DOI 10.1007/s00233-016-9814-9 R E S E A R C H A RT I C L E
Ends of semigroups
S. Craik1 · R. Gray2 · V. Kilibarda3 · J. D. Mitchell1 · N. Ruškuc1
Received: 3 September 2014 / Accepted: 4 July 2016
© The Author(s) 2016. This article is published with open access at Springerlink.com
Abstract We define the notion of the partial order of ends of the Cayley graph of a semigroup. We prove that the structure of the ends of a semigroup is invariant under change of finite generating set and at the same time is inherited by subsemigroups and extensions of finite Rees index. We prove an analogue of Hopf’s Theorem, stating that an infinite group has 1, 2 or infinitely many ends, for left cancellative semigroups and that the cardinality of the set of ends is invariant in subsemigroups and extension of finite Green index in left cancellative semigroups.
Keywords Digraph·Ends·Cayley graph
Communicated by Victoria Gould.
B
S. Craik[email protected] R. Gray
[email protected] V. Kilibarda
[email protected] J. D. Mitchell
[email protected] N. Ruškuc
1 Mathematical Institute, University of St Andrews, St Andrews KY16 9SS, UK 2 School of Mathematics, University of East Anglia, Norwich NR4 7TJ, UK
1 Introduction
The study of ends in group theory has been extensive and has had widespread influence. Stallings’ Theorem characterising groups with more than one end has been used in such varied topics as distance-transitive graphs [13], groups with context-free word problem [14], pursuit-evasion problems in infinite graphs [17] and to describe accessible groups [5]. This paper follows the trend of relating geometric properties of semigroups to algebraic properties, see for example [8,16] and [11]. We consider the notions of ends for a semigroup and try to recover some basic theorems from the theory of ends of groups. In the case of a finitely generated group, the ends of the group are defined to be the ends of the Cayley graph of G with respect to the generating set A. This definition is invariant under change of finite generating setA. For a full definition for ends of graphs see [4]. In particular, the paper will focus on generalising the following theorems for semigroups.
Theorem 1.1 [9, Satz IV]Let G be an infinite finitely generated group and let H be a subgroup of finite index. Then the number of ends of H is the same as the number of ends of G.
Theorem 1.2 [9, Satz II]An infinite finitely generated group has 1, 2 or2ℵ0 ends.
In this paper we consider the definition of the ends of a digraph introduced by Zuther in [21] and apply it to the left and right Cayley graphs of a semigroup.
In [10] Jackson and Kilibarda introduce a definition for the number of ends of a semigroup. Their definition is based on the number of ends of the underlying undirected Cayley graph associated with a finitely generating semigroup. Jackson and Kilibarda prove that the number of ends of a semigroup is invariant under change of finite generating set and provide examples of semigroups withnends in the left Cayley graph andmends in the right Cayley graph for any prescribed positive integersnandm. In a follow up paper [12] some of the authors of this paper, together with Malcev, used the definition of Jackson and Kilibarda to investigate the relationship between the number of ends when considering subsemigroups of finite index. It was demonstrated that the number of ends was invariant under taking subsemigroups of finite Rees index. Also, the number of ends was shown to be invariant under taking subsemigroups of finite Green index when restricted to the class of cancellative semigroups. These notions of index are defined later in the paper.
We argue that although there are many ways to generalise the notion of ends to a semigroup; by preserving the notion of direction there is a greater chance of interre-lating the algebraic structure and the ends.
In this paper we will use the notationNto denote the set of positive integers and
N0to denote the set of non-negative integers.
Letbe any set and let ⊆×. We will refer toas adigraphon, the elements ofas theverticesof, and the elements ofasedges. Awalk inis just a (finite or infinite) sequence (v0, v1, . . .) of (not necessarily distinct) vertices such that (vi, vi+1) ∈ for alli. An anti-walk is a sequence(v0, v1, . . .)of (not
necessarily distinct) vertices such that(vi+1, vi)∈for alli. For a walk or anti-walk
α=(α0, α1, . . .)inwe may write{α}for the set{α0, α1, . . .}of vertices which occur inα. Apathinis just a walk consisting of distinct vertices. Ifα=(α0, α1, . . . , αn)
is a walk in , then thelength of α isn and it is straightforward to verify thatα contains a path fromα0toαn. A rayinis just an infinite path(v0, v1, . . .) such
that(vi, vi+1)∈for alli and ananti-rayis an infinite path(v0, v1, . . .)such that
(vi+1, vi) ∈ for all i. Ifα = (α0, α1, . . . , αm) andβ = (β0, β1, . . . , βn) are
arbitrary finite sequences of vertices from, then we denote byαβthe sequence (α0, α1, . . . , αm, β0, β1, . . . , βn).
The out-degreeof a vertex αin a digraph is just|{β ∈ : (α, β) ∈ }|. A digraphisout-locally finiteif every vertex has finite out-degree.
If ⊆, then∩(×)is theinduced subdigraphofon. Ifand are infinite subsets of, then we writeif there exist infinitely many disjoint paths (including paths of length 0) inwith initial vertex belonging toand final vertex belonging to. It is straightforward to verify thatis reflexive on infinite subsets ofbut not necessarily transitive, symmetric, or anti-symmetric. However, it was shown in [21] that ifis restricted to the set of rays and anti-rays on, then it is transitive, and hence a preorder. Letrandrbe rays or anti-rays. Ifrrandrr, then we writer ≈ r. It follows that ≈is an equivalence relation andinduces a partial order on≈-classes of rays and anti-rays. As such we refer to rays and anti-rays as beingequivalentif they belong to the same≈-class; andinequivalentotherwise. We denote this poset by, and as in [21] we refer to≈-classes of rays and anti-rays as theendsof. Note that an equivalence class can contain both rays and anti-rays.
