Automorphisms of partial endomorphism semigroups
By J. Ara ´ujo, V. H. Fernandes, M. M. Jesus, V. Maltcev and J. D. Mitchell
Abstract. In this paper we propose a general recipe for calculating the automor-phism groups of semigroups consisting of partial endomorautomor-phisms of relational structures with a singlem-ary relation for anym∈Nover a finite set.
We use this recipe to determine the automorphism groups of the following semi-groups: the full transformation semigroup, the partial transformation semigroup, and the symmetric inverse semigroup, and their wreath products, partial endomorphisms of partially ordered sets, the full spectrum of semigroups of partial mappings preserving or reversing a linear or circular order. We also determine the automorphism groups of the so-called Madhaven semigroups as an application of the methods developed herein.
1. Introduction and the Main Result
A number of works in the literature are dedicated to calculating the auto-morphism groups of certain transformation semigroups. In an earlier paper [3], a method for calculating automorphism groups of some such objects is given. In this paper we prove a more general result and use it to find the automorphism group of several well-known transformation semigroups. In order to state our main result we must recall some definitions and introduce some notation.
We assume throughout the paper thatΩis a finite set. We denote the semi-group of all partial mappings onΩunder composition of functions byPΩ, the semigroup of total mappings onΩbyTΩ, and the group of permutations onΩ bySΩ. IfΩ = {1,2, . . . , m}, then we abbreviatePΩ,TΩ, andSΩtoPm,Tm, and
Sm, respectively. IfU is a subsemigroup ofPΩ, then we letAut(U)denote the group of automorphisms ofU. The group ofinner automorphismsofU is defined
Mathematics Subject Classification: 20M10, 20M20.
as
Inn(U) ={φa :f 7→a−1f a | a∈SΩanda−1U a=U}.
The imageof f ∈ PΩ is denoted byim(f) and thedomainof f bydom(f). A mappingf ∈PΩis called aconstant with valueαifβf = αfor allβ ∈ dom(f). For the sake of convenience, we will assume that the empty mapping∅is also a constant.
Letm∈N. Then anm-ary relationρonΩis just a subset of
Ωm={(α1, α2, . . . , αm) | α1, α2, . . . , αm∈Ω}.
Ifρis anm-ary relation, then define
ρ′={(α1, α2, . . . , αm)∈ρ | αi6=αjifi6=j}.
Letρandσbem-ary relations onΩ. We say that a subsemigroupU ofPΩacts transitively fromρ′ toσ′ if for all(α1, . . . , αm) ∈ ρ′ and(β1, . . . , βm) ∈ σ′ there existsf ∈U with
(α1f, . . . , αmf) = (β1, . . . , βm).
If U is a group of permutations and ρ′ = σ′ = (Ωm)′, then our defini-tion of transitivity is just the usual definidefini-tion ofm-transitivity for permutation groups. If (α1, . . . , αm) ∈ Ωm andf ∈ U, then we denote(α1f, . . . , α
mf)by (α1, . . . , αm)f. The monoid ofpartial endomorphisms ofρis
PEnd(ρ) ={f ∈PΩ | (α1, . . . , αm)∈ρ∩(dom(f))mimplies(α1, . . . , αm)f ∈ρ}.
Ifρis a binary relation onΩ, then ananti-automorphismofρis an element
f ∈ SΩ such that (α, β) ∈ ρimplies(β, α)f ∈ ρ. Since Ωis finite, the set of automorphisms and anti-automorphisms ofρforms a group. The following def-inition can be thought of as a generalization of this notion to relations of higher arity. LetHbe a subgroup ofSm. Then we define
AutH(ρ) ={f ∈SΩ | (∀(α1, . . . , αm)∈ρ)(∃t∈H)((α1t, . . . , αmt)f ∈ρ)}.
Again sinceΩis finite, AutH(ρ)is a group. The group of automorphisms and anti-automorphisms of a binary relationρmentioned above is denotedAutS2(ρ)
using this notation. We also require the following definition:
N(ρ, H) ={(α1, . . . , αm)∈(Ωm)′ | (α1t, . . . , αmt)6∈ρfor allt∈H}.
Theorem 1.1. Letρbe anm-ary relation on a finite setΩfor somem ∈N,
letH be a subgroup ofSm, and letUis a subsemigroup ofPEnd(ρ)such that (i) U contains a constant idempotent with valueαfor allα∈Ω;
(ii) U acts transitively fromN(ρ, H)∪ρ′toρ′.
ThenAut(U) ={φa:f 7→a−1f a | a∈AutH(ρ′)anda−1U a=U }. Moreover,
ifa, b∈AutH(ρ′)such thata6=bandφ
a, φb∈Aut(U), thenφa6=φb.
We prove Theorem 1.1 in Section 2. In Section 3, we derive several corol-laries of Theorem 1.1, and discuss whether it is possible to weaken its hypoth-esis and still obtain its conclusion. In Sections 4 and 5, we use the main theo-rem and its corollaries to determine the automorphism groups of the following semigroups: the full transformation semigroup, the partial transformation semi-group, and the symmetric inverse semisemi-group, and their wreath products, the partial endomorphisms of finite partially ordered sets, the full spectrum of semi-groups of partial mappings preserving or reversing a linear or circular order. In Section 6 we determine the automorphism groups of the Madhaven semigroups as an application of the methods developed in the proof of Theorem 1.1.
