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-
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Ph.D. hesis, Nijmegen (1983)
Ma ´ıa Cumplido
Ins i u o de Ma em´a icas de la Uni e sidad de Se illa (IMUS), Depa amen o de
´
Algeb a
Uni e sidad de Se illa
Se ille
Spain
e-mail: [email p o ec ed]
Recei ed: Sep embe 7, 2021.
Re ised: Janua y 29, 2022.
Accep ed: Augus 6, 2022.