Cancellative orders

David Easdown and Victoria Gould

December 1995

School of Mathematics and Statistics Department of Mathematics University of Sydney University of York Sydney NSW 2006 Heslington, York YO1 5DD Australia England

Abstract

A subsemigroup S of a semigroup Q is a left order in Q and Q is a semigroup of left quotients of S if every q ∈ Q can be written as q = a^{∗}b for some a, b ∈ S, where a^{∗} denotes the inverse of a in a subgroup of Q, and if, in addition, every square-cancellable element of S lies in a subgroup of Q. Perhaps surprisingly, a semigroup, even a commutative cancellative semigroup, can have non-isomorphic semigroups of left quotients. We show that if S is a cancellative left order in Q then Q is completely regular and the D-classes of Q are left groups. The semigroup S is right reversible and its group of left quotients is the minimum semigroup of left quotients of S.

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The authors are grateful to the ARC for its generous financial support.

1 Introduction

If S is a right reversible, cancellative semigroup, then as shown by Ore [CP], S has a group G of left quotients . That is, S is a subsemigroup of G and any element of G can be written as a^{−1}b where a and b are elements of S. Inspired by methods of both classical ring and semigroup theory, Fountain and Petrich in 1982 [FP] extended the notion of group of left quotients of a semigroup S to that of semigroup of left quotients of S. Their approach, which we adopt here, is to consider inverses of elements in any subgroup of a semigroup and not just the group of units (which may not exist).

Let S be a subsemigroup of a semigroup Q. Then Q is a semigroup of left quotients of S if every q ∈ Q can be written as q = a^{∗}b for some a, b ∈ S, where a^{∗} denotes the inverse of a in a subgroup of Q and if, in addition, every element of S satisfying a weak cancellation condition known as square cancellability lies in a subgroup of Q. Clearly if S has a group of left quotients G then G is also a semigroup of left quotients of S. If Q is a semigroup of left quotients of S then, borrowing terminology from ring theory, we also say that S is a left order in Q. Semigroups of right quotients and right orders are defined dually. If S is both a left order and a right order in Q then S is an order in Q and Q is a semigroup of quotients of S.

Unlike the case for groups and rings, a semigroup can have non-isomorphic semigroups of left quotients [FP]. Indeed a commutative cancellative semigroup can be an order in a non-trivial semilattice of groups, as well as, of course, an order in a group [G]. The thrust of many earlier papers was to consider semigroups that are (left) orders in semigroups in a well known class, for example the class of completely 0-simple semigroups, and in particular to choose classes such that semigroups of (left) quotients taken from the class are unique up to isomorphism.

Here we take a different approach, fixing a property of the orders under consideration. Specifically, we consider semigroups that are cancellative left orders. It is easy to see that such a semigroup S must be right reversible, and thus have a group of left quotients. On the other hand we show that if S is a left order in Q then Q is completely regular (a union of groups).

Many left orders naturally satisfy an additional property called straightness. Straight left orders in completely regular semigroups have been characterised by Smith in [P]. Left orders in completely regular semigroups need not be straight, however cancellative left orders must be. This enables us to use the results of [P] to describe all semigroups of left quotients of a given cancellative left order.

After some preliminaries in Section 2, we study in Section 3 cancellative left orders. We show that if S is such an order in Q, then Q is completely regular. Further, the D-classes of Q are left groups, that is, direct products of groups with left zero semigroups, and S is straight and local in Q. The group of left quotients of S is, in an established sense, the minimum semigroup of left quotients of S. If S is commutative then any semigroup of left quotients must be commutative [EG], so that if in addition S is cancellative, any semigroup of left quotients of S is a semilattice of commutative groups.

In the final section we specialise the result of [P] charactersising straight left orders in completely regular semigroups, to the case where the left order is cancellative. This enables us to present an example of a semigroup that is a left order in a free group on countably many generators and which is also a left order in a countable chain of left groups.

We would like to thank P.N. Anh for suggesting the topic of this paper.

2 Preliminaries

We assume the reader has a basic knowledge of algebraic semigroup theory, as may be found in the early chapters of [CP] or [H]. As far as possible we follow the notation and terminology of these references. However we deviate from the norm in denoting by a^{∗} the inverse of an element a of a semigroup Q in a subgroup containing a. Thus if a^{∗} exists, the H-class H_{a} is a subgroup of Q and a^{∗} is the inverse of a in H_{a}.