2. Proof of Theorem 1.1
We prove Theorem 1.1 in the following sequence of lemmas, some of them belonging to the folklore of this topic and included here for the sake of complete-ness.
Lemma 2.1. LetV be a subsemigroup of constants inPΩsuch that for allα∈ Ωthere exists a constant idempotent inV with valueα. ThenAut(V) = Inn(V).
PROOF. The inclusionInn(V)≤Aut(V)is obvious.
Letφ∈Aut(V). If∅ ∈V, then clearly∅φ=∅. Iff ∈V \ {∅}, then denote by
αfthe unique element ofim(f). Letf, g∈V\ {∅}withαf =αg. By assumption, there exists an idempotent constanth∈V withαh=αf =αg. Thenf h=fand
gh=gand so(f h)φ=f φhφ6=∅. It follows that
αf φ=α(f h)φ=αf φhφ =αhφ=αgφhφ=α(gh)φ=αgφ.
Henceαf φ =αgφ for allf, g ∈ V \ {∅}withαf = αg. Therefore the mapping
a: Ω−→Ωdefined byαf 7→αf φis a well-defined bijection, i.e. a∈SΩ.
Hence for allα∈dom(f φ) = dom(a−1f a)we have
(α)(f φ) =αf φ= (αf)a= (α)a−1f a.
Thusf φ=a−1f afor allf ∈V and soφ∈Inn(V).
Lemma 2.2. Let U be a subsemigroup ofPΩ such that for allα ∈ Ωthere exists a constant idempotent inUwith valueα. ThenAut(U) = Inn(U).
PROOF. LetV denote the set of all constants inU and letVα ={f ∈ V | im(f) ⊆ {α} } for everyα ∈ Ω. It is easy to checkVα is a minimal left ideal ofU andVαcontains an idempotent. Consequently,V =∪α∈ΩVαis a subsemi-group ofU. Conversely, letI⊆Ube a minimal left ideal such thatIcontains an idempotentf whereαf =αfor someα∈Ω.ThenVα=Vαf ⊆Iand the mini-mality ofIimpliesVα=I. As an automorphisms sends every minimal left ideal containing an idempotent into a minimal left ideal containing an idempotent, it follows thatV φ=V for allφ∈Aut(U).
Let φ ∈ Aut(U). Then we will prove that φ ∈ Inn(U). SinceAut(V) = Inn(V), by Lemma 2.1 and the preceding paragraph, it follows thatφ|V ∈Inn(V). As in the proof of Lemma 2.1, for allf ∈V\{∅}, we denote byαfthe unique element inim(f). Sinceφ|V ∈Aut(V) = Inn(V), there existsa∈SΩsuch that
f φ|V =a−1f aandαfa=αf φ=αf φ|V
for allf ∈V.
Let f ∈ V and g ∈ U be arbitrary. Then, as in the proof of Lemma 2.1,
αf ∈ dom(g)if and only if (αf)a = αf φ ∈ dom(gφ). It follows thatdom(gφ) equalsdom(a−1ga). Moreover,
(αfa)(gφ) = (αf φ)(gφ) =αf φgφ=α(f g)φ= (αf g)a= (αfg)a
and sogφ:αfa7→(αfg)afor allαf ∈dom(g). That is,gφ:α7→(α)a−1gafor all
α∈dom(gφ), as required. Finally,a−1U a=U follows by the assumption thatφ
is an automorphism. Thusφis an inner automorphism.
PROOF OFTHEOREM1.1. From Lemma 2.2, we have thatAut(U) = Inn(U).
It remains to prove thatInn(U) ={φa:f 7→a−1f a| a∈AutH(ρ′)anda−1U a=
U}.
Leta∈SΩ\AutH(ρ′). We will prove thata−1U a6=U. From the definition
Case 1.there existt∈H and(β1, . . . , βm)∈ρ′with(β1
t, . . . , βmt)a ∈ρ. Let (α1, . . . , αm) ∈ ρ′ such that(α1
t, . . . , αmt)a 6∈ ρ. SinceU is transitive fromρ′toρ′, there existsf ∈U such that(β1, . . . , βm)f= (α1, . . . , αm). Then we have(β1t, . . . , βmt)f = (α1t, . . . , αmt)and so
[(β1t, . . . , βmt)a]a
−1f a
= (β1t, . . . , βmt)f a= (α1t, . . . , αmt)a 6∈ρ.
Hencea−1f a6∈PEnd(ρ)and, in particular,a−1f a6∈U, as required.
Case 2.for allt∈Hand for all(β1, . . . , βm)∈ρ′we have(β1t, . . . , βmt)a 6∈ρ. By assumption,(β1, . . . , βm)a ∈N(ρ, H)for all(β1, . . . , βm)∈ ρ′. Lett∈H and(β1, . . . , βm)∈ρ′. Assume that(β1
t, . . . , βmt)a
−1
∈ρ′. Then
[(β1t, . . . , βmt)a
−1
]a= (β1
t, . . . , βmt)∈N(ρ, H).