In this paper the direction ofis the reverse of that in [21]. That is, for raysr1 andr2, if there are infinitely many disjoint paths with initial vertexr1and final vertex inr2then the notation in [21] would ber1 r2 whereas in this paper the notation would ber2r1. The reason for this alteration is to reflect that the set of vertices that can be reached by a path fromr1contains the set of vertices that can be reached by a path from a cofinite set of vertices ofr2. Readers should note that by reversing the direction ofthe properties of anti-symmetry and transitivity are preserved.
topological and graph-theoretical definitions remain the same. For a detailed proof of this correspondence see [4, Sect. 8.5].
We finish this section with some technical results which will be needed later.
Lemma 1.3 Letbe a digraph onand letα =(α0, α1, . . .)be an infinite walk (or anti-walk) insuch that every vertex ofαoccurs only finitely many times. Then
αcontains a rayr(or anti-ray, respectively) such thatrhas infinitely many disjoint paths to every infinite subset of{α0, α1, . . .}, and every infinite subset of{α0, α1, . . .} has infinitely many disjoint paths tor.
Proof We prove that α contains a ray in the case thatα is a walk; an analogous argument proves thatαcontains an anti-ray in the case that it is an anti-walk.
Leta(0)=1 and for everyi ≥1 definea(i)=max{j ∈N:αj =αa(i−1)} +1, i.e.αa(i)−1is the last appearance ofαa(i−1)inα. We will show that
r=(αa(0), αa(1), . . .)
is the required ray. Since (αi, αi+1) ∈ for alli, in particular,(αa(i−1), αa(i)) =
(αa(i)−1, αa(i))∈.Henceris an infinite walk whereαa(i) =αa(j)for alli,j ∈N
such thati = jand soris a ray.
Let be any infinite subset of{α0, α1, . . .}. If infinitely many elements in are vertices ofr, thenrand r. If only finitely many elements ofbelong tor, then\ {r}and\ {r}and so we may assume without loss of generality thatcontains no elements in{αa(0), αa(1), . . .}.
We define infinitely many disjoint paths fromtorby induction. Letb(0)∈N
be any number such thatαb(0) ∈. Then there existsk(0)∈ Nsuch thata(k(0))≤
b(0) < a(k(0)+1)andβ0 := (αa(k(0)), αa(k(0))+1, . . . , αb(0), . . . , αa(k(0)+1))is a walk from αa(k(0)) inr to αa(k(0)+1) inr via αb(0) ∈ . Since every finite walk contains a path, we conclude that there is a path contained inβ0from a vertex ofrto αb(0)∈and a path back fromαb(0)to a vertex ofr.
Suppose that we have defined the valuesb(0), . . . ,b(i−1),k(0), . . . ,k(i−1)∈N and finite walksβ0,β1, . . . ,βi−1for some i ≥ 1. Choosek(i),b(i) ∈ Nso that
b(i) > a(k(i)), αj does not equal any vertex in any of β0,β1, . . . ,βi−1 for all
j>a(k(i)), andαb(i)∈. Then we define
βi =(αa(k(i)), αa(k(i))+1, . . . , αb(i), . . . , αa(k(i)+1)).
By construction, ifi = j, thenβi andβj are disjoint and so we have infinitely
many disjoint paths (contained in theβi) fromrtoand back, as required.
Lemma 1.4 Letbe an out-locally finite digraph on a set X and letw0,w1, . . .be finite walks of bounded length inwith distinct final vertices. Then every vertex in the sequencew0w1 . . .occurs only finitely many times.
i ∈N. But the final vertices of thewi are distinct and soB is infinite, contradicting
the assumption thatis out-locally finite.
Lemma 1.5 Letbe an out-locally finite digraph on, let ⊆be infinite and letα0∈such that there is a path fromα0to everyβ ∈. Then there exists a rayr
instarting atα0such thatr.
Proof We constructrrecursively. Start by setting0 :=and letP0be a set con-taining precisely one pathqβ fromα0toβ for allβ ∈0. Then, sinceα0has finite out-degree and there is a path inP0fromα0to everyβ∈0, there exists a vertexγ0 such that(α0, γ0)∈ and there is a pathqβ ∈ P0fromα0viaγ0to everyβ in the infinite subset1⊆0.
Letβ1∈1be fixed and also fix a path
p1=(δ1=α0, δ2=γ0, δ3, . . . , δn−1, δn=β1).
Let P1 = {qβ ∈ P0 : β ∈ 1}. Ifβ ∈ 1 is arbitrary andqβ ∈ P1, then there existsi(β)∈ Nsuch thatδi(β)is the last vertex belonging to both the paths p1and
qβ. The numberi(β)exists since, in particular, both paths go throughγ0. For each β ∈1we have 2≤i(β)≤ n. If{β ∈1:i(β)=m}is finite for all 2 ≤m ≤n
then 1 would be finite. Hence, there exists m ∈ N such that 2 ≤ m ≤ n and 2 = {β ∈ 1 : i(β) =m}is infinite. Setα1 = δm. Sincem ≥ 2,α1 = α0and,
by construction, there is a path from α1 to every elementβ of the infinite set 2 (consisting of the vertices betweenα1andβ inqβ ∈ P1) such that the only vertex in
p1and this path isα1. SetP2to the set of paths fromαtoβ∈2from the previous sentence.
We may repeat the above processad infinitumto obtain for alli>0:βi+1∈2i+1 and a pathpi+1∈ P2i+1fromαi toβi+1, anαi+1inpi+1, an infinite2i+2⊆2i+1 and an infinite set P2i+2of paths fromαi+1to every element of2i+2such that the only vertex inpi+1and any path inP2i+2isαi+1.
Hence there is a walkrcontaining{αi :i ∈ N}consisting of the vertices on the
pathspi+1betweenαi andαi+1. In fact, by construction, the only vertex on both pi
andpi+1isαi+1, and so the walkris a ray. Moreover, there are infinitely many paths
fromr to consisting of the remaining vertices on pi+1betweenαi+1 andβi+1.