Recall that a semigroup S is right reversible if Sa ∩ Sb ≠ ∅ for all elements a, b of S. As shown in Theorem 1.24 of [CP], a semigroup S has a group G of left quotients if and only if it is right reversible and cancellative; transparently, all elements of S then lie in a subgroup of G. If a semigroup S has a semigroup of left quotients Q then only the square cancellable elements of S lie in subgroups of Q.

To define square cancellability, we make use of the generalisations L^{∗}, R^{∗} and H^{∗} of Green’s relations L, R and H. The relation L^{∗} is defined on a semigroup S by the rule that aL^{∗}b if and only if

ax = ay if and only if bx = by

for all x, y ∈ S^{1}. Lemma 1.1 of [F] explains that aL^{∗}b if and only if aLb in some oversemigroup of S. The relation R^{∗} is defined dually and we put H^{∗} = L^{∗} ∩R^{∗}. An element a of a semigroup S is square cancellable if aH^{∗}a^{2}. Being square cancellable is clearly a necessary condition for an element of S to lie in a subgroup of some oversemigroup. The definition of (left, right) semigroup of quotients insists that all square cancellable elements lie in subgroups of any semigroup of (left, right) quotients. If S is cancellative, then all elements of S are square cancellative and so if S is a left order in Q, all elements of S lie in subgroups of Q. P.N. Anh asked the natural question whether in this case Q must be completely regular, which we answer in the affirmitive in the next section.

In addition to Green’s relations L and R it is useful for our purposes to consider the preorders ≤_{L} and ≤_{R} with which they are associated. If a and b are elements of a semigroup Q, then a≤_{L}b if and only if Q^{1}a ⊆ Q^{1}b. Clearly ≤_{L} is a preorder right compatible with multiplication with associated equivalence relation L. The preorder ≤_{R} is defined dually. Where S is a (left) order in Q, the relations L, R, ≤_{L} and ≤_{R} will always refer to relations on Q.

Observe that if a and b are elements of a semigroup Q and aHa^{2}, then if b≤_{R}a we have aa^{∗}b = b and so bLa^{∗}b. Now if S is a left order in Q then any q ∈ Q can be written as q = c^{∗}d for some c, d ∈ S. Then q = (c^{∗})^{2}cd = a^{∗}b where a = c^{2}, b = cd and b≤_{R}a in Q. Thus qLb and S has non-empty intersection with every L-class of Q. If S has the additional property that any q ∈ Q can be written as q = a^{∗}b where a, b ∈ S and aRb in Q, then we say that S is a straight left order in Q. Note that in this case bRa^{∗}aRa^{∗}b = q so that bHq in Q and S has non-empty intersection with every H-class of Q. Moreover S is local in Q [G2], that is, S ∩ H is a left order in H for each group H-class H of Q. Many examples of straight left orders are known, indeed if S is a left order in Q and Q is a regular semigroup on which H is a congruence, then S is straight in Q [G2]. If S is a left order in a completely regular semigroup, then S need not be straight (see Example 4.2). However, we show in Theorem 3.3 that if S is a cancellative left order, then S must be straight.

3 Cancellative left orders

Our first aim is to show that if S is a cancellative left order in Q, then Q must be completely regular. As a preliminary we have

Lemma 3.1 Let S be a cancellative left order in Q. Suppose that a, b ∈ S and b≤_{R}a. Then b≤_{L}a.

Proof Since S is cancellative, a^{∗} exists in Q and ba^{∗} = c^{∗}d for some c, d ∈ S. By a previous remark, we can assume that d≤_{R}c in Q. Now ba = ba^{∗}a^{2} = c^{∗}da^{2} and then cba = cc^{∗}da^{2} = da^{2}; cancelling in S gives that cb = da. Further, b lies in a subgroup of Q so that bHb^{2}≤_{R}baRba^{∗}≤_{R}b. Thus bRba^{∗} = c^{∗}d≤_{R}c and so b = c^{∗}cb = c^{∗}da and b≤_{L}a in Q.

As a consequence of this lemma, if S is a cancellative left order in Q, then R∩(S ×S) = H∩(S ×S); we show later that S intersects every H-class of Q, whence R = H in Q. Note that, as in the lemma, if p and q are elements of a semigroup Q where pHp^{2} and p≤_{R}q, then pRpq. We use this fact repeatedly.

Proposition 3.2 Let S be a cancellative left order in Q. Then Q is completely regular.

