arXiv:1712.06727v2 [] 8 Feb 2018

On parabolic subgroups of Artin–Tits groups
of spherical type
Marı́a Cumplido, Volker Gebhardt, Juan González-Meneses
and Bert Wiest∗
February 8, 2018

Abstract
We show that, in an Artin–Tits group of spherical type, the intersection of two parabolic subgroups is a parabolic subgroup. Moreover, we
show that the set of parabolic subgroups forms a lattice with respect to
inclusion. This extends to all Artin–Tits groups of spherical type a result
that was previously known for braid groups.
To obtain the above results, we show that every element in an Artin–
Tits group of spherical type admits a unique minimal parabolic subgroup
containing it. Also, the subgroup associated to an element coincides with
the subgroup associated to any of its powers or roots. As a consequence,
if an element belongs to a parabolic subgroup, all its roots belong to the
same parabolic subgroup.
We define the simplicial complex of irreducible parabolic subgroups,
and we propose it as the analogue, in Artin–Tits groups of spherical type,
of the celebrated complex of curves which is an important tool in braid
groups, and more generally in mapping class groups. We conjecture that
the complex of irreducible parabolic subgroups is δ-hyperbolic.

1

Introduction

Artin–Tits groups are a natural generalization of braid groups from the algebraic
point of view: In the same way that the braid group can be obtained from
the presentation of the symmetric group with transpositions as generators by
dropping the order relations for the generators, other Coxeter groups give rise
∗ Partially supported by a PhD contract founded by University of Rennes 1, Spanish
Projects MTM2013-44233-P, MTM2016-76453-C2-1-P, FEDER, the French-Spanish mobility programme “Mérimée 2015”, and Western Sydney University. Part of this work was done
during a visit of the third author to Western Sydney University, and visits of the first, third
and fourth authors to the University of Seville and the University of Rennes I.

1

to more general Artin–Tits groups. If the underlying Coxeter group is finite,
the resulting Artin–Tits group is said to be of spherical type.
Artin–Tits groups of spherical type share many properties with braid groups.
For instance, they admit a particular algebraic structure, called Garside structure, which allows to define normal forms, solve the word and conjugacy problems, and prove some other algebraic properties (like torsion-freeness).
However, some properties of braid groups are proved using topological or geometrical techniques, since a braid group can be seen as the fundamental group
of a configuration space, and also as a mapping class group of a punctured disc.
As one cannot replicate these topological or geometrical techniques in other
Artin–Tits groups, they must be replaced by algebraic arguments, if one tries
to extend properties of braid groups to all Artin–Tits groups of spherical type.
In this paper we will deal with parabolic subgroups of Artin–Tits groups. A
parabolic subgroup is by definition the conjugate of a subgroup generated by a
subset of the standard generators. The irreducible parabolic subgroups, as we
will see, are a natural algebraic analogue of the non-degenerate simple closed
curves in the punctured disc.
The simple closed curves, in turn, are the building blocks that form the wellknown complex of curves. The properties of this complex, and the way in
which the braid group acts on it, allow to use geometric arguments to prove
important results about braid groups (Nielsen–Thurston classification, structure
of centralizers, etc.). Similarly, the set of irreducible parabolic subgroups also
forms a simplicial complex, which we call the complex of parabolic subgroups. We
conjecture that the geometric properties of this complex, and the way in which
an Artin–Tits group acts on it, will allow to extend many of the mentioned
properties to all Artin–Tits groups of spherical type. The results in the present
paper should help to lay the foundations for this study.
In this paper we will show the following:
Theorem 1.1. (see Proposition 7.2) Let AS be an Artin–Tits group of spherical
type, and let α ∈ AS . There is a unique parabolic subgroup Pα which is minimal
(by inclusion) among all parabolic subgroups containing α.
We will also show that the parabolic subgroup associated to an element coincides
with the parabolic subgroup associated to any of its powers and roots:
Theorem 8.2. Let AS be an Artin–Tits group of spherical type. If α ∈ AS and
m is a nonzero integer, then Pαm = Pα .
The above result will have an interesting consequence:
Corollary 8.3. Let AS be an Artin–Tits group of spherical type. If α belongs
to a parabolic subgroup P , and β ∈ AS is such that β m = α for some nonzero
integer m, then β ∈ P .
2

Finally, we will use Theorem 1.1 to show the following results, which describe
the structure of the set of parabolic subgroups with respect to the partial order
determined by inclusion.
Theorem 9.5. Let P and Q be two parabolic subgroups of an Artin–Tits group
AS of spherical type. Then P ∩ Q is also a parabolic subgroup.
Theorem 10.3. The set of parabolic subgroups of an Artin–Tits group of spherical type is a lattice with respect to the partial order determined by inclusion.

2

Complex of irreducible parabolic subgroups

An Artin–Tits group AS is a group generated by a finite set S, that admits
a presentation with at most one relation for each pair of generators s, t ∈ S
of the form sts · · · = tst · · · , where the length (which may be even or odd) of
the word on the left hand side is equal to the length of the word on the right
hand side. We will denote this length m(s, t), and we will say, as usual, that
m(s, t) = ∞ if there is no relation involving s and t. We can also assume that
m(s, t) > 1, otherwise we could just remove one generator. The two sides of
the relation sts · · · = tst · · · are expressions for the least common multiple of
the generators s and t with respect to the prefix and suffix partial orders (cf.
Section 3).
The given presentation of AS is usually described by a labeled graph ΓS (called
the Coxeter graph of AS ) whose vertices are the elements of S, in which there
is an edge joining two vertices s and t if and only if m(s, t) > 2. If m(s, t) > 3
we write m(s, t) as a label on the corresponding edge.
Given an Artin–Tits group AS , where S is the standard set of generators and ΓS
is the associated Coxeter graph, we say that AS is irreducible if ΓS is connected.
For a subset X of S, the subgroup AX of AS generated by X is called a standard
parabolic subgroup of AS ; it is isomorphic to the Artin–Tits group associated to
the subgraph ΓX of ΓS spanned by X [3, 10]. A subgroup P of AS is a parabolic
subgroup of AS if it is conjugate to a standard parabolic subgroup of AS .
If one adds the relations s2 = 1 for all s ∈ S to the standard presentation of
an Artin–Tits group AS , one obtains its corresponding Coxeter group WS . If
WS is a finite group, then AS is said to be of spherical type. Artin–Tits groups
of spherical type are completely classified [4], and they are known to admit a
Garside structure, as we will see in Section 3.
The main example of an Artin–Tits group of spherical type is the braid group on
n strands, Bn , which is generated by n − 1 elements, σ1 , . . . , σn−1 , with defining
relations σi σj = σj σi (if |i − j| > 1) and σi σj σi = σj σi σj (if |i − j| = 1). Its
corresponding Coxeter group is the symmetric group of n elements, Σn .

3

The braid group Bn can be seen as the group of orientation-preserving automorphisms of the n-times punctured disc Dn , fixing the boundary pointwise,
up to isotopy. That is, Bn is the mapping class group of Dn . As a consequence,
Bn acts naturally on the set of isotopy classes of non-degenerate, simple closed
curves in Dn (here non-degenerate means that the curve encloses more than one
and less than n punctures; simple means that the curve does not cross itself).
The (isotopy classes of) curves form a simplicial complex, called the complex of
curves, as follows: A simplex of dimension d is a set of d + 1 (isotopy classes
of) curves which admit a realisation consisting of d + 1 mutually disjoint curves.
The 1-skeleton of the complex of curves is called the graph of curves, in which
the vertices are the (isotopy classes of) curves, and two vertices are connected
by an edge if and only if the corresponding isotopy classes of curves can be
represented by two disjoint curves.
Our goal is to extend the notion of complex of curves from the braid group Bn to
all Artin–Tits groups of spherical type. Hence, we need to find some algebraic
objects which can be defined for all Artin–Tits groups of spherical type, and
which correspond to isotopy classes of non-degenerate simple closed curves in the
case of Bn . We claim that the irreducible parabolic subgroups are such objects.
We can define a correspondence:




Isotopy classes of


Irreducible parabolic
non-degenerate simple
ϕ:
−→
subgroups of Bn


closed curves in Dn

so that, for a curve C, its image ϕ(C) is the set of braids that can be represented
by an automorphism of Dn whose support is enclosed by C.
Let us see that the image of ϕ consists of irreducible parabolic subgroups. Suppose that Dn is represented as a subset of the complex plane, whose boundary
is a circle, and its punctures correspond to the real numbers 1, . . . , n. Let Cm
be a circle enclosing the punctures 1, . . . , m where 1 < m < n. Then ϕ(Cm ) is
a subgroup of Bn which is naturally isomorphic to Bm . Actually, it is the standard parabolic subgroup AXm , where Xm = {σ1 , . . . , σm−1 }. More generally, if
we consider a non-degenerate simple closed curve C in Dn , we can always find
an automorphism α of Dn such that α(C) = Cm for some 1 < m < n. This
automorphism α represents a braid, and it follows by construction that ϕ(C) is
precisely the irreducible parabolic subgroup αAXm α−1 .
It is clear that ϕ is surjective as, given an irreducible parabolic subgroup P =
αAX α−1 , we can take a circle C ′ in Dn enclosing the (consecutive) punctures
which involve the generators in X, and it follows that ϕ(α−1 (C ′ )) = P . The
injectivity of ϕ will be shown later as a consequence of Lemma 2.1.
Therefore, instead of talking about curves, in an Artin–Tits group AS of spherical type we will talk about irreducible parabolic subgroups. The group AS acts
(from the right) on the set of parabolic subgroups by conjugation. This action
corresponds to the action of braids on isotopy classes of non-degenerate simple
closed curves.
4

Now we need to translate the notion of curves being “disjoint” (the adjacency
condition in the complex of curves) to the algebraic setting. It is worth mentioning that disjoint curves do not necessarily correspond to irreducible parabolic
subgroups with trivial intersection. Indeed, two disjoint nested curves correspond to two subgroups with nontrivial intersection, one containing the other.
Conversely, two irreducible parabolic subgroups with trivial intersection may
correspond to non-disjoint curves. (For instance, the curves corresponding to
the two cyclic subgroups of Bn generated by σ1 and σ2 respectively intersect.)
One simple algebraic translation of the notion of “disjoint curves” is the following: Two distinct irreducible parabolic subgroups P and Q of Bn correspond to
disjoint curves if and only if one of the following conditions is satisfied:
1. P ( Q.
2. Q ( P .
3. P ∩ Q = {1} and pq = qp for every p ∈ P , q ∈ Q.
This can be deduced easily, using geometrical arguments, but it will also follow
from the forthcoming results in this section. Hence, we can say that two irreducible parabolic subgroups P and Q in an Artin–Tits group AS are adjacent
(similarly to what we say in the complex of curves) if P and Q satisfy one of
the three conditions above, that is, if either one is contained in the other, or
they have trivial intersection and commute.
However, this characterization is not completely satisfactory, as it contains three
different cases. Fortunately, one can find a much simpler equivalent characterization by considering a special element for each parabolic subgroup, as we will
now see.
If P is an irreducible parabolic subgroup of an Artin–Tits group of spherical
type, we saw that P is itself an irreducible Artin–Tits group of spherical type.
Hence, the center of P is cyclic, generated by an element zP . To be precise,
there are two possible elements (zP and zP−1 ) which could be taken as generators
of the center of P , but we take the unique one which is conjugate to a positive
element of AS (an element which is a product of standard generators). Hence,
the element zP is determined by P . We shall see in Lemma 2.1 (case x = 1),
that conversely, the element zP determines the subgroup P .
If P is standard, that is, if P = AX for some X ⊆ S, we will write zX = zAX . It
turns out that, using the standard Garside structure of AS , one has zX = (∆X )e ,
where ∆X is the least common multiple of the elements of X and e ∈ {1, 2}.
Namely, e = 1 if ∆X is central in AX , and e = 2 otherwise [3].
If a standard parabolic subgroup AX is not irreducible, that is, if ΓX is not
connected, then X = X1 ⊔ · · · ⊔ Xr , where each ΓXi corresponds to a connected
component of ΓX , and AX = AX1 ×· · ·×AXr . In this case we can also define ∆X
as the least common multiple of the elements of X, and it is clear that (∆X )e
is central in AX for either e = 1 or e = 2. We will define zX = ∆X if ∆X is
5

