Ends of Semigroups
aa r X i v : . [ m a t h . G R ] S e p ENDS OF SEMIGROUPS
S. CRAIK, R. GRAY, V. KILIBARDA, J. D. MITCHELL, AND N. RUˇSKUC
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 finitegenerating set and at the same time is inherited by subsemigroups and extensions of finite Reesindex. We prove an analogue of Hopf’s Theorem, stating that a group has 1, 2 or infinitely manyends, for left cancellative semigroups and that the cardinality of the set of ends is invariant insubsemigroups and extension of finite Green index in left cancellative semigroups. 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 top-ics as distance-transitive graphs [9], groups with context-free word problem [10], pursuit-evasionproblems in infinite graphs [13] and to describe accessible groups [3]. This paper follows the trendof relating geometric properties of Cayley graphs of semigroups to algebraic properties, see forexample [5], [12] and [8]. We consider the notions of ends for a semigroup and try to recover somebasic theorems from the theory of ends of groups.In this paper we consider a definition for ends of digraphs introduced by Zuther in [16] andapply it to the left and right Cayley graphs of a semigroup. In [7] Jackson and Kilibarda introducea notion of ends for semigroups which is based on the ends of the underlying undirected graph ofthe Cayley graph. They prove that the number of ends of a semigroup is invariant under change offinite generating set and provide examples of semigroups with n ends in the left Cayley graph and m ends in the right Cayley graph for any prescribed n, m ∈ N . We argue that although there aremany ways to generalise the notion of ends to a semigroup; by preserving the notion of directionthere is a greater chance of interrelating the algebraic structure and the ends.In the remainder of this section we introduce the relevant definitions and technical resultsrequired to prove the main theorems of this paper. In Section 2, we prove that the structureof the ends of a semigroup is invariant under change of finite generating set and at the sametime is inherited by subsemigroups and extensions of finite Rees index. In Section 3, we provean analogue of Hopf’s Theorem, stating that a group has 1, 2 or infinitely many ends, for leftcancellative semigroups and that the cardinality of the set of ends is invariant in subsemigroupsand extensions of finite Green index in left cancellative semigroups.Let Ω be any set and let Γ ⊆ Ω × Ω. We will refer to Γ as a digraph on Ω, the elements ofΩ as the vertices of Γ, and the elements of Γ as edges . A walk in Γ is just a (finite or infinite)sequence ( v , v , . . . ) of (not necessarily distinct) vertices such that ( v i , v i +1 ) ∈ Γ for all i . An anti-walk is a sequence ( v , v , . . . ) of (not necessarily distinct) vertices such that ( v i +1 , v i ) ∈ Γ forall i . A path in Γ is just a walk consisting of distinct vertices. If α = ( α , α , . . . , α n ) is a walkin Γ, then the length of α is n and it is straightforward to verify that α contains a path from α to α n . A ray in Γ is just an infinite path ( v , v , . . . ) such that ( v i , v i +1 ) ∈ Γ for all i and an anti-ray is an infinite path ( v , v , . . . ) such that ( v i +1 , v i ) ∈ Γ for all i . If α = ( α , α , . . . , α m )and β = ( β , β , . . . , β n ) are arbitrary finite sequences of elements from Ω, then we denote by α a β the sequence ( α , α , . . . , α m , β , β , . . . , β n ).The out-degree of a vertex α in a digraph Γ is just |{ β ∈ Ω : ( α, β ) ∈ Γ }| . A digraph Γ is out-locally finite if every vertex has finite out-degree.If Σ ⊆ Ω, then Γ ∩ (Σ × Σ) is the induced subdigraph of Γ on Σ. If Σ and Σ ′ are infinite subsetsof Ω, then we write Σ ′ Σ if there exist infinitely many disjoint paths (including paths of length
0) in Γ with initial vertex belonging to Σ ′ and final vertex belonging to Σ. It is straightforwardto verify that is reflexive on infinite subsets of Ω but not necessarily transitive, symmetric, oranti-symmetric. However, it was shown in [16] that if is restricted to the set of rays and anti-rayson Γ, then it is transitive, and hence a preorder. If Σ ′ Σ and Σ Σ ′ , then we write Σ ≈ Σ ′ .It follows that ≈ is an equivalence relation and induces a partial order on ≈ -classes of rays andanti-rays. As such we refer to rays as being equivalent if they belong to the same ≈ -class; and inequivalent otherwise. We denote this poset by ΩΓ, and we refer to ≈ -classes of rays as the ends of Γ. Lemma 1.1.