Proof Let q ∈ Q and write q = a^{∗}b for some a, b ∈ S with b≤_{R}a. Since aa^{∗}b = b, b^{∗}aa^{∗}bb^{∗}a = b^{∗}a and a^{∗}bb^{∗}aa^{∗}b = a^{∗}b so that a^{∗}bb^{∗}a is idempotent, and a^{∗}bRa^{∗}bb^{∗}aLb^{∗}a. By the comment preceding the proposition we have bRba in Q and so by Lemma 3.1, bLba. Hence

a^{∗}bLbLbaLb^{∗}aLa^{∗}bb^{∗}a.

This gives that a^{∗}bHa^{∗}bb^{∗}a and so q = a^{∗}b lies in a subgroup of Q.

A completely regular semigroup has a decomposition as a semilattice of completely simple semigroups [H]. A completely simple semigroup is a left group if it is a single L-class, that is, R = H. It is easy to see that a completely simple semigroup is a left group if and only if it is isomorphic to the direct product of a left zero semigroup with a group.

Theorem 3.3 Let S be a cancellative left order in Q. Then

(i) Q is a semilattice of left groups; (ii) S is local and straight in Q.

Proof We know from Proposition 3.2 that Q is completely regular. Let p, q ∈ Q be D-related elements and write p = a^{∗}b, q = c^{∗}d for some a, b, c, d ∈ S with b≤_{R}a and d≤_{R}c. Since bLpDqLd we have that b and d belong to the completely simple semigroup D_{p}. Hence bRbdLd and Lemma 3.1 gives that bLbdLd so that pLq and D_{p} is a left group.

To prove (ii) we first show that S has non-empty intersection with every H-class of Q. Note that in Proposition 3.2 it is not necessarily the case that a^{∗}bHb, as illustrated by Example 4.6.

Let e be any idempotent of Q. Then e = u^{∗}v for some u, v ∈ S with v≤_{R}u. From previous comments, eLv and vuRv in Q; from Lemma 3.1, vuHv. Now e = u^{∗}v gives that ue = v and so v^{2} = vue. But vuLe so that v^{2} = vu and cancelling in S gives v = u. Thus e = u^{∗}uHu and S ∩ H_{e} ≠ ∅.

Now let q ∈ Q and choose c ∈ S ∩ H_{q}. Then cq = a^{∗}b for some a, b ∈ S with b≤_{R}a so that also q≤_{R}a. Since Q is a semilattice of left groups, H = R on Q and it follows that if h, k ∈ Q and h≤_{R}k, then hkHh in Q. Here we have q = c^{∗}a^{∗}b and c^{∗}≤_{R}a^{∗} so that c^{∗}Hc^{∗}a^{∗} and q = c^{∗}a^{∗}c^{∗}cb. Arguments that are now routine give that c^{∗}a^{∗}c^{∗}HcacHc and c^{∗}a^{∗}c^{∗} = (cac)^{∗}. Further, bLcqHc so that cb, cac ∈ H_{q} and q = (cac)^{∗}cb. This shows that S is local in Q. Since Q is completely regular, certainly S is also straight in Q.

Note that in the above theorem, we cannot deduce (ii) immediately from (i), as H need not be a congruence on a semilattice of left groups.

Corollary 3.4 Let S be a cancellative order in Q. Then Q is a semilattice of groups.

We give examples in the next section to show that if S is a cancellative left order in Q, then the semilattice decomposition of Q can be non-trivial and we can have D ≠ H in Q.

Certainly the above corollary applies in the case where S is a commutative cancellative left order in Q, for by Corollary 3.2 of [EG], Q is also commutative. We remark that if S is a cancellative left order in Q then for S and Q to be commutative it is enough to assume that the H-classes of Q are commutative groups. For in this case, if u, v ∈ S and uLv in Q we have by Theorem 3.3 that uHuv and so by hypothesis, u(uv) = (uv)u. Cancelling in S gives uv = vu and uHv in Q. Thus Q is a semilattice of commutative groups so that Q (and S) are commutative.

In the final result of this section we show that any cancellative left order S is right reversible so that by Ore’s theorem, S has a group G of left quotients. In fact G is the minimum semigroup of quotients of S in the sense that if S is a left order in Q, then G is the image of Q under a homomorphism which restricts to the identity on S.

Proposition 3.5 Let S be a cancellative left order. Then S is right reversible and the group G of left quotients of S is the minimum semigroup of left quotients of S.