central in AX , and zX = ∆2X otherwise. Notice that zX is a central element
in AX , but it is not a generator of the center of AX if AX is reducible, as in
this case the center of AX is not cyclic: it is the direct product of the centers
of each component, so it is isomorphic to Zr .
Now suppose that P is a parabolic subgroup, so P = αAX α−1 for some X ⊆ S.
We define zP = αzX α−1 . This element zP is well defined: if P = αAX α−1 =
βAY β −1 , then we have β −1 αAX α−1 β = AY , so β −1 αzX α−1 β = zY hence
αzX α−1 = βzY β −1 . Using results by Godelle [16] one can deduce the following:
Lemma 2.1. [5, Lemma 34] Let P and Q be two parabolic subgroups of AS .
Then, for every x ∈ AS , one has x−1 P x = Q if and only if x−1 zP x = zQ .
It follows from the above lemma that, if we want to study elements which
conjugate a parabolic subgroup P to another parabolic subgroup Q, we can
replace P and Q with the elements zP and zQ . It will be much easier to work
with elements than to work with subgroups. Moreover, in the case in which
P = Q, the above result reads:
NAS (P ) = ZAS (zP ),
that is, the normalizer of P in AS equals the centralizer of the element zP in AS .
Remark. We can use Lemma 2.1 to prove that the correspondence ϕ is injective.
In the case of braid groups, if C is a non-degenerate simple closed curve, and P =
ϕ(C) is its corresponding irreducible parabolic subgroup, the central element zP
is either a conjugate of a standard generator (if C encloses two punctures) or
the Dehn twist along the curve C (if C encloses more than two punctures).
Hence, if C1 and C2 are such that ϕ(C1 ) = ϕ(C2 ) then either zP or zP2 is the
Dehn twist along C1 and also the Dehn twist along C2 . Two Dehn twists along
non-degenerate curves correspond to the same mapping class if and only if their
corresponding curves are isotopic [13, Fact 3.6], hence C1 and C2 are isotopic,
showing that ϕ is injective.
Lemma 2.1 allows us to simplify the adjacency condition for irreducible parabolic
subgroups, using the special central elements zP :
Theorem 2.2. Let P and Q be two distinct irreducible parabolic subgroups of
an Artin–Tits group AS of spherical type. Then zP zQ = zQ zP holds if and only
if one of the following three conditions are satisfied:
1. P ( Q.
2. Q ( P .
3. P ∩ Q = {1} and xy = yx for every x ∈ P and y ∈ Q.
The proof of this result will be postponed to Section 11, as it uses some ingredients which will be introduced along the paper.
6

Remark. Consider two isotopy classes of non-degenerate simple closed curves C1
and C2 in the disc Dn , and their corresponding parabolic subgroups P = ϕ(C1 )
and Q = ϕ(C2 ) of Bn . Then C1 and C2 can be realized to be disjoint if and
only if their corresponding Dehn twists commute [13, Fact 3.9]. It is known
that the centralizer of a generator σi is equal to the centralizer of σi2 . Hence C1
and C2 can be realized to be disjoint if and only if zP and zQ commute, which
is equivalent, by Theorem 2.2, to the three conditions in its statement.
We can finally extend the notion of complex of curves to all Artin–Tits groups
of spherical type, replacing curves with irreducible parabolic subgroups.
Definition 2.3. Let AS be an Artin–Tits group of spherical type. We define
the complex of irreducible parabolic subgroups as a simplicial complex in which a
simplex of dimension d is a set {P0 , . . . , Pd } of irreducible parabolic subgroups
such that zPi commutes with zPj for every 0 ≤ i, j ≤ d.
As it happens with the complex of curves in the punctured disc (or in any other
surface), we can define a distance in the complex of irreducible parabolic subgroups, which is determined by the distance in the 1-skeleton, imposing all edges
to have length 1. Notice that the action of AS on the above complex (by conjugation of the parabolic subgroups) is an action by isometries, as conjugation
preserves commutation of elements.
We believe that this complex can be an important tool to study properties of
Artin–Tits groups of spherical type. One important result, that would allow to
extend many properties of braid groups to Artin–Tits groups of spherical type,
is the following:
Conjecture 2.4. The complex of irreducible parabolic subgroups of an Artin–
Tits group of spherical type is δ–hyperbolic.

3

Results from Garside theory for Artin–Tits
groups of spherical type

In this section, we will recall some results from Garside theory that will be
needed. In order to simplify the exposition, we present the material in the
context of Artin–Tits groups of spherical type that is relevant for the paper,
instead of in full generality. For details, we refer to [6, 7, 9, 12].
Let AS be an Artin–Tits group of spherical type. Its monoid of positive elements (the monoid generated by the elements of S) will be denoted A+
S . The
group AS forms a lattice with respect to the prefix order 4, where a 4 b if and
only if a−1 b ∈ A+
S . We will denote by ∧ and ∨ the meet and join operations,
respectively, in this lattice. The Garside element of AS is ∆S = s1 ∨ · · · ∨ sn ,
where S = {s1 , . . . , sn }. One can similarly define the suffix order <, where b < a
if and only if ba−1 ∈ A+
S . In general, a 4 b is not equivalent to b < a.
7

If AS is irreducible (that is, if the defining Coxeter graph ΓS is connected),
then either ∆S or ∆2S generates the center of AS [3, 10]. Actually, conjugation
by ∆S induces a permutation of S. We will denote τS (x) = ∆−1
S x∆S for every
element x ∈ AS ; notice that the automorphism τS has either order 1 or order 2.
The triple (AS , A+
S , ∆S ) determines the so called classical Garside structure of
the group AS .
The simple elements are the positive prefixes of ∆S , which coincide with the
positive suffixes of ∆S . In an Artin–Tits group AS of spherical type, a simple
element is a positive element that is square-free, that is, that cannot be written
as a positive word with two consecutive equal letters [3, 10]. For a simple
element s, we define its right complement as ∂S (s) = s−1 ∆S . Notice that ∂S
is a bijection of the set of simple elements. Notice also that ∂S2 (s) = τS (s) for
every simple element s.
The left normal form of an element α ∈ AS is the unique decomposition of the
form α = ∆pS α1 · · · αr , where p ∈ Z, r ≥ 0, every αi is a non-trivial simple
element different from ∆S , and ∂S (αi ) ∧ αi+1 = 1 for i = 1, . . . , r − 1.
The numbers p and r are called the infimum and the canonical length of α,
denoted inf S (α) and ℓS (α), respectively. The supremum of α is supS (α) =
inf S (α) + ℓS (α). By [11], we know that inf S (α) and supS (α) are, respectively,
the maximum and minimum integers p and q such that ∆pS 4 α 4 ∆qS , or
equivalently ∆qS < α < ∆pS , holds.
There is another normal form which will be important for us. The mixed normal
form, also called the negative-positive normal form, or np-normal form, is the
−1
decomposition of an element α as α = x−1
s · · · x1 y1 · · · yt , where x = x1 · · · xs
and y = y1 · · · yt are positive elements written in left normal form (here some of
the initial factors of either x or y can be equal to ∆), such that x ∧ y = 1 (that
is, there is no possible cancellation in x−1 y).
This decomposition is unique, and it is closely related to the left normal form
of α. Indeed, if x 6= 1 and y 6= 1 then inf S (x) = inf S (y) = 0 holds; otherwise
there would be cancellations. In this case, if one writes xi−1 = ∂(xi )∆S−1 for
i = 1, . . . , s, and then collects all appearances of ∆−1
on the left, one gets
S
−i
α = ∆−s
x
e
·
·
·
x
e
y
·
·
·
y
,
where
x
e
=
τ
(∂(x
)).
The
latter
is precisely the left
s
1 1
t
i
i
S
normal form of α. If x is trivial, then α = y1 · · · yt where the first p factors could
be equal to ∆S , so the left normal form is α = ∆pS yp+1 · · · yt . If y is trivial,
−1
then α = x−1
s · · · x1 where some (say k) of the rightmost factors could be equal
−1
es · · · x
ek+1 .
to ∆S . The left normal form of α in this case would be α = ∆−s
S x
Notice that if x 6= 1 then inf S (α) = −s, and if y 6= 1 then supS (α) = t.

The np-normal form can be computed from any decomposition of α as α = β −1 γ,
where β and γ are positive elements: One just needs to cancel δ = β ∧ γ in the
middle. That is, write β = δx and γ = δy; then α = β −1 γ = x−1 δ −1 δy = x−1 y,
where no more cancellation is possible. Then compute the left normal forms
of x and y and the np-normal form will be computed.
The mixed normal form is very useful to detect whether an element belongs
8

to a proper standard parabolic subgroup. If α ∈ AX , with X ⊆ S, and
−1
α = x−1
s · · · x1 y1 · · · yt is the np-normal form of α in AX , then the simple factors x1 , . . . , xs , y1 , . . . , yt ∈ AX are also simple elements in AS such that x1 · · · xs
and y1 · · · ys are in left normal form in AS . Hence, the above is also the npnormal form of α in AS . Therefore, given α ∈ AS we will have α ∈ AX if and
only if all factors in its np-normal form belong to AX . It follows that if x 6= 1,
inf S (α) = inf X (α) = −s, and if y 6= 1, supS (α) = supX (α) = t.
We finish this section with an important observation: The Artin–Tits group AS
N
admits other Garside structures (AS , A+
S , ∆S ), which are obtained by replacing
the Garside element ∆S with some non-trivial positive power ∆N
S . To see this,
recall that ∆pS 4 α 4 ∆qS is equivalent to ∆qS < α < ∆pS for any p, q ∈ Z, so the
N
positive prefixes of ∆N
S coincide with the positive suffixes of ∆S , and note that
N
one has s 4 ∆S 4 ∆N
for
any
s
∈
S,
so
the
divisors
of
∆
generate
AS .
S
S
The simple elements with respect to this Garside structure are the positive
prefixes of ∆N
S (which are no longer square-free, in general, if N > 1). The npN
normal form of an element with respect to the Garside structure (AS , A+
S , ∆S )
can be obtained from that of the Garside structure (AS , A+
,
∆
)
by
grouping
S
S
together the simple factors of the positive and the negative parts of the latter
in groups of N , ”padding“ the outermost groups with copies of the identity
element as necessary.
−1
Suppose that α = x−1
s · · · x1 y1 · · · yt is in np-normal form, for the classical
Garside structure of AS . If we take N ≥ max(s, t), it follows that x = x1 · · · xs 4
N
∆N
S and y = y1 · · · yt 4 ∆S . This means that x and y are simple elements with
respect to the Garside structure in which ∆N
S is the Garside element. Therefore,
for every α ∈ AS , we can consider a Garside structure of AS such that the npnormal form of α is x−1 y, with x and y being simple elements.

4

Other results for Artin–Tits groups of spherical type

This section focuses on some further properties of Artin–Tits groups of spherical
type that we will need. The properties listed in this section are specific to Artin–
Tits groups and do not directly extend to Garside groups in general.
Definition 4.1. Let AS be an Artin–Tits group of spherical type. Given X ⊆ S
and t ∈ S, we define the positive element
rX,t = ∆−1
X ∆X∪{t} .
If t ∈
/ X, this positive element coincides with the elementary ribbon dX,t defined
in [16]. If t ∈ X we just have rX,t = 1 while dX,t = ∆X .
Lemma 4.2. If AS is an Artin–Tits group of spherical type, X ⊆ S and t ∈ S,
−1
then there is a subset Y ⊆ X ∪ {t} such that rX,t
XrX,t = Y holds.
9