Sincet−1 ∈ H, it follows that(β1, . . . , βm) 6∈ ρ′, a contradiction. Hence for all
t∈Hand for all(β1, . . . , βm)∈ρ′we have that(β1
t, . . . , βmt)a
−1
6∈ρ.
If(β1, . . . , βm)∈ρ′, then sinceUis transitive fromN(ρ, H)toρ′, there exists
f ∈U such that[(β1, . . . , βm)a]f = (β1, . . . , βm). Then
(β1, . . . , βm)af a−1
= (β1, . . . , βm)a−1
6∈ρ.
Thusaf a−16∈U and soa−1f a6∈U, as required.
Leta, b∈AutH(ρ′)such thata6=bandφ
a, φb∈Aut(U). It remains to prove thatφa 6=φb. But there existsα∈Ωsuch that(α)a6= (α)band soφaandφbdiffer
on any constant idempotent inUwith valueα.
3. Corollaries and Examples
Note that ifρin Theorem 1.1 satisfiesρ′ = ∅and U is a subsemigroup of PEnd(ρ)satisfying (i) in Theorem 1.1, then Theorem 1.1 offers no new informa-tion regarding Aut(U). That is, as there are no tuples of distinct elements in
ρ,U vacuously acts transitively fromN(ρ, H)∪ρ′ toρ′. Moreover, in this case AutH(ρ′) =SΩand so Theorem 1.1 reasserts that
Aut(U) ={φa :f 7→a−1f a | a∈SΩanda−1U a=U}= Inn(U).
Corollary 3.1. Letρbe anm-ary relation on a finite setΩfor somem ∈N,
letH a subgroup ofSm, and letU ∈ {PEnd(ρ),End(ρ),IEnd(ρ)}such that the
following hold
(i) U contains a constant idempotent with valueαfor allα∈Ω;
(ii) U acts transitively fromρ′∪N(ρ, H)toρ′;
(iii) AutH(ρ′) = AutH(ρ).
ThenAut(U)∼= AutH(ρ).
PROOF. We prove the corollary in the case thatU = End(ρ), the remaining
cases can be proved analogously.
It follows from Theorem 1.1 and (iii) in the hypothesis of the corollary that
Aut(U) ={φa :f 7→a−1f a | a∈AutH(ρ)anda−1U a=U }
and that distinct elements inAutH(ρ)induce distinct automorphisms ofU. Leta∈ AutH(ρ). We will prove thata−1U a =U. Letf ∈End(ρ)be
arbi-trary and let(α1, . . . , αm)∈ ρ. Sincea ∈AutH(ρ), there existst ∈H such that (α1t−1, . . . , αmt−1)
a−1
∈ρ. Hence, sincef ∈U,(α1t−1, . . . , αmt−1)
a−1f
∈ρand so (α1, . . . , αm)a−1f a
∈ρ.Hencea−1f a∈Uand soa−1U a=U, as required.
It follows thatAut(U) = {φa : f 7→ a−1f a | a ∈ AutH(ρ)}. LetF : AutH(ρ)−→ Aut(U)be defined by(a)F =φa. Then, since distinct elements in AutH(ρ)induce distinct automorphisms ofU andAut(U)is finite,F is a bijec-tion. It is straightforward to verify thatF is a homomorphism, and the corollary
follows.
Corollary 3.2. Letρbe a reflexive binary relation on a finite setΩand letH be a subgroup ofS2. ThenAutH(ρ′) = AutH(ρ).
PROOF. Sinceρ′ ⊆ ρ, it follows thatAutH(ρ) ≤ AutH(ρ′). To prove the converse, leta ∈AutH(ρ′)and let(α, β)∈ ρbe arbitrary. Ifα6=β, then either (α, β)a ∈ ρ′ ⊆ ρor(β, α)a ∈ ρ′ ⊆ ρ, as required. Ifα = β, then, since ρis reflexive, (α, β)a ∈ ρand(β, α)a ∈ ρ. Hencea ∈ AutH(ρ)and soAutH(ρ′) =
AutH(ρ).
Example 3.3. LetΩ ={1,2,3,4,5}, letH =S5, and let
ρ={(1,2,3,4,5),(2,1,3,4,5)} ∪ {(α1, α2, . . . , α5) | {α1, α2, . . . , α5}={3,4} }
∪ {(α, α, α, α, α) | α∈ {1, . . . ,5} }.
ThenN(ρ, H) =∅,Aut(ρ′) =S
{1,2} = Aut(ρ),S{1,2}×S{3,4,5}is a subgroup of AutH(ρ′), and every element ofAutH(ρ)stabilizes{3,4}setwise. In particular, AutH(ρ′)6= AutH(ρ).
LetU = Aut(ρ)∪ {f ∈ T5 | im(f) = {3,4}or|im(f)|= 1}. ThenU is a subsemigroup ofPEnd(ρ)and, sinceAut(ρ)is transitive fromρ′ toρ′,U is also transitive fromρ′toρ′. Let
f = 1 2 3 4 5
3 3 3 3 4 !
∈U.
Then(3,4,3,4,3)(4 5)f(4 5) = (3,5,3,5,3) 6∈ ρ. Hence(4 5)f(4 5)6∈ U and so the
element(4 5)∈AutH(ρ′)does not induce an inner automorphism ofU.