Let Γ be a digraph on Ω and let α = ( α , α , . . . ) be an infinite walk (or anti-walk)in Γ such that every vertex of α occurs only finitely many times. Then α contains a ray r (oranti-ray, respectively) such that r is has infinitely many disjoint paths to every infinite subset of { α , α , . . . } and every infinite subset of { α , α , . . . } has infinitely many disjoint paths to r .Proof. We prove that α contains a ray in the case that α is a walk; an analogous argument provesthat α contains an anti-ray in the case that it is an anti-walk.Let a (0) = 1 and for every i ≥ a ( i ) = max { j ∈ N : α j = α a ( i − } + 1, i.e. α a ( i ) − isthe last appearance of α a ( i − in α . We will show that r = ( α a (0) , α a (1) , . . . )is the required ray. Since ( α i , α i +1 ) ∈ Γ for all i , in particular, ( α a ( i − , α a ( i ) ) = ( α a ( i ) − , α a ( i ) ) ∈ Γ . Hence r is an infinite walk where α a ( i ) = α a ( j ) for all i, j ∈ N such that i = j and so r is a ray.Let Σ be any infinite subset of { α , α , . . . } . If infinitely many elements in Σ are vertices of r ,then r is equivalent to Σ. If only finitely many elements of Σ belong to r , then Σ \{ α a (0) , α a (1) , . . . } is equivalent to Σ and so we may assume without loss of generality that Σ contains no elementsin { α a (0) , α a (1) , . . . } .We define infinitely many disjoint paths from Σ to r by induction. Let b (0) ∈ N be any numbersuch that α b (0) ∈ Σ. Then there exists k (0) ∈ N such that a ( k (0)) < b (0) < a ( k (0) + 1) and β := ( α a ( k (0)) , α a ( k (0))+1 , . . . , a b (0) , . . . , α a ( k (0)+1) ) is a walk from α a ( k (0)) in r to α a ( k (0))+1 in r via a b (0) ∈ Σ. Since every finite walk contains a path, we conclude that there is a path containedin β from a vertex of r to a b (0) ∈ Σ and a path back from a b (0) to a vertex of r .Suppose that we have defined b (0) , . . . , b ( i − , k (0) , . . . , k ( i − ∈ N and finite walks β , β , . . . , β i − for some i ≥
1. Choose k ( i ) , b ( i ) ∈ N so that b ( i ) > a ( k ( i )), α j does not equal any vertex in anyof β , β , . . . , β i − for all j > a ( k ( i )), and α b ( i ) ∈ Σ. Then we define β i = ( α a ( k ( i )) , α a ( k ( i ))+1 , . . . , α b ( i ) , . . . , α a ( k ( i )+1) ) . By construction, if i = j , then β i and β j are disjoint and so we have infinitely many disjointpaths (contained in the β i ) from r to Σ and back, as required. (cid:3) Lemma 1.2.
Let Γ be an out-locally finite digraph on a set X and let w , w , . . . be finite walksof bounded length in Γ with distinct final vertices. Then every vertex in the sequence w a w a · · · occurs only finitely many times.Proof. Let K ∈ N be bound on the lengths of w , w , . . . . If a vertex v occurs in infinitely manyof w , w , . . . , then the set B of vertices that can be reached from v by a path of length at most K contains the final vertex of w i for infinitely many i ∈ N . But the final vertices of the w i aredistinct and so B is infinite, contradicting the assumption that Γ is out-locally finite. (cid:3) Lemma 1.3.
Let Γ be an out-locally finite digraph on Ω , let Σ ⊆ Ω be infinite and let α ∈ Ω such that there is a path from α to every β ∈ Σ . Then there exists a ray r in Γ starting at α such that Σ r .Proof. We construct r recursively. Start by setting Σ := Σ and let P be a set containing preciselyone path q β from α to β for all β ∈ Σ . Then, since α has finite out-degree and there is a pathin P from α to every β ∈ Σ , there exists a vertex γ such that ( α , γ ) ∈ Γ and there is a path q β ∈ P from α via γ to every β in the infinite subset Σ ⊆ Σ . NDS OF SEMIGROUPS 3
Let β ∈ Σ be fixed and also fix a path p = ( δ = α , δ = γ , δ , . . . , δ n − , δ n = β ) . Let P = { q β ∈ P : β ∈ Σ } . If β ∈ Σ is arbitrary and q β ∈ P , then there exists i ( β ) ∈ N suchthat δ i ( β ) is the last vertex belonging to both the paths p and q β . The number i ( β ) exists since,in particular, both paths go through γ . By the pigeonhole principle, there exists m ∈ N such that2 ≤ m ≤ n and Σ = { β ∈ Σ : i ( β ) = m } is infinite. Set α = δ m . Since m ≥ α = α and, byconstruction, there is a path from α to every element β of the infinite set Σ (consisting of thevertices between α and β in q β ∈ P ) such that the only vertex in p and this path is α . Set P to the set of paths from α to β ∈ Σ from the previous sentence.We may repeat the above process ad infinitum to obtain for all i > β i +1 ∈ Σ i +1 and a path p i +1 ∈ P i +1 from α i to β i +1 , an α i +1 in p i +1 , an infinite Σ i +2 ⊆ Σ i +1 and an infinite set P i +2 of paths from α i +1 to every element of Σ i +2 such that the only vertex in p i +1 and any path in P i +2 is α i +1 .Hence there is a walk r containing { α i : i ∈ N } consisting of the vertices on the paths p i +1 between α i and α i +1 . In fact, by construction, the only vertex on both p i and p i +1 is α i +1 , andso the walk r is a ray. Moreover, there are infinitely many paths from r to Σ consisting of theremaining vertices on p i +1 between α i +1 and β i +1 . Again by construction the only vertex on both p i and p i +1 is α i +1 and so the paths from α i +1 ∈ r to β i +1 ∈ Σ are disjoint. (cid:3) The ends of a semigroup