Proof. This follows from the definition, as the automorphisms τX and τX∪{t}
permute the elements of X and X ∪ {t}, respectively.
Lemma 4.3. If AS is an Artin–Tits group of spherical type, X ⊆ S and t ∈ S,
then the element rX,t can be characterized by the following property:
∆X ∨ t = ∆X rX,t
Proof. This follows immediately from the definition of ∆X , which is the least
common multiple of the elements in X, and the definition of ∆X∪{t} = ∆X rX,t ,
which is the least common multiple of the elements in X ∪ {t}.
Lemma 4.4. If AS is an Artin–Tits group of spherical type, X ( S, and
t ∈ S \ X, then the following hold:
(a) t 4 rX,t
(b) If s ∈ S with s 4 rX,t , then s = t.
Proof. We start by proving (b). Recall that the set of prefixes of a Garside
element with the classical Garside structure coincides with its set of suffixes. If
s ∈ S with s 4 rX,t , then one has s 4 rX,t 4 ∆X∪{t} , so s ∈ X ∪ {t}. On
the other hand, ∆X rX,t is a simple element by Lemma 4.3, hence square-free,
and ∆X < u for all u ∈ X, so s ∈
/ X and part (b) is shown. Turning to the
proof of (a), as t ∈
/ X, one has t 64 ∆X and thus rX,t 6= 1 by Lemma 4.3. This
means that rX,t must start with some letter, which by part (b) must necessarily
be t.
We define the support of a positive element α ∈ AS , denoted supp(α), as the
set of generators s ∈ S that appear in any positive word representing α. This
is well defined by two reasons: Firstly, two positive words represent the same
element in AS if and only if one can transform the former into the latter by
repeatedly applying the relations in the presentation of AS , that is, if and only
if the elements are the same in the monoid A+
S defined by the same presentation
as AS [18]. Secondly, due to the form of the relations in the presentation of A+
S,
applying a relation to a word does not modify the set of generators that appear,
so all words representing an element of A+
S involve the same set of generators.
For a not necessarily positive element α ∈ AS , we define its support, using the
np-normal form α = x−1 y, as supp(α) = supp(x) ∪ supp(y).
Lemma 4.5. Let α be a simple element (with respect to the usual Garside
structure) in an Artin–Tits group AS of spherical type. Let t, s ∈ S. Then:

t 64 α
⇒ αs = tα
t 4 αs

10

Proof. As the relations defining A+
S are homogeneous, the word length of α is
well defined. We proceed by induction on the length of α. If α = 1 the result is
trivially true, so suppose α 6= 1 and that the result is true for shorter elements.
Let a ∈ S such that a 4 α. There is a relation in AS of the form atat · · · =
tata · · · , where the words on each side have length m = m(a, t). Let us denote
by ρi the i-th letter of the word atat · · · , for i = 1, . . . , m. That is, ρi = a if i
is odd, and ρi = t if i is even. Recall that ρ1 · · · ρm = a ∨ t. Also, ρ1 · · · ρm =
tρ1 · · · ρm−1 .
We have t 4 αs and a 4 α 4 αs, so a ∨ t = ρ1 · · · ρm 4 αs. Notice that
ρ1 · · · ρm = a ∨ t 64 α (as t 64 α), but ρ1 = a 4 α. Hence, there is some k, where
0 < k < m, such that ρ1 · · · ρk 4 α and ρ1 · · · ρk+1 64 α.
Write α = ρ1 · · · ρk α0 . Then ρk+1 64 α0 , but ρk+1 4 α0 s. By induction hypothesis, α0 s = ρk+1 α0 . Hence αs = ρ1 · · · ρk+1 α0 , where k + 1 ≤ m.
We claim that k + 1 = m. Otherwise, as ρ1 · · · ρm 4 αs = ρ1 · · · ρk+1 α0 , we
would have ρk+2 4 α0 , and then ρ1 · · · ρk ρk+2 4 α, which is not possible as
ρk = ρk+2 and α is simple (thus square-free).
Hence k + 1 = m and αs = ρ1 · · · ρm α0 = tρ1 · · · ρm−1 α0 = tα, as we wanted to
show.
Remark. The above result is not true if α is not simple (with the usual Garside
structure). As an example, consider the Artin–Tits group ha, b | abab = babai,
and the elements α = aaba and t = s = b. We have b 64 α, but
αb = aabab = ababa = babaa,
so b 4 αb, but αb 6= bα as α = aaba 6= abaa.
We end this section with an important property concerning the central elements zP .
Lemma 4.6. Let P and Q be parabolic subgroups of an Artin–Tits group AS
of spherical type. Then the following are equivalent:
1. zP zQ = zQ zP
2. (zP )m (zQ )n = (zQ )n (zP )m for some n, m 6= 0
3. (zP )m (zQ )n = (zQ )n (zP )m for all n, m 6= 0
Proof. If zP and zQ commute, it is clear that (zP )m and (zQ )n commute for
every n, m 6= 0. Conversely, suppose that (zP )m and (zQ )n commute for some
n, m 6= 0.
By Godelle [16, Proposition 2.1], if X, Y ⊆ S and m 6= 0, then u−1 (zX )m u ∈ AY
holds if and only if u = vy with y ∈ AY and v −1 Xv = Y .

11

Hence, if u−1 (zX )m u = (zX )m we can take Y = X, so y commutes with zX
and v induces a permutation of X, which implies that v −1 AX v = AX and then
v −1 zX v = zX . Therefore u−1 zX u = y −1 v −1 (zX )vy = y −1 zX y = zX .
Now recall that (zP )m (zQ )n = (zQ )n (zP )m . Since α−1 P α = AX for some
α ∈ AS and some X ⊆ S, we can conjugate the above equality by α to obtain
(zX )m (α−1 zQ α)n = (α−1 zQ α)n (zX )m , which by the argument in the previous
paragraph implies that zX (α−1 zQ α)n = (α−1 zQ α)n zX . Conjugating back, we
get zP (zQ )n = (zQ )n zP .
Now take β ∈ AS such that β −1 Qβ = AY for some Y ⊆ S. Conjugating
the equality from the previous paragraph by β, we obtain (β −1 zP β)(zY )n =
(zY )n (β −1 zP β), which implies (β −1 zP β)zY = zY (β −1 zP β). Conjugating back,
we finally obtain zP zQ = zQ zP .

5

Cyclings, twisted cyclings and summit sets

Let α ∈ AS be an element whose left normal form is α = ∆pS α1 · · · αr with r > 0.
The initial factor of α is the simple element ιS (α) = τS−p (α1 ). Thus, α =
ιS (α)∆pS α2 · · · αr . The cycling of α is the conjugate of α by its initial factor,
that is,
cS (α) = ∆pS α2 · · · αr ιS (α).
This expression is not necessarily in left normal form, so in order to apply a
new cycling one must first compute the left normal form of cS (α), to know the
new conjugating element ιS (cS (α)). If r = 0, that is, if α = ∆pS , we just define
cS (∆pS ) = ∆pS .
The twisted cycling of α is defined as e
cS (α) = τS−1 (cS (α)). It is the conjugate of
−1
α by ιS (α)∆S , which is the inverse of a simple element. (One can also think
of e
cS as a left-conjugation by a simple element.) Notice that the conjugating
element is
−(p+1)
ιS (α)∆S−1 = ∆pS α1 ∆S
.
The following lemma tells us that twisted cycling is actually more natural than
cycling from the point of view of the mixed normal form. Notice that its
proof also works when using the alternative Garside structure with Garside
element ∆n , for some n > 1.
Lemma 5.1. If α = x−1 y = xs−1 · · · x−1
1 y1 · · · yt is the np-normal form of an
element in AS , and x 6= 1, then the conjugating element for twisted cycling is
precisely x−1
s . Hence,
−1
−1
e
cS (α) = xs−1
· · · x−1
1 y1 · · · yt xs .

Proof. We have seen that the conjugating element for the twisted cycling of α
s−1
is ∆−s
S α1 ∆S , where α1 is the first non-∆S factor in its left normal form, that
is, α1 = x
es = τS−s (∂S (xs )) = ∆sS ∂S (xs )∆−s
S .
12

−s s−1
s
−1
Thus, the conjugating element is ∆−s
= ∂S (xs )∆−1
S ∆S ∂S (xs )∆S ∆S
S = xs .

If α = ∆pS α1 · · · αr is in left normal form and r > 0, the decycling of α is the
conjugate of α by α−1
r , that is,
dS (α) = αr ∆p α1 · · · αr−1 .
If r = 0, that is, if α = ∆pS , we define dS (∆pS ) = ∆pS . Cyclings (or twisted
cyclings) and decyclings are used to compute some important finite subsets of
the conjugacy class of an element, which we will refer to with the common name
of summit sets.
Definition 5.2. Given α ∈ AS , we denote by C + (α) the set of positive conjugates of α. Notice that this set is always finite, and it could be empty.
Definition 5.3. [11] Given α ∈ AS , the Super Summit Set of α, denoted
SSS(α), is the set of all conjugates of α with maximal infimum and minimal
supremum. Equivalently, SSS(α) is the set of conjugates of α with minimal
canonical length.
Definition 5.4. [15] Given α ∈ AS , the Ultra Summit Set of α, denoted
U SS(α), is the set of all elements β ∈ SSS(α) such that ckS (β) = β for some
k > 0.
It is easy to deduce from the definitions that cS and τS commute; indeed, for
any β ∈ AS , the conjugating elements for the conjugations β 7→ cS (τS (β)) and
β 7→ τS (cS (β)) are identical. Therefore, ckS (β) = β for some k > 0 holds if and
only if e
ctS (β) = β for some t > 0. This means that twisted cycling can be used
to define the Ultra Summit Set.
We will also need two other types of summit sets:

Definition 5.5. [17] Let α be an element in a Garside group G. The Reduced
Super Summit Set of α is
RSSS(α) = {x ∈ αG : ckS (x) = x = dtS (x) for some k, t > 0}.
As above, one can use twisted cycling instead of cycling to define RSSS(α).
Definition 5.6. [1] Let α be an element in a Garside group G. The stable ultra
summit set of α is SU (α) = {x ∈ αG : xm ∈ U SS(xm ) ∀m ∈ Z}.
It is well known [11, 15] that when applying iterated cycling, starting with
α ∈ AS , one obtains an element α′ whose infimum is maximal in its conjugacy
class. Then, by applying iterated decycling to α′ one obtains an element α′′
whose supremum is minimal in its conjugacy class, so α′′ ∈ SSS(α). Finally,
when applying iterated cycling to α′′ until the first repeated element α′′′ is

13

obtained, one has α′′′ ∈ U SS(α). If one then applies iterated decycling to α′′′
until the first repeated element, one obtains α
b ∈ RSSS(α).

In order to conjugate α
b to SU (α), as explained in [1], one just need to apply the
conjugating elements for iterated cycling or decycling of suitable powers of α
b. It
follows that all the above summit sets are nonempty (and finite), and moreover
we have the following:
Lemma 5.7. Let α be an element in a Garside group G with Garside element ∆,
and let I be either SSS(α), or U SS(α), or RSSS(α), or SU (α). One can
conjugate α to β ∈ I by a sequence of conjugations:
α = α0 → α1 → · · · → αm = β
where, for i = 0, . . . , m − 1, the conjugating element from αi to αi+1 has the
form αpi ∆q ∧ ∆r for some integers p, q and r.
Proof. We just need to show that the conjugating elements, for either cycling
or decycling, of a power xk of an element x has the form xp ∆q ∧ ∆r for some
integers p, q and r.
If xk is a power of ∆, the conjugating element is trivial, so the result holds.
Otherwise, suppose that ∆m x1 · · · xn is the left normal form of xk . Then
xk ∆−m = τ −m (x1 ) · · · τ −m (xn ), where the latter decomposition is in left normal form. Hence, xk ∆−m ∧ ∆ = τ −m (x1 ) = ι(xk ). So the conjugating element
for cycling has the desired form.
−k
On the other hand, ∆m+n−1 ∧ xk = ∆m x1 · · · xn−1 . So x−1
(∆m+n−1 ∧
n = x
k
−k m+n−1
−1
x )=x ∆
∧1. Since xn is the conjugating element for decycling of xk ,
the result follows.

The sets C + (α), SSS(α), U SS(α), RSSS(α) and SU (α) share a common
important property, which is usually called convexity. Recall that β x means
x−1 βx and that ∧ denotes the meet operation in the lattice associated to the
prefix partial order 4.
Lemma 5.8. [14, Propositions 4.8 and 4.12], [15, Theorem 1.18] Let α be an
element in a Garside group G and let I be either C + (α), or SSS(α), or U SS(α),
or RSSS(α), or SU (α). If α, αx , αy ∈ I then αx∧y ∈ I.
Remark. Convexity of SU (α) is not shown in [1], but it follows immediately
from convexity of U SS(α).
Remark. Convexity in RSSS(α) also follows from convexity in U SS(α), after
the following observation: An element x belongs to a closed orbit under decycling if and only if x−1 belongs to a closed orbit under twisted cycling, hence to a
closed orbit under cycling. Actually, the conjugating element for twisted cycling
of x−1 equals the conjugating element for decycling of x. Hence the elements
in RSSS(α) are those elements x, conjugate to α, such that both x and x−1
belong to closed orbits under cycling. This implies the convexity in RSSS(α).
Note that it also implies that SU (α) ⊆ RSSS(α).
14

The set I is usually obtained by computing the directed graph GI , whose vertices are the elements of I, and whose arrows correspond to minimal positive
conjugators. That is, there is an arrow labelled x starting from a vertex u and
finishing at a vertex v if and only if:
1. x ∈ A+
S;
2. ux = v; and
3. uy ∈
/ I for every non-trivial proper prefix y of x.
Thanks to the convexity property, one can see that the graph GI is finite and
connected, and that the label of each arrow is a simple element. This is why that
graph can be computed starting with a single vertex, iteratively conjugating the
known vertices by simple elements, until no new elements of I are obtained.
n
We will need to use different Garside structures (AS , A+
S , ∆S ) for distinct positive values of n. To distinguish the precise Garside structure we are using, we
will write In instead of I for a given summit set. For instance, given n ≥ 1 and
an element α ∈ AS , we will write SSSn (α), U SSn (α), RSSSn (α) and SUn (α)
to refer to the super summit set, the ultra summit set, the reduced super summit set and the stable ultra summit set, respectively, of α with respect to the
n
structure (AS , A+
S , ∆S ). Notice that all those sets are finite sets consisting of
conjugates of α. Also, the set C + (α) is independent of the Garside structure
under consideration, so Cn+ (α) = C + (α) for every n > 0.