Again by construction the only vertex on both pi and pi+1isαi+1and so the paths fromαi+1∈rtoβi+1∈are disjoint.
2 The ends of a semigroup
Throughout this section, we letSbe a finitely generated semigroup and letAbe any finite generating set forS. Theright Cayley graphr(S,A)ofSwith respect toAis
the directed graph with vertex setSand edges(s,sa)∈r(S,A)for alls∈ Sand for
alla∈ A. We refer toaas alabelof the edge(s,sa). Theleft Cayley graphl(S,A)
is defined dually.
l(S,A) =r(S∗,A). Therefore to understand the end structure of semigroups, it
suffices to study right Cayley graphs only.
We require the following lemma to prove the results in this section.
Lemma 2.1 Let S be a semigroup, let T be a subsemigroup of S generated by a finite set A, and let s ∈S. Suppose that|T,s \T| =n∈N. Then there exists N∈Nsuch that for all b1,b2, . . . ,bn ∈ A∪ {s}if s,sb1, . . . ,sb1. . .bn are distinct, then there
exists i≤n and a1,a2, . . . ,aj ∈ A such that j ≤N and sb1. . .bi =a1. . .aj. Proof Let X := {sc1c2. . .ci ∈ T : cj ∈ A∪ {s}, 1 ≤ i ≤ n}. Then X is finite and so there existsN ∈Nsuch that every element ofX can be given as a product of elements of Aof length at most N. By the pigeonhole principle, there existsi such thatsb1. . .bi ∈Tand hencesb1. . .bi ∈ X. It follows that there exista1, . . . ,aj ∈ A
such that j ≤Nandsb1. . .bi =a1. . .aj, as required.
Proposition 2.2 Let S be a semigroup, let T be a subsemigroup of S generated by a finite set A, and let s∈ S. IfT,s \T is finite, thenr(T,A)is isomorphic (as a partially ordered set) tor(T,s,A∪ {s}).
Proof For the sake of brevity, we denoter(T,s,A∪{s})by. We useto denote
the preorder defined above on the rays and anti-rays of. We prove the proposition in two steps. The first step is to show that every ray or anti-ray inis equivalent to a ray or anti-ray with vertices inT and edges labelled by elements ofA. The second step is to show that ifrandrare rays or anti-rays with vertices inT, edges labelled by elements ofA, andrr, then there exist infinitely many disjoint paths fromrto rwith edges labelled by elements of A. So, the first step ensures that every endωof , contains a ray or anti-rayrωwith vertices inT and edges labelled by elements of
A. The second step implies that the mapping :−→r(T,A)defined so that
(ω)equals the end ofr(T,A)containingrω is an isomorphism. We only give
the proof of these steps for rays, an analogous argument can be used for anti-rays. LetU = T,s \T, letn = |U|, and letr =(x,xb1,xb1b2, . . .)be a ray in for someb1,b2, . . . ∈ A∪ {s}andx ∈ T,s. SinceU is finite, only finitely many elements ofrcan lie inU, and so we may assume without loss of generality that
x,xb1,xb1b2, . . .∈T. Ifbi =sfor alli, then there is nothing to prove.
Ifbk is the first occurrence of s in{b1,b2, . . .}, then since the vertices of rare distinct so are the elementsbk,bkbk+1, . . . ,bkbk+1. . .bk+n. Hence by Lemma2.1
there existi,N ∈Nanda1,a2, . . . ,aj ∈ Asuch that j ≤ Nandbkbk+1. . .bk+i = sbk+1. . .bk+i =a1. . .aj. Hence
w0=(xb1. . .bk−1,xb1. . .bk−1a1, . . . ,xb1. . .bk−1a1. . .aj)
is a walk inwith vertices inT and edges labelled by elements of A. We repeatedly apply Lemma2.1to successive occurrences ofsin{b1,b2, . . .}to obtain finite walks w1,w2, . . .with vertices inT and edges labelled by elements of A. The length ofwi
in somewi occurs only once, sinceris a ray. Hence every vertex ofwoccurs only
finitely many times, and so by Lemma1.3there is a subrayrofwsuch thatr≈r, as required.
For the second step of the proof, letrandrbe rays inwith vertices inT, edges labelled by elements ofA, andrr. Sincerr, there exist infinitely many disjoint paths infromrtor. We may assume without loss of generality that there are at leastnvertices in each of these paths after the last occurrence ofsas an edge label. Hence, by repeatedly applying Lemma2.1, there existsN ∈ Nand infinitely many paths fromrtorlabelled by elements of A. Moreover, there is a path of length at mostN from every element in one of the new paths to some element in the original path it was obtained from by applying Lemma2.1. If infinitely many of these new paths are disjoint, then there is nothing to prove. Otherwise infinitely many of these paths have non-empty intersection with a finite subset ofT, and so infinitely many paths contain some fixed elementt ∈ T. Hence there are paths of length at mostN
fromtto infinitely many vertices in the original paths, which contradicts the out-local
finiteness of.
Corollary 2.3 Let S be a finitely generated semigroup and let A and B be any finite generating sets for S. Thenr(S,A)is isomorphic (as a partially ordered set) to
r(S,B).
Proof It suffices to show thatr(S,A)is isomorphic tor(S,A∪{s})for anys∈ S, since thenr(S,A)is isomorphic tor(S,A∪B)is isomorphic tor(S,B),
as required. CertainlySis a finitely generated subsemigroup ofSsuch thatS,s\Sis finite, and so it follows by Proposition2.2thatr(S,A)is isomorphic tor(S,A∪
{s}), as required.
Following from Corollary2.3we defineS=r(S,A)for any finite generating
setAofS. We refer toSas theends of S.