Proof Let S be a left order in Q and let a, b ∈ S. Then a^{∗} exists in Q and ba^{∗} = c^{∗}d for some c, d ∈ S with d≤_{R}c in Q. This equation yields cba = da^{2} and we cancel in S to obtain cb = da. Thus S is right reversible and so by Ore’s theorem, S has a group G of left quotients.

Suppose now that S is a left order in P. Trivially, if h, k are elements of S that are R-related in P, then they are R-related in G. Since S is straight in both P and G, an adaptation of Corollary 4.3 of [G4] (see [G5]) gives immediately that the mapping a^{∗}b ↦ a^{−1}b where a, b ∈ S and b≤_{R}a in P is a homomorphism from P onto G which restricts to the identity on S. Thus G is the minimum semigroup of left quotients of S.

4 Semilattices of cancellative left orders

In the previous section we showed that if S is a cancellative left order in Q, then S is straight in Q and Q is completely regular, indeed, Q is a semilattice of left groups.

Straight left orders in completely regular semigroups are investigated in depth in [P]. If S is a straight left order in Q, where Q is a semilattice Y of completely simple semigroups Q_{α}, α ∈ Y, then it is easy to see that putting S_{α} = S ∩ Q_{α}, α ∈ Y, each S_{α} is a left order in Q_{α} and S is a semilattice Y of semigroups S_{α}, α ∈ Y.

On the other hand, given a semigroup S that is a semilattice Y of semigroups S_{α}, α ∈ Y, where each S_{α} is a left order in a completely simple semigroup Q_{α}, under what conditions is S a straight left order in Q, where Q is a semilattice Y of semigroups Q_{α}, α ∈ Y? It is not difficult to see this is equivalent to being able to define an associative multiplication on ⋃{Q_{α} : α ∈ Y } which extends the multiplication on S and each Q_{α}. This question is answered in Theorem III 1.2 of [P], which we quote below. In fact we quote a slightly stronger result, the proof of which is implicit in [P]; it is also somewhat reworded. The statement involves the relation R′ where R′ = ⋃{R′_{α} : α ∈ Y } and for each α ∈ Y, R′_{α} = R^{Q_{α}} ∩ (S_{α} × S_{α}).

Theorem 4.1 Let S be a semilattice Y of semigroups S_{α}, α ∈ Y, where for each α ∈ Y, S_{α} is a left order in a completely simple semigroup Q_{α}. Put Q = ⋃{Q_{α} : α ∈ Y }. Then Q is a semilattice of semigroups Q_{α}, α ∈ Y, if and only if conditions (I) to (VI) hold. In this case, S is a straight left order in Q. (I) For any a ∈ S, if u, v ∈ S and uR′at for some t ∈ S^{1}, then uvR′as for some s ∈ S^{1}. (II) The relation R′ is contained in R^{∗}. (III) If a, b ∈ S_{α}, aS_{α} ∩ bS_{α} ≠ ∅ and abx = aby for some x, y ∈ S^{1}, then bx = by. (IV) The relation R′ is a left congruence. (V) If a^{2}x, a^{2}y ∈ S and a^{2}xR′a^{2}y, then axR′ay. (VI) If x ∈ S_{α}, z ∈ S_{β} then there exist u, v ∈ S_{αβ} with uR′vR′xz and uxz = vz^{2}.

The example below shows that a left order in a completely regular semigroup Q need not be straight. This Q is also an example of a completely regular semigroup for which R = H but H is not a congruence.

Example 4.2 .

Let V be the infinite cyclic group generated by an element a and let U be the subsemigroup of V consisting of the positive powers of a. Let G be a group and put P = Z × G, where Z is regarded as a left zero semigroup, so that P is a left group. Now let Q = P ∪ V and define a multiplication on Q extending that on P and V by

a^{i}(j, g) = (i + j, g), (j, g)a^{i} = (j, g)

for all a^{i} ∈ V and (j, g) ∈ P. It is easy to see that under this multiplication Q is a completely regular semigroup. Indeed, Q is a chain of the group V and the left group P so that in particular, H = R on Q. Note that H is not a congruence on Q since aHa^{−1} but

a(0, g) = (1, g) H̸(−1, g) = a^{−1}(0, g).

Put S = (N × G) ∪ U. Clearly S is a subsemigroup of Q and it is not difficult to show that S is a left order in Q. Since S does not intersect every H-class of Q, S cannot be straight in Q.

We are interested here in applying Theorem 4.1 to cancellative left orders. From Theorem 3.3, cancellative left orders are always straight. Also from that theorem, if S is a cancellative left order in a completely simple semigroup Q, then Q is a left group.