The following example shows that it is not true that ifa ∈ AutH(ρ′)and
a−1U a=U, thena∈AutH(ρ).
Example 3.4. Letρbe the relation from Example 3.3 and letV = Aut(ρ)∪ {f ∈T5 | fis constant}. Then, as in Example 3.3,V is transitive onρ′. However, (4 5)∈AutH(ρ′)and(4 5)V(4 5) =V but(4 5)6∈AutH(ρ), as required.
4. Applications I - Transformation semigroups
In this section we apply Theorem 1.1 to determine the automorphism groups of several well-known transformation semigroups, defined below. Some of the results contained in this section are well-known and included here only to il-lustrate how Theorem 1.1 can be used. Recall thatTΩ, PΩ, andIΩdenote the monoids of all total mappings, all partial mappings, and all partial injective map-pings of the finite setΩ, respectively. As above, ifΩ ={1,2, . . . , m}, then we may writeTm,Pm, orIminstead ofTΩ,PΩ, orIΩ, respectively.
Corollary 4.1. LetU ∈ {PΩ, TΩ, IΩ}whereΩis a finite set. ThenAut(U)∼=
PROOF. Letρ= Ω×Ωand letH≤S2be arbitrary. Then
PEnd(ρ) =PΩ,End(ρ) =TΩ, and IEnd(ρ) =IΩ.
In any case, U contains a constant idempotent with valueαfor allα ∈ Ωand so part (i) of the hypothesis of Corollary 3.1 is satisfied. Now,N(ρ, H) =∅and
ρ′ = (Ω×Ω)\ {(α, α) | α ∈ Ω}. It is clear thatU is transitive fromρ′ to
ρ′ and so part (ii) of Corollary 3.1 is satisfied. Finally, ρ is a reflexive binary relation and so by Corollary 3.2, AutH(ρ′) = AutH(ρ). Thus, by Corollary 3.1,
Aut(U)∼= AutH(ρ′) =SΩ, as required.
Next, we find the automorphisms of some wreath products of transforma-tion semigroups. IfS andT are semigroups acting or partially acting on setsΩ andΣ, then thewreath productofSandT denotedS≀T is the setS×TΩwhere TΩdenotes the set of mappings or partial mappings fromΩtoT, respectively,
with multiplication
(s, f)(t, g) = (st, f ⋆sg)
where (α)sg = (αs)g for allα ∈ Ω and where(α)f ⋆sg = (α)f ·(α)sg. The semigroupS≀T acts onΩ×Σas follows:(α, σ)(s,f)= (αs,(σ)(αf)). For further
details about wreath products see [15] or [18].
Corollary 4.2. LetU ∈ {Pm≀Pn, Tm≀Tn, Im≀In}. ThenAut(U)∼=Sm≀Sn.
PROOF. LetΩbe a set of sizemnand letρbe an equivalence relation with
m classes each of size n. ThenPEnd(ρ) = Pm≀Pn, End(ρ) = Tm≀Tn, and IEnd(ρ) = Im≀In(see [4, Lemma 2.1] for a proof). It is clear thatU contains a constant idempotent with valueαfor allα∈Ω.
Let H be the trivial subgroup ofS2. Then, sinceρis symmetric,N(ρ, H) contains all the pairs of distinct elements in Ω2\ρ′. We must prove thatU is transitive from the pairs of distinct entries inΩ2toρ′.
Let(α, β)∈Ω2and(γ, δ)∈ρsuch thatα6=βandγ6=δ. Then
f = α β
γ δ !
is an element ofIEnd(ρ), and so ofPEnd(ρ), such that(α, β)f = (γ, δ).Letg : Ω−→ Ωbe any total mapping with image{γ, δ}such thatαg =γandβg =δ. Theng∈End(ρ)and again(α, β)g= (γ, δ).
5. Applications II – Ordered sets
In this section, we consider the automorphisms groups of semigroups of order-preserving partial mappings of a partially ordered set, and some related semigroups.
Theorem 5.1. LetΩbe a finite set, letρbe partial order onΩ, and letU ∈ {PEnd(ρ),End(ρ),IEnd(ρ)}. ThenAut(U)is isomorphic to the group of auto-morphisms and anti-autoauto-morphisms of≤.
PROOF. Ifα, β ∈Ωsuch that(α, β)6∈ ρand(β, α)6∈ ρ, then we will write αkβ. Ananti-chainis any subsetΣofΩwhereαkβfor allα, β∈Σ.
Sinceρis reflexive,PEnd(ρ),End(ρ), andIEnd(ρ)contain a constant idem-potent with valueαfor allα∈Ω. IfH =S2, then
N(ρ, H) ={(α, β)∈Ω2 | αkβ}.
We will prove thatUis transitive fromN(ρ, H)∪ρ′toρ′.
If α, β, γ, δ ∈ Ωsuch that (β, α) 6∈ ρand (γ, δ) ∈ ρ, then the mapping f
from the proof of Corollary 4.2 is an element ofIEnd(ρ)andPEnd(ρ)such that (α, β)f = (γ, δ).
