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Journal of Algebra
www.elsevier.com/locate/jalgebra
Conjugacy in inverse semigroups
João Araújoa,b,∗, Michael Kinyonc,b, Janusz Koniecznyd
a DepartamentodeMatemática,CentrodeMatemáticaeAplicaçoes(CMA), FaculdadedeCiênciaseTecnologia,UniversidadeNovadeLisboa,2829–516 Caparica,Portugal
b CEMAT-CIÊNCIASUniversidadedeLisboa,Portugal
cDepartmentofMathematics,UniversityofDenver,Denver,CO80208,USA
d DepartmentofMathematics,UniversityofMaryWashington,Fredericksburg, VA 22401,USA
a r t i c l e i n f o a bs t r a c t
Article history:
Received6October2018 Availableonline31May2019 CommunicatedbyVolodymyr Mazorchuk
MSC:
20M10 20M18 20M05 20M20
Keywords:
Inversesemigroups Conjugacy
Symmetricinversesemigroups Freeinversesemigroups McAllisterP-semigroups Factorizableinversemonoids Cliffordsemigroups Bicyclicmonoid
Stableinversesemigroups
InagroupG,elementsaandb areconjugate ifthereexists g ∈Gsuch that g−1ag=b.Thisconjugacyrelation,which plays animportantroleingrouptheory,canbeextendedin anaturalwaytoinversesemigroups:forelementsaandbin an inversesemigroup S, ais conjugate to b,which wewill writeasa∼ib,ifthereexistsg∈S1suchthatg−1ag=band gbg−1=a.Thepurposeofthispaperistostudytheconjugacy
∼iinseveralclassesofinversesemigroups:symmetricinverse semigroups, McAllister P-semigroups, factorizable inverse monoids, Clifford semigroups, the bicyclic monoid, stable inversesemigroups,andfreeinversesemigroups.
©2019PublishedbyElsevierInc.
* Correspondingauthor.
E-mailaddresses:jj.araujo@fct.unl.pt(J. Araújo),mkinyon@du.edu(M. Kinyon),jkoniecz@umw.edu (J. Konieczny).
https://doi.org/10.1016/j.jalgebra.2019.05.022 0021-8693/©2019PublishedbyElsevierInc.
1. Introduction
Theconjugacyrelation ∼G inagroupGisdefinedas follows: fora,b∈G, a∼Gb if thereexists g∈Gsuchthatg−1ag=b andb=gag−1.
A semigroup S is said to be inverse if for every x ∈ S, there exists exactly one x−1∈S suchthatx=xx−1xandx−1=x−1xx−1.Thus onecanextendthedefinition ofgroupconjugacyverbatimtoinversesemigroups.(AsusualS1denotesthesemigroup S extendedbyanidentity elementwhennoneispresent.)
Definition 1.1.Let S be an inverse semigroup. Elements a,b ∈ S are said to be i-conjugate,denoted a∼ib, ifthere exists g∈ S1 suchthatg−1ag =b and gbg−1 =a.
Inshort,
a∼ib ⇐⇒ ∃g∈S1 (g−1ag=b and gbg−1=a). (1.1) Wecalltherelation∼i i-conjugacy(“i” for“inverse”).
Atfirst glance,thisnotionofconjugacyforinversesemigroupsseemssimultaneously bothnaturaland naive:naturalbecauseitisan obviousway toextend∼G formallyto inversesemigroups,andnaivebecauseonemightnotinitiallyexpectaformalextension to exhibit much structure. But surprisingly, it turns out thatthis conjugacy coincides with onethat Mark Sapir considered the best notionfor inverse semigroups. The aim of this paper is to carry outan in depthstudy of ∼i, and to show that its naturality goesfarbeyonditsdefinition;∼i is,infact,ashintedbySapir,ahighlystructuredand interestingnotionof conjugacy.
Ourfirstthreeresults(Section2)generalizetheknowntheoremsforpermutationson aset X to partialinjectivetransformationsonX.
(1) Elementsαand β of thesymmetric inversesemigroupI(X) are i-conjugateif and onlyiftheyhavethesamecycle-chain-raytype(Theorem2.10).
(2) IfX is afinite set with n elements,then I(X) has n
r=0p(r)p(n−r) i-conjugacy classes(Theorem2.16).
(3) IfX isaninfiniteset with|X|=ℵε, thenI(X) hasκℵ0 i-conjugacyclasses,where κ=ℵ0+|ε|(Theorem 2.18).
Thefollowingresults(Sections3,4,5and6)concernotherimportantclassesofinverse semigroups.
(4) If S = P(G,X,Y) is a McAlister P-semigroup, then for all (A,g),(B,h) ∈ S, (A,g)∼i(B,h) if and only if there exists (C,k) ∈ S such that (i) A = kB = C∧gC∧Aand (ii)g=khk−1 (Theorem3.1).
(5) IfS is aninverse semigroup,then for alla,b ∈S, theset of all g ∈S1 such that g−1ag = b and gbg−1 = a is upward closed in the natural partial order on S1 (Theorem4.1).
(6) IfS isafactorizable inversemonoid,thentwo elementsare∼i-relatedifandonly iftheyareconjugateunderaunitelement.(Corollary4.2).