Corollary 4.3 Let S be a cancellative semigroup, and suppose that S is a semilattice Y of semigroups S_{α}, α ∈ Y, where for each α ∈ Y, S_{α} is a left order in a left group Q_{α}. Put Q = ⋃{Q_{α} : α ∈ Y }.

Then Q is a semilattice of semigroups Q_{α}, α ∈ Y, if and only if conditions (I), (IV) and (V) hold. In this case, S is a straight left order in Q.

Proof Suppose that conditions (I), (IV) and (V) hold. Since R^{∗} is universal on a cancellative semigroup, it is immediate that conditions (II) and (III) also hold. Let x ∈ S_{α} and z ∈ S_{β}. Then zR′z^{3} so that as R′ is a left congruence, xzR′xz^{3}. Thus in Q_{αβ}, xzHxz^{3} and as S_{αβ} is a local left order in Q_{αβ}, there exist u, w ∈ S_{αβ} ∩ H_{xz} with uxz = wxz^{3}. Put v = wxz so that v ∈ S_{αβ} ∩ H_{xz}, uxz = vz^{2} and uR′vR′xz. Thus condition (VI) holds. The corollary is now immediate from Theorem 4.1.

To apply Corollary 4.3 we need to know when a cancellative semigroup S is a left order in a left group. By Proposition 3.5, S must be right reversible.

Proposition 4.4 Let S be a right reversible, cancellative semigroup. Then S is a left order in a left group Q = L×G, where L is a left zero semigroup and G is a group, if and only if S is the disjoint union of right ideals R_{ℓ}, ℓ ∈ L. Moreover, if this holds then for any a, b ∈ S, aRb in Q if and only if a, b ∈ R_{ℓ} for some ℓ ∈ L.

Proof It is possible to deduce this result directly from the characterisations of left orders in completely simple semigroups given in [AFM] or [G3], however we give a short direct argument.

If S is a left order in Q = L × G then clearly S intersects every H-class of Q, so that putting

R_{ℓ} = S ∩ {(ℓ, g) : g ∈ G}, ℓ ∈ L,

we have that S is a disjoint union of right ideals R_{ℓ}, ℓ ∈ L.

Conversely, suppose that S is a disjoint union of right ideals R_{ℓ}, ℓ ∈ L. Let G be the group of left quotients of S. It is easy to see that each R_{ℓ} is a left order in G. Put Q = L × G and define ϕ : S → Q by sϕ = (ℓ, s), where s ∈ R_{ℓ}. Then ϕ is an embedding of S into Q such that Sϕ is a left order in Q. Moreover, if s, t ∈ S, then sϕ R tϕ in Q if and only if a, b ∈ R_{ℓ} for some ℓ ∈ L.

Example 4.5 .

Let G be the group semidirect product given by

G = {(a, b) : a, b ∈ Q, b ≠ 0}

with multiplication

(a, b)(c, d) = (a + bc, bd)

so that (0, 1) is the identity of G and (a, b)^{−1} = (−a/b, 1/b). Put

S = {(a, b) : a, b ∈ Z, b ≠ 0}.

Then S is a subsemigroup, indeed a left order in G. Let n be a positive integer. For any (a, n), (c, d) ∈ S,

(a, n)(c, d) = (a + nc, nd)

and a + nc ≡ a modulo n. Thus (0, n), (1, n), ..., (n − 1, n) generate disjoint right ideals. Putting

T = ⋃{(r, n)S : 0 ≤ r ≤ n − 1}

we have that T satisfies the conditions of Proposition 4.3, where L = {0, 1, ..., n − 1}.

Fitting together Theorem 3.3, Corollary 4.3 and Proposition 4.4 we have the following picture for a cancellative semigroup S.

If S is a left order in Q, then Q is a semilattice Y of left groups Q_{α}, α ∈ Y. Putting S_{α} = S ∩ Q_{α}, α ∈ Y, we have that S_{α} is a left order in Q_{α}. Write Q_{α} = L_{α} × G_{α} where L_{α} is a left zero semigroup and G_{α} is a group. For ℓ_{α} ∈ L_{α} put R_{ℓ_{α}}^{α} = S_{α} ∩ {(ℓ_{α}, g_{α}) : g_{α} ∈ G_{α}}. Then for each α ∈ Y, S_{α} is right reversible and is a disjoint union of right (S_{α}-)ideals R_{ℓ_{α}}^{α}, ℓ_{α} ∈ L_{α}. Defining the relation R′ on S by the rule that for elements a, b of S we have aR′b if and only if a, b ∈ R_{ℓ_{α}}^{α} for some α ∈ Y and ℓ_{α} ∈ L_{α}, we have that R′ = R^{Q} ∩ (S × S) and R′ satisfies conditions (I), (IV) and (V).