It remains to prove thatU is transitive fromN(ρ, H)∪ρ′ toρ′ whenU = End(ρ). LetΣbe a subset ofΩ. Then we define
Σ∧={β ∈Ω | (∀α∈Σ) (α, β)∈ρ′orβ kα}
and
Σ∨={β ∈Ω | (∀α∈Σ) (β, α)∈ρ′orβ kα}
IfΣis a maximal (with respect to containment) anti-chain inρ, then the setsΣ∧, Σ, andΣ∨are disjoint and their union is the whole ofΩ.
Letα, β, γ, δ ∈Ωsuch thatα6=β,(β, α)6∈ ρ, and(γ, δ)∈ρ′. If(α, β) ∈ρ, then sinceΩis finite, there exists a maximal antichainΣinρsuch thatα∈Σ. It follows thatβ ∈Σ∧. Letf : Ω−→Ωbe defined by
ǫf = (
γ ǫ∈Σ∪Σ∨
δ ǫ∈Σ∧.
If (α, β) 6∈ ρ, then as in the previous case, sinceΩis finite, there exists a maximal antichainΣinρsuch thatα, β∈Σ. Letf : Ω−→Ωbe defined by
ǫf = (
γ ǫ∈(Σ\ {β})∪Σ∨
δ ǫ∈Σ∧∪ {β}.
Thenf ∈End(ρ)and(α, β)f = (γ, δ).
In any case,U is transitive fromN(ρ, H)∪ρ′toρ′. Thus, by Corollaries 3.1
and 3.2,Aut(U)∼= AutS2(ρ), as required.
Ifρis the usual total order of{1, . . . , n}, thenEnd(ρ),PEnd(ρ), andIEnd(ρ) are usually denotedOn,POn,POIn(the semigroups of total, partial, and partial injective order-preserving mappings of the chain , respectively). These monoids have been extensively studied, for example see [1, 2, 8, 13, 14].
The following is an immediate corollary of Theorem 5.1.
Corollary 5.2. IfU ∈ {POn,On,POIn}, then
Aut(U) =hf 7→f(1n)(2n−1)···(⌊n/2⌋⌈n/2⌉+ 1)i ∼=C2
where⌊n/2⌋is the greatest integer less thann/2,⌈n/2⌉is the least integer greater thann/2andC2denotes the cyclic group of order2.
Let ODn, let PODn, and let PODIn be the monoids of all total, partial, and partial injective order-preserving and order-reversing mappings of the chain 1 ≤2 ≤ · · · ≤n, respectively. Again these monoids appear in several papers in the literature, for example see [9, 10, 11]. The automorphism groups of these monoids are given in the theorem below. However, the theorem is not a direct corollary of any of the preceding theorems, as these semigroups are not defined as the partial endomorphisms of a relation.
We require the following well-known combinatorial fact in order find the automorphism groups ofODn,PODn, andPODIn.
Lemma 5.3. Ifp, q ∈ N, then in any sequence of distinct natural numbers
of length pq+ 1, there exists a strictly increasing subsequence of lengthpor a strictly decreasing subsequence of lengthq.
PROOF. We prove the theorem in the case thatU = ODn, the other cases follow by an analogous argument.
Letρbe the ternary relation on{1, . . . , n}defined by
ρ={(α1, α2, α3) | α1≤α2≤α3orα1≥α2≥α3}.
ThenODn≤End(ρ).
It is straightforward to verify thatU is transitive fromρ′∪N(ρ, S3) = ρ′ to
ρ′, andUcontains a constant idempotent with valueαfor allα∈Ω. We will prove thatAut(U)∼= Aut(ρ)and
Aut(ρ) =h(1n)(2n−1)· · ·(⌊n/2⌋ ⌈n/2⌉+ 1)i ∼=C2.
We begin by showing thatAutS3(ρ
′) = AutS
3(ρ). SinceAutS3(ρ) ≤ AutS3(ρ
′)
always holds, it suffices to show thatAutS3(ρ
′)≤AutS
3(ρ).
Leta∈AutS3(ρ
′)and let(α1, α2, α3)∈ρ. Then there exists(β1, β2, β3)∈ρ′
such that{α1, α2, α3} ⊆ {β1, β2, β3}. Sincea∈AutS3(ρ
′), there existst∈S3such
that(β1t, β2t, β3t)a∈ρ. Hence(α1t, α2t, α3t)a∈ρand soa∈AutS3(ρ).
It follows from Corollary 3.1 thatAut(U) ∼= AutS3(ρ). The next step is to
prove thatAutS3(ρ) = Aut(ρ).
Clearly,Aut(ρ)≤AutS3(ρ). To prove the converse inclusion, take arbitrary
f ∈ AutS3(ρ). Then there exists t ∈ S3 such that (α1t, α2t, α3t)
f ∈ ρ for all
(α1, α2, α3) ∈ ρ. By Lemma 5.3, the sequence1f,2f, . . . ,10f contains a subse-quence α1f, α2f, α3f whereα1 < α2 < α3 andα1f < α2f < α3forα1f > α2f > α3f.Hence either(α1, α2, α3)f ∈ρor(α3, α2, α1)f ∈ρ. This implies that
tis the identity permutation id ort= (1 3). In either case, by the definition ofρ,
f ∈Aut(ρ). ThusAut(ρ) = AutS3(ρ).
The final step in the proof is to show that
Aut(ρ) =h(1n)(2n−1)· · ·(⌊n/2⌋ ⌈n/2⌉+ 1)i.