We will now see that in a summit set I there is always some element which
belongs to In for every n > 1.
Definition 5.9. Let α ∈ T
AS and let I be either SSS, or U SS, or RSSS, or SU .
Then we define I∞ (α) = n≥1 In (α).
Proposition 5.10. For every α ∈ AS , the set I∞ (α) is nonempty.

Proof. We write In for In (α) for n ∈ N ∪ {∞}. For every N ≥ 1, let I≤N =
TN
n=1 In . We will show that I≤N 6= ∅ by induction on N .

If N = 1 then I≤N = I1 , which is known to be nonempty. We can then assume
that N > 1 and that there is an element β ∈ I≤(N −1) . Using the Garside
N
structure (AS , A+
S , ∆S ), we can conjugate β to an element γ ∈ IN by applying
some suitable conjugations
β = β0 → β1 → · · · → βm = γ.
By Lemma 5.7, for every i = 0, . . . , m − 1 the conjugating element from βi to
q
Nr
βi+1 has the form βip ∆N
for some integers p, q and r. Recall that
S ∧ ∆S
β ∈ I≤(N −1) . We claim that, if we apply such a conjugating element to β, the
resulting element β1 will still belong to I≤(N −1) . Repeating the argument m
times, it will follow that γ ∈ I≤(N −1) ∩ IN = I≤N , so I≤N 6= ∅ as we wanted to
show.
15

p

Nq

Nq

But notice that, if β ∈ In for some n, then β β ∆S = β ∆S = τSN q (β) ∈ In , and
Nr
also β ∆S = τSN r (β) ∈ In . By Lemma 5.8, it follows that β1 ∈ In . Hence, if
β ∈ I≤(N −1) then β1 ∈ I≤(N −1) , which shows the claim.
Therefore, we have shown that I≤N 6= ∅ for every N ≥ 1. As the set I1 is finite,
and we have a monotonic chain
I1 = I≤1 ⊇ I≤2 ⊇ I≤3 ⊇ · · · ,
this chain must stabilize for some N ≥ 1, so I∞ = I≤N 6= ∅.
We finish this section by recalling an important tool for studying properties of
the Ultra Summit Set: The transport map.
Definition 5.11. [15] Let α be an element of a Garside group G and v, w ∈
U SS(α). Let x ∈ AS be an element conjugating v to w, that is, x−1 vx = w.
For every i ≥ 0, denote v (i) = ciS (v) and w(i) = ciS (w). The transport of x at v
is the element x(1) = ιS (v)−1 x ιS (w), which conjugates v (1) to w(1) .
We can define iteratively x(i) as the transport of x(i−1) at v (i−1) . That is,
the i-th transport of x at v, denoted x(i) , is the following conjugating element
from v (i) to w(i) :

−1 

x(i) = ι(v)ι(v (1) ) · · · ι(v (i−1) )
x ι(w)ι(w(1) ) · · · ι(w(i−1) ) .
Lemma 5.12. [15, Lemma 2.6] Let α be an element of a Garside group G and
v, w ∈ U SS(α). For every conjugating element x such that x−1 vx = w there
exists some integer N > 0 such that v (N ) = v, w(N ) = w and x(N ) = x.
We remark that in [15] it is assumed that x is a positive element, but this is not a
constraint: Multiplying x by a suitable central power of the Garside element ∆,
we can always assume that x is positive.
Lemma 5.12 can be rewritten in the following way:
Lemma 5.13. Let α be an element in a Garside group G, and let v, w ∈
U SS(α). Let m and n be the lengths of the orbits under cycling of v and w,
respectively. Denote Ct (v) (resp. Ct (w)) the product of t consecutive conjugating
elements for cycling, starting with v (resp. starting with w). Then, for every x
such that x−1 vx = w, there is a positive common multiple N of m and n such
that x−1 CkN (v) x = CkN (w) for every k > 0.
Proof. By Lemma 5.12, there is some N > 0 such that v (N ) = v, w(N ) = w and
x(N ) = x. The first property implies that N is a multiple of m. The second one,
that N is a multiple of n. Finally, by definition of transport, the third property
just means x−1 CN (v)x = CN (w).
Now notice that, as N is a multiple of the length m of the orbit of v under
cycling, one has CkN (v) = CN (v)k for every k > 0. In the same way, CkN (w) =
CN (w)k for every k > 0. Therefore x−1 CkN (v)x = x−1 CN (v)k x = CN (w)k =
CkN (w).
16

Corollary 5.14. Let α be an element in a Garside group G, and let v, w ∈
et (v) (resp. C
et (w)) the product of t consecU SS(α). For every t > 0, denote C
utive conjugating elements for twisted cycling, starting with v (resp. starting
with w). Then, for every x such that x−1 vx = w, there is a positive integer M
eM (v) x = C
eM (w), where C
eM (v) commutes with v and C
eM (w)
such that x−1 C
commutes with w.
et (v) = Ct (v)∆−t
Proof. By definition of cycling and twisted cycling, we have C
−t
e
and Ct (w) = Ct (w)∆ for every t > 0.

We know from Lemma 5.13 that there is some positive integer N such that
x−1 CkN (v) x = CkN (w) for every k > 0. If we take k big enough so that ∆kN
is central, and we denote M = kN , we have:
eM (v) x = x−1 CM (v)∆−M x = x−1 CM (v) x∆−M = CM (w)∆−M = C
eM (w).
x−1 C

Moreover, from Lemma 5.13 we know that CM (v) commutes with v, hence
eM (v) = CM (v)∆−M also commutes with v. In the same way, C
eM (w) comC
mutes with w.

6

Positive conjugates of elements in a parabolic
subgroup

Suppose that an element α belongs to a proper parabolic subgroup P ( AS
and has a positive conjugate. We will show in this section that then all positive
conjugates of α belong to a proper standard parabolic subgroup, determined
by their corresponding supports.
Moreover, the support of a positive conjugate of α will be shown to be preserved
by conjugation, in the sense that if α′ and α′′ are two positive conjugates of α,
whose respective supports are X and Y , then every element conjugating α′ to
α′′ will also conjugate AX to AY .
This will allow us to define a special parabolic subgroup associated to α, which
will be the smallest parabolic subgroup (by inclusion) containing α.
Lemma 6.1. If α ∈ AX ( AS where X ( S, then cS (α) ∈ AY , where either
Y = X or Y = τS (X).
−1
Proof. Let x−1 y = x−1
s · · · x1 y1 · · · yt be the np-normal form of α in AS . As
this is precisely the np-normal form of α in AX , one has that supp(x) ⊆ X and
supp(y) ⊆ X.

Suppose that x = 1, so y = y1 · · · yt . It is clear that y1 6= ∆S , as X ( S. Then
cS (α) = αy1 = y2 · · · yt y1 , hence cS (α) ∈ AX . Notice that if y1 6= ∆X or if α
is a power of ∆X , then one has cS (α) = cX (α), but otherwise we may have
cS (α) 6= cX (α).
17

Now suppose that x 6= 1. We saw in Lemma 5.1 that
−1
−1
e
cS (α) = x−1
s−1 · · · x1 y1 · · · yt xs ,

hence e
cS (α) = τS−1 (cS (α)) ∈ AX , which implies that cS (α) ∈ AτS (X) . Notice
that in this case cS (α) is not necessarily equal to cX (α), but e
cS (α) = e
cX (α).
Lemma 6.2. If α ∈ AX ( AS where X ( S, then dS (α) ∈ AY , where either
Y = X or Y = τS (X).

Proof. It is obvious from the conjugating elements for decycling and twisted
cycling that for every element α, one has dS (α) = (e
cS (α−1 ))−1 . Now α ∈ AX ,
−1
−1
so α
∈ AX and then e
cS (α ) ∈ AY , where either Y = X or Y = τS (X).
Therefore, dS (α) = (e
cS (α−1 ))−1 ∈ AY .
Thanks to Lemma 6.1 and Lemma 6.2, we see that if α ∈ AX , then we obtain
α′′ ∈ AX ∩ SSS(α), where α′′ is obtained from α by iterated cycling and decycling (and possibly conjugating by ∆S at the end). Notice that if α is conjugate
to a positive element, then α′′ will be positive (as the infimum of α′′ is maximal
in its conjugacy class).
Let us suppose that α′′ ∈ AX is positive. We want to study the graph GC + (α) .
We already know that some vertex α′′ ∈ C + (α) belongs to a proper standard
parabolic subgroup. Let us see that this is the case for all elements in C + (α).
Proposition 6.3. Let v ∈ C + (α) with supp(v) = X ( S. Then the label of
every arrow in GC + (α) starting at v either belongs to AX , or is a letter t ∈ S
that commutes with every letter in X, or is equal to rX,t for some letter t ∈ S
adjacent to X in ΓS .
Proof. Let x be the label of an arrow in GC + (α) starting at v. Let t ∈ S be such
that t 4 x. We distinguish three cases.
Case 1: Suppose that t ∈ X. As we know that ∆X conjugates v to a positive
element τX (v), the convexity property implies that ∆X ∧ x also conjugates v
to a positive element. As x is minimal with this property, ∆X ∧ x must be
either trivial or equal to x. However, ∆X ∧ x cannot be trivial as t 4 ∆X ∧ x.
Therefore, ∆X ∧ x = x, which is equivalent to x 4 ∆X . Hence x ∈ AX .
Case 2: Suppose that t ∈
/ X and t is not adjacent to X. Then t commutes
with all letters of X, which means that v t = v is a positive element. Hence, by
minimality of x, we have x = t.
Case 3: Suppose that t ∈
/ X and t is adjacent to X. We must show x = rX,t .
We will determine x algorithmically, starting with t 4 x, and iteratively adding
letters until we obtain the whole conjugating element x.
Write v = v1 · · · vr , where vi ∈ X for i = 1, . . . , r, and let c0 be such that
t 4 c0 4 x and c0 4 rX,t . Consider the following diagram, in which every two
paths with the same initial and final vertices represent the same element:
18

v1

c0

v2

/
c1

u1



vr

/
/

cr−1

c2

/

u2

/

/

ur

/

cr

/

Starting with c0 and v1 , the elements c1 and u1 are defined by the condition
c0 ∨ v1 = c0 u1 = v1 c1 , corresponding to the first square. Then c2 and u2 are
determined by the condition c1 ∨ v2 = c1 u2 = v2 c2 corresponding to the second
square, and so on.
Consider the following claims that will be proven later:
Claim 1: c0 ∨ (v1 · · · vi ) = (v1 · · · vi )ci for every 1 ≤ i ≤ r.
Claim 2: t 4 c0 4 c1 4 c2 4 · · · 4 cr 4 x and cr 4 rX,t .
These claims give us a procedure to compute x: We start with c0 = t = x0 , and
compute c1 , c2 , . . . , cr =: x1 . If x1 is longer than x0 , we start the process again,
this time taking c0 = x1 , and compute c1 , . . . , cr =: x2 . We keep going while xi
is longer than xi−1 . As all obtained elements are prefixes of x, it follows that
the process will stop and we have xk−1 = xk for some k. On the other hand,
Claim 1 implies that xk 4 vxk , so v xk is a positive element. Now notice that
xk 4 x. Hence, by minimality of x, it follows that xk = x.
At this point of the iteration, something interesting will happen. We can substitute c0 = x in the above diagram and obtain:
v1

x

v2

/



x
u1

/

vr

/

x

/

u2

/

x

/

/

x
ur

/

This happens because each vertical arrow is a prefix of the following one, and the
first and last arrows coincide, hence all vertical arrows must be the same. Then
vi x = xui for every i = 1, . . . , r. As the relations defining AS are homogeneous,
it follows that each ui is a single letter, and thus x conjugates the whole set X to
a set Y ( S (since supp(v) = X). But then x−1 ∆X x is a positive element, that
is, x 4 ∆X x. Hence t 4 x 4 ∆X x. By Lemma 4.3, ∆X rX,t = ∆X ∨ t 4 ∆X x,
so we have that rX,t 4 x. By minimality of x, we thus have x = rX,t , as we
wanted to show.
The proofs of Claim 1 and Claim 2 remain to be done:
Proof of Claim 1: We will show by induction on i that c0 ∨ (v1 · · · vi ) =
(v1 · · · vi )ci . This is true for i = 1 by definition of c1 , so assume that i > 1
and that the claim is true for i − 1. Let then c0 ∨ (v1 · · · vi ) = v1 · · · vi d. We
have
(v1 · · · vi−1 )ci−1 = c0 ∨ (v1 · · · vi−1 ) 4 c0 ∨ (v1 · · · vi ) = (v1 · · · vi )d,
19