(7) IfGisagroup,thenforanyHa,Kbinthecosetmonoid CM(G),Ha∼iKb ifand only if there exists g ∈ G such that g−1Hag = Kb and gKbg−1 = Ha (Corol- lary4.4).
(8) Elementsaand b inaCliffordsemigroup S are i-conjugateif andonly ifaand b aregroupconjugateinsomesubgroup ofS (Theorem5.7).
(9) IfS is aninverse semigroup,then S isa Cliffordsemigroupif and onlyif notwo differentidempotentsofS arei-conjugate(Theorem5.8).
(10) InthebicyclicmonoidB,i-conjugacycoincideswiththeminimalgroupcongruence (Theorem6.1).
(11) AninversesemigroupS isstableifandonlyifi-conjugacyandthenaturalpartial orderintersecttriviallyas relationsinS×S (Theorem6.3).
Thenextfourresults(Section7)concernfreeinversesemigroups.
(12) Elements wτ and vτ of the free inverse semigroup FI(X) are i-conjugate if and only if their birooted word trees (Tw,αw,βw) and (Tv,αv,βv) are equaland w(Π(αw,αv))=w(Π(βw,βv)) (Theorem7.9).
(13) Every i-conjugacyclassinFI(X) isfinite(Corollary7.7).
(14) Thenumberofelementsofi-conjugacyclassofwτ∈ FI(X) isequaltothenumber ofverticesofacertainwordsubtreeof(Tw,αw,βw) (Corollary7.11).
(15) The relation∼i inFI(X) isdecidable(Corollary 7.15).
We givenow aquickoverview ofthestate oftheart regardingnotionsof conjugacy forsemigroups.Asrecalledabove,theconjugacyrelation∼GinagroupGisdefinedby a∼Gb ⇐⇒ ∃g∈G (g−1ag=b and gbg−1=a). (1.2) (Infact,∼G istraditionallydefinedlesssymmetrically,butthesymmetricform of(1.2) follows since g−1ag=b if and onlyif gbg−1 =a.) This definitiondoes notmake sense in generalsemigroups,so conjugacy hasbeen generalizedto semigroupsinavariety of ways.
For a monoid S with identity element 1, let U(S) denote its group of units. Unit conjugacy inS ismodeledon∼G ingroupsby
a∼ub ⇐⇒ ∃g∈U(S)(g−1ag=b and gbg−1=a). (1.3) (Wewillwrite“∼”withvarioussubscripts forpossible definitionsofconjugacyinsemi- groups.Inthiscase,thesubscriptustandsfor“unit.”)See,forinstance,[14,15].However,
∼udoesnotmakesenseinasemigroupwithoutanidentityelement.Requiringg∈U(S1) inunit conjugacy does nothelp forarbitrary semigroups becauseU(S1)= {1}if S is notamonoid.
ConjugacyinagroupGcan,ofcourse,berewrittenwithoutusinginverses:a∼Gb∈G areconjugateifandonlyifthereexistsg∈Gsuchthatag=gb.Usingthisformulation, left conjugacy∼lhasbeendefinedforasemigroupS [24,33,34]:
a∼lb ⇐⇒ ∃g∈S1 ag=gb. (1.4)
InageneralsemigroupS,therelation∼lisreflexiveandtransitive,butnotsymmetric.
Inaddition, ifS hasazero,then ∼l isthe universalrelation S×S,so ∼l is notuseful forsuchsemigroups.
The relation ∼l, however, is an equivalence on any free semigroup. Lallement [16]
definedtwoelements ofafreesemigrouptobe conjugateiftheyarerelatedby∼l,and thenshowedthat∼l isequalto thefollowing relationinafreesemigroupS:
a∼pb ⇐⇒ ∃u,v∈S1 (a=uv and b=vu). (1.5) Ina generalsemigroup S, ∼p = ∼l and infact, therelation ∼p is reflexiveand sym- metric,but notnecessarily transitive.Kudryavtseva and Mazorchuk[14,15] considered thetransitiveclosure∼∗p of∼p asaconjugacyrelationinageneralsemigroup.(Seealso [9].)
Otto[24] studiedtherelations∼l and∼p inthemonoidsS presented byfiniteThue systems, and then symmetrized ∼l to give yet another definition of conjugacy insuch anS:
a∼ob ⇐⇒ ∃g,h∈S1 (ag=gb and bh=ha). (1.6) Therelation∼o isanequivalence relationinanarbitrarysemigroupS,but,again,it is theuniversalrelationforany semigroupwith zero.
Thisdeficiencyof∼owasremediedin[2],wherethefollowingrelationwasdefinedon anarbitrarysemigroupS:
a∼cb ⇐⇒ ∃g∈P1(a) ∃h∈P1(b)(ag=gb and bh=ha), (1.7) where for a = 0, P(a) = {g ∈ S : ∀m∈S1(ma = 0 ⇒ (ma)g = 0)}, P(0) = {0}, and P1(a)=P(a)∪ {1}.(See[2,§2] for amotivation forthisdefinition.)Therelation∼c is anequivalenceonS,itdoesnotreducetoS×S ifS hasazero,and itisequalto∼o if S doesnothaveazero.
In2018,thethirdauthor [13] definedaconjugacy∼nonanysemigroupS by a∼nb ⇐⇒ ∃g,h∈S1 (ag=gb, bh=ha, hag=b, and gbh=a). (1.8)
The relation∼n isan equivalencerelation onany semigroupand it doesnotreduceto S×SifShasazero.Infact,itisthesmallestofallconjugaciesdefineduptothispoint forgeneralsemigroups.
