ORDERS IN POWER SEMIGROUPS

December 1995

David Easdown Victoria Gould

School of Mathematics and Statistics Department of Mathematics University of Sydney University of York

Heslington

NSW 2006 York YO1 5DD

Australia England

Abstract

We consider the Fountain-Petrich notion of a semigroup of quotients. If Q is a semigroup of quotients of S we also say that S is an order in Q. We concentrate here on orders in the restricted power semigroup P(Q) of a semigroup Q, where P(Q) consists of all nonempty finite subsets of Q under the natural multiplication. Our first result shows that if S is a commutative cancellative semigroup with group of quotients G, then P(S) is an order in P(G). In the latter part of the paper we give necessary and sufficient conditions for P(S) to be an order in P(Q), where Q is a semilattice of torsion-free commutative groups and S is an order in Q.

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

1 Introduction

In this paper we consider examples of orders in restricted power semigroups, where for any semigroup S the restricted power semigroup P(S) is given by P(S) = {X ⊆ S : 1 ≤ |X| < ℵ_{0}} with multiplication XY = {xy : x ∈ X, y ∈ Y } for all X, Y ∈ P(S). We use the notion of order introduced by Fountain and Petrich in [2] which first appears in the form used here in [3]. If S is a subsemigroup of Q then S is an order in Q and Q is a semigroup of quotients of S if any q ∈ Q can be written as q = a^{∗}b = cd^{∗} where a, b, c, d ∈ S and a^{∗}(d^{∗}) is the inverse of a(d) in a subgroup of Q, and in addition, all elements of S satisfying a weak cancellability condition called square-cancellability lie in a subgroup of Q.

It is clear that the concept of a semigroup of quotients extends that of the group of quotients G of a commutative cancellative semigroup S. Our first result shows that for such an S and G, the restricted power semigroup P(S) is an order in P(G).

In the latter part of the paper we turn our attention to orders in a semigroup Q which is a semilattice Y of commutative groups G_{α}, α ∈ Y. To handle the idempotents of P(Q) we make the further assumption that the groups G_{α}, α ∈ Y, are torsion-free. We find a necessary and sufficient condition for an order S in such a semigroup Q to have the property that P(S) is an order in P(Q). In fact we prove a slightly stronger result. We say that a subsemigroup S of Q is a weak order in Q if any q ∈ Q can be written as q = a^{∗}b = cd^{∗} where a, b, c, d ∈ S and a^{∗}(d^{∗}) is the inverse of a(d) in a subgroup of Q. Proposition 4.1 gives a necessary and sufficient condition on S such that P(S) is a weak order in P(Q), where S is an order in Q and Q is a semilattice of torsion-free commutative groups. We then show that for such an S and Q, if P(S) is a weak order in P(Q) then necessarily P(S) is an order in P(Q).

Section 2 consists of some preliminary definitions and results concerning orders and Green’s relations in certain restricted power semigroups. Section 3 considers restricted power semigroups of orders in commutative groups. In our last section we turn our attention to restricted power semigroups of orders in semilattices of commutative torsion-free groups, and prove the results mentioned in the previous paragraph.

We comment that if S is an order in Q then S^{0} (the semigroup S with a zero adjoined) is clearly an order in Q^{0}. Moreover it is easy to see that if T is an order in Q^{0} then T = S^{0} where S is an order in Q. Including the empty set in P(S) would correspond to considering P(S)^{0}. Thus excluding the empty set from P(S) does not affect our results in any essential way, it is merely convenient.

2 Preliminaries

We assume the reader has a basic knowledge of semigroup theory as in the early chapters of [1] or [4]. Any undefined notation or concepts may be found in these references. We deviate from standard notation in denoting by a^{∗} the group inverse, where it exists, of an element a of a semigroup Q. That is, a^{∗} exists if and only if a lies in a subgroup of Q, and the inverse of a in this subgroup is a^{∗}. By a famous result of Green [4, Theorem II 2.5], a^{∗} exists if and only if a is related to its square by the relation H. Moreover, where a^{∗} exists it is unique, being the inverse of a in the group H-class H_{a}.

Let a be an element of a semigroup S. Then a is square-cancellable if for all x, y ∈ S^{1}

xa^{2} = ya^{2} implies xa = ya

and

a^{2}x = a^{2}y implies ax = ay.

It is clear that if a lies in a subgroup of an oversemigroup of S, then a is square-cancellable. We insist that if S is an order in Q, then all such elements lie in subgroups of Q. The set of square-cancellable elements of S is denoted by S(S).