and thus
(v1 · · · vi−1 )vi ci = (v1 · · · vi−1 )vi ∨ (v1 · · · vi−1 )ci−1 4 (v1 · · · vi−1 )vi d.
This implies that ci 4 d. But (v1 · · · vi )ci is a common multiple of c0 and v1 · · · vi
by construction, so d = ci and the claim is shown.
Proof of Claim 2: By hypothesis, we have that c0 4 rX,t . We will first show by
induction on i that ci 4 rX,t for every i ≥ 0. Suppose that ci−1 4 rX,t for some i.
r
Since X rX,t = Y for some Y ( S by Lemma 4.2, it follows that vi X,t is positive,
that is, rX,t 4 vi rX,t . Hence ci−1 4 vi rX,t and finally vi ci = ci−1 ∨ vi 4 vi rX,t ,
so ci 4 rX,t .
Secondly, we show that ci−1 4 ci for i ≥ 1: Since vi is a single letter belonging to X, and t is the only possible initial letter of rX,t by Lemma 4.4, the
assumption t ∈
/ X implies that vi 64 ci−1 , hence ui 6= 1. Write the word ci−1 ui
as ci−1 u′i s, where s is a single letter. Since ci−1 ui = ci−1 ∨ vi , it follows that
vi 64 ci−1 u′i and vi 4 ci−1 u′i s. By Lemma 4.5 (notice that ci−1 is a simple
element with respect to the usual Garside structure, as it is a prefix of rX,t ,
hence ci−1 u′i is simple as it is a prefix of ci−1 ∨ vi , which is also simple), we
obtain that ci−1 u′i s = vi ci−1 u′i . That is, vi ci = ci−1 ui = ci−1 u′i s = vi ci−1 u′i ,
which implies that ci−1 4 ci .
Finally, we prove that cr 4 x: As x−1 vx is positive, we have that c0 4 x 4 vx,
and also vc0 4 vx. Hence, c0 ∨vc0 4 vx. But notice that by Claim 1, c0 ∨v = vcr ,
so vcr = c0 ∨ v 4 c0 ∨ vc0 4 vx. Hence cr 4 x.
Example 6.4. In Figure 1 we can see the graph GC + (α) for α = σ1 σ2 in the
braid group on 5 strands (the Artin–Tits group of type A4 ). We see the 6
vertices corresponding to the positive conjugates of σ1 σ2 , and the three kinds
of arrows explained in Proposition 6.3. For instance, the arrows starting from
σ1 σ2 are labeled σ1 (type 1), σ4 (type 2) and σ3 σ2 σ1 (type 3).
σ3 σ2 σ1

σ4

σ4 σ3 σ2

σ1 σ2

σ2 σ3
σ1 σ2 σ3

σ1

σ2

σ2 σ3 σ4
σ2

σ3

σ3 σ2 σ1
σ4

σ2 σ1

σ1

σ3 σ4
σ3

σ4

σ4 σ3 σ2
σ3 σ2

σ1 σ2 σ3

σ4 σ3
σ2 σ3 σ4

σ1

Figure 1: The graph GC + (α) for α = σ1 σ2 in the braid group on 5 strands.
Corollary 6.5. Let α ∈ AS be a non-trivial element that belongs to a proper
parabolic subgroup, and is conjugate to a positive element. Then all positive
20

conjugates of α belong to a proper standard parabolic subgroup. Moreover, if v
and w are positive conjugates of α with supp(v) = X and supp(w) = Y , then
for every x ∈ AS such that x−1 vx = w, one has x−1 zX x = zY (and hence
x−1 AX x = AY ).
Proof. With the given hypothesis, we know that we can conjugate α to a positive
element v ∈ AX , where X ( S: First compute a conjugate of α in a proper
standard parabolic subgroup, then apply iterated twisted cycling and decycling
until a super summit conjugate is obtained; the latter will be positive and, by
Lemma 6.1 and Lemma 6.2, contained in a proper standard parabolic subgroup.
By Proposition 6.3, there are three types of arrows in GC + (α) starting at v. In
each of these three cases, consider the target vertex v x of an arrow with label x
starting at v:
1. x ∈ AX . In this case v x ∈ AX .
2. x = t ∈
/ X where t is not adjacent to X. In this case v x = v ∈ AX .
3. x = rX,t , where t ∈
/ X is adjacent to X. Then x−1 Xx = Z for some
proper subset Z ( S. Hence v x ∈ AZ .
Therefore, in every case, v x belongs to a proper standard parabolic subgroup.
We can apply the same argument to every vertex of GC + (α) and, since GC + (α)
is connected, it follows that all vertices in GC + (α) belong to a proper standard
parabolic subgroup.
Now denote X = supp(v), let x be the label of an arrow in GC + (α) starting at v,
and let Y = supp(v x ).
If x ∈ AX , we have Y = supp(v x ) ⊆ X. If we had Y ( X, then v x would be
a positive element belonging to a proper standard parabolic subgroup of AX .
Hence, taking AX as the global Artin–Tits group, all positive conjugates of v x
in AX would belong to a proper standard parabolic subgroup of AX , which is
is not the case, as v itself does not satisfy that property. Hence Y = X, so
supp(v x ) = X. Moreover, as x ∈ AX , we have x−1 zX x = zX .
If x = t ∈
/ X where t is not adjacent to X, then x commutes with all letters
of X, so we have v x = v, whence Y = X, and also x−1 zX x = zX .
Finally, if x = rX,t where t ∈
/ X is adjacent to X, then x−1 Xx = Z for some
Z ( S. As v contains all letters of X, it follows that v x contains all letters of Z,
that is, Z = Y . Therefore x−1 Xx = Y , which implies that x−1 AX x = AY and
hence x−1 zX x = zY .
Applying this argument to all arrows in GC + (α) , it follows that the label of any
arrow starting at a vertex u0 and ending at a vertex u1 conjugates zsupp(u0 )
to zsupp(u1 ) . This can be extended to paths in GC + (α) : If a path goes from u0
to uk , the element associated to the path conjugates zsupp(u0 ) to zsupp(uk ) .

21

Now suppose that v and w are two positive conjugates of α, where supp(v) = X
and supp(w) = Y , and suppose that x ∈ AS is such that x−1 vx = w. Then v
and w are vertices of GC + (α) . Up to multiplying x by a central power of ∆S ,
we can assume that x is positive. Decomposing x as a product of minimal
conjugators, it follows that x is the element associated to a path in GC + (α)
starting at v and finishing at w. Therefore, x−1 zX x = zY , as we wanted to
show.
Example 6.6. Consider again the situation from Example 6.4 and Figure 1.
We see that all positive conjugates of σ1 σ2 belong to a proper standard parabolic
subgroup, namely either to hσ1 , σ2 i, or to hσ2 , σ3 i, or to hσ3 , σ4 i. We can also
check that the labels of the arrows in the graph GC + (α) conjugate the central elements of the (minimal) standard parabolic subgroups containing the respective
conjugates as expected; cf. Figure 2.
σ1
σ4

σ2

σ3

σ3 σ2 σ1
2

σ4 σ3 σ2
2

(σ1 σ2 σ1 )

σ1 σ2 σ3
σ2

(σ3 σ4 σ3 )2

(σ2 σ3 σ2 )

σ1

σ2 σ3 σ4
σ3

σ4

Figure 2: The action of the conjugating elements from Figure 1 on the central elements z. of the (minimal) standard parabolic subgroups containing the
positive conjugates of σ1 σ2 .
The result from Corollary 6.5 allows to introduce an important concept:
Definition 6.7. Let α ∈ AS be conjugate to a positive element α′ = β −1 αβ ∈
′
A+
S . Let X = supp(α ). We define the parabolic subgroup associated to α as the
subgroup Pα = βAX β −1 .
Proposition 6.8. Under the above assumptions, the parabolic subgroup Pα is
well defined, and it is the smallest parabolic subgroup (by inclusion) containing α.
Proof. Suppose that α′′ = γ −1 αγ is another positive conjugate of α, and denote
−1
Y = supp(α′′ ). We must show that βAX β −1 = γAY γ −1 . We have (α′ )β γ =
α′′ . If α belongs to a proper parabolic subgroup, we can apply Corollary 6.5,
so both AX and AY are proper standard parabolic subgroups, the conjugating
element β −1 γ maps zX to zY , and then it maps AX to AY by Lemma 2.1.
Therefore βAX β −1 = γAY γ −1 , that is, Pα is well defined. If α does not belong
to a proper parabolic subgroup, none of its conjugates does, hence AX = AY =
AS . We then have βAX β −1 = βAS β −1 = AS = γAS γ −1 = γAY γ −1 , so Pα is
also well defined (and equal to AS ) in this case.
22

Now let P be a parabolic subgroup containing α. Let x ∈ AS be such that
x−1 P x is standard, that is, x−1 P x = AZ for some Z ⊆ S. Then x−1 αx ∈ AZ
and we can obtain another conjugate α′ = y −1 x−1 αxy ∈ AZ ∩ SSS(α), where
y ∈ AZ , by iterated twisted cycling and iterated decycling. Since α is conjugate
to a positive element, all elements in SSS(α) are positive, so α′ is positive.
Hence, if we denote X = supp(α′ ), we have AX ⊆ AZ . Conjugating back,
we have Pα = xyAX y −1 x−1 ⊆ xyAZ y −1 x−1 = xAZ x−1 = P . Hence Pα is
contained in any parabolic subgroup containing α, as we wanted to show.

7

Parabolic subgroup associated to an element

In the previous section we defined a parabolic subgroup associated to a given
element α ∈ AS , provided that α is conjugate to a positive element. In this
section we will see how to extend this definition to every element α ∈ AS . That
is, we will see that for every element α ∈ AS there is a parabolic subgroup Pα
which is the smallest parabolic subgroup (by inclusion) containing α.
Instead of using the positive conjugates of α (which may not exist), we will define
Pα by using one of the summit sets we defined earlier: RSSS∞ (α). We will see
that the support of the elements in RSSS∞ (α) is also preserved by conjugation,
so it can be used to define the smallest parabolic subgroup containing α.
Definition 7.1. Let α ∈ AS . Let α′ ∈ RSSS∞ (α), where α′ = β −1 αβ. If
we denote Z = supp(α′ ), we define the parabolic subgroup associated to α as
Pα = βAZ β −1 .
The next result shows Theorem 1.1.
Proposition 7.2. Under the above assumptions, the parabolic subgroup Pα is
well defined, and it is the smallest parabolic subgroup (by inclusion) containing α.
Proof. Write α′ = x−1 y in np-normal form, and recall that Z = supp(α′ ) =
supp(x) ∪ supp(y). If α does not belong to a proper parabolic subgroup, none of
its conjugates does, hence supp(v) = S for every v ∈ RSSS∞ (α), so Pα = AS
is well defined, and it is indeed the smallest parabolic subgroup containing α.
Hence we can assume that α belongs to some proper parabolic subgroup.
If α is conjugate to a positive element, then all elements in SSS(α) will be
positive. As RSSS∞ (α) ⊆ RSSS(α) ⊆ SSS(α), it follows that α′ is positive
and Z = supp(α′ ). Therefore the above definition of Pα coincides with the
definition we gave in the previous section. Hence, by Proposition 6.8, Pα is well
defined and it is the smallest parabolic subgroup containing α.
Suppose that α−1 is conjugate to a positive element. As one has β ∈ SSS(α)
if and only if β −1 ∈ SSS(α−1 ), the inverse of every element of RSSS(α) ⊆
SSS(α) is positive. In this case Z = supp((α′ )−1 ), and Pα coincides with the
23

definition of Pα−1 in the previous section. Hence Pα is well defined, and it is the
smallest parabolic subgroup containing α−1 , thus the smallest one containing α.
We can then assume that α belongs to some proper parabolic subgroup, and
−1
that the np-normal form of α′ has the form α′ = x−1 y = x−1
s · · · x1 y1 · · · yt ,
with s, t > 0 (that is, x, y 6= 1).
N
Let N = max(s, t), and let us use the Garside structure (AS , A+
S , ∆S ). With
′
respect to this structure, x and y are simple elements and, as α ∈ RSSS∞ (α),
both α′ and (α′ )−1 belong to their respective ultra summit sets.

In order to show that Pα is well defined, let α′′ = γ −1 αγ ∈ RSSS∞ (α). We
have to show that every element g conjugating α′ to α′′ , conjugates AZ to AU ,
where Z = supp(α′ ) and U = supp(α′′ ). We will show this by constructing
positive elements with supports Z and U , respectively, which are also conjugate
by g; then the claim follows by Corollary 6.5.
N
′
−1
As we are using the Garside structure (AS , A+
y
S , ∆S ), we have that α = x
′′
in np-normal form, where x and y are simple elements, and also α = u−1 v
in np-normal form, where u and v are simple elements, since both α′ and α′′
belong to SSSN (α), so they have the same infimum and supremum.