Conversely, suppose that S is a semilattice Y of semigroups S_{α} where for each α ∈ Y, S_{α} is right reversible and is a disjoint union of right (S_{α}-)ideals R_{ℓ_{α}}^{α}, ℓ_{α} ∈ L_{α}. Suppose also that the relation R′ satisfies conditions (I), (IV) and (V), where R′ is defined by the rule that aR′b if and only if a, b ∈ R_{ℓ_{α}}^{α} for some α ∈ Y and ℓ_{α} ∈ L_{α}. Then each S_{α} is a left order in a left group Q_{α}, α ∈ Y, where for a, b ∈ S_{α}, aRb in Q_{α} if and only if aR′b. Further, Q = ⋃{Q_{α} : α ∈ Y } is a semilattice of semigroups Q_{α}, α ∈ Y and S is a straight left order in Q.

In each case the relation R′ is the restriction to S of the relation R in Q. Indeed a semigroup of left quotients Q of S is uniquely determined in the following sense by R′. If Q_{1} and Q_{2} are semigroups of left quotients of S then by Theorem 3.3 S is straight in Q_{1} and Q_{2}. Corollary 4.3 of [G4] yields that Q_{1} is isomorphic to Q_{2} under an isomorphism which restricts to the identity on S if and only if R^{Q_{1}} ∩ (S × S) = R^{Q_{2}} ∩ (S × S).

We end this paper by presenting an example of a cancellative left order S that is also a left order in a chain of left groups.

Example 4.6 .

Let F = \{a_{i}, a_{i}^{−1} : i ∈ N\}^{∗} be a free monoid and write \overline{w} for the reduced form of any w ∈ F, so that G = \{\overline{w} : w ∈ F \} is the free group on \{a_{i} : i ∈ N\} under the multiplication \overline{w_{1}} · \overline{w_{2}} = \overline{w_{1}w_{2}} and ^{−} : F → G is a homomorphism.

For each i ∈ N we define a homomorphism Δ_{i} from F to the integers under addition by

a_{i}Δ_{i} = 1, a_{i}^{−1}Δ_{i} = −1, a_{j}Δ_{i} = a_{j}^{−1}Δ_{i} = 0, i ≠ j.

Call a word w ∈ F i-positive if w ≠ 1 and for all u, v ∈ F

u ≠ 1, w = uv implies that uΔ_{i} > 0.

Put T_{0} = {1} and for i ∈ N

T_{i} = {a_{1}, ..., a_{i}, a_{1}^{−1}, ..., a_{i}^{−1}}^{∗}.

Now for i ∈ N let

U_{i} = {t_{i−1}t_{i} : t_{i−1} ∈ T_{i−1}, t_{i} ∈ T_{i}, t_{i} i-positive}.

Claim 1 For each i, j ∈ N, U_{i}U_{j} ⊆ U_{max{i,j}} so that U = ⋃{U_{i} : i ∈ N} is a subsemigroup of F. Proof Let u ∈ U_{i}, v ∈ U_{j} so that u = s_{i−1}s_{i}, v = t_{j−1}t_{j} where s_{i−1} ∈ T_{i−1}, s_{i} ∈ T_{i}, t_{j−1} ∈ T_{j−1}, t_{j} ∈ T_{j}, s_{i} is i-positive and t_{j} is j-positive.

If i < j then s_{i−1}s_{i}t_{j−1} ∈ T_{j−1} so that uv ∈ U_{j}. Suppose now that i ≥ j so that s_{i}t_{j−1}t_{j} ∈ T_{i}. Let x ≠ 1 be a prefix of s_{i}t_{j−1}t_{j}. If x is a prefix of s_{i} then certainly xΔ_{i} > 0. If x = s_{i}y where y is a prefix of T_{j−1}, then

xΔ_{i} = s_{i}Δ_{i} + yΔ_{i} = sΔ_{i} > 0.

If x = s_{i}t_{j−1}z where z is a prefix of t_{j}, then as above if j < i we have xΔ_{i} = s_{i}Δ_{i} > 0; on the other hand if j = i we have

xΔ_{i} = sΔ_{i} + t_{i−1}Δ_{i} + zΔ_{i} = s_{i}Δ_{i} + zΔ_{i} > 0.