Wepoint outthateachoftherelations(1.3)–(1.8) reduces togroupconjugacywhen S isagroup.However,assuming werequireconjugacyto beanequivalence relationon generalsemigroups,only∼∗p,∼o,∼c,and∼ncanprovidepossibledefinitionsofconjugacy.
Wehave
∼n ⊆ ∼∗p ⊆ ∼o and∼n ⊆ ∼c ⊆ ∼o,
and,withrespecttoinclusion,∼∗pand∼carenotcomparable[13,Prop. 2.3].Fordetailed comparisonandanalysis,invariousclassesofsemigroups,oftheconjugacies∼∗p,∼o,∼c, and alsotrace(character) conjugacy∼tr definedforepigroups,see[1].
A notionofconjugacyforinversesemigroupsequivalenttoour∼i hasappearedelse- where.Infact,partofourmotivationforthepresentstudywasaMathOverflowpost by Sapir[28], inwhichheclaimed thatthefollowingis thebestnotionof conjugacyin inversesemigroups:fora,binaninversesemigroupS,aisconjugatetob ifthereexists t∈S1 suchthat
t−1at=b, a·tt−1=tt−1·a=a, and b·t−1t=t−1t·b=b. (1.9) Sapir notesthatthisnotionofconjugacyisimplicitintheworkofYamamura [32].Itis easyto showthatSapir’srelationcoincideswith∼i (Proposition1.3).
Oftheconjugacies∼∗p,∼o,∼c,and∼ndefinedforanarbitrarysemigroupS,only∼n
reduces to ∼i ifS is an inversesemigroup [13, Thm. 2.6].Observe alsothat ifS is an inversemonoid,
∼u ⊆ ∼i. (1.10)
This inclusion is generally proper, but we will see that equality holds in factorizable inversemonoids(Corollary 4.2).
We conclude this introduction with three general results about i-conjugacy. In an inversesemigroupS,both ofthefollowingidentitieshold:forallx,y∈S,
(x−1)−1=x and (xy)−1=y−1x−1. From these,thefollowing iseasytosee.
Lemma 1.2.The conjugacy∼i isan equivalence relationinany inversesemigroup.
Proof. LetSbeaninversesemigroup.Then∼i isreflexive(since1∈S1)andsymmetric (since(g−1)−1=gforeveryg∈S).Foralla,b,c∈S,g,h∈S1,ifg−1ag=b,gbg−1=a, h−1bh=c, and hch−1 =b,then (gh)−1·a·gh=c and gh·c·(gh)−1=a. Thus∼i is also transitive. 2
ForaninversesemigroupS,theequivalenceclassofa∈S withrespectto∼i willbe calledthei-conjugacy classofaanddenotedby[a]∼i.
Thefollowingpropositionshows that(1.1) isequivalent toSapir’s formulation(1.9), andalsoshowsthespecificconnectionbetweentheconjugacies∼iand∼o.Foraninverse semigroupS,a,b∈S andg∈S1,weconsider thefollowingequations.
(i) g−1ag=b (ii) gbg−1=a (iii) ag=gb (iv) bg−1=g−1a
(v) a·gg−1=a (vi) gg−1·a=a (vii) b·g−1g=b (viii) g−1g·b=b
Proposition 1.3.Let S be an inverse semigroup. Fora,b∈S andg ∈S1, thefollowing setsof conditions areequivalent andeach setimplies allof(i)–(viii).
(a) {(i), (ii)}(that is,a∼ib);
(b) {(i), (v), (vi)}; (c) {(iii), (v), (viii)}; (d) {(ii), (vii), (viii)}; (e) {(iv), (vi), (vii)}.
Proof. (a) =⇒ (b): a·gg−1 = gbg−1gg−1 = gbg−1 = a and gg−1·a = gg−1gbg−1 = gbg−1=a.
(b) =⇒(c):ag=gg−1·ag=gband g−1g·b=g−1g·g−1ag=g−1ag=b.
(c) =⇒(a):g−1ag=g−1g·b=b andgbg−1=a·gg−1=a.
The cycle of implications (a) =⇒ (d) =⇒ (e) =⇒ (a) follows from the cycle al- ready proven by exchanging the roles of a and b and replacing g with g−1 (since (g−1)−1=g). 2
Finally, we characterize one of the two extreme cases for i-conjugacy on an inverse semigroup S, namely where ∼i is the universal relation S ×S. In Theorem 5.9 we willconsider the oppositeextreme,where ∼i isthe identityrelation (equality).Similar discussionsforothernotionsofconjugacycanbe foundin[1].
Foran inversesemigroupS, we denote byE(S) thesemilattice ofidempotents of S [10,p. 146].
Theorem1.4. LetS beaninverse semigroup.Then∼i istheuniversalrelationS×S if andonly ifS isasingleton.