Letf ∈Aut(ρ). We start by proving that1f ∈ {1, n}. Assume the contrary, that is,1f 6∈ {1, n}. Ifnf 6= 1, then(1,1f−1, n)∈ρbut(1,1f−1, n)f = (1f,1, nf)∈/ρ, a contradiction. Therefore nf = 1. But (1, nf−1, n) ∈ ρand (1, nf−1, n)f = (1f, n, nf) = (1f, n,1) ∈/ ρsince1f 6= n, a contradiction. Hence we have that 1f ∈ {1, n}.
If1f = 1andα, β ∈ {1, . . . , n}such that α < β, then(1, α, β) ∈ ρand so (1, αf, βf)∈ρ. Henceαf < βfand sof is the identity on{1,2, . . . , n}.
A finite sequence(α1, α2, . . . , αm)of natural numbers is calledcyclicif there existsk ≥ 1 such thatα1gk ≤ α2gk ≤ · · · ≤ αmgk, whereg = (1 2· · ·m). A
partial mapping with domain and range contained in{1,2, . . . , n}is orientation-preservingif the image of every cyclic sequence indom(f)is a cyclic sequence.
Let OPn, let POPn, and letPOPIn be the monoids of all total, partial, and partial injective orientation-preserving mappings, respectively, of the set {1,2, . . . , n}. Some references from the literature concerning these monoids are [6, 7, 9, 10, 12, 17].
Theorem 5.5. Let U ∈ {POPn,OPn,POPIn}. ThenAut(U) = hf 7→
f(1 2···n), f 7→ f(1n)(2n−1)···(⌊n/2⌋ ⌈n/2⌉+1)i ∼= D2
n where D2n denotes the
di-hedral group with2nelements.
PROOF. We prove the theorem only in the case thatU = OPn, the other cases follow by analogous arguments. Letρdenote the set of all cyclic sequences of length3 over{1,2, . . . , n} and letH = h(1 3)i ≤ S3. ThenOPn ≤End(ρ) and N(ρ, H) = ∅. We will prove thatOPn acts transitively fromρ′ toρ′. Let (i, j, k),(i′, j′, k′)∈ρ′be arbitrary. Then eitheri < j < k,j < k < iork < i < j. Ifi < j < k, then define
f= 1 · · · i−1 i i+ 1 · · · j−1 j j+ 1 · · · n
i′ · · · i′ i′ j′ · · · j′ j′ k′ · · · k′
! .
Ifj < k < i, then define
f= 1 · · · j−1 j j+ 1 · · · k−1 k k+ 1 · · · n
j′ · · · j′ j′ k′ · · · k′ k′ i′ · · · i′
! .
Ifk < i < j, then define
f= 1 · · · k−1 k k+ 1 · · · i−1 i i+ 1 · · · n
k′ · · · k′ k′ i′ · · · i′ i′ j′ · · · j′
! .
In any case,f ∈ OPnand(i, j, k)f= (i′, j′, k′).
Next, we verify thatAutH(ρ′) = AutH(ρ). As always, we haveAutH(ρ)≤ AutH(ρ′). Letf ∈AutH(ρ′)and let(α1, α2, α3)∈ρbe arbitrary. Then there exists
Therefore, by Corollary 3.1,Aut(OPn)∼= AutH(ρ)and it remains to prove that AutH(ρ) = h(1 2 · · ·n),(1n)(2n−1)· · ·(⌊n/2⌋ ⌈n/2⌉+ 1)i. First of all (1 2 · · ·n)and(1n)(2n−1)· · ·(⌊n/2⌋ ⌈n/2⌉+ 1)are inAutH(ρ).
Letf ∈AutH(ρ). Then there existst∈ {id,(1 3)}such that(α1, α2, α3)∈ρ
implies (α1t, α2t, α3t)f ∈ ρ. By postmultiplying by the appropriate power of (1 2 · · ·n), if necessary, we may assume that1f = 1.
Lett =id. Then for anyαandβwith1 < α < β≤nwe have(1, α, β)∈ρ
and so (1, αf, βf) ∈ ρ which implies thatαf < βf. Hence f is the identity automorphism fromAutH(ρ).
Lett= (1 3). Then for anyαandβwith1< α < β≤nwe have(1, α, β)∈ρ
and so(βf, αf,1) ∈ ρwhich implies thatβf < αf. Thereforef = (1n)(2n− 1)· · ·(⌊n/2⌋ ⌈n/2⌉+ 1)∈AutH(ρ)and the proof is concluded.
A finite sequence(α1, α2, . . . , αm)of natural numbers is calledanti-cyclicif there existsk≥1such thatα1gk ≥α2gk ≥ · · · ≥αmgk, whereg= (1 2 · · ·m). A
partial mapping with domain and range contained in{1,2, . . . , n}is orientation-reversing if the image of every cyclic sequence in dom(f) is an anti-cyclic se-quence.
LetORn, letPORn, and letPORInbe the monoids of all total, partial, and partial injective orientation-preserving and reversing mappings, respectively, of the set{1,2,· · ·, n}. For further details of the known results concerning these monoids see [5, 9, 10, 12].