Let then g be an element such that g −1 α′ g = α′′ , and recall that α′ , α′′ ∈
RSSS∞ (α) ⊆ U SS∞ (α) ⊆ U SSN (α). By Corollary 5.14, there is a positive
eM (α′ )g = C
eM (α′′ ), where C
eM (α′ ) (resp. C
eM (α′′ )) is
integer M such that g −1 C
the product of the conjugating elements for M consecutive twisted cyclings of
N
α′ (resp. α′′ ) with respect to the Garside structure (AS , A+
S , ∆S ).
eM (α′ ) is the inverse of a positive element, say w1 (α′ ) :=
Hence, by Lemma 5.1, C
eM (α′ )−1 . Recall that α′ ∈ AZ , hence the factors composing C
eM (α′ ) belong to
C
+
′
′′
e
AZ , and then w1 (α ) ∈ AZ . In the same way, CM (α ) is the inverse of a positive
eM (α′′ )−1 ∈ A+ . We then have g −1 w1 (α′ )g = w1 (α′′ ),
element, say w1 (α′′ ) := C
U
+
′
where w1 (α ) ∈ AZ and w1 (α′′ ) ∈ A+
U.

Now, from g −1 α′ g = α′′ we obtain g −1 (α′ )−1 g = (α′′ )−1 . Since α′ , α′′ ∈
RSSS∞ (α), it follows that (α′ )−1 , (α′′ )−1 ∈ U SS∞ (α−1 ) ⊆ U SSN (α−1 ). Thus,
we can apply Corollary 5.14 and obtain that there is a positive integer T such
eT ((α′ )−1 )g = C
eT ((α′′ )−1 ) in the same way as above. If we denote
that g −1 C
eT ((α′ )−1 )−1 and w2 (α′′ ) = C
eT ((α′′ )−1 )−1 , we have g −1 w2 (α′ )g =
w2 (α′ ) = C
′′
w2 (α′′ ), where w2 (α′ ) ∈ A+
and
w
(α
)
∈ A+
2
Z
U.

+
′′
′′
′′
Let us denote w(α′ ) = w1 (α′ )w2 (α′ ) ∈ A+
Z and w(α ) = w1 (α )w2 (α ) ∈ AU .
By construction, g −1 w(α′ )g = w(α′′ ). We will now show that supp(w(α′ )) = Z
and supp(w(α′ )) = U .

Notice that the conjugating element for twisted cycling of α′ = x−1 y using the
N
−1
Garside structure (AS , A+
. By Lemma 5.1, x is a suffix of w1 (α′ ).
S , ∆S ) is x
On the other hand, the conjugating element for twisted cycling of (α′ )−1 = y −1 x
is y −1 . Hence, y is a suffix of w2 (α′ ). This implies that Z = supp(α′ ) =
supp(x)∪supp(y) ⊆ supp(w1 (α′ ))∪supp(w2 (α′ )) = supp(w(α′ )) ⊆ Z. Therefore
supp(w(α′ )) = Z. In the same way it follows that supp(w(α′′ )) = U .
24

We can then apply Corollary 6.5, since g −1 w(α′ )g = w(α′′ ), to conclude that
g −1 AZ g = AU , as we wanted to show. This means that Pα is well defined, as
taking g = β −1 γ (a conjugating element from α′ to α′′ ) one has:
βAZ β −1 = βgAU g −1 β −1 = γAU γ −1 ,
hence one can equally use either α′ or α′′ to define Pα .
Now let us prove that Pα is the smallest parabolic subgroup (for inclusion)
containing α. Suppose that P is a parabolic subgroup containing α, and let a ∈
AS be an element such that a−1 P a = AX is standard. Then a−1 αa ∈ AX . We
can now apply iterated twisted cyclings and iterated decyclings to this element
(in all needed Garside structures), so that the resulting element, say α
b, belongs
to RSSS∞ (α). The product of all conjugating elements, call it b, will belong to
AX , so we will have α
b = b−1 a−1 αab ∈ AX ∩ RSSS∞ (α).
Let Y = supp(b
α). By definition, we have Pα = abAY b−1 a−1 . But on the other
hand, as α
b ∈ AX , all letters in the np-normal form of α
b belong to AX . Hence
AY ⊆ AX , and we finally have:
Pα = abAY b−1 a−1 ⊆ abAX b−1 a−1 = aAX a−1 = P.
Therefore, Pα is contained in every parabolic subgroup containing α, as we
wanted to show.

8

Parabolic subgroups, powers and roots

In this section we will see that the parabolic subgroup Pα associated to an element α ∈ AS behaves as expected under conjugation, taking powers and taking
roots. The behavior under conjugation follows directly from the definition:
Lemma 8.1. For every α, x ∈ AS , one has Px−1 αx = x−1 Pα x.
Proof. Let α′ = β −1 αβ ∈ RSSS∞ (α). Then α′ = β −1 x(x−1 αx)x−1 β.
If X = supp(α′ ), by definition we have Pα = βAX β −1 , and also
Px−1 αx = x−1 βAX β −1 x = x−1 Pα x.

The behavior of Pα when taking powers or roots is not so easy, but it is also as
expected:
Theorem 8.2. Let AS be an Artin–Tits group of spherical type. If α ∈ AS and
m is a nonzero integer, then Pαm = Pα .

25

Proof. By Lemma 8.1, we can conjugate α to assume that α ∈ RSSS∞ (α). We
can further conjugate α by the conjugating elements for iterated twisted cycling
and iterated decycling of its m-th power (for all needed Garside structures), in
order to take αm to RSSS∞ (αm ). By Lemma 5.7, these conjugating elements
are the meet of two elements which conjugate α to other elements in RSSS∞ (α).
Hence, by the convexity of RSSS∞ (α), they maintain α in RSSS∞ (α). In
summary, up to conjugacy we can assume that α ∈ RSSS∞ (α) and αm ∈
RSSS∞ (αm ). We can also assume that m is positive, as Pα−m = Pαm by
definition.
Under these assumptions, the associated parabolic subgroups of α and αm will
be determined by their corresponding supports. Hence, we must show that
supp(α) = supp(αm ).
If either α or α−1 is positive, the result is clear. We can then assume that this
is not the case, that is, we have α = x−1
1 y1 in np-normal form with x1 , y1 6= 1.
Moreover, using a suitable Garside structure, we can assume that x1 and y1 are
simple elements.
We will now use the following result [1, Theorem 2.9]: If α ∈ U SS(α), inf(α) =
p, ℓ(α) > 1 and m ≥ 1, one has
αm ∆−mp ∧ ∆m = Cm (α).

(1)

In other words, if we consider the positive element αm ∆−mp , and compute the
first m factors in its left normal form (including any ∆ factors), the product
of these m factors equals the product of the first m conjugating elements for
iterated cyclings of α.
N
We point out that we are using a Garside structure (AS , A+
S , ∆S ) such that
−1
α = x1 y1 , where x1 and y1 are nontrivial simple elements. So the ∆ in (1)
means ∆N
S is our case. Also, we remark that α satisfies the hypotheses for (1),
as α ∈ RSSS∞ (α) ⊆ U SS∞ (α) ⊆ U SSN (α), and the left normal form of α is
∆−1 x
e1 y1 , so inf(α) = −1 and ℓ(α) = 2.

We will restate the above result from [1] in terms of np-normal forms. It turns
out that the statement will become much nicer.

−1
−1
Write αm = x−1
1 y1 x1 y1 · · · x1 y1 (see Figure 3). Notice that applying a twisted
−1
cycling to α means to conjugate it by x−1
1 , and one obtains α2 = y1 x1 , whose
−1
np-normal form will be of the form α2 = x2 y2 . We then have:
−1
−1
−1
αm = x−1
1 (x2 y2 x2 y2 · · · x2 y2 )y1

Now we can apply twisted cycling to α2 (conjugating it by x−1
2 ), and we obtain
α3 = x3−1 y3 . Then we see that:
−1 −1
−1
αm = x−1
1 x2 (x3 y3 · · · x3 y3 )y2 y1 .

Repeating the process m times, we finally obtain:
−1
−1
αm = x−1
1 x2 · · · xm ym · · · y2 y1 .

26

O
x1
y1

/O

O
x2

x1
y2

y1

/O

O
x3

/O

x2
y3

y2

/O

O
x4

x1

x3
y4

/

y1

/O
x2

/O

y3

/

x1
y2

/

y1

/

−1 −1 −1 −1
4
Figure 3: How to transform α4 = (x−1
1 y1 ) into x1 x2 x3 x4 y4 y3 y2 y1 .
−1
Around each square, the product xi yi is the np-normal form of yi−1 x−1
i−1 .

That is, we have written αm as a product of a negative times a positive element,
where the negative one is the product of the first m conjugating elements for
iterated twisted cycling of α. We will see that [1, Theorem 2.9] is equivalent to
−1
−1
the fact that no cancellation occurs between x−1
1 x2 · · · xm and ym · · · y2 y1 .
Indeed, on the one hand, as p = −1, we have the element αm ∆−pm = αm ∆m ,
which is:
−1
m
αm ∆m = x−1
1 · · · xm ym · · · y1 ∆
2
m−1 −1 m m
= (x1−1 ∆)(∆−1 x−1
xm ∆ )τ (ym · · · y1 ).
2 ∆ ) · · · (∆

On the other hand, for i = 1, . . . , m, we have αi = e
ci−1 (α) = τ −i+1 ◦ ci−1 (α), so
i−1
i−1
i−1
−1 i−1
c (α) = τ (αi ) = τ (xi ) τ (yi ). This means that the i-th conjugating
i
element for iterated cycling of α is τ i−1 (xi )−1 ∆ = ∆i−1 x−1
i ∆ .
Hence, by [1, Theorem 2.9] the product of the first m factors of the left normal
−1 −1 2
m
form of αm ∆m is equal to (x−1
x2 ∆ ) · · · (∆m−1 x−1
m ∆ ), that is, to
1 ∆)(∆
−1 −1
−1 m
x1 x2 · · · xm ∆ . In other words:
−1
−1 m
αm ∆m ∧ ∆m = x−1
1 x2 · · · xm ∆ .

As the greatest common prefix is preserved by multiplication on the right by
any power of ∆, we obtain:
−1
−1
αm ∧ 1 = x−1
1 x2 · · · xm .

But the biggest common prefix of an element and 1 is precisely the negative
part of its np-normal form, so this shows that there is no cancellation between
−1
−1
x−1
1 x2 · · · xm and ym · · · y2 y1 , as we claimed.
27

But then:
supp(αm ) = supp(xm · · · x1 ) ∪ supp(ym · · · y1 )
⊇ supp(x1 ) ∪ supp(y1 ) = supp(α).
Since it is clear that supp(αm ) ⊆ supp(α) (as no new letters can appear when
computing the np-normal form of αm starting with m copies of the np-normal
form of α), we finally have supp(αm ) = supp(α), as we wanted to show.
This behavior of Pα with respect to taking powers or roots allows us to show an
interesting consequence: All roots of an element in a parabolic subgroup belong
to the same parabolic subgroup.
Corollary 8.3. Let AS be an Artin–Tits group of spherical type. If α belongs
to a parabolic subgroup P , and β ∈ AS is such that β m = α for some nonzero
integer m, then β ∈ P .
Proof. Since α is a power of β, Theorem 8.2 tells us that Pα = Pβ . Now α ∈ P
implies Pα ⊆ P , as Pα is the minimal parabolic subgroup containing α. Hence
β ∈ Pβ = Pα ⊆ P .