Proof. Suppose ∼i is universal. For all e ∈ E(S) and g ∈ S, we have g−1eg ∈ E(S), andsoeveryelementofSisanidempotent,thatis,Sisasemilattice.Nowfore,f ∈S, let g ∈ S be given such that g−1eg = f and gf g−1 = e. Since g−1 = g, we have f =geg = egg= eg and so e =gf g =f gg =f g = (eg)g =eg =f. Therefore S has onlyoneelement.Theconverseis trivial. 2
2. Conjugacyinsymmetricinversesemigroups
For a nonempty set X (finite or infinite), denote by I(X) the symmetric inverse semigroup onX,thatis,thesemigroupof partialinjectivetransformations onX under composition. The semigroup I(X) is universalfor the class of inverse semigroups(see [25] and[10,Ch. 5])sinceeveryinversesemigroupcanbe embeddedinsomeI(X) [10, Thm. 5.1.7]. This is analogous to thefact thatevery groupcanbe embeddedin some symmetricgroupSym(X) ofpermutationsonasetX.ThesemigroupI(X) hasSym(X) asitsgroupofunitsandcontainsazero(theemptytransformation,whichwewilldenote by0).
In this section, we will describe i-conjugacy in I(X) and its ideals, and count the i-conjugacyclassesinI(X) forboth finiteandinfiniteX.
2.1. Cycle-chain-raydecomposition ofelements of I(X)
The cycle decomposition of a permutation can be extended to the cycle-chain-ray decompositionofapartialinjectivetransformation(see[12]).
We will write functions on the right and compose from left to right; that is, for f :A→B andg :B →C, we willwrite xf,rather thanf(x), andx(f g),ratherthan g(f(x)). Letα∈ I(X).We denote thedomain of αbydom(α) andthe image ofα by im(α).Theuniondom(α)∪im(α) willbecalledthespanofαanddenotedspan(α).We saythatαandβ inI(X) arecompletelydisjointifspan(α)∩span(β)=∅.Forx,y∈X, we writex→α y ifx∈dom(α) andxα=y.
Definition 2.1.Let M be a set of pairwise completely disjoint elements of I(X). The join of the elements of M, denoted
γ∈Mγ, is the element of I(X) whose domain is
γ∈Mdom(γ) andwhosevaluesaredefinedby x(γ∈M
γ) =xγ0,
where γ0 is the (unique) element of M such that x ∈ dom(γ0). If M = ∅, we define
γ∈Mγ to be 0 (the zeroinI(X)).If M={γ1,γ2,. . . ,γk}is finite,wemaywrite the
joinasγ1γ2 · · · γk.
Definition 2.2.Let . . . ,x−2,x−1,x0,x1,x2,. . . be pairwise distinct elements of X. The following elementsofI(X) willbe calledbasic partialinjectivetransformationsonX.
• A cycle of lengthk (k ≥1),written(x0x1. . . xk−1),is an element δ ∈ I(X) with dom(δ)={x0,x1,. . . ,xk−1},xiδ=xi+1 forall0≤i< k−1,andxk−1δ=x0.
• A chain of length k (k ≥ 1), written [x0x1. . . xk], is an element θ ∈ I(X) with dom(θ)={x0,x1,. . . ,xk−1}and xiθ=xi+1 forall0≤i≤k−1.
• A double ray, written . . . x−1x0x1. . ., is an element ω ∈ I(X) with dom(ω) = {. . . ,x−1,x0,x1,. . .}andxiω=xi+1 foralli.
• A right ray, written [x0x1x2. . ., is an element υ ∈ I(X) with dom(υ) = {x0,x1,x2,. . .}andxiυ=xi+1 foralli≥0.
• A left ray, written . . . x2x1x0], is an element λ ∈ I(X) with dom(λ) = {x1,x2,x3,. . .}andxiλ=xi−1 foralli>0.
Byaraywewill meanadouble,right,or leftray.
Wenotethefollowing.
• Thespanofabasicpartialinjectivetransformationisexhibitedbythenotation.For example,thespanoftherightray[123. . .is{1,2,3,. . .}.
• The left bracket in “η = [x. . .” indicates that x ∈/ im(η); while the right bracket in“η =. . . x]” indicatesthatx∈/dom(η).For example, forthe chainθ = [1234], dom(θ)={1,2,3}andim(θ)={2,3,4}.
• A cycle (x0x1. . . xk−1) differs from the corresponding cycle in the symmetric group of permutations on X in that the former is undefined for every x ∈ (X \ {x0,x1,. . . ,xk−1}),whilethelatterfixeseverysuchx.
Thefollowingdecompositionwasprovedin[12,Prop. 2.4].
Proposition 2.3.Let α∈ I(X) with α= 0. Thenthere exist unique sets:Δα of cycles, Θα of chains, Ωα of double rays, Υα of right rays, and Λα of left rays such that the transformationsin Δα∪Θα∪Ωα∪Υα∪Λα are pairwisecompletelydisjointand
α=δ
∈Δα
δθ
∈Θα
θω
∈Ωα
ωυ
∈Υα
υλ
∈Λα
λ. (2.1)
Wewill call thejoin(2.1) the cycle-chain-ray decomposition of α. If η∈Δα∪Θα∪ Ωα∪Υα∪Λα,we willsaythatη is contained inα(orthatαcontains η).Ifα= 0, we setΔα= Θα= Ωα= Υα= Λα=∅.Wenotethefollowing.
• If α ∈ Sym(X), then α =
δ∈Δαδ
ω∈Ωαω (since Θα = Υα = Λα =∅), which
corresponds totheusualcycledecompositionofapermutation[29,1.3.4].