It is easy to see that adjoining an identity to a semigroup will not change the cardinality of the set of ends, ie|S| = |S1|. For a formal proof of this see Corollary 2.4. Hence, when considering the cardinality of the set of ends of a semigroup we may always assume that the semigroup is a monoid. In addition when considering the ends of an infinite finitely generated group G (under Zuther’s definition) we have from Corollary2.3that for any finite semigroup generating setAforGthe set of ends of r(S,A)is isomorphic to the set of ends ofr(S,A∪A−1). As the generating set A∪A−1is closed under taking inverses the digraphr(S,A∪A−1)is in fact a graph.
Also, as the degree of each vertex in the graph is at most 2|A|the graph is locally finite. Hence, when considering the cardinality of the set of ends of a group we are justified in using Hopf’s theorems.
Note that if S is an infinite finitely generated group, then it follows by Hopf’s Theorem [9, Satz II] that the ends ofSform an anti-chain with 1, 2, or 2ℵ0 elements.
In Sect.5we give examples of finitely generated semigroups with any finite number orℵ0ends (Examples5.5,5.3). Any group with 2ℵ0 group ends will also have 2ℵ0
ends as a semigroup. It is easy to see that the free monoid on two generators will have 2ℵ0ends as all pairs of rays are incomparable. It is not known whether, in the absence
S hasκ elements whereℵ0< κ <2ℵ0. The question of which posets can occur as
the partial order of endsSof some finitely generated semigroupSis unresolved. IfSis a semigroup andT is a subsemigroup ofS, then theRees indexofT inSis just|S\T| +1.
Corollary 2.4 Let S be a finitely generated semigroup and let T be a subsemigroup of S of finite Rees index. Then the partial orderS of the ends of S is isomorphic to the partial orderT of the ends of T .
Proof SinceSis finitely generated, it follows by [18, Theorem 1.1], thatT is finitely generated. Let A be any finite generating set forT and lets ∈ S\T be arbitrary. ThenT,s \T ⊆ S\T and soT,s \T is finite. It follows from Proposition2.2 thatr(T,A)is isomorphicr(T,s,A∪ {s}), and henceT is isomorphic to
T,s. SinceT T,s ≤ S, by repeating this process (at most|S\T|times) we
have shown thatSis isomorphic toT.
3 The number of ends of a left cancellative semigroup
In this section we prove that the right Cayley graph of a left cancellative semigroup can only have a restricted number of ends, unlike the general case (see Proposition 5.5).
A semigroupSisleft cancellativeifx=ywheneverax =aywherea,x,y∈S.
Right cancellativeis defined analogously. A semigroup iscancellativeif it is both left and right cancellative.
A left or right cancellative monoid contains only one idempotent (the identity). A left cancellative semigroup contains at most one idempotent in everyL-class, and the analogous statement holds for right cancellative semigroups. The structure of a cancellative semigroupSis straightforward to describe: eitherSisR- trivial orSis a monoid with group of unitsG, everyR-class is of the formx G, and everyL-class is of the formG xfor somex∈S(see for example [15]). We start this section by giving an analogous description of the structure of a left cancellative semigroup. It is possible to deduce these results from [19] although they are not couched in this notation, and so we include a proof for the sake of completeness.
Aright groupis the direct product of a groupGand right zero semigroupE.
Theorem 3.1 [3, Theorem 1.27]A semigroup is a right group if and only if it is left cancellative andR-simple.
Proposition 3.2 Let S be a left cancellative semigroup and let U be the set of regular elements in S. Then:
(i) S\U is an ideal (in the case when S is a group it is empty);
(ii) if U is non-empty, then U is a right group;
(iii) if x ∈S has non-trivialR-class Rx, then Rx =xU ;
Proof Letx,y∈Sand assume thatx yis a regular element. It follows that there exists
z∈Ssuch thatx yzx y=x y. By cancelling we see thatyzx y=yand henceymust be a regular element. Fromyzx y=ywe must haveyzx yzx =yzx, again by cancelling we seex yzx=xand hencexis also a regular element. This means thatS\Uis an ideal ofS.
For the second part we assume thatUis non-empty. IfScontains a regular element then it contains an idempotent. Let eand f be idempotents in S. Thene2f = e f
and f2e = f eand so, by cancelling,e f = f and f e = e. ThuseRf and, since every regular element isR-related to an idempotent, the regular elements of S are contained in a singleR-class of S. Ifx and yare regular, then there exists x ∈ S
such thatx xx = x and, sinceyRx, there existsz ∈ S such thatyz = x. Hence
x yzx y=x xx y=x yand sox yis regular. Hence,Uis a subsemigroup ofSand by part oneS\Uis an ideal and soU isR-simple. It follows from Theorem3.1thatU
is a right group
We now prove parts three and four together. Letx∈Sand assumeUis non-empty. Lete∈Ube an idempotent. Then as all elements inUareRrelatedxU is contained within anR-class, say R. Let y be an element of R distinct from xe. Then there existss,t ∈ S such thatxes =yandyt =xe. It follows thatxest =xeand hence
xest st = xest. By cancelling(st)2= stis an idempotent and as S\U is an ideal
s,t ∈U. This means thatR=xU. Ifxlies in a non-trivialR-class then there exists
yRxsuch thaty =xand there existss,t∈ Ssuch thatxs=yandyt=x. Then as before we see thatstis an idempotent andx=xstsox∈xU.
Lemma 3.3 A left cancellative semigroup S has either oneR-class or infinitely many R-classes.
Proof We show that either S is regular or (x,x2) /∈ Rfor somex ∈ S. Suppose that(x,x2)∈ Rfor allx ∈ S. Then there existss ∈ S1such that x2s = x. Hence
x2st =xtand so(xs)t =t for allt ∈ S. Hencexsis a left identity forSand soxs
is an idempotent andxRxs. ThusSis regular and so by [3, Exercise 1.11.4] has only oneR-class. If there existsx ∈ Ssuch that(x,x2) /∈ R, then(xi,xj) /∈ Rfor all
i,j∈Nsuch thati = j. HenceShas infinitely manyR-classes.