Thus s_{i}t_{j−1}t_{j} is i-positive and uv ∈ U_{i}.

For i ∈ N let

S_{i} = \overline{U_{i}} = \{\overline{u_{i}} : u_{i} ∈ U_{i}\}

and put

S = ⋃{S_{i} : i ∈ N}.

By Claim 1, S is a subsemigroup of G. We shall show that S is the left order for which we are searching.

Claim 2 For all words w ∈ F, if w is i-positive, then \overline{w} is i-positive. Proof Suppose that w = w_{0}a_{j}^{ϵ}a_{j}^{−ϵ}w_{1} is i-positive, where ϵ ∈ {1, −1}. Note that w_{0}w_{1} ≠ 1. Let v ≠ 1 be a prefix of w_{0}w_{1}. If v is a prefix of w_{0} then clearly vΔ_{i} > 0. Otherwise, v = w_{0}z for some prefix z of w_{1}. Then

vΔ_{i} = w_{0}Δ_{i} + zΔ_{i} = w_{0}Δ_{i} + a_{j}^{ϵ}Δ_{i} + a_{j}^{−ϵ}Δ_{i} + zΔ_{i} = (w_{0}a_{j}^{ϵ}a_{j}^{−ϵ}z)Δ_{i} > 0.

Thus w_{0}w_{1} is i-positive and the claim follows by induction.

For a word w ∈ F we denote by c(w) the content of w, that is, c(w) is the set of letters occurring in w.

Claim 3 Each element of S has a unique representation as \overline{s_{i−1}} \overline{s_{i}} (multiplication in F) where s_{i−1} ∈ T_{i−1}, s_{i} ∈ T_{i} and s_{i} is i-positive. Proof If s ∈ S then by definition, s = \overline{s_{i−1}s_{i}} for some s_{i−1} ∈ T_{i−1} and ipositive s_{i} ∈ T_{i}. By Claim 2, \overline{s_{i}} is i-positive, so that \overline{s_{i}} = a_{i}x for some x ∈ F. But a_{i}^{−1} ∉ c(\overline{s_{i−1}}), whence s = \overline{s_{i−1}} \overline{s_{i}}.

Suppose now that \overline{s_{i−1}} \overline{s_{i}} = \overline{t_{j−1}} \overline{t_{j}} where s_{i−1} ∈ T_{i−1}, s_{i} ∈ T_{i} is i-positive, t_{j−1} ∈ T_{j−1}, t_{j} ∈ T_{j} is j-positive. As above, \overline{s_{i}} = a_{i}x and \overline{t_{j}} = a_{j}y for some x, y ∈ F. Thus a_{i} ∈ c(\overline{t_{j−1}} \overline{t_{j}}) ⊆ \{a_{1}, ..., a_{j}, a_{1}^{−1}, ..., a_{j}^{−1}\} and so i ≤ j. Dually, j ≤ i and so i = j. Now \overline{s_{i−1}}a_{i}x = \overline{t_{i−1}}a_{i}y gives \overline{s_{i−1}} = \overline{t_{i−1}} and x = y, since a_{i} ∉ c(\overline{s_{i−1}}) ∪ c(\overline{t_{i−1}}). The claim follows.

We say that an element s of S is in normal form if s = \overline{s_{i−1}} \overline{s_{i}} where s_{i−1} ∈ T_{i−1}, s_{i} ∈ T_{i} and s_{i} is i-positive.. Claim 3 says that each element of S has a unique normal form. Moreover it follows also that if i ≠ j then S_{i} ∩ S_{j} = ∅. Together with Claim 1 this gives

Claim 4 The semigroup S is a chain N of semigroups S_{i}, i ∈ N.

Consider now the product s · t of two elements s = \overline{s_{i−1}} \overline{s_{i}} and t = \overline{t_{j−1}} \overline{t_{j}} of S in normal form. We have that

s · t = (\overline{s_{i−1}} \overline{s_{i}}) · (\overline{t_{j−1}} \overline{t_{j}}) = \overline{s_{i−1}s_{i}} · \overline{t_{j−1}t_{j}} = \overline{s_{i−1}s_{i}t_{j−1}t_{j}}.

¿From the proof of Claim 1, if i < j then

s · t = \overline{s_{i−1}s_{i}t_{j−1}} \overline{t_{j}}

is in normal form, whilst if i ≥ j then

s · t = \overline{s_{i−1}} \overline{s_{i}t_{j−1}t_{j}}

is in normal form.