• Ifdom(α)=X,thenα=
δ∈Δαδ
ω∈Ωαω
υ∈Υαυ(sinceΘα= Λα=∅),which
corresponds tothedecompositiongivenin[18].
• IfX isfinite,thenα=
δ∈Δαδ
θ∈Θαθ (sinceΩα= Υα= Λα=∅), whichisthe
decompositiongivenin[20,Thm. 3.2].
Forexample,ifX ={1,2,3,4,5,6,7,8,9},then α=
1 2 3 4 5 6 7 8 9
3 6 − 5 9 8 − 2 −
∈ I(X)
written incycle-chain decomposition (norays since X is finite) is α= (268)[13]
[459].The following β is an exampleof an element of I(Z) written incycle-chain-ray decomposition:
β = (2 4)[6 8 10] . . .−6 −4−2−1 −3−5. . . [1 5 9 13 . . . . . .15 11 7 3].
2.2. Characterizationof∼i in I(X)
Wewillnowcharacterize∼iinI(X) usingthecycle-chain-raydecompositionofpartial injectivetransformations.
Notation 2.4.We will fix an element ∈/ X. For α∈ I(X) and x ∈ X, we will write xα=ifandonlyifx∈/dom(α).Wewillalsoassumethatα=.Withthisnotation, it will make sense to write xα = yβ or xα = yβ (α,β ∈ I(X), x,y ∈ X) even when x∈/dom(α) or y /∈dom(β).
Lemma 2.5.Let α,β,τ ∈ I(X) and suppose τ−1ατ =β and τ βτ−1 =α.Then for all x,y∈X:
(1) span(α)⊆dom(τ) andspan(β)⊆im(τ);
(2) if x→α y thenxτ →β yτ;
(3) if x∈/dom(α)andx∈dom(τ),then xτ /∈dom(β);
(4) if x∈/im(α)andx∈dom(τ),thenxτ /∈im(β).
Proof. By Proposition 1.3, α = τ(τ−1α) and α = (ατ)τ−1. Thus dom(α) ⊆ dom(τ) and im(α)⊆im(τ−1)= dom(τ),andsospan(α)⊆dom(τ).Bytheforegoingargument, span(β)⊆dom(τ−1)= im(τ).Wehaveproved (1).
To prove(2), letx→α y. Since ατ =τ β (by Proposition1.3),(xτ)β = (xα)τ =yτ. Since yτ =by(1),itfollows thatxτ →β yτ.
To prove(3),letx∈/dom(α) and x∈dom(τ).Then (xτ)β = (xα)τ =τ =.Thus (xτ)β=,thatis, xτ /∈dom(β).
Toprove(4),letx∈/im(α) andx∈dom(τ).Supposetothecontrarythatxτ ∈im(β).
Then zβ=xτ forsomez∈X. ByProposition1.3,βτ−1=τ−1α. Thusx= (xτ)τ−1= (zβ)τ−1= (zτ−1)α,andsox∈im(α),whichisacontradiction.Hencexτ /∈im(β). 2 Definition 2.6.Let . . . ,x−1,x0,x1,. . . be pairwise distinct elements of X. Let δ = (x0. . . xk−1), θ = [x0x1. . . xk], ω = . . . x−1x0x1. . ., υ = [x0x1x2. . ., and λ = . . . x2x1x0]. Foranyη∈ {δ,θ,ω,υ,λ}andanyτ ∈ I(X) suchthatspan(η)⊆dom(τ), we defineητ∗ tobe η inwhich eachxi hasbeen replacedwithxiτ. Sinceτ is injective, ητ∗isacycleoflengthk[chainoflengthk,double ray,rightray,leftray]ifη isacycle of lengthk [chainof lengthk,doubleray,rightray,leftray]. Forexample,
δτ∗ = (x0τ x1τ . . . xk−1τ) andλτ∗=. . . x2τ x1τ x0τ].
Notation2.7.For0=α∈ I(X),letΔαbe thesetofcyclesandΘαbethesetofchains thatoccur inthe cycle-chain-raydecomposition of α(see (2.1)).For k≥1, wedenote byΔkα theset of cyclesinΔα oflengthk,and byΘkα theset ofchainsinΘαof length k.Ifα= 0, wesetΔkα= Θkα=∅.
Forafunctionf :A→B andA0⊆A,A0f ={af :a∈A0}denotestheimageofA0 underf.
Proposition 2.8.Let α,β,τ ∈ I(X) be such that τ−1ατ =β and τ βτ−1 =α.Then for every k≥1,Δkατ∗= Δkβ,Θkατ∗ = Θkβ,Ωατ∗= Ωβ,Υατ∗ = Υβ,andΛατ∗= Λβ. Proof. Let k ≥ 1. Let δ = (x0x1. . . xk−1) ∈ Δkα. Then δτ∗ = (x0τ x1τ . . . xk−1τ).
We havex0 →α x1 → · · ·α →α xk−1 →α x0, and so x0τ →β x1τ → · · ·β →β xk−1τ →β x0τ by Lemma2.5.Thusδτ∗∈Δkβ.WehaveprovedthatΔkατ∗⊆Δkβ.Letσ= (y0y1. . . yk−1)∈ Δkβ. Bythe foregoing argument, σ(τ−1)∗ = (y0τ−1y1τ−1. . . yk−1τ−1) ∈ Δkα. Further, (σ(τ−1)∗)τ∗ = (y0τ−1τ y1τ−1τ . . . yk−1τ−1τ) = (y0y1. . . yk−1) = σ. It follows that Δkατ∗= Δkβ.