Corollary 3.4 If S has infinitely manyR-classes at least one of which is infinite, then it has infinitely many infiniteR-classes.
Proof Since there is at least one infiniteR-class inS, thatR-class is of the formyU
for somey ∈ S, and|yU| = |U|by left cancellativity, it follows thatU is infinite. From the proof of Lemma 3.3, there exists x ∈ S such that (xi,xj) /∈ Rfor all
i,j∈Nsuch thati = j. By Proposition3.2,xiUis anR-class ofSfor alli ∈Nand |xiU| = |U|and, in particular,xiUis infinite for alli ∈N. It suffices to show that the setsxiUare disjoint. Suppose to the contrary thatxiU∩xjU = ∅for somei,j ∈N
withi < j. Then, by left cancellativity, xj−iU ∩U = ∅and so xj−i ∈ U since
S\U is an ideal. Therefore infinitely many powers ofx, namely,xj−i,x2j−2i, . . .,
areR-related, contradicting our assumption.
Corollary 3.5 Let G be a finitely generated group and let E be a finite right zero semigroup. Then|(G×E)| = |G|.
Lemma 3.6 Let S be a finitely generated left cancellative semigroup with no infinite R-classes. If the Cayley graph of S with respect to a finite generating set contains a rayrand there is an s ∈S such that there are paths from infinitely many points inr
to s, thenS is infinite.
Proof LetAbe a finite generating set forS, letr=(r0,r1, . . .)be a ray inr(S,A)
and lets∈ Ssuch that there are infinitely many paths fromrtos. We may writer0as a producta1. . .anof generators inA.
Assume, seeking a contradiction, that S has finitely many ends. Since S is left cancellative,ri = (si,sia1, . . . ,sia1. . .an =sir0,sir1, . . .)is a ray for alli ∈ N. Thus, by assumption, there existi,j ∈Nsuch thati < j andri ≈rj. Again using
the left cancellativity ofS, it follows thatrj−i ≈rand so there is a path fromsj−irk
torl for somek,l ∈N. There is a path fromstosj−irk and hence torl. But in this case,rlRrl+1R. . .and soShas an infiniteR-class, which is a contradiction.
The main results of this section are given below.
Theorem 3.7 Let S be an infinite finitely generated left cancellative semigroup. Then
|S| =1,2or|S| ≥ ℵ0.
Proof IfShas only oneR-class, then by [3, Theorem 1.27] it follows thatS∼=G×E
whereGis a group andEis a right zero semigroup. Since Sis finitely generated, it follows thatGis finitely generated and Eis finite. Hence, by Corollary3.5,|S| =
|G|and by Hopf’s Theorem [9, Satz I],|G| =1,2 or 2ℵ0.
Suppose that S has more than oneR-class. Then Lemma3.3implies that S has infinitely manyR-classes. IfScontains an infiniteR-class then then by Corollary3.4 Scontains infinitely many infiniteR-classes. AsS is finitely generated every vertex has finite out-degree. Letrbe an element of an infiniteR-classR. Then there exists a path fromrto infinitely many elements ofR. By Lemma1.5, there exists a ray starting atrthat has paths to infinitely many elements ofR. Asr ∈ R, it follows that all the elements of the ray must also be inRand hence each infiniteR-class contains a ray. Furthermore, none of these rays can be equivalent as theR-classes are distinct. Thus |S| ≥ ℵ0.
Assume that S has no infiniteR-class. Letdenote the Cayley graph of S with respect to some finite generating setAforS. Ifcontains a rayrand there is ans∈S
such that there are paths from infinitely many points inrtos, thenS is infinite by Lemma3.6. Ifcontains an anti-rayr, then there existsa ∈ Asuch that infinitely many of the elements inrare of the format for somet ∈S. In particular, there is a path fromato everyatinrand so by Lemma1.5there exists a rayrsuch thatrr. But then there are paths from infinitely many of the vertices inrto any fixed element inr, and soSis infinite by Lemma3.6.
Suppose that the Cayley graph of S does not have the property of Lemma 3.6. Seeking a contradiction assume thatShas finitely many ends, and letr1,r2, . . . ,rnbe
ri pass throughF. By assumption there exists elementsinr1such that there are no
paths fromsto any element ofFand hence to any element in anyri. SinceSis left
cancellative,sr1andsr2are rays. Ifsr1≈riorsr2≈ri, then there is a path froms
toriand soi =1. In particular,sr1≈sr2and so, sinceSis left cancellative,r1≈r2,
a contradiction. We have shown thatSeither has 1 or infinitely many ends.
Corollary 3.8 Let S be an infinite finitely generated cancellative semigroup that is not a group. Then|S| =1or|S| ≥ ℵ0.
Proof SinceS is cancellative, it is certainly left cancellative and so|S| =1,2, or |S| ≥ ℵ0by Theorem3.7. If|S| =2, then from the proof of Theorem3.7,Shas
only oneR-class and hence is a group.
As mentioned above, it is not known what cardinalities S can have, even for restricted types of semigroups such as those which are left cancellative. We prove that Shas cardinality 2ℵ0for a particular type of cancellative semigroup. Ore’s Theorem
(see for instance [3, Theorem 1.23]) states that if a cancellative semigroupSsatisfies the condition thats S∩t S = ∅for alls,t ∈SthenScan be embedded in a group.
Theorem 3.9 A finitely generated cancellative semigroup which cannot be embedded in a group has2ℵ0ends.