Throughout this section, we let S be a finitely generated semigroup and let A be any finitegenerating set for S . The right Cayley graph Γ r ( S, A ) of S with respect to A is the directed graphwith vertex set S and edges ( s, sa ) ∈ Γ r ( S, A ) for all s ∈ S and for all a ∈ A . We refer to a as the label of the edge ( s, sa ). The left Cayley graph Γ l ( S, A ) is defined dually.If S is a semigroup, then the dual S ∗ of S is just the set S with multiplication ∗ defined by x ∗ y = yx for all x, y ∈ S . It follows directly from the definition that Γ l ( S, A ) = Γ r ( S ∗ , A ).Therefore to understand the end structure of a semigroup it suffices to study right Cayley graphsonly.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 |h T, s i \ T | = n ∈ N . Then there exists N ∈ N such that for all b , b , . . . , b n ∈ A ∪ { s } if s, sb , . . . , sb · · · b n are distinct, then there exists i ≤ n and a , a , . . . , a j ∈ A such that j ≤ N and sb · · · b i = a · · · a j .Proof. Let X := { sc c . . . c i ∈ T : c j ∈ A ∪ { s } , ≤ i ≤ n } . Then X is finite and so there exists N ∈ N such that every element of X can be given as a product of elements of A of length at most N . By the pigeonhole principle, there exists i such that sb · · · b i ∈ T and hence sb · · · b i ∈ X . Itfollows that there exist a , . . . , a j ∈ A such that j ≤ N and sb · · · b i = a · · · a j , as required. (cid:3) 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 . If h T, s i \ T is finite, then ΩΓ r ( T, A ) is isomorphic (as a partially ordered set)to ΩΓ r ( h T, s i , A ∪ { s } ) .Proof. For the sake of brevity, we denote Γ r ( h T, s i , A ∪{ s } ) by Γ. We use to denote the preorderdefined above on the rays and anti-rays of Γ. We prove the proposition in two steps. The first stepis to show that every ray or anti-ray in Γ is equivalent to a ray or anti-ray with vertices in T andedges labelled by elements of A . The second step is to show that if r and r ′ are rays or anti-rayswith vertices in T , edges labelled by elements of A , and r r ′ , then there exist infinitely manydisjoint paths from r ′ to r with edges labelled by elements of A . So, the first step ensures thatevery end ω of Γ, contains a ray or anti-ray r ω with vertices in T and edges labelled by elements of A . The second step implies that the mapping Ψ : ΩΓ −→ ΩΓ r ( T, A ) defined so that Ψ( ω ) equalsthe end of ΩΓ r ( T, A ) containing r ω is an isomorphism. We only give the proof of these steps forrays, an analogous argument can be used for anti-rays. S. CRAIK, R. GRAY, V. KILIBARDA, J. D. MITCHELL, AND N. RUˇSKUC
Let U = h T, s i \ T , let n = | U | , and let r = ( x, xb , xb b , . . . ) be a ray in Γ for some b , b , . . . ∈ A ∪ { s } and x ∈ h T, s i . Since U is finite, only finitely many elements of r can lie in U , and so wemay assume without loss of generality that x, xb , xb b , . . . ∈ T . If b i = s for all i , then there isnothing to prove.If b k is the first occurrence of s in { b , b , . . . } , then since the vertices of r are distinct soare the elements b k , b k b k +1 , . . . , b k b k +1 · · · b k + n . Hence by Lemma 2.1 there exist i, N ∈ N and a , a , . . . , a j ∈ A such that j ≤ N and b k b k +1 · · · b k + i = sb k +1 · · · b k + i = a · · · a j . Hence w = ( xb · · · b k − , xb · · · b k − a , . . . , xb · · · b k − a · · · a j )is a walk in Γ with vertices in T and edges labelled by elements of A . We repeatedly apply Lemma2.1 to successive occurrences of s in { b , b , . . . } to obtain finite walks w , w , . . . with vertices in T and edges labelled by elements of A . The length of w i is bounded by N for all i ∈ N and thefinal vertices are distinct, and hence by Lemma 1.2 every vertex in the sequence w a w a · · · occursonly finitely many times. Let w be the walk obtained by replacing the subpaths of r by the w i .Every vertex of w not in some w i occurs only once, since r is a ray. Hence every vertex of w occurs only finitely many times, and so by Lemma 1.1 there is a subray r ′ of w such that r ≈ r ′ ,as required.For the second step of the proof, let r and r ′ be rays in Γ with vertices in T , edges labelled byelements of A , and r r ′ . Since r r ′ , there exist infinitely many disjoint paths in Γ from r ′ to r . We may assume without loss of generality that there are at least n vertices in each of thesepaths after the last occurrence of s as an edge label. Hence, by repeatedly applying Lemma 2.1,there exists N ∈ N and infinitely many paths from r ′ to r labelled by elements of A . Moreover,there is a path of length at most N from every element in one of the new paths to some elementin the original path it was obtained from by applying Lemma 2.1. If infinitely many of these newpaths are disjoint, then there is nothing to prove. Otherwise infinitely many of these paths havenon-empty intersection with a finite subset of T , and so infinitely many paths contain some fixedelement t ∈ T . Hence there are path of length at most N from t to infinitely many vertices in theoriginal paths, which contradicts the out-local finiteness of Γ. (cid:3) Corollary 2.3.
Let S be a finitely generated semigroup and let A and B be any finite generatingsets for S . Then ΩΓ r ( S, A ) is isomorphic (as a partially ordered set) to ΩΓ r ( S, B ) .Proof. It suffices to show that ΩΓ r ( S, A ) is isomorphic to ΩΓ r ( S, A ∪{ s } ) for any s ∈ S , since thenΩΓ r ( S, A ) is isomorphic to ΩΓ r ( S, A ∪ B ) is isomorphic to ΩΓ r ( S, B ), as required. Certainly S isa finitely generated subsemigroup of S such that h S, s i \ S is finite, and so it follows by Proposition2.2 that ΩΓ r ( S, A ) is isomorphic to ΩΓ r ( S, A ∪ { s } ), as required. (cid:3) Following from Corollary 2.3 we define Ω S = ΩΓ r ( S, A ) for any finite generating set A of S .We refer to Ω S as the ends of S .Note that if S is a finitely generated group, then it follows by Hopf’s Theorem [6, Satz II ] thatthe ends of S form an anti-chain with 1, 2, or 2 ℵ elements.In section 5 we give examples of finitely generated semigroups with any finite number or ℵ ends (Examples 5.5, 5.3). Any group with 2 ℵ group ends will also have 2 ℵ ends as a semigroup.It is easy to see that the free monoid on two generators will have 2 ℵ ends as all pairs of rays areincomparable. It is not known whether, in the absence of the Continuum Hypothesis, there existsa finitely generated semigroup S such that Ω S has κ elements where ℵ < κ < ℵ . The questionof which posets can occur as the partial order of ends Ω S of some finitely generated semigroup S is unresolved.If S is a semigroup and T is a subsemigroup of S , then the Rees index of T in S is just | S \ T | +1. Corollary 2.4.