Let θ = [x0x1. . . xk] ∈ Θkα. Then θτ∗ = [x0τ x1τ . . . xkτ]. We have x0
→α x1
→α
· · · →α xk, and so x0τ →β x1τ → · · ·β →β xkτ by Lemma 2.5. Also by Lemma 2.5, x0τ /∈ im(β) (since x0 ∈/ im(α)) and xkτ /∈ dom(β) (since xk ∈/ dom(α)). Thus θτ∗ ∈ Θkβ. We have proved that Θkατ∗ ⊆ Θkβ. Let η = (y0y1. . . yk−1) ∈ Θkβ. By the foregoing argument, η(τ−1)∗ = [y0τ−1y1τ−1. . . ykτ−1] ∈ Θkα. Further, (η(τ−1)∗)τ∗ = [y0τ−1τ y1τ−1τ . . . ykτ−1τ]= [y0y1. . . yk]=η.It followsthatΘkατ∗= Θkβ.
Theproofsoftheremainingequalitiesaresimilar. 2 Definition2.9.Letα∈ I(X).Thesequence
|Δ1α|,|Δ2α|,|Δ3α|, . . .;|Θ1α|,|Θ2α|,|Θ3α|, . . .;|Ωα|,|Υα|,|Λα|
(indexedbytheelementsoftheordinal2ω+ 3)willbecalledthecycle-chain-raytype of α. Thisnotiongeneralizesthecycletypeofapermutation[5,p. 126].
Thecycle-chain-raytypeofαiscompletelydeterminedbytheformofthecycle-chain- raydecompositionofα.Theform isobtainedfrom thedecompositionbyomittingeach occurrenceofthesymbol“”andreplacingeachelementofX bysomegenericsymbol, say“∗.”Forexample,α= (268)[13][459] hastheform(∗ ∗ ∗)[∗∗][∗ ∗ ∗], and
β= (2 4)[6 8 10] . . .−6−4−2 −1−3−5. . . [1 5 9 13. . . . . .15 11 7 3]
hastheform (∗∗)[∗ ∗ ∗]. . .∗ ∗ ∗. . .[∗∗ ∗. . .. . .∗ ∗∗].
It is wellknownthattwo elements ofthesymmetric groupSym(X) areconjugate if andonlyiftheyhavethesamecycletype[5,Prop. 11,p. 126].Thefollowingdescription of thei-conjugacyinthesymmetricinversesemigroupI(X) generalizesthisresult.
Theorem 2.10.Elements αand β of I(X) are i-conjugate if and only if they have the same cycle-chain-ray type.
Proof. Letα,β ∈ I(X).Supposeα∼iβ,thatis,thereisτ∈ I(X) suchthatτ−1ατ =β and τ βτ−1=α.Thenαandβ havethesametypebyProposition2.8andthefactthat τ∗ restrictedtoanyset from{Δkα:k≥1}∪ {Θkα:k≥1}∪ {Ωα,Υα,Λα}isinjective.
Conversely, suppose α and β have the same cycle-chain-ray type. Then for every k ≥1,there arebijections fk : Δkα →Δkβ, gk : Θkα →Θkβ,h: Ωα →Ωβ, i: Υα →Υβ, and j : Λα → Λβ. For all δ ∈ Δkα, θ ∈ Θkα, ω ∈ Ωα, υ ∈ Υα, and λ ∈ Λα, we define τ on span(δ)∪span(θ)∪span(ω)∪span(υ)∪span(λ) in such a way that δτ∗ = δfk, θτ∗=θgk,ωτ∗=ωh,υτ∗=υi,andλτ∗=λj.Notethatthisdefinesaninjectiveτwith dom(τ)= span(α) andim(τ)= span(β).
Let x∈ X. We will prove thatx(τ−1ατ) =xβ. If x∈/ span(β) thenx∈/ dom(τ−1) (since dom(τ−1) = im(τ) = span(β)), and so x(τ−1ατ) = (ατ) = and xβ = . Suppose x∈ (im(β)\dom(β)). Then there is ξ = . . . x] that is either a chain or left ray contained inβ. Bythe definitionof τ, there is η =. . . z] that is eithera chainor left ray contained inα with zτ = x. Then x(τ−1ατ) = z(ατ) =τ = and xβ = . Finally,supposex∈dom(β).Thenthereisξ=. . . xy . . .thatisabasicpartialinjective transformation contained inβ.By thedefinition ofτ, there isη =. . . z w . . . thatis a basic partial injective transformation contained in αwith zτ = x and wτ = y. Then x(τ−1ατ)=z(ατ)=wτ =yand xβ=y.
Wehaveprovedthatτ−1ατ =β.Bythe sameargument,appliedtoτ−1,β,αinstead of τ,α,β,wehaveτ βτ−1=α.Henceα∼iβ. 2
Theorem 2.10also followsfrom [13, Cor. 5.2] andthefactthat∼i = ∼n ininverse semigroups (see Section1).However, the proof in[13] is notdirect sinceit relies on a characterization of∼n insubsemigroups ofthe semigroupP(X) of allpartial transfor- mationsonX.