Proof LetSbe a cancellative semigroup that cannot be embedded in a group. AsSis not group-embeddable there existss,t ∈Ssuch thats S∩t S= ∅. Firstly we show that
all elements of{s,t}∗ are distinct. Letu =u1u2. . .un, v = v1, v2. . . vm ∈ {s,t}∗
and assumeu =Sv, without loss of generality we assume the length ofuis less than
or equal to the length ofv. Ifuis a prefix ofvthenu =v=uv. It follows thatvis a left identity for all elements ofS. The first letter ofvis (without loss of generality)
sand hencet x =vt x ∈s Sfor allx ∈S. Ifuis not a prefix ofvthen there exists a positioni ≤nsuch thatuj =vj for all j <ibutui =vj. Asuj =vj for all j <i
andu =S v it follows by left-cancellativity thatui. . .un =vi. . . vm andui = vj,
however,s S∩t S= ∅a contradiction.
We now show that foru, v∈ {s,t}∗we havev∈u Sif and only ifuis a prefix of v. Clearly ifu is a prefix ofv thenv ∈ u S. With the aim of getting a contradiction assume thatu = u1u2. . .un is not a prefix of v = v1v2. . . vm butv ∈ u S. This
means there exists x ∈ S such thatux = v. Asu is not a prefix ofv there exists 1≤i ≤nsuch thatuj =vj for all j <i butui =vj. But then by left-cancellativity ui. . .unx=Svi. . . vm. Then as{ui, vi} = {s,t}it follows thats S∩t S = ∅.
Combining these facts gives a copy of the free semigroup on two generators as a subsemigroup ofS and there can be no paths between elements, this meansShas at least 2ℵ0 ends. This is also the maximum possible number of ends so|S| =2ℵ0.
4 Subsemigroups of finite Green index
In [20] Wallace introduced a generalisation of Green’s relations involving subsemi-groups. LetSbe a semigroup, letT be a subsemigroup ofSand letx,y∈S. We say
yt =x. We may defineLT,HT andDT analogously. As with Green’sR,L,Hand
DrelationsRT,LT,HT andDT are all equivalence relations onS.
It follows from Proposition3.2that ifT is a subsemigroup of a left cancellative semigroupS, thenRT =RV whereV is the right group of regular elements ofT.
LetSbe a semigroup and letT be a subsemigroup ofS. We sayT is of finiteGreen indexinSifS\T has finitely manyHT-classes. We say theGreen indexofT inSis the number ofHT-classes inS\T plus one.
Let Sbe a semigroup and let T be a subsemigroup of finite Green index. It was shown in [2] thatS is finitely generated if and only ifT is finitely generated. IfT is a submonoid of a left-cancellative monoidSandT has finite Green index inS, then, since the complement of the group of units is an ideal, the group of units ofT has finite index in the group of units ofS.
Lemma 4.1 If S=G×E is a right group, G is infinite and T is a subsemigroup of finite Green index then T =K ×E is a right group where K is of finite index in G.
Proof One can seeS has only oneRS-class, therefore theHS-classes of S are the
LS-classes ofS. As(g,e)·(h,f)=(gh,f)we see thatLS-classes are of the form G× {e}for eache∈E.
IfT contains no elements of the form(g,e)for some fixede∈ E then theRT
-class of each(h,e)must be trivial. This follows as(h,e)(g,f)can only be of the form(hg,f)where f = eand then there exists no element(g, f)∈ T such that (hg, f)(g, f)=(k,e)asT contains no elements of the form(g,e).
For eache ∈ E we letKe be those elements g ∈ G such that(g,e) ∈ T. We now show each Ke contains 1G. Lete ∈ E. One can see Ke is a subsemigroup of G as in particular(g, f)(h,e) = (gh,e)so KfKe ⊆ Ke. It is easy to see that a
subsemigroup of finite Rees index inG is equal toG so we may assumeG \Ke
is infinite. AsHT-classes are contained inHS-classes and asG\Ke is infinite we must have at least one non-trivialHT-class containing distinct elements(g,e), (g,e)
withg,g ∈/ Ke. As these elements areHT-related they areRT-related and hence there exists(h, f), (h, f, )∈T such that(g,e)(h, f)=(g,e)and(g,e)(h, f)=
(g,e). This means f = f=eand furthermore thatghh=g. It followshh=1G
is an element ofKe. Hence,Ke ⊆Kf for alle, f ∈EsoKe=Kf for alle, f ∈ E.
We call this semigroupK.
AsK ×Ehas finite Green index inG×Eit must follow thatKhas finite Green index in G. It was shown in [6, Corollary 34] that ifK is a subsemigroup of finite Green index in a groupGthenKis a subgroup ofGwith finite group index.
Lemma 4.2 Let S be a semigroup generated by A and let T be a subsemigroup of S generated by B with Green index n∈N. If s∈S and a1,a2, . . . ,am+k∈ A such that the number ofRT-classes containing any of sa1a2. . .am+1,sa1a2. . .am+2, . . . ,sa1a2 . . .am+k is at least n, then there exist i > m and b1,b2, . . . ,bj ∈ B such that sa1. . .ai =b1. . .bj.
Theorem 4.3 Let S be an infinite finitely generated left cancellative semigroup and let T be a subsemigroup of S of finite Green index. Then|S| = |T|.
Proof IfSis right simple, thenS∼=G×Efor some finitely generated groupGand
Eis a finite right zero semigroup. SinceT has finite Green index inS, it follows that
T ∼= H×Ewhere His a subgroup of finite index inG. In other words,T is a right group and so by Corollary3.5|T| = |H|and|S| = |G|. From Hopf’s theorem
[9, Satz IV] we know|G| = |H|, hence,|S| = |T|.
LetUbe the right group of regular elements inS. SinceSis finitely generated, it follows thatT is finitely generated. LetAandBbe finite generating sets forSandT, respectively, such that B ⊆ A. SinceS\U is an ideal,U is also finitely generated. Hence, as T is also left cancellative, the right group of regular elements V of T is finitely generated. It follows by Proposition3.2thatRT =RV, and soV has finite Green index inU.