Let S be a finitely generated semigroup and let T be a subsemigroup of S of finiteRees index. Then the partial order Ω S of the ends of S is isomorphic to the partial order Ω T ofthe ends of T .Proof. Since S is finitely generated, it follows by [14, Theorem 1.1], that T is finitely generated.Let A be any finite generating set for T and let s ∈ S \ T be arbitrary. Then h T, s i\ T ⊆ S \ T and so NDS OF SEMIGROUPS 5 h T, s i\ T is finite. It follows from Proposition 2.2 that ΩΓ r ( T, A ) is isomorphic ΩΓ r ( h T, s i , A ∪{ s } ),and hence Ω T is isomorphic to Ω h T, s i . Since T (cid:12) h T, s i ≤ S , by repeating this process (at most | S \ T | times) we have shown that Ω S is isomorphic to Ω T . (cid:3) The number of ends of a left cancellative semigroup
In this section we prove that left cancellative semigroups can only have a restricted number ofends, unlike the general case (See Proposition 5.5).A semigroup S is left cancellative if x = y whenever ax = ay where a, x, y ∈ S . Right cancellative is defined analogously. A semigroup is cancellative if it is both left and right cancellative.A left or right cancellative monoid contains only one idempotent (the identity). A left cancella-tive semigroup contains at most one idempotent in every L -class, and the analogous statementholds for right cancellative semigroups. The structure of a cancellative semigroup S is straightfor-ward to describe: either S is R - trivial or S is a monoid with group of units G , every R -class is ofthe form xG , and every L -class is of the form Gx for some x ∈ S (see for example [11]). We startthis section by giving an analogous description of the structure of a left cancellative semigroup. Itis possible to deduce these results from [15] although they are not couched in this notation, andso we include a proof for the sake of completeness.A right group is the direct product of a group G and right zero semigroup E . Theorem 3.1. [2, Theorem 1.27]
A semigroup is a right group if and only if it is left cancellativeand R -simple. Proposition 3.2.
Let S be a left cancellative semigroup and let U be the set of regular elementsin 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-trivial R -class R x , then R x = xU ; (iv) if x ∈ S is arbitrary and U is non-empty, then xU is an R -class of S (not necessarilycontaining x ).Proof. Let x, y ∈ S and assume that xy is a regular element. It follows that there exists z ∈ S such that xyzxy = xy . By cancelling we see that yzxy = y and hence y must be a regular element.From yzxy = y we must have yzxyzx = yzx , again by cancelling we see xyzx = x and hence x isalso a regular element. This means that S \ U is an ideal of S .For the second part we assume that U is non-empty. If S contains a regular element then itcontains an idempotent. Let e and f be idempotents in S . Then e f = ef and f e = f e andso, by cancelling, ef = f and f e = e . Thus e R f and, since every regular element is R -related toan idempotent, the regular elements of S are contained in a single R -class of S . If x and y areregular, then there exists x ′ ∈ S such that xx ′ x = x and, since y R x ′ , there exists z ∈ S such that yz = x ′ . Hence xyzxy = xx ′ xy = xy and so xy is regular. Hence, U is a subsemigroup of S andby part one S \ U is an ideal and so U is R -simple. It follows from Theorem 3.1 that U is a rightgroupWe now proof parts three and four together. Let x ∈ S and assume U is non-empty. Let e ∈ U be an idempotent. Then as all elements in U are R related xU is contained within an R -class,say R . Let y be an element of R distinct from xe . Then there exists s, t ∈ S such that xes = y and yt = xe . It follows that xest = xe and hence xestst = xest . By cancelling ( st ) = st is anidempotent and as S \ U is an ideal s, t ∈ U . This means that R = xU . If x lies in a non-trivial R -class then there exists y R x such that y = x and there exists s, t ∈ S such that xs = y and yt = x . Then as before we see that st is an idempotent and x = xst so x ∈ xU . (cid:3) Lemma 3.3.
A left cancellative semigroup S has either one R -class or infinitely many R -classes.Proof. We show that either S is regular or ( x, x )
6∈ R for some x ∈ S . Suppose that ( x, x ) ∈ R for all x ∈ S . Then there exists s ∈ S such that x s = x . Hence x st = xt and so ( xs ) t = t for all t ∈ S . Hence xs is a left identity for S and so xs is an idempotent and x R xs . Thus S is regular S. CRAIK, R. GRAY, V. KILIBARDA, J. D. MITCHELL, AND N. RUˇSKUC and so by [2, Exercise 1.11.4] has only one R -class. If there exists x ∈ S such that ( x, x )
6∈ R ,then ( x i , x j )
6∈ R for all i, j ∈ N such that i = j . Hence S has infinitely many R -classes. (cid:3) Corollary 3.4. If S has infinitely many R -classes at least one of which is infinite, then it hasinfinitely many infinite R -classes.Proof. Since there is at least one infinite R -class in S , that R -class is of the form yU for some y ∈ S , and | yU | = | U | by left cancellativity, it follows that U is infinite. From the proof of Lemma3.3, there exists x ∈ S such that ( x i , x j )
6∈ R for all i, j ∈ N such that i = j . By Proposition 3.2, x i U is an R -class of S for all i ∈ N and | x i U | = | U | and, in particular, x i U is infinite for all i ∈ N .It suffices to show that the sets x i U are disjoint. Suppose to the contrary that x i U ∩ x j U = ∅ for some i, j ∈ N with i < j . Then, by left cancellativity, x j − i U ∩ U = ∅ and so x j − i ∈ U since S \ U is an ideal. Therefore infinitely many powers of x , namely, x j − i , x j − i , . . . , are R -related,contradicting our assumption. (cid:3) Right groups are a special case of Rees matrix semigroups where | I | = 1 and the multiplicationmatrix P consists of identity elements. Hence as a corollary to Proposition 5.5 below we have. Corollary 3.5.