9

Intersection of parabolic subgroups

In this section we will show one of the main results of this paper: The intersection of two parabolic subgroups in an Artin–Tits group of spherical type AS is
also a parabolic subgroup.
We will use the parabolic subgroup Pα associated to an element α ∈ AS , but
we will also need some technical results explaining how the left normal form of
a positive element, in which some factors equal ∆X for some X ⊆ S, behaves
when it is multiplied by another element:
Lemma 9.1. Let X ⊆ S be nonempty, and let α ∈ A+
S be a simple element such
that ∆X α is simple. Then there is Y ⊆ S and a decomposition α = ρβ in A+
S,
such that Xρ = ρY and the left normal form of ∆X (∆X α) is (ρ∆Y )(∆Y β).
Proof. We proceed by induction on the length |α| of α as a word over S. If
|α| = 0 then ∆X (∆X α) = ∆X ∆X , which is already in left normal form, and
the result holds taking Y = X and ρ = β = 1. Suppose then that |α| > 0 and
the result is true when α is shorter.
Let Z ⊆ S be the set of initial letters of ∆X α. That is, Z = {σi ∈ S : σi 4
∆X α}. It is clear that X ⊆ Z, as every letter of X is a prefix of ∆X .
Notice that the set of final letters of ∆X is precisely X. Hence, if Z = X
the decomposition ∆X (∆X α) is already in left normal form, so the result holds
taking X = Y , ρ = 1 and β = α.
28

Suppose on the contrary that there exists some t ∈ Z which is not in X. Then
t 4 ∆X α so t ∨ ∆X = ∆X rX,t 4 ∆X α. Cancelling ∆X from the left we have
rX,t 4 α, so α = rX,t α1 .
We know that XrX,t = rX,t T for some T ( S. Hence:
∆X (∆X α) = ∆X (∆X rX,t α1 ) = (rX,t ∆T )(∆T α1 ).
Now consider the element ∆T (∆T α1 ). As α1 is shorter than α, we can apply
the induction hypothesis to obtain that α1 = ρ1 β, where T ρ1 = ρ1 Y for some
Y ( S, and the left normal form of ∆T (∆T α1 ) is (ρ1 ∆Y )(∆Y β).
We claim that the left normal form of ∆X (∆X α) is (ρ∆Y )(∆Y β), where ρ =
rX,t ρ1 . First, we have:
∆X (∆X α) = ∆X (∆X rX,t ρ1 β) = (rX,t ∆T )(∆T ρ1 β)
= (rX,t ρ1 ∆Y )(∆Y β) = (ρ∆Y )(∆Y β).
Next, we see that ρ∆Y = ∆X ρ is simple, as it is a prefix of ∆X α which is
simple. Finally, the set of final letters of ρ∆Y contains the set of final letters of
its suffix ρ1 ∆Y , which in turn contains the set of initial letters of ∆Y β (as the
product (ρ1 ∆Y )(∆Y β) is in left normal form). So the claim holds.
Now it just remains to notice that Xρ = XrX,t ρ1 = rX,t T ρ1 = rX,t ρ1 Y = ρY ,
to finish the proof.
Lemma 9.2. Let X ⊆ S be nonempty, and let m > r > 0. Let α ∈ A+
S be
such that sup(α) = r, and let x1 x2 · · · xm+r be the left normal form of (∆X )m α
(where some of the the first factors can be equal to ∆S and some of the last
factors can be trivial). Then there is Y ⊆ S and a decomposition α = ρβ in
A+
S , such that Xρ = ρY and:
1. x1 · · · xr = (∆X )r ρ = ρ (∆Y )r .
2. xi = ∆Y for i = r + 1, . . . , m − 1.
3. xm · · · xm+r = ∆Y β.
Proof. Suppose first that α is a simple element, so r = 1. By the domino rule
[7, Definition III 1.57, Proposition V 1.52] (see also [8, Lemma 1.32]), the left
normal form of (∆X )m α is computed as follows:
∆X

∆X

/
y2

x1



x2

/

∆X

/

ym−2

y3

/

x3

∆X

/

/

/

xm−1

ym−1

/

xm

∆X

/

/

ym

/

xm+1

α

/

where the xi and the yi are defined from right to left, in the following way: First,
the left normal form of ∆X α is ym xm+1 (here xm+1 could be trivial). Then the
29

left normal form of ∆X ym is ym−1 xm , and so on. Around each square, the
down-right path represents the left normal form of the right-down path.
By construction, ∆X 4 ym . Hence, by the previous lemma, the left normal
form of ∆X ym is (ρ∆Y )(∆Y β1 ), for some Y ⊆ S and some ρ which conjugates
X to Y . Hence ym−1 = ρ∆Y = ∆X ρ, and xm = ∆Y β1 .
Now, if some yk = ∆X ρ, as it is clear that the left normal form of ∆X (∆X ρ) =
(ρ∆Y )(∆Y ), it follows that yk−1 = ρ∆Y = ∆X ρ and xk = ∆Y . Therefore, the
above diagram is actually as follows:
∆X

ρ∆Y

∆X

/



ρ∆Y

∆Y

/

∆X

/

ρ∆Y

/

∆Y

/

ρ∆Y

/

/

∆Y

∆X

∆X

/
ym

ρ∆Y

/

∆Y β1

/

/

xm+1

α

/

By construction, α = ρβ1 xm+1 . This shows the result for r = 1.
The case r > 1 follows from the above one and the domino rule. If α = α1 · · · αr
in left normal form, the left normal form of (∆X )m α is computed by completing
the squares from the diagram in Figure 4 (row by row, from right to left), where
the down-right path is the left normal form of the right-down path, around each
square.
By construction, the subsets X = Y0 , Y1 , · · · , Yr = Y of S and the elements
ρ1 , . . . , ρr satisfy Yi−1 ρi = ρi Yi . This implies that the first r factors in the
normal form of (∆X )m α are:
(ρ1 ∆Y1 )(ρ2 ∆Y2 ) · · · (ρr ∆Yr ) = (∆X )r ρ = ρ(∆Y )r ,
where ρ = ρ1 · · · ρr . Moreover, xi = ∆Y for i = r + 1, . . . , m − 1, as we can see
in the diagram. And finally we see that xm = ∆Y βr and
(∆X )m α = ρ(∆Y )m−1 xm · · · xm+r = (∆X )m−1 ρxm · · · xm+r .
Therefore ρxm · · · xm+r = ∆X α = ∆X ρβ = ρ∆Y β. Hence xm · · · xm+r = ∆Y β,
as we wanted to show.
Therefore, if we multiply a big power (∆X )m by some element α which is a
product of r simple factors, the normal form of the result still has m − r − 1
factors of the form ∆Y for some Y .
In the forthcoming result, we will need a special procedure to compare elements
in AS . For that purpose, we introduce the following:
Definition 9.3. For every element γ ∈ AS we will define an integer ϕ(γ) as
follows: Conjugate γ to γ ′ ∈ RSSS∞ (γ). Let U = supp(γ ′ ). Then ϕ(γ) = |∆U |,
the length of the element ∆U as a word in the standard generators.

30

∆X

ρ1 ∆Y1

∆X

/

ρ1 ∆Y1
∆Y1


ρ2 ∆Y2

∆Y2

∆X

/

ρ1 ∆Y1

/

ρ2 ∆Y2



/

∆Y1

ρ1 ∆Y1

/

/

ρ2 ∆Y2

/

∆Y2

∆X

/

∆Y1

ρ1 ∆Y1

/

ρ2 ∆Y2

/

/

/

∆Y2

∆Y1

∆Y2

∆X

/

ρ1 ∆Y1

/

ρ2 ∆Y2

/

∆X

∆Y1

ρ1 ∆Y1

/

ρ2 ∆Y2

/

∆Y2

∆X

/

∆Y1

∆X

/

/

ρ1 ∆Y1

/

∆Y1 β1

α1

/

/

ρ2 ∆Y2

/

∆Y2 β2

α2

/

/

/

/

/

/

31

/

/

∆Yr−1


ρr ∆Yr

/



ρr ∆Yr

∆Yr

∆Yr−1

ρr ∆Yr

/

∆Yr

∆Yr−1

/

ρr ∆Yr

/

/

∆Yr

∆Yr−1

/

∆Yr−1 βr−1

/

αr

ρr ∆Yr

/

∆Yr βr

/

xm+1

/

xm+2

/

Figure 4: Computing the left normal form of (∆X )m α in the proof of Lemma 9.2.

/

xm+r

/

Proposition 9.4. The integer ϕ(γ) is well defined. Moreover, if γ is conjugate
to a positive element, then ϕ(γ) = |∆X |, where X = supp(β) for any positive
element β conjugate to γ.
Proof. Suppose that γ ′ , γ ′′ ∈ RSSS∞ (γ), and let U = supp(γ ′ ) and V =
supp(γ ′′ ). Then Pγ ′ = AU and Pγ ′′ = AV , and every element x conjugating γ ′
to γ ′′ must also conjugate AU to AV . Hence x−1 zU x = zV , by Lemma 2.1.
This implies that |zU | = |zV |. Moreover, as AU is conjugate to AV , we have
zU = ∆eU and zV = ∆eV for the same e ∈ {1, 2}. Therefore |∆U | = |∆V |, which
proves that ϕ(γ) is well defined.
On the other hand, if γ is conjugate to a positive element β, and X = supp(β),
then Pβ = AX . We can then apply the same argument as above to γ ′ and β, to
obtain that ϕ(γ) = |∆X |.
We can finally show one of the main results in this paper:
Theorem 9.5. Let P and Q be two parabolic subgroups of an Artin–Tits group
AS of spherical type. Then P ∩ Q is also a parabolic subgroup.
Proof. If either P or Q is equal to AS or to {1}, the result is trivially true.
Hence we can assume that both subgroups are proper parabolic subgroups.
If P ∩ Q = {1} the result holds. Hence we will assume that there exists some
nontrivial element α ∈ P ∩ Q. We take α such that ϕ(α) is maximal (notice
that ϕ(α) is bounded above by |∆S |).
Let Pα be the parabolic subgroup associated to α. By Theorem 1.1, we know
that Pα ⊆ P , and also Pα ⊆ Q, so Pα ⊆ P ∩ Q. Moreover, up to conjugating
Pα , P and Q by the same suitable element, we can assume that Pα is standard,
so Pα = AZ ⊆ P ∩ Q for some Z ( S. Notice that ∆Z ∈ Pα ⊆ P ∩ Q.
We will show that P ∩ Q = Pα , that is, P ∩ Q = AZ .
Take any element w ∈ P ∩ Q. In order to show that w ∈ AZ , we will consider
its associated parabolic subgroup Pw , which we will denote by T . By the above
arguments, we have T ⊆ P ∩ Q and, in particular, zT ∈ P ∩ Q. Notice that
T is conjugate to AX for some X ( S, hence zT is conjugate (by the same
conjugating element) to the positive element zX . Since the support of zX is X,
it follows that PzX = AX , and conjugating back we have PzT = T . Therefore,
if we show that zT ∈ AZ , this will imply that T ⊆ AZ and then w ∈ AZ , as
desired.
We then need to show that zT ∈ AZ . Let a−1 b be the np-normal form of zT .
We will now construct an infinite family of elements in P ∩ Q, using zT and
∆Z . For every m > 0, consider βm = zT (∆Z )m = a−1 b(∆Z )m . By construction
βm ∈ P ∩ Q for every m > 0.
Suppose that a = a1 · · · ar and b = b1 · · · bs are the left normal forms of a and

32

b, respectively. Then
m
βm = ar−1 · · · a−1
1 b1 · · · bs (∆Z ) .

The np-normal form of βm is computed by making all possible cancellations
in the middle of the above expression. As all the involved factors are simple
elements, it follows that inf(βm ) ≥ −r and sup(βm ) ≤ s + m.
Recall that ϕ(α) = |∆Z | is maximal among the elements in P ∩Q. Let us denote
n = ϕ(α) = |∆Z |. Now, for every m > 0, choose some βem ∈ RSSS∞ (βm ).

Claim There is M > 0 such that βem is positive for all m > M .
Let Um = supp(βem ). We know by maximality of ϕ(α) that |∆Um | ≤ n. So the
length of each simple element in the normal form of βem must be at most n.
e
Let x−1
As βem ∈ RSSS∞ (βm ) ⊆
m ym be the the np-normal form of βm .
SSS(βm ), it follows that xm is a positive element formed by at most r simple elements, and ym is a positive element formed by at most s + m simple
elements.
Given some m > 0, suppose that none of the factors in the left normal form of ym
is equal to ∆Um . This means that the length of each factor of ym is at most n−1,
so |ym | ≤ (n − 1)(s + m), that is, |ym | ≤ (n − 1)m + k where k is a constant
independent of m. Hence the exponent sum of βem as a product of standard
generators and their inverses is s(βem ) = |ym | − |xm | ≤ |ym | ≤ (n − 1)m + k.
But this exponent sum is invariant under conjugation, hence s(βem ) = s(βm ) =
|b| − |a| + nm. That is, s(βem ) = nm + K for some constant K independent of
m. We then have nm + K ≤ (n − 1)m + k, that is, m ≤ k − K.

If we denote M = k − K, it follows that for every m > M , the left normal form
of ym has some factor equal to ∆Um . Recall that Um = supp(βem ), so βem ∈ AUm .
This means that the left normal form of ym starts with ∆Um . But there cannot
be cancellations between x−1
m and ym , so this implies that xm = 1. Therefore,
βem is positive for all m > M . This proves the claim.