Suppose X is finite with |X| = n and let α ∈ I(X). Then α contains no rays, no cycles of lengthgreater thann, andno chainsof lengthgreaterthan n−1.Therefore, thecycle-chain-raytypeofαcanbewrittenas
|Δ1α|,|Δ2α|, . . . ,|Δnα|;|Θ1α|,|Θ2α|, . . . ,|Θnα−1|. (2.2) Wewillreferto(2.2) asthecycle-chaintypeofα.ByTheorem2.10,forallα,β∈ I(X), α∼iβ ⇐⇒ |Δ1α|, . . . ,|Δnα|;|Θ1α|, . . . ,|Θnα−1|=|Δ1β|, . . . ,|Δnβ|;|Θ1β|, . . . ,|Θnβ−1|. (2.3)
Supposeα∈ I(X) hasafinitedomain.Thenαdoesnotcontainanyrays.Therefore, wewill referto thecycle-chain-raytypeofαasthecycle-chaintypeof αeven whenX isinfinite.
By(1.1),αand β inI(X) are i-conjugateifand only ifthere exists τ ∈ I(X) such thatτ−1ατ =β and τ βτ−1 =α. If X is finite, we can replace τ with a permutation onX.
Proposition2.11.LetX beafiniteset,andletα,β ∈ I(X).Thenthefollowingconditions areequivalent:
(i) αandβ are i-conjugate;
(ii) αandβ have thesamecycle-chaintype;
(iii) thereexistsσ∈Sym(X)suchthat σ−1ασ=β.
Proof. Conditions (i)and (ii)are equivalentby Theorem 2.10, and (iii)clearly implies (i).It remainsto show that(i)implies(iii). Suppose (i)holds, thatis,τ−1ατ =β and τ βτ−1=αforsomeτ ∈ I(X).ByProposition2.8,τ mapsspan(α) onto span(β).Thus
|span(α)|=|span(β)|,andso,sinceX isfinite,|X\span(α)|=|X\span(β)|.Wefixa bijectionf :X\span(α)→X\span(β) anddefine σ:X →X by
xσ=
xτ ifx∈span(α), xf ifx∈(X\span(α)).
Clearly, σ ∈ Sym(X). Let x ∈ X. If x ∈/ span(β), then xσ−1 ∈/ span(α), and so x(σ−1ασ)=σ==xβ. Supposex∈(im(β)\dom(β)).Then xτ−1 =xσ−1 (bythe definitionof σ) and xτ−1 ∈/ dom(α) (by Lemma 2.5). Thus, x(σ−1ασ) =x(τ−1ασ)= σ = = xβ. Suppose x∈ dom(β). Then xτ−1 =xσ−1 and xτ−1 ∈ dom(α). Hence, (xτ−1)α∈im(α),andso((xτ−1)α)τ = ((xτ−1)α)σ.Therefore,x(σ−1ασ)=x(τ−1ατ)= xβ.
Wehaveprovedthatx(σ−1ασ)=xβ forallx∈X,andso (i)implies(iii). 2 Theequivalence of (ii) and (iii) is stated in[4, p. 120]. Proposition2.11 isnot true foraninfinitesetX.LetX={1,2,3,. . .}andconsiderα= [234. . .andβ= [123. . . inI(X). Then α and β are i-conjugate by Theorem 2.10. Note that1 ∈/ dom(α) and dom(β)=X.Thus,byLemma2.5(3),ifτ ∈ I(X) issuchthatτ−1ατ =βandτ βτ−1= α, then1∈/dom(τ).Consequently,(iii)isnotsatisfied.
2.3. Conjugacy intheidealsof I(X)
Wehavealreadydealtwithi-conjugacyinI(X) (Theorem2.10).Here,wewilldescribe i-conjugacy in an arbitrary proper (that is, different from I(X)) ideal of I(X). For α ∈ I(X), the rank of α is the cardinality of im(α). Since α is injective, we have
rank(α)=|im(α)|=|dom(α)|.Foracardinalrwith 0< r≤ |X|,letJr={α∈ I(X): rank(α)< r}.Then theset {Jr: 0< r≤ |X|}consistsof allproperidealsofI(X) [19].
Theorem 2.12.Let Jr be a proper ideal of I(X), where r is finite, and let α,β ∈ Jr. Then α andβ are i-conjugate in Jr if andonly if theyhave the same cycle-chaintype and |span(α)|< r.
Proof. Suppose α∼i β in Jr. Then α∼iβ in I(X), and so α and β have the same cycle-chain type by Theorem 2.10. Let τ ∈ Jr such that τ−1ατ = β and τ βτ−1 = α.
Then, byLemma2.5, span(α)⊆dom(τ),andso|span(α)|≤ |dom(τ)|= rank(τ)< r.
Conversely,suppose thatαand β havethe samecycle-chaintypeand|span(α)|< r.
Then α∼iβ inI(X) by Theorem 2.10. Inthe proof of Theorem 2.10, weconstructed τ ∈ I(X) such that dom(τ)= span(α), τ−1ατ =β, and τ βτ−1 =α. Since rank(τ) =
|dom(τ)|=|span(α)|< r,wehaveτ∈Jr,andsoα∼iβ inJr. 2 Wenote thatforallα,β ∈Jr,wherer isfinite,
|span(α)|= rank(α) + the number of chains inα,
and that if α and β have the same cycle-chain type, then rank(α) = rank(β) and
|span(α)|=|span(β)|.