Let G be a finitely generated group and let E be a finite right zero semigroup.Then | Ω( G × E ) | = | Ω 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 any finite generating set contains a ray r and there is an s ∈ S such that there are paths from infinitely many points in r to s , then Ω S is infinite.Proof. Let A be any finite generating set for S and let r = ( r , r , . . . ) be a ray in Γ r ( S, A ). Wemay write r as a product a · · · a n of generators in A .Assume, seeking a contradiction, that S has finitely many ends. Since S is left cancellative, r i = ( s i , s i a , . . . , s i a · · · a n = s i r , s i r , . . . ) is a ray for all i ∈ N . Thus, by assumption, thereexist i, j ∈ N such that i < j and r i ≈ r j . Again using the left cancellativity of S , it follows that r j − i ≈ r and so there is a path from s j − i r k to r l for some k, l ∈ N . There is a path from s to s j − i r k and hence to r l . But in this case, r l R r l +1 R · · · and so S has an infinite R -class, which isa contradiction. (cid:3) The main results of this section are given below.
Theorem 3.7.
Let S be an infinite finitely generated left cancellative semigroup. Then | Ω S | = 1 , or | Ω S | ≥ ℵ .Proof. If S has only one R -class, then by [2, Theorem 1.27] it follows that S ∼ = G × E where G isa group and E is a right zero semigroup. Since S is finitely generated, it follows that G is finitelygenerated and E is finite. Hence, by Proposition 3.5, | Ω S | = | Ω G | and by Hopf’s Theorem [6,Satz I], | Ω G | = 1 , ℵ .Suppose that S has more than one R -class. Then Lemma 3.3 implies that S has infinitely many R -classes. If S contains an infinite R -class then then by Corollary 3.4 S contains infinitely manyinfinite R -classes. By Konig’s lemma each infinite R -class contains a ray, none of these rays canbe equivalent as the R -classes are distinct. Thus | Ω S | ≥ ℵ .Assume that S has no infinite R -class. Let Γ denote the Cayley group of S with respect tosome finite generating set A for S . If Γ contains a ray r and there is an s ∈ S such that there arepaths from infinitely many points in r to s , then Ω S is infinite by Lemma 3.6. If Γ contains ananti-ray r , then there exists a ∈ A such that infinitely many of the elements in r are of the form at for some t ∈ S . In particular, there is a path from a to every at in r and so by Lemma 1.3 thereexists a ray r ′ such that r r ′ . But then there are paths from infinitely many of the vertices in r ′ to any fixed element in r , and so Ω S is infinite by Lemma 3.6.Suppose that the Cayley graph of S does not have the property of Lemma 3.6. Seeking acontradiction assume that S has finitely many ends, and let r , r , . . . , r n be rays belonging indistinct ends such that the end containing r is minimal with respect to . Since r i r , there NDS OF SEMIGROUPS 7 exists a finite F ⊆ S such that all paths from r to every r i pass through F . By assumption thereexists element s in r such that there are no paths from s to any element of F and hence to anyelement in any r i . Since S is left cancellative, s r and s r are rays. If s r ≈ r i or s r ≈ r i , thenthere is a path from s to r i and so i = 1. In particular, s r ≈ s r and so, since S is left cancellative, r ≈ r , a contradiction. We have shown that S either has 1 or infinitely many ends. (cid:3) Corollary 3.8.
Let S be an infinite finitely generated cancellative semigroup that is not a group.Then | Ω S | = 1 or | Ω S | ≥ ℵ .Proof. Since S is cancellative, it is certainly left cancellative and so | Ω S | = 1 ,
2, or | Ω S | ≥ ℵ byTheorem 3.7. If | Ω S | = 2, then from the proof of Theorem 3.7, S has only one R -class and henceis a group. (cid:3) As mentioned above, it is not known what cardinalities Ω S can have, even for restricted typesof semigroups such as those which are left cancellative. We prove that Ω S has cardinality 2 ℵ fora particular type of cancellative semigroup. Ore’s Theorem (see for instance [2, Theorem 1.23])states that if a cancellative semigroup S satisfies the condition that sS ∩ tS = ∅ for all s, t ∈ S then S can be embedded in a group. Theorem 3.9.
