)do i he connec ed componen ΓYo ΓY∪{ }con aining is wis able hen Z=Δ −1 YYΔY; T=(Δ −1 YY1ΔY,Δ−1 YY2ΔY,...,Δ−1 YYmΔY); i T/∈C hen C=C∪{T}; i Z=X2and T∈ D hen D=D∪{T}; o (Y1,Y 2,...,Y m)∈Cdo Y:= Y1∪Y2∪···∪Ym; o ∈Adj(Y)do i he connec ed componen ΓYo ΓY∪{ }con aining is wis able hen Z=Δ −1 YYΔY T=(Δ −1 YY1ΔY,Δ−1 YY2ΔY,...,Δ−1 YYmΔY) i T/∈C hen C=C∪{T}; i Z=X2and T∈ D hen e u n “AXis no conjugacy s able”; e u n “AXis conjugacy s able”; MJOM The Conjugacy S abili y P oblem Page 21 o 22 237 Acknowledgemen s The idea o w i ing his pape came o me while doing a collabo a ion wi h Alexand e Ma in, o whom I am e y g a e ul o he yea I spen in Ed- inbu gh wo king unde his supe ision. Thanks o Yago An ol´ın o use ul discussions abou basics on amalgama ed ee p oduc s. Thanks o Juan Gonz´alez-Meneses o eading his pape , his sugges ions and nume ous help- ul con e sa ions. I also e y much app ecia e he commen s and ema ks made by he e e ee o his a icle. Funding Open Access unding p o ided hanks o he CRUE-CSIC ag ee- men wi h Sp inge Na u e. Funding was p o ided by Andalusian Minis y o Economy and Knowledge and he Ope a ional P og am FEDER 2014-2020 (G an no. US-1263032). Minis e io de Ciencia e Inno aci´on o Spain (G an no. PID2020-117971GB-C21). Open Access. This a icle is licensed unde a C ea i e Commons A ibu ion 4.0 In e na ional License, which pe mi s use, sha ing, adap a ion, dis ibu ion and e- p oduc ion in any medium o o ma , as long as you gi e app op ia e c edi o he o iginal au ho (s) and he sou ce, p o ide a link o he C ea i e Commons licence, and indica e i changes we e made. The images o o he hi d pa y ma- e ial in his a icle a e included in he a icle’s C ea i e Commons licence, unless indica ed o he wise in a c edi line o he ma e ial. I ma e ial is no included in he a icle’s C ea i e Commons licence and you in ended use is no pe mi ed by s a u o y egula ion o exceeds he pe mi ed use, you will need o ob ain pe mis- sion di ec ly om he copy igh holde . To iew a copy o his licence, isi h p:// c ea i ecommons.o g/licenses/by/4.0/. Publishe ’s No e Sp inge Na u e emains neu al wi h ega d o ju isdic- ional claims in published maps and ins i u ional affilia ions. Re e ences [1] Al obelli, J.A.: The wo d p oblem o A in g oups o FC ype. J. Pu e Appl. Algeb a 129, 1–22 (1998) [2] A in, E.: Theo y o B aids. Ann. Ma h. 2(48), 101–126 (1947) [3] Blu s ein, M.A.: Pa abolic subg oups o wo-dimensional A in g oups and sys olic-by- unc ion complexes. To appea in Bull. Lond. Ma h. 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