As an example, let X = {1,. . . ,8} and consider α = (12)[34][567] and β = (59)[16][387] in I(X). Then α,β ∈ J6 but they are not i-conjugate in J6 since
|span(α)|= 7>6.Note,however,thatα∼iβ inJ8.
If r isinfinite, then theconjugacy ∼i inJr is therestriction of ∼i inI(X), thatis, forallα,β∈Jr,α∼iβ inJr ifandonlyifα∼iβ inI(X).
Theorem 2.13. LetJr be a proper idealof I(X), where r isinfinite, and let α,β ∈Jr. Then α and β are i-conjugate in Jr if and only if they have the same cycle-chain-ray type.
Proof. Ifα∼iβ inJr,thenα∼iβinI(X),andsoαandβ havethesamecycle-chain-ray typebyTheorem2.10.Conversely,supposethatαandβ havethe samecycle-chain-ray type.Thenα∼iβinI(X) byTheorem2.10.IntheproofofTheorem2.10,weconstructed τ ∈ I(X) such thatdom(τ)= span(α), τ−1ατ =β, and τ βτ−1 =α. Since span(α)= dom(α)∪im(α),wehave|span(α)|≤ |dom(α)|+|im(α)|= rank(α)+rank(α)< r+r=r (sincer is infinite). Thus rank(τ) =|dom(τ)| =|span(α)|< r. Henceτ ∈ Jr, and so α∼iβ inJr. 2
2.4. Numberof conjugacyclassesinI(X)
Wewillnowcount thei-conjugacyclassesinI(X).Ofcourse,wewillhavetodistin- guish betweenthefiniteandinfiniteX.
Letnbeapositiveinteger.Recall thatapartition ofnisasequencen1,n2,. . . ,ns of positive integers such that n1 ≤ n2 ≤ . . . ≤ ns and n1 +n2+· · ·+ns = n. We denote by p(n) the number of partitions of n and define p(0) to be 1. For example, n= 4 hasfivepartitions:1,1,1,1,1,1,2,1,3,2,2,and 4; sop(4)= 5.Denote byQ(n) thesetofsequences(i1,k1),. . . ,(iu,ku)ofpairsofpositiveintegerssuchthat k1 < k2 < . . . < ku and i1k1+i2k2+· · ·+iuku = n. There is an obvious one-to-one correspondencebetweenthesetofpartitionsofnandthesetQ(n),so|Q(n)|=p(n).For example,thepartition1,1,2,2,2,2,5of15 correspondsto(2,1),(4,2),(1,5)∈Q(15).
WedefineQ(0) tobe (0,0).
Notation2.14.LetX beafinitesetwith|X|=n.Theneveryα∈ I(X) canbeexpressed uniquely as a join α = σαηα, where σα is either 0 or a join of cycles, and ηα is either 0 or a join of chains. In other words, σα =
δ∈Δαδ and ηα =
θ∈Θαθ. For
example,ifα= (268)[13][459],thenσα= (268) andηα= [13][459].Notethat
|span(σα)|=n
k=1k|Δkα|and|span(ηα)|=n−1
k=1(k+ 1)|Θkα|.
Let C = {[α]∼i : α ∈ I(X)} be the set of i-conjugacy classes of I(X). For r ∈ {0,1,. . . ,n},denotebyCr thefollowing subsetofC:
Cr={[α]∼i ∈C:|span(σα)|=r}. (2.4) ByTheorem2.10,eachCris welldefined(ifα∼iβ then|span(σα)|=|span(σβ)|)and C0,C1,. . . ,Cn arepairwisedisjoint.
Lemma 2.15.LetX be a finite set with n elements, let r∈ {0,1,. . . ,n}, and letCr be thesetdefined by(2.4).Then|Cr|=p(r)p(n−r).
Proof. Let[α]∼i∈ Cr. LetK={k∈ {1,. . . ,n}: Δkα=∅}.WriteK={k1,k2,. . . ,ku} withk1 < k2 < . . . < ku (u= 0 if K=∅).Forp∈ {1,. . . ,u},letip=|Δkαp|. By(2.3), the sequence (i1,k1),. . . ,(iu,ku) (which we define to be (0,0) if K = ∅) does not dependonthechoiceofarepresentativein[α]∼i and
i1k1+· · ·+iuku= n
k=1
k|Δkα|=|span(σα)|=r. (2.5)
Let L = {l ∈ {1,. . . ,n} : l ≥ 2 andΘlα−1 = ∅ or l = 1 and X \span(α) = ∅}. (The reasonwe includel when Θl−1α =∅is thatthere arel pointsinthe spanof eachchain [x0x1. . . xl−1] fromΘlα−1;andweinclude1 whenX\span(α)=∅becauseX\span(α) consists of single points.) Write L = {l1,l2,. . . ,lv} with l1 < l2 < . . . < lv (v = 0 if L=∅).Forq∈ {1,. . . ,v}, letjq=|Θlαq−1|(iflq≥2) andjq=|X\span(α)|(iflq= 1).
By(2.3),thesequence(j1,l1),. . . ,(jv,lv)(whichwedefinetobe(0,0)ifL=∅)does notdependonthechoiceofarepresentative in[α]∼i and