A cancellative semigroup which cannot be embedded in a group has ℵ ends.Proof. Let S be a cancellative semigroup that cannot be embedded in a group. As S is not group-embeddable there exists s, t ∈ S such that sS ∩ tS = ∅ . Firstly we show that all elements of { s, t } ∗ are distinct. Let u = u u . . . u n , v = v , v . . . v m ∈ { s, t } ∗ and assume u = S v , without loss ofgenerality we assume the length of u is less than or equal to the length of v . If u is a prefix of v then u = v = uv ′ . It follows that v ′ is a left identity for all elements of S . The first letter of v ′ is (without loss of generality) s and hence tx = v ′ tx ∈ sS for all x ∈ S . If u is not a prefix of v then there exists a position i ≤ n such that u j = v j for all j < i but u i = v j . As u j = v j for all j < i and u = S v it follows by left-cancellativity that u i . . . u n = v i . . . v m and u i = v j , however, sS ∩ tS = ∅ a contradiction.We now show that for u, v ∈ { s, t } ∗ we have v ∈ uS if and only if u is a prefix of v . Clearly if u is a prefix of v then v ∈ uS . With the aim of getting a contradiction assume that u = u u . . . u n is not a prefix of v = v v . . . v m but v ∈ uS . This means there exists x ∈ S such that ux = v . As u is not a prefix of v there exists 1 ≤ i ≤ n such that u j = v j for all j < i but u i = v j . But thenby left-cancellativity u i . . . u n x = S v i . . . v m . Then as { u i , v i } = { s, t } it follows that sS ∩ tS = ∅ .Combining these facts gives a copy of the free semigroup on two generators as a subsemigroupof S and there can be no paths between elements, this means S has at least 2 ℵ ends. This is alsothe maximum possible number of ends so | Ω S | = 2 ℵ . (cid:3) Subsemigroups of finite Green index
It follows from Proposition 3.2 that if T is a subsemigroup of a left cancellative semigroup S ,then R T = R V where V is the right group of regular elements of T .Let S be a semigroup and let T be a subsemigroup of finite Green index. It was shown in[1] that S is finitely generated if and only if T is finitely generated. If T is a submonoid of aleft-cancellative monoid S and T has finite Green index in S , then, since the complement is anideal, the group of units of T has finite index in the group of units of S . Lemma 4.1. If S = G × E is a right group, G is infinite and T is a subsemigroup of finite Greenindex then T = H × E is a right group where H is of finite index in G .Proof. One can see S has only one R S -class, therefore the H S -classes of S are the L S -classes of S . As ( g, e ) · ( h, f ) = ( gh, f ) we see that L S -classes are of the form G × { e } for each e ∈ E .If T contains no elements of the form ( g, e ) for some fixed e ∈ E then the R T -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 = e and thenthere exists no element ( g ′ , f ′ ) ∈ T such that ( hg, f )( g ′ , f ′ ) = ( h, e ) as T contains no elements ofthe form ( g, e ).For each e ∈ E we let H e be those elements h ∈ G such that ( h, e ) ∈ T . We now showeach H e contains 1 G . Let e ∈ E . One can see H e is a subsemigroup of G as in particular S. CRAIK, R. GRAY, V. KILIBARDA, J. D. MITCHELL, AND N. RUˇSKUC ( g, f )( h, e ) = ( gh, e ) so H f H e ⊆ H e . It is easy to see that a subsemigroup of finite Rees index in G is equal to G so we may assume G \ H e is infinite. As H T -classes are contained in H S -classes andas G \ H e is infinite we must have at least one non-trivial H T -class containing distinct elements( g, e ) , ( g ′ , e ) with g, g ′ H e . As these elements are H T -related they are R T -related and hencethere exists ( h, f ) , ( h ′ , f, ) ∈ T such that ( g, e )( h, f ) = ( g ′ , e ) and ( g ′ , e )( h ′ , f ′ ) = ( g, e ). Thismeans f = f ′ = e and furthermore that ghh ′ = g . It follows hh ′ = 1 G is an element of H e . Hence, H e ⊆ H f for all e, f ∈ E so H e = H f for all e, f ∈ E . We call this semigroup H .As H × E has finite Green index in G × E it must follow that H has finite Green index in G .It was shown in [4, Corollary 34] that if H is a subsemigroup of finite index in a group G then H is a subgroup of G with finite group index. (cid:3) 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 a , a , . . . , a m + k ∈ A such that the number of R T -classescontaining any of sa a · · · a m +1 , sa a · · · a m +2 , . . . , sa a · · · a m + k is at least n , then there exist i > m and b , b , . . . , b j ∈ B such that sa · · · a i = b · · · b j .Proof. If sa a · · · a m +1 , sa a · · · a m +2 , . . . , sa a · · · a m + k contains elements from n R T -classesthen sa a · · · a m + i is an element of T for some i . Any element of T can be expressed over B andhence there exists b , b . . . b j ∈ B such that sa a · · · a m + i = b b · · · b j . (cid:3) Theorem 4.3.