Suppose thatS has more than oneR-class. Then, by Lemma3.3,Shas infinitely manyR-classes. IfS has no infiniteR-classes, then sinceRT-classes are contained inRS-classes, it follows thatT has finite Rees index in S and so by Corollary2.4, the theorem follows. We now consider the case that S has infinitely many infinite
R-classes. By Proposition3.5,U either has 1, 2, or 2ℵ0 ends.
IfUhas 2ℵ0 ends, then, sinceS\Uis an ideal,Shas 2ℵ0ends. SinceV has finite
Green index inUandUis a right group,V has 2ℵ0 ends and soT has 2ℵ0 ends also.
Suppose thatUhas 1 or 2 ends. ThenSandT have at leastℵ0ends, since every pair of infiniteR-classes contain a pair of inequivalent rays. Let(S)be the set of ends of
Scontaining a ray that has non-empty intersection with infinitely manyRS-classes. By [21, Lemma 2.8], ifωis an end ofr(S,A), then every ray inωis contained in a
strongly connected component or intersects infinitely many strongly connected com-ponents (but not both). Since strongly connected comcom-ponents ofr(S,A)are precisely RS-classes, it follows that|S| = max{ℵ0,|(S)|}and|T| =max{ℵ0,|(T)|}.
We conclude the proof by showing that|(S)| = |(T)|.
Letrbe a ray or anti-ray inr(S,A)that has non-empty intersection with infinitely
manyRS-classes. Since everyRS-class is a union ofRT-classes,rhas non-empty intersection with infinitely manyRT-classes. Since there are only finitely manyRT -classes inS\T, we may assume without loss of generality that the elements inrare inT. Letnbe the number ofRT-classes inS\T and let(xc1,xc1c2, . . . ,xc1. . .cm) be a subpath ofr that has non-empty intersection withn+1, RT-classes. By left cancellativity, the path(c1,c1c2, . . . ,c1. . .cm)has non-empty intersection with at leastn+1RT-classes also. It follows that there existsisuch thatc1. . .ci ∈T. Hence
c1. . .ci is a productb1b2. . .bj of elements in the generating set B for T.
Recur-sively replacing every such path (xc1,xc1c2, . . . ,xc1. . .ci) by the corresponding walk(xb1,xb1b2, . . . ,xb1. . .bj)we obtain a walkw= (w0, w1, . . .)inr(T,B)
that has non-empty intersection with infinitely manyRT-classes contained inT. If
i < jandwiRTwj, thenwiRTwi+1RT. . .RTwj. Butwhas non-empty intersection
with infinitely manyRT-classes and so every vertex ofwoccurs only finitely many times. Hence, by Lemma1.3,wis equivalent to a ray or anti-ray inr(T,B).
Letr1be a ray or anti-ray and letr2be a ray or anti-ray inr(T,B)such thatr1and
tor2inr(T,B), then clearlyr1is equivalent tor2inr(S,A). Suppose thatr1is
equivalent tor2inr(S,A). In this case, there are infinitely many disjoint paths from
r1tor2and vice versa. By repeatedly applying Lemma4.2, there exist infinitely many paths fromr1tor2labelled by elements of B. If infinitely many of these paths are disjoint, then the proof is complete. Otherwise infinitely many of these paths have non-empty intersection with a finite subset ofS, and so infinitely many paths contain some fixed elements ∈ S. But then there exists a path froms to element inr2 and a path from that vertex to an element in r1, and so infinitely many elements inr1 areRS-related, a contradiction. We have shown that for all rays or anti-raysr1and r2inr(T,B)such thatr1andr2have non-empty intersection with infinitely many RS-classes, r1 is equivalent tor2in
r(T,B)if and only if they are equivalent in
r(S,A). Therefore|(S)| = |(T)|, as required.
5 Examples
In this section we give several examples of finitely generated semigroups S and describeSfor these examples.
The following example shows that unlike in the groups case it is possible for a left cancellative semigroup to haveℵ0ends.
Example 5.1 The semigroup N0 × N0 under componentwise addition has ℵ0 ends. For the sake of brevity we use to denote the Cayley graph r(N0 ×
N0,{(0,0), (0,1), (1,0)}). We show that any ray inis equivalent to one of
((i,0), (i,1), (i,2), . . .), ((0,i), (1,i), (2,i), . . .)or ((0,0), (1,0), (1,1), (2,1), (2,2) . . .)
for each i ∈ N0. We first note that there are no anti-rays in . Any ray either contains finitely many elements in the first component of its vertices, finitely many elements in the second component of its vertices or infinitely many distinct elements in both components. In the first case as elements are eventually of the form (i,j)
for some fixedithe ray is equivalent to((i,0), (i,1), (i,2), . . .). Equivalently if the
ray has finitely many elements in the second component of its vertices then it will be equivalent to some((0,i), (1,i), (2,i), . . .). In the case that the ray has infinitely
many distinct elements in both components then for any element(i,j)wherei < j
there is a path from(i,i)to(i,j)to(j,j)and we see that the ray is equivalent to ((0,0), (1,0), (1,1), (2,1), (2,2), . . .).
The following example demonstrates the existence of anti-rays which are not equiv-alent to any ray. It also shows that it is possible to have anti-rays in a semigroup with trivialR-classes.
1 a a2 a3 a4 a5
b ab a2b a3b a4b a5b
ba ba2
ba3
ba4
ba5
b2 ab2 a2b2 a3b2 a4b2 a5b2
b2a
b2a2
b2a3
b2a4
b2a5
b3 ab3 a2b3 a3b3 a4b3 a5b3
b3a
b3a2
b3a3
b3a4
b3a5
Fig. 1 A portion of the right Cayley graph ofa,b|aba=bfrom Example5.2,edgeslabelled byaare represented withsolid linesand those labelled bybwithdashed lines
Fig. 2 A portion of the right Cayley graph of the semigroupa × {0,1}from Example5.3. Note that a line without a direction is used here to represent a pair ofinverse edges
The following example demonstrates that in general a subsemigroup of finite Green index may have a different number of ends from the original semigroup.