We then know that, if m > M , the element βm is conjugate to a positive element.
The good news is that one can conjugate βm to a positive element βbm , using
a conjugating element cm whose length is bounded by a constant independent
of m. Indeed, one just needs to apply iterated cycling to βm until its infimum
becomes non-negative. Since inf(βm ) ≥ −r, one just needs to increase the
infimum at most r times, and we know from [2] that this can be done with
at most r|∆S | − r cyclings. Hence, we can take sup(cm ) ≤ r|∆S | − r, say
sup(cm ) ≤ N , a number which is independent of m.
−1 −1
We then have βbm = c−1
b(∆Z )m cm ∈ A+
m βm c m = c m a
S . We will now try to
b
describe the support of the positive element βm .

Consider the element (∆Z )m cm . By Lemma 9.2, if m is big enough we can
decompose cm = ρm dm so that ∆Z ρm = ρm ∆Ym (actually Zρm = ρm Ym ) for
some Ym ( S, and the left normal form of (∆Z )m cm finishes with m − N − 1
33

copies of ∆Ym followed by some factors whose product equals ∆Ym dm . Notice
that making m big enough, we can have as many copies of ∆Ym as desired.
−1
The negative part of βbm as it is written above is c−1
, which is the product
m a
b
of at most N + r inverses of simple factors. Since βm is positive, this negative
part must cancel completely with the positive part. Namely, it cancels with the
first N + r simple factors in the left normal form of b(∆Z )m cm .
But the first N +r simple factors of b(∆Z )m cm are a prefix of b multiplied by the
first N + r factors of (∆Z )m cm . Recall that we can take m big enough so that
(∆Z )m cm has as many copies of ∆Ym as desired. Hence, for m big enough we
can decompose b(∆Z )m cm = A∆Ym dm , where A is a positive element containing
−1
enough simple factors to absorb c−1
completely. That is, A = acm B for some
m a
positive element B. It follows that βbm = B∆Ym dm .
Now recall that ρ−1
m ∆Z ρm = ∆Ym , hence |∆Ym | = n. On the other hand,
b
since βm is positive, its support determines ϕ(βm ). Hence, if U = supp(βbm ),
we have |∆U | ≤ n. As Ym ⊆ U , it follows that n = |∆Ym | ≤ |∆U | ≤ n, so
|∆Ym | = |∆U | = n and then Ym = U = supp(βbm ). This implies in particular
that dm ∈ AYm .
b
But now recall that βbm = c−1
m βm cm , where βm is positive. Hence the min−1
−1
= ρm dm AYm d−1
imal parabolic subgroup containing βm is cm AYm cm
m ρm =
−1
ρm AYm ρm = AZ .

Therefore, βm ∈ AZ for some m big enough. Since βm = zT (∆Z )m , it follows
that zT ∈ AZ , as we wanted to show.

10

The lattice of parabolic subgroups

We finish this paper with an interesting simple consequence of the main result in
the previous section: The set of parabolic subgroups forms a lattice for inclusion.
Proposition 10.1. Let AS be an Artin–Tits group of spherical type and let P
be the set of parabolic subgroups of AS . If π is a predicate on P such that the
conjunction of π(P ) and π(Q) implies π(P ∩ Q) for any P, Q ∈ P, then the
set Pπ = {P ∈ P : π(P )} contains a unique minimal element with respect to
inclusion, namely
\
P.
P ∈Pπ

Proof. First note that for P, Q ∈ P, we have P ∩ Q ∈ P by Theorem 9.5, so
π(P ∩ Q) is defined.
The set Pπ is partially ordered by inclusion. We will show that
\
R=
P
P ∈Pπ

34

is the unique minimal element in Pπ . It is clear by definition that R is contained
in every element of Pπ , hence it just remains to show that R is an element of Pπ .
Notice that the set P of parabolic subgroups in AS is a countable set, as every
element P ∈ P can be determined by a subset X ⊆ S and an element α ∈ AS
such that α−1 P α = AX . As there are a finite number of subsets of S and a
countable number of elements in AS , it follows that P is countable.
Therefore, the set Pπ is also countable, and we can enumerate its elements:
Pπ = {Pi : i ∈ N}. Now let
n
\
Pi .
Tn =
i=0

By Theorem 9.5 and the assumption on π, the intersection of any finite number
of elements of Pπ is contained in Pπ , so we have Tn ∈ Pπ for all n ≥ 0.
We then have the following descending chain of elements of Pπ
T0 ⊇ T1 ⊇ T2 ⊇ · · ·
where the intersection of all the parabolic subgroups in this chain equals R.
We finish the proof by noticing that in AS there cannot be an infinite chain of
distinct nested parabolic subgroups, as if α−1 AX α ( β −1 AY β then |X| < |Y |.
Hence, there can be at most |S| + 1 distinct nested parabolic subgroups in any
chain. Therefore, there exists N ≥ 0 such that TN = TN +k for every k > 0, and
then
N
∞
\
\
Pi ,
Ti = TN =
R=
i=0

i=0

so R is an element of Pπ .
Example 10.2. Let AS be an Artin–Tits group of spherical type and α ∈ AS .
Applying Proposition 10.1 with the predicate π(P ) = (α ∈ P ), we see that the
minimal parabolic subgroup Pα containing α that was defined in Proposition 7.2
is the intersection of all parabolic subgroups containing α.
Theorem 10.3. The set of parabolic subgroups of an Artin–Tits group of spherical type is a lattice with respect to the partial order determined by inclusion.
Proof. Let AS be an Artin–Tits group of spherical type, and let P be the set
of parabolic subgroups of AS . This set is partially ordered by inclusion. Now
assume that P, Q ∈ P are given.
By Theorem 9.5, P ∩ Q is the unique maximal parabolic subgroup among those
parabolic subgroups contained in both P and Q.
Applying Proposition 10.1 with the predicate π(T ) = (P ∪ Q ⊆ T ) shows that
there is a unique minimal parabolic subgroup among those parabolic subgroups
containing both P and Q.

35

11

Adjacency in the complex of irreducible parabolic subgroups

We postponed to this section the proof of the following result, which characterizes the pairs of adjacent subgroups in the complex of irreducible parabolic
subgroups.
Theorem 2.2. Let P and Q be two distinct irreducible parabolic subgroups of
an Artin–Tits group AS of spherical type. Then zP zQ = zQ zP holds if and only
if one of the following three conditions are satisfied:
1. P ( Q.
2. Q ( P .
3. P ∩ Q = {1} and xy = yx for every x ∈ P and y ∈ Q.
Proof. If P ( Q then zP ∈ Q and zQ is central in Q, so both elements commute.
Similarly, if Q ( P then zP and zQ commute. Also, if the third condition is
satisfied every element of P commutes with every element of Q, so zP and zQ
commute.
Conversely, assume that zP and zQ commute. We can assume {1} =
6 P ( AS
and {1} 6= Q ( AS , as otherwise either Condition 1 or Condition 2 holds. We are
going to prove a result which is slightly stronger than what is required: We shall
show that P and Q can be simultaneously conjugated to standard irreducible
parabolic subgroups AX and AY (for some subsets X, Y ⊆ S); moreover, one of
the following holds:
1. X ( Y .
2. Y ( X.
3. X ∩ Y = ∅, and all elements of X commute with all elements of Y .
Notice that the four properties listed in the statement of Theorem 2.2 are preserved by conjugation. Hence, up to conjugation, we can assume that P is
standard.
We decompose zQ in pn-normal form. Namely, zQ = ab−1 where a and b are
positive elements such that a ∧ b = 1 (where ∧ means the greatest common
suffix in AS ). The suffix order is preserved by right multiplication, so we can
right-multiply the above equation by b−1 , and we have (ab−1 ) ∧ 1 = b−1 , that
is, zQ ∧ 1 = b−1 .
−1
Now zQ commutes with zP , and 1 also commutes with zP . Hence zQ zP zQ
and
−1
1zP (1) are positive elements (we are assuming that P is standard, so zP is
positive). It follows by convexity (Lemma 5.8 applied to the suffix order) that
(zQ ∧ 1)zP (zQ ∧ 1)−1 is positive, that is, b−1 zP b is positive.

36

But we know from Corollary 6.5 (see also [5, Corollary 40]) that each positive
conjugate of zP is the generator of the center of a standard irreducible parabolic
subgroup. That is, b−1 zP b = zX for some X ( S.
On the other hand, it is shown in [5, Theorem 4] that if zQ = ab−1 is in
pn-normal form, then b is the minimal standardizer of Q, that is, b is the smallest positive element which conjugates Q to a standard parabolic subgroup, so
b−1 zQ b = zY for some Y ( S.
Therefore, when conjugating both zP and zQ by b, we obtain elements zX
and zY , generators of the centers of standard parabolic subgroups. We can
then assume, up to conjugacy, that P and Q are both standard irreducible
parabolic subgroups.
Now we will need the following:
Claim: Let AS be an arbitrary Artin–Tits group. Let s0 , . . . , sk ∈ S be standard generators such that si does not commute with si+1 , and si 6= si+2 for
every i. If an element α ∈ A+
S is represented by a positive word w which contains the subsequence s0 s1 · · · sk , then all positive words representing α contain
the same subsequence.
Proof of the claim: It is known [18] that Artin monoids inject in their groups.
This implies that every positive word representing α is obtained from w after
a finite sequence transformations, each one replacing a subword sts · · · (having m(s, t) letters) with tst · · · (also having m(s, t) letters). It suffices to show
that the word obtained from w after a single transformation contains the subsequence s0 · · · sk . If m(s, t) = 2, the transformation replaces st with ts. But
the subword st can intersect the subsequence s0 · · · sk in at most one letter
(as si and si+1 do not commute for every i), hence the subsequence survives
after the transformation. If m(s, t) ≥ 3 then the subword sts · · · intersects the
subsequence s0 · · · sk of w in at most two consecutive letters (as si 6= si+2 for
every i). This intersection is either (si , si+1 ) = (s, t) or (si , si+1 ) = (t, s). In either case, the subsequence survives after the transformation, as tst · · · contains
both possible subsequences. This shows the claim.
Recall that we are assuming that P and Q are distinct nonempty proper standard parabolic subgroups of AS , so P = AX and Q = AY , where X, Y ( S
and ΓX and ΓY are connected graphs. We are also assuming that zP and zQ
(that is, zX and zY ) commute. We will further assume that none of the three
conditions in the statement holds, and we will arrive at a contradiction.
If X ∩ Y = ∅, Condition 3 not being satisfied implies the existence of a ∈ X and
b ∈ Y that are adjacent in ΓS .
Otherwise, as Condition 1 is not satisfied and ΓX is connected, there exist
a ∈ X \ Y and s1 ∈ X ∩ Y that are adjacent in ΓS . Moreover, as Condition 2
is not satisfied and ΓY is connected, there are b ∈ Y \ X and a simple path
s1 , s2 , . . . , sk = b in ΓY .
In either case, we have a path a = s0 , s1 , . . . , sk = b satisfying the hypothesis of
37

the above claim, where a ∈ X \ Y , b ∈ Y \ X, and s1 , . . . , sk ∈ Y .
Now consider the element zP zQ , which is equal to zX zY . It is a positive element,
and any representative of zX involves the letter a. On the other hand, let us
denote Ai = {s1 , . . . , si } for i = 1, . . . , k. Then ∆Ak 4 ∆Y 4 zY , and we have
a decomposition
∆Ak = ∆A1 rA1 ,s2 rA2 ,s3 · · · rAk−1 ,sk
where the product of the i leftmost factors is precisely ∆Ai , for i = 1, . . . , k.
Now notice that s1 = ∆A1 and recall that si is the first letter of rAi−1 ,si (by
Lemma 4.4) for i = 2, . . . , k. Therefore, the sequence s0 , . . . , sk is a subsequence
of zX zY .
From the above claim, it follows that every positive word representing zX zY
must contain s0 , . . . , sk as a subsequence. Now choose a representative of zY zX
which is the concatenation of a word representing zY and a word representing zX . In such a representative, each instance of the letter b appears to the left
of each instance of the letter a. Therefore, this word does not contain s0 , . . . , sk
as a subsequence. Hence zX zY 6= zY zX , that is, zP zQ 6= zQ zP . The latter is a
contradiction which finishes the proof.

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Marı́a Cumplido.
maria.cumplidocabello@univ-rennes1.fr
Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes (France).
& Depto. de Álgebra. Instituto de Matemáticas (IMUS).
Universidad de Sevilla. Av. Reina Mercedes s/n, 41012 Sevilla (Spain).
Volker Gebhardt.
v.gebhardt@westernsydney.edu.au
Western Sydney University
Centre for Research in Mathematics
Locked Bag 1797, Penrith NSW 2751, Australia
Juan González-Meneses.
meneses@us.es
Depto. de Álgebra. Instituto de Matemáticas (IMUS).
Universidad de Sevilla. Av. Reina Mercedes s/n, 41012 Sevilla (Spain).
39

Bert Wiest.
bertold.wiest@univ-rennes1.fr
Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes (France).

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