Theorem 5.6. Letn≥17andU ∈ {PORn,ORn,PORIn}. ThenAut(U) = hf 7→f(1 2···n), f7→f(1n)(2n−1)···(⌊n/2⌋ ⌈n/2⌉+1)i ∼=D2
n.
PROOF. We prove the theorem in the case thatU = ORn, the remaining cases can be proved analogously. Letρbe the set of cyclic sequencers of length 3 and letσbe the set of quadruples−→α = (α1, α2, α3, α4)such that there exists
t∈ {id,(1 3)}and for all subsequences(β1, β2, β3)of−→α we have that
(β1t, β2t, β3t)∈ρ.
Now we will prove thatORn ≤End(σ). Letf ∈ ORn and(α1, α2, α3, α4)∈σ. Sincef ∈ ORn, there existss ∈ {id,(1 3)}such that(γ1s, γ2s, γ3s)f ∈ ρfor all (γ1, γ2, γ3)∈ρ. By definition ofσ, there existst∈ {id,(1 3)}such that for all sub-sequences(β1, β2, β3)of(α1, α2, α3, α4)we have that(β1t, β2t, β3t)∈ρ. Therefore by definition ofσ, we have that(α1, α2, α3, α4)f lies inσwith the correspondent permutationst∈ {id,(1 3)}.
cyclic and anti-cyclic sequences of length 4 lie in σ. Let (α1, α2, α3, α4) ∈ σ′. Let the correspondent permutation tfor this sequence be id. Then any subse-quence(β1, β2, β3)from(α1, α2, α3, α4)is cyclic. This means that ifαiis the min-imum among α1,α2,α3 andα4, then since(αi, αi+1, αi+2)and(αi, αi+1, αi+3)
are cyclic, we haveαi< αi+1< αi+2< αi+3(here we worked modulo4).
There-fore(α1, α2, α3, α4)is cyclic. Analogously, it can be proved that in the case when
t= (1 3),(α1, α2, α3, α4)must be anti-cyclic.
LetH =S4. ThenN(σ, H) =∅. Now we will prove thatU acts transitively fromσ′toσ′. Let(α1, α2, α3, α4)and(β1, β2, β3, β4)be any fromσ′, i.e. any cyclic or anti-cyclic sequences of length4each of4distinct elements. There existp, q≥ 1such thatα1gp < α2gp < α3gp < α4gpandβ1gq < β2gq < β3gq < β4gq, where
g= (1 2 · · ·n). There exists an order-preserving functionf ∈ Onwhich extends
αigp 7→ βigq for alli ∈ {1,2,3,4}. Then(α1, α2, α3, α4)g pf g−q
= (β1, β2, β3, β4) and sincef, g∈ ORn,gpf g−q ∈ ORn.
Next, we will prove thatAutH(σ′) = AutH(σ) = Aut(σ).
The first step is to show thatAutH(σ′) = Aut(σ′). Letf ∈AutH(σ′). Then there existst∈S4such that(α1t, α2t, α3t, α4t)f ∈σ′for all(α1, α2, α3, α4)∈σ′. By Lemma 5.3, the sequence1f, . . . ,17f contains a subsequenceβ1f,β2f,β3f,
β4f where eitherβ1f < β2f < β3f < β4f orβ1f > β2f > β3f > β4f. Therefore
t ∈ {id,(1 4)(2 3)}. Since σ′ consists of all cyclic and ant-cyclic sequences of length4 of distinct4 elements, we have thatf ∈ Aut(σ′). HenceAutH(σ′) ≤ Aut(σ′). As observed above the converse inequality always holds, and so this step is concluded. ThatAutH(σ) = Aut(σ)follows by a similar argument.
Next, we prove that Aut(σ′) = Aut(σ). Again it suffices to prove that Aut(σ′) ≤ Aut(σ). So, let f ∈ Aut(σ′) and −→α = (α1, α2, α3, α4) ∈ σ. If |{α1, α2, α3, α4}|= 4, then−→αf ∈ σsincef ∈ Aut(σ′). If|{α1, α2, α3, α4}|<3, then−→αf ∈σ, asσcontains all quadruples containing at most 2 distinct elements. Let, finally, |{α1, α2, α3, α4}| = 3. Among all4 subsequences of−→α of length3 there will be two subsequences of distinct elements and these two subsequences must be equal. If this common sequence of distinct elements is(β1, β2, β3)then it is cyclic or anti-cyclic. Then there exists a cyclic or anti-cyclic sequence−→β with 4 distinct elements−→β with subsequence (β1, β2, β3). As we showed earlier, it follows that −→β ∈ σ′. Hence by definition(β1, β2, β3)f ∈ ρor(β3, β2, β1)f ∈ ρ. Let(γ1, γ2, γ3)be any of the2remaining subsequences of−→α of length3. Then |{γ1, γ2, γ3}| ≤ 2and so(γ1, γ2, γ3)f ∈ ρand(γ3, γ2, γ1)f ∈ ρ. Therefore again −
→αf ∈σ. Thus, we proved thatf ∈Aut(σ), as required.
The inclusionh(1 2· · ·n),(1n)(2n−1)· · · i ≤ Aut(σ)is obvious. Letf ∈ Aut(σ). By post multiplying by a power of(1 2· · ·n), if necessary, we may as-sume that1f = 1. We will prove thatf ∈ {id,(2n)(3n−1)· · · }, this will com-plete the proof.