Now fix i ∈ N and consider S_{i}. For t_{i−1} ∈ T_{i−1} put

Ri ; \overline{t_{i−1}} = \{\overline{t_{i−1}} \overline{t_{i}} : t_{i} ∈ T_{i}, t_{i} is i-positive\}.

By the above, each Ri ; \overline{t_{i−1}} is a right ideal of S_{i} and by the uniqueness of normal forms,

S = ⋃\{Ri ; \overline{t_{i−1}} : \overline{t_{i−1}} ∈ \overline{T_{i−1}}\}

where the union is disjoint.

Claim 5 For i ∈ N the semigroup S_{i} is a left order in \overline{T_{i}}. Proof Let g ∈ \overline{T_{i}}, so that g = \overline{w} for some w ∈ T_{i}. Let N = 1 + max{|vΔ_{i}| : v is a prefix of w}. Then a_{i}^{N}, a_{i}^{N}w are i-positive elements of T_{i} so that a_{i}^{N}, a_{i}^{N}w ∈ U_{i} and

g = \overline{w} = \overline{a_{i}^{−N}a_{i}^{N}w} = \overline{a_{i}^{−N}} · \overline{a_{i}^{N}w} = \overline{a_{i}^{N}}^{−1} · \overline{a_{i}^{N}w}

as required.

The next claim is now immediate.

Claim 6 The semigroup S is a left order in G.

¿From Claim 5 and Proposition 4.4 it follows that for each i ∈ N, S_{i} is a left order in a left group Q_{i}. Moreover, if \overline{s_{i−1}} \overline{s_{i}}, \overline{t_{i−1}} \overline{t_{i}} ∈ S_{i} are in normal form, then \overline{s_{i−1}} \overline{s_{i}}R\overline{t_{i−1}} \overline{t_{i}} in Q_{i} if and only if \overline{s_{i−1}} = \overline{t_{i−1}}.

We wish to apply Corollary 4.3 to give that S is a left order in Q, where Q is a chain N of the semigroups Q_{i}, i ∈ N. The relation R′ in question is given by the rule that if \overline{s_{i−1}} \overline{s_{i}}, \overline{t_{j−1}} \overline{t_{j}} ∈ S are in normal form, then they are R′-related if and only if i = j and \overline{s_{i−1}} = \overline{t_{j−1}}. It is straightforward to show that conditions (I), (IV) and (V) hold for R′. Corollary 4.3 now gives our result.

Claim 7 The semigroup S is a left order in Q, where Q is a chain N of the left groups Q_{i}, i ∈ N.

Note that the R-classes of each Q_{i} are indexed by \overline{T_{i−1}}, so that Q_{1} is a group, but for i > 1, Q_{i} has infinitely many R-classes.

References

[AFM] P.N. Ánh, J.B. Fountain and L. Márki, ‘Subsemigroups and left orders in completely 0-simple semigroups’, submitted.

[CP] A.H. Clifford and G.B. Preston, The algebraic theory of semigroups Vol. I, American Math. Soc. 1961.

[EG] D. Easdown and V.A.R. Gould, ‘Commutative orders’, Proc. Royal Soc. Edinburgh, to appear.

[F] J.B. Fountain, ‘Abundant semigroups’, Proc. London Math. Soc. 44 (1982), 103-129.

[FP] J.B. Fountain and M. Petrich, ‘Completely 0-simple semigroups of quotients’, J. Algebra 101 (1986), 365-402.

[G1] V.A.R. Gould, ‘Clifford semigroups of left quotients’, Glasgow Math. J. 28 (1986), 181-191.

[G2] V.A.R.Gould, ‘Orders in semigroups’, Contributions to general algebra 5, Proceedings of the Salzburg conference May 29-June 1 1986, Verlag Hölder-Pichler-Tempsky, Vienna 1987.

[G3] V.A.R. Gould, ‘Left orders in regular H-semigroups II’, Glasgow Math. J. 32 (1990), 95-108.

[G4] V.A.R. Gould, ‘Semigroups of left quotients - the uniqueness problem’, Proc. Edinburgh Math. Soc. 35 (1992), 213-226.

[G5] V.A.R. Gould, ‘Semigroups of left quotients - the homomorphism problem’, in preparation.

[H] J.M. Howie, An introduction to semigroup theory, Academic Press 1976.

[P] P.M. Smith, ‘Orders in completely regular semigroups’, D. Phil. thesis, University of York 1990.