Let S be a finitely generated left cancellative semigroup and let T be a subsemigroupof S of finite Green index. Then | Ω S | = | Ω T | .Proof. If S is right simple, then S ∼ = G × E for some finitely generated group G and E is a finiteright zero semigroup. Since T has finite Green index in S , it follows that T ∼ = H × E where H is asubgroup of finite index in G . In other words, T is a right group and so | Ω T | = | Ω H | = | Ω G | = | Ω S | by Lemma 3.5.Let U be the right group of regular elements in S . Since S is finitely generated, it follows that T is finitely generated. Let A and B be finite generating sets for S and T , respectively, such that B ⊆ A . Since S \ 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 Proposition 3.2 that R T = R V , and so V has finite Green index in U .Suppose that S has more than one R -class. Then, by Lemma 3.3, S has infinitely many R -classes. If S has no infinite R -classes, then since R T -classes are contained in R S -classes, it followsthat T has finite Rees index in S and so by Corollary 2.4, the theorem follows. We now considerthe case that S has infinitely many infinite R -classes. By Proposition 3.5, U either has 1, 2, or2 ℵ ends.If U has 2 ℵ ends, then, since S \ U is an ideal, S has 2 ℵ ends. Since V has finite Green indexin U and U is a right group, V has 2 ℵ ends and so T has 2 ℵ ends also.Suppose that U has 1 or 2 ends. Then S and T have at least ℵ ends, since every pair of infinite R -classes contain a pair of inequivalent rays. Let Σ( S ) be the set of ends of S containing a ray thathas non-empty intersection with infinitely many R S -classes. By [16, Lemma 2.8], if ω is an end ofΓ r ( S, A ), then every ray in ω is contained in a strongly connected component or intersects infinitelymany strongly connected components (but not both). Since connected components of Γ r ( S, A )are precisely R S -classes, it follows that | Ω S | = max {ℵ , | Σ( S ) |} and | Ω T | = max {ℵ , | Σ( T ) |} . Weconclude the proof by showing that | Σ( S ) | = | Σ( T ) | .Let r be a ray or anti-ray in Γ r ( S, A ) that has non-empty intersection with infinitely many R S -classes. Since every R S -class is a union of R T -classes, r has non-empty intersection with infinitelymany R T -classes. Since there are only finitely many R T -classes in S \ T , we may assume withoutloss of generality that the elements in r are in T . Let n be the number of R T -classes in S \ T and let ( xc , xc c , . . . , xc · · · c m ) be a subpath of r that has non-empty intersection with n + 1, R T -classes. By left cancellativity, the path ( c , c c , . . . , c · · · c m ) has non-empty intersectionwith at least n + 1 R T -classes also. It follows that there exists i such that c · · · c i ∈ T . Hence c · · · c i is a product b b · · · b j of elements in the generating set B for T . Recursively replacingevery such path ( xc , xc c , . . . , xc · · · c i ) by the corresponding walk ( xb , xb b , . . . , xb · · · b j ) we NDS OF SEMIGROUPS 9 obtain a walk w = ( w , w , . . . ) in Γ r ( T, B ) that has non-empty intersection with infinitely many R T -classes contained in T . If i < j and w i R T w j , then w i R T w i +1 R T · · · R T w j . But w has non-empty intersection with infinitely many R T -classes and so every vertex of w occurs only finitelymany times. Hence, by Lemma 1.1, w is equivalent to a ray or anti-ray in Γ r ( T, B ).Let r be a ray or anti-ray and let r be a ray or anti-ray in Γ r ( T, B ) such that r and r have non-empty intersection with infinitely many R S -classes. If r is equivalent to r in Γ r ( T, B ),then clearly r is equivalent to r in Γ r ( S, A ). Suppose that r is equivalent to r in Γ r ( S, A ).In this case, there are infinitely many disjoint paths from r to r and vice versa. By repeatedlyapplying Lemma 4.2, there exist infinitely many paths from r to r labelled by elements of B . Ifinfinitely many of these paths are disjoint, then the proof is complete. Otherwise infinitely manyof these paths have non-empty intersection with a finite subset of S , and so infinitely many pathscontain some fixed element s ∈ S . But then there exists a path from s to element in r and apath from that vertex to an element in r , and so infinitely many elements in r are R S -related, acontradiction. We have shown that for all rays or anti-rays r and r in Γ r ( T, B ) such that r and r have non-empty intersection with infinitely many R S -classes, r is equivalent to r in Γ r ( T, B )if and only if they are equivalent in Γ r ( S, A ). Therefore | Σ( S ) | = | Σ( T ) | , as required. (cid:3) Examples
In this section we give several examples of finitely generated semigroups S and describe Ω S forthese examples.The following example shows that unlike in the groups case it is possible for a left cancellativesemigroup to have ℵ ends. Example 5.1.
The semigroup N × N under componentwise addition has ℵ ends. For the sakeof brevity we use Γ to denote the Cayley graph Γ r ( N × N , { (0 , , (1 , } ). We show that any rayin Γ is equivalent to one of(( i, , ( i, , ( i, , . . . ) , ((0 , i ) , (1 , i ) , (2 , i ) , . . . ) or ((0 , , (1 , , (1 , , (2 , , (2 , . . . )for each i ∈ N . We first note that there are no anti-rays in Γ. Any ray either contains finitely manyelements in the first component of its vertices, finitely many elements in the second component ofits vertices or infinitely many distinct elements in both components. In the first case as elementsare eventually of the form ( i, j ) for some fixed i the ray is equivalent to (( i, , ( i, , ( i, , . . . ).Equivalently if the ray has finitely many elements in the second component of its vertices thenit will be equivalent to some ((0 , i ) , (1 , i ) , (2 , i ) , . . . ). In the case that the ray has infinitely manydistinct elements in both components then for any element ( i, j ) where i < j there is a path from( i, i ) to ( i, j ) to ( j, j ) and we see that the ray is equivalent to ((0 , , (1 , , (1 , , (2 , , (2 , , . . . ).The following example demonstrates the existence of anti-rays which are not equivalent to anyray. It also shows that it is possible to have anti-rays in a semigroup with trivial R -classes. Example 5.2.
Let M be the monoid h a, b | aba = b i . It is easy to check that aba −→ b and b a −→ ab is a complete rewriting system. In a similar way to Example 5.1 we can show thatthis monoid has ℵ ends.The following example demonstrates that in general a subsemigroup of finite Green index mayhave a different number of ends from the original semigroup. Example 5.3.
Let { , } be the semigroup with the usual multiplication (of real numbers).Consider the semigroup Z × Z × { , } . Then T = Z × Z × { } is a subsemigroup and Z × Z × { } is an H T -class in the complement. Hence Z × Z × { } has finite Green index in Z × Z × { , } .However, by inspection we see Z × Z ×{ , } has 2 ends corresponding to Z × Z ×{ } and Z × Z ×{ } ,however, Z × Z × { } has only 1 end. For a diagram of a portion of the right Cayley graph of Z × Z × { , } see Figure 2.Following Theorem 4.3 one might question whether for a left cancellative semigroup it is possibleto show that the end poset of a subsemigroup of finite Green index is isomorphic to the end posetof the semigroup. The following example answers this in the negative. Figure 1.