First we prove that2f ∈ {2, n}. Assume that this is not the case. Then either
nf 6=nornf =n. If the former is true, then(1,2, nf−1, n)∈σ. But
(1,2, nf−1, n)f = (1,2f, n, nf)∈/ σ,
a contradiction. In the latter case,(1,2,2f−1, n)∈σbut
(1,2,2f−1, n)f = (1,2f,2, n)∈/ σ,
a contradiction.
Hence2f ∈ {2, n}and by symmetrynf ∈ {2, n}. By post multiplying by (2n)(3n−1)· · ·, if necessary, we deduce that 2f = nand nf = 2. Now, if 2 < α < β < n, then(2, α, β, n)f = (n, αf, βf,2) ∈ σand soαf > βf. This implies thatf = (2n)(3n−1)· · · concluding the proof.
6. Further remarks
So far we applied our main result, Theorem 1.1, to describe the automor-phisms of semigroups satisfying the conditions of the theorem. In this section we provide an example of a transformation semigroup which does not satisfy the conditions of Theorem 1.1, but still, using the developed machinery, we are able to calculate its automorphism group. This semigroup was first introduced by Madhaven [16] and is constructed as follows. Letρbe an equivalence relation on a finite setΩ. A partial mappingf ∈ PEnd(ρ)is named aρ-mappingif the following hold:
(1) (x, y)∈ρimpliesxf =yf for allx, y∈dom(f); (2) (xf, yf)∈ρimplies(x, y)∈ρfor allx, y∈dom(f);
(3) ifxρdenotes the equivalence class ofxinρ, thenxρ∩dom(f)6=∅implies
xρ⊆dom(f)for allx∈Ω.
The collection of all ρ-mappings forms a semigroup denoted by Pρ(Ω). This semigroup is regular and satisfies the law
Moreover, the semigroups Pρ(Ω) play a role analogous to the role of SΩ for groups, for the class of regular semigroups which satisfy this law.
ObviouslyPρ(Ω)≤PEnd(ρ)satisfies condition (i) of Theorem 1.1. However condition (ii) is not satisfied. In fact, for any(α, β)∈ρ′there are no(δ, γ)∈Ω×Ω and f ∈ Pρ(Ω) with (δ, γ)f = (α, β). Thus we cannot apply Theorem 1.1 to determine the automorphism group ofPρ(Ω). Nonetheless, we can determine the automorphism group ofPρ(Ω).
Theorem 6.1. Letρbe an equivalence relation on a finite setΩand letPρ(Ω)
be the Madhaven semigroup ofρ. ThenAut(Pρ(Ω)∼= Aut(ρ).
PROOF. By Lemma 2.2, it follows that
Aut(Pρ(Ω)) ={φa :f 7→a−1f a | a∈SΩanda−1Pρ(Ω)a=Pρ(Ω)}.
We prove that the permutationsa∈SΩsuch thata−1Pρ(Ω)a=Pρ(Ω)are exactly
the elements ofAut(ρ). It is straightforward to verify that iff ∈ Aut(ρ), then
f−1Pρ(Ω)f =Pρ(Ω). Letf ∈SΩ\Aut(ρ)and suppose thatf−1Pρ(Ω)f =Pρ(Ω).
Then there exists(α, β)∈ρsuch that(α, β)f∈/ ρ.
Letgbe a partial mapping which sendsαρtoαf and wheredom(g) = αρ. Theng ∈ Pρ(Ω) and sof−1gf ∈ Pρ(Ω). But(αf)f−1gf = (αg)f = αf2 and
(βf)f−1gf = (βg)f =αf2, a contradiction since(αf, βf)∈/ ρ.
We have shown thatAut(Pρ(Ω)) ={ψa :f 7→a−1f a | a∈Aut(ρ)}.Since
Pρ(Ω)contains a constant idempotent with valueαfor allα∈Ω, it follows, by the arguments at the end of the proofs of Theorem 1.1 and Corollary 3.1, that
Aut(Pρ(Ω))∼= Aut(ρ).
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DEPARTAMENTO DE MATEMTICA DA UNIVERSIDADE ABERTA RUA DA ESCOLA POLIT ´ECNICA, 147, 1269-001 LISBOA, PORTUGAL
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CENTRO DE LGEBRA DA UNIVERSIDADE DE LISBOA AV. PROF. GAMA PINTO, 2, 1649-003 LISBOA, PORTUGAL
E-mail: [email protected]
DEPARTAMENTO DE MATEMTICA DA FACULDADE DE CINCIAS E TECNOLOGIA DA UNIVERSI-DADE NOVA DE LISBOA QUINTA DA TORRE, 2829-516 CAPARICA, PORTUGAL
E-mail: [email protected]
MATHEMATICAL INSTITUTE, ST ANDREWS UNIVERSITY, NORTH HAUGH, ST ANDREWS, FIFE, UK, KY16 9SS
E-mail: [email protected]
MATHEMATICAL INSTITUTE, ST ANDREWS UNIVERSITY, NORTH HAUGH, ST ANDREWS, FIFE, UK, KY16 9SS