Example 5.3 Let{0,1}be the semigroup with the usual multiplication (of real num-bers) andG=Gr pabe an infinite cyclic group. Consider the semigroupa×{0,1} with generating set{(a,0), (a,1), (a−1,0), (a−1,1)}. ThenT = a × {1}is a sub-semigroup anda × {0}is anHT-class in the complement. Hencea × {1}has finite Green index ina × {0,1}. However, by inspection we seea × {0,1}has 4 ends corresponding to the two ends ofa × {1}anda × {0}, however,a × {1}has only 2 ends. For a diagram of a portion of the right Cayley graph ofa × {0,1}see Fig.2.
Following Theorem4.3one might question whether for a left cancellative semi-group it is possible to show that the end poset of a subsemisemi-group of finite Green index is isomorphic to the end poset of the semigroup. The following example answers this in the negative.
Example 5.4 Consider the semigroupS=Z×Z×N0under componentwise addition. The subsemigroupT =Z×Z×(N0\ {1})is of finite Green index as the complement consists of 1HT-class. One can see thatShasℵ0ends corresponding to eachZ×Z×{i}
and to{0} × {0} ×N0. In the poset of ends ofSany two elements are comparable. Either by inspection or by Theorem4.3we see thatT also hasℵ0ends. However, there are no paths fromZ×Z× {2}toZ×Z× {3}or vice versa and hence the ends in these components cannot be comparable.
The following proposition describes the left and right end posets of Rees matrix semigroups. As a corollary we see that for anyn,m∈Nthere exists a semigroup with
[image:15.439.73.367.53.164.2] [image:15.439.109.330.201.251.2]Recall a Rees matrix semigroupM[G;I, ;P]has elementsI×G×whereG
is a group andIandare index sets. Multiplication is defined by(i,g, λ)(j,h, μ)=
(i,gpλjh, μ)whereP =(pλj)λ∈,j∈I is a|| × |I|matrix overG.
Proposition 5.5 If S is the Rees matrix semigroup M[G;I, ;P] where I =
{i1,i2, . . . ,in}and = {λ1, λ2, . . . λm}, G is a finitely generated group and P is a m×n matrix with entries in G then the right ends of S form an anti-chain of size n· |G|and the left ends of S form an anti-chain of size m· |G|.
Proof LetXbe a finite semigroup generating set forGcontaining 1G and let
A=
(i,pμ,−1jx, λ)|x∈ X, λ, μ∈,i,j∈ I
.
ClearlyAis a finite generating set forS.
Letibe the induced subgraph ofr(S,A)on the vertices{i}×G×and letλ,i
be the subgraph ofiwith vertices{i} ×G× {λ}and edges with labels(i,pλ,−1ix, λ).
As(i,g, λ)(j,h, μ) =(i,gpλ,jh, μ)note thatr(S,A)is the disjoint union of the
i. This means thatr(S,A)isn incomparable copies ofi. As all ends inG
are incomparable it suffices to show thati is isomorphic toGfor alli∈ I.
We first note that for a fixedλ∈,λ,i is isomorphic tor(G,X). We now prove
that any ray ini is equivalent to a ray inλ,i, the proof for anti-rays is analogous.
Letr=((i,g0, λj0), (i,g1, λj1) . . .)be a ray and letr=((i,g0, λ), (i,g1, λ) . . .)be
a sequence inλ,i. We show that there is an infinite walkwinλ,i containingrin
which every vertex appears finitely often.
We constructwby concatenating the shortest paths inλ,i between each(i,gk, λ)
and(i,gk+1, λ). These shortest paths exist because λ,i is isomorphic to(G,X).
Next we show that there is a global bound on the lengths of these shortest paths. If (i,g, μ) = (i,h, ν)(j,p−ξ,1kx, π) then it followsμ = π andg = hpν,jpξ,−1kx. This
means the shortest path inλ,i between any consecutive elements ofris of length
less than or equal to K =max{|pμ,jp−ν,1k|X : j,k∈ I, μ, ν ∈} +1. Asris a ray
it follows there are at most||repetitions of vertices inr. Every vertex ofwhas a path of length less thanK to a vertex ofrand asλ,i is out-locally finite this means
that if some vertexvappears infinitely often inwthen infinitely many elements ofr can be reached fromv by a path of length less than or equal to K. But each vertex inrappears at most||times so any infinite set of elements ofrcontains infinitely many vertices, a contradiction. By Lemma1.3,wcontains a rayswith infinitely many disjoint paths fromsto and fromrand hence to and fromr.
This means any ray ini is equivalent to a ray inλ,i. To complete the proof we
must now verify that if we have raysr1andr2inλ,i such thatr1r2thenr1r2
ini. Letr1andr2be incomparable rays in λ,i. As the rays are incomparable in
λ,i there exists a finite set F = {(i, f1, λ), . . . , (i, fm, λ)}such that all paths from
r1tor2inλ,i pass through F. For any edge((i,g, μ), (i,gpμ,jp−ν,1kx, ξ))we have
a wordw=x1x2. . .xpoverX of minimal length such thatw =G pμ,jpν,−1kxand a
corresponding path
(i,g, μ), (i,g, λ), (i,gx1, λ), . . . , (i,gx1x2. . .xp, λ), (i,gpμ,jpν,−1x, ξ)
This means that any path inihas a corresponding walk inλ,isuch that any point
on the walk has a path of length less thanK +2 to a vertex on the path ini. This
means any pathπfromr1tor2inihas a corresponding walk inλ,iand this must
pass throughFand henceπmust contain an element that can be reached fromFby a path of length less than or equal toK +2. Asi is out-locally finite there are only
finitely many such elements sor1r2.
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