A portion of the right Cayley graph of h a, b | aba = b i from Example5.2, edges labelled by a are represented with solid lines and those labelled by b with dashed lines. Figure 2.
A portion of the right Cayley graph of the semigroup defined by thepresentation Z × { , } from Example 5.3. Example 5.4.
Consider the semigroup S = Z × Z × N under componentwise addition. Thesubsemigroup T = Z × Z × ( N \ { } ) is of finite Green index as the complement consists of 1 H T -class. One can see that S has ℵ ends corresponding to each Z × Z × { i } and to { } × { } × N .In the poset of ends of S any two elements are comparable. Either by inspection or by Theorem4.3 we see that T also has ℵ ends. However, the are no paths from Z × Z × { } to Z × Z × { } 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 any n, m ∈ N there exists a semigroup with n left ends and m rightends.Recall a Rees matrix semigroup M [ G ; I, Λ; P ] has elements I × G × Λ where G is a groupand I and Λ are index sets. Multiplication is defined by ( i, g, λ )( j, h, µ ) = ( i, gp λj h, µ ) where P = ( p λj ) λ ∈ Λ ,j ∈ I is a | Λ | × | I | matrix over G . Proposition 5.5. If S is the Rees matrix semigroup M [ G ; I, Λ; P ] where I = { i , i , . . . , i n } and Λ = { λ , λ , . . . λ m } , 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 · | Ω G | and the left ends of S form ananti-chain of size m · | Ω G | .Proof. Let X be a finite semigroup generating set for G containing 1 G and let A = { ( i, p − µ,j x, λ ) | x ∈ X, λ, µ ∈ Λ , i, j ∈ I } . Clearly A is a finite generating set for S .Let Γ i be the induced subgraph of Γ r ( S, A ) on the vertices { i } × G × Λ and let Γ i,λ be thesubgraph of Γ i with vertices { i } × G × { λ } and edges with labels ( i, p − λ,i x, λ ). As ( i, g, λ )( j, h, µ ) =( i, gp λ,j h, µ ) note that Γ r ( S, A ) is the disjoint union of the Γ i . This means that ΩΓ r ( S, A ) is n incomparable copies of ΩΓ i . As all ends in Ω G are incomparable it suffices to show that ΩΓ i isisomorphic to Ω G for all i ∈ I .We first note that for a fixed λ ∈ Λ, Γ i,λ is isomorphic to Γ r ( G, X ). We now prove thatany ray in Γ i is equivalent to a ray in Γ i,λ , the proof for anti-rays is analogous. Let r = NDS OF SEMIGROUPS 11 (( i, g , λ j ) , ( i, g , λ j ) . . . ) be a ray and let r ′ = (( i, g , λ ) , ( i, g , λ ) . . . ) be a sequence in Γ i,λ .We show that there is an infinite walk w in Γ i,λ containing r ′ in which every vertex appearsfinitely often.We construct w by concatenating the shortest paths in Γ i,λ between each ( i, g k , λ ) and ( i, g k +1 , λ ),these shortest paths exist because Γ i,λ is isomorphic to Γ( G, X ). Next we show that there is aglobal bound on the lengths of these shortest paths. If ( i, g, µ ) = ( i, h, ν )( j, p − ξ,k x, π ) then it follows µ = π and g = hp ν,j p − ξ,k x . This means the shortest path in Γ i,λ between any consecutive elementsof r ′ is of length less than K = max {| p j,µ p − k,ν | X : j, k ∈ I, µ, ν ∈ Λ } + 1. As r is a ray it followsthere are at most | Λ | repetitions of vertices in r ′ . Every vertex of w has a path of length lessthan K to a vertex of r ′ and as Γ i,λ is out-locally finite this means that if some vertex v appearsinfinitely often in w then infinitely many elements of r ′ can be reached from v by a path of lengthless than or equal to K . But each vertex in r ′ appears at most | Λ | times so any infinite set ofelements of r ′ contains infinitely many vertices, a contradiction. By Lemma 1.1, w contains a ray s with infinitely many disjoint paths from s to and from r ′ and hence to and from r .This means any ray in Γ i is equivalent to a ray in Γ i,λ , to complete the proof we must nowverify that if we have rays r and r in Γ i,λ such that r r then r r in Γ i . Let r and r be incomparable rays in Γ i,λ , as the rays are incomparable in Γ i,λ there exists a finite set F = { ( i, f , λ ) , . . . , ( i, f m , λ ) } such that all paths from r to r in Γ i,λ pass through F . For anyedge (( i, g, µ ) , ( i, gp µ,j p − ν,k x, ξ )) we have a word w = x x . . . x p over X of minimal length suchthat w = G p µ,j p − ν,k x and a corresponding path(( i, g, µ ) , ( i, g, λ ) , ( i, gx , λ ) , . . . , ( i, gx x · · · x p , λ ) , ( i, gp µ,j p − ν,k x, ξ )) . This means that any path in Γ i has a corresponding walk in Γ i,λ such that any point on thewalk has a path of length less than K + 2 to a vertex on the path in Γ i . This means any path π from r to r in Γ i has a corresponding walk in Γ i,λ and this must pass through F and hence π must be contain an element that can be reached from F by a path of length less than or equal to K + 2. As Γ i is out-locally finite there are only finitely many such elements so r r . (cid:3) References [1] A. J. Cain, R. Gray and N. Rˇuskuc, Green index in semigroups: generators, presentations and automaticstructures,
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