Let Q be a semilattice Y of commutative groups G_{α}, α ∈ Y. We follow the usual practice of abbreviating {g}H where g ∈ Q and H ∈ P(Q) by gH. Further, we make the notational convention that e_{α} denotes the idempotent of G_{α}, α ∈ Y and also e_{i} denotes the idempotent of G_{α_{i}}, α_{i} ∈ Y, i ∈ N. For X ∈ P(Q) and α ∈ Y put

X_{α} = X ∩ G_{α}

and

supp X = {α ∈ Y : X_{α} ≠ ∅}, max X = max {supp X}.

Lemma 2.1 For H, K ∈ P(Q), if H H K then supp H = supp K and |H_{α}| = |K_{α}| for all α ∈ Y.

Proof Suppose that H H K. Then H = K (in which case the result is clearly true) or there exist S, T ∈ P(Q) with H = SK and K = T H. Let α ∈ max H. Since H_{α} ⊆ SK there exists β ∈ max K with α ≤ β. But K_{β} ⊆ T H so β ≤ γ for some γ ∈ max H. Thus α = β = γ so that α ∈ max K. With the dual argument this gives that max H = max K.

Now let δ ∈ supp H. Then δ ≤ α for some α ∈ max K. Since K_{α} ⊆ T H there exists ϵ ∈ supp T with α ≤ ϵ. Let t ∈ T_{ϵ}, then tH_{δ} ⊆ K_{δ} so δ ∈ supp K; hence supp H = supp K. Further, |H_{δ}| = |tH_{δ}| ≤ |K_{δ}| and with the dual we obtain that |H_{δ}| = |K_{δ}|.

The following corollaries are now easy to see.

Corollary 2.2 [5] Let G be a commutative group. Then for any H, K ∈ P(G), H H K if and only if H = gK for some g ∈ G.

Corollary 2.3 [5] Let G be a commutative group. The following conditions are equivalent for H ∈ P(G):

(i) H H H^{2}; (ii) H^{2} = gH for some g ∈ G; (iii) H^{2} = hH for any h ∈ H; (iv) |H^{2}| = |H|.

Corollary 2.4 [5] Let G be a commutative group. Then E ∈ P(G) is idempotent if and only if E is a finite subgroup. Moreover if H ∈ P(G) then H H E if and only if H is a coset of E.

We now return to the case where Q is a semilattice Y of commutative groups G_{α}, α ∈ Y. To adequately describe the idempotents of P(Q) and the elements of P(Q) lying in group H-classes, we make henceforth the additional assumption that each G_{α} is torsion-free. Clearly then the idempotents of P(Q) are the finite subsemigroups of E(Q). By Lemma 2.1, if H ∈ P(Q) and H H H^{2} then |H_{α}| = 1 for all α ∈ supp H. Also supp H must be a subsemilattice of Y. We write H = {h_{1}, ..., h_{n}} where |H| = n and h_{i} ∈ G_{α_{i}}, 1 ≤ i ≤ n. Then H H E where E = {e_{1}, ..., e_{n}} and e_{i} = e_{i}^{2} ∈ G_{α_{i}}, 1 ≤ i ≤ n. Further, if i, j ∈ {1, ..., n} and α_{i} ≥ α_{j} then from EH = H we have e_{j}h_{i} = h_{j}. This leads us to the following definition.

An element H of P(Q) is balanced if |H_{α}| = 1 for all α ∈ supp H and α, β ∈ supp H with α ≥ β implies e_{β}h_{α} = h_{β} where {h_{α}} = H_{α} and {h_{β}} = H_{β}.

Lemma 2.5 An element H of P(Q) is in a subgroup of P(Q) if and only if supp H is a subsemilattice of Y and H is balanced. In this case, writing H = {h_{1}, ..., h_{n}} where supp H = {α_{1}, ..., α_{n}} and H_{α_{i}} = {h_{i}}, 1 ≤ i ≤ n, then H^{∗} = {h_{1}^{∗}, ..., h_{n}^{∗}} and HH^{∗} = {e_{1}, ..., e_{n}}.

Proof We have seen that if H ∈ P(Q) and H H H^{2} then supp H is a subsemilattice of Y and H is balanced.

Conversely, suppose these conditions hold. Put H = {h_{1}, ..., h_{n}} where supp H = {α_{1}, ..., α_{n}} and {h_{i}} = H_{α_{i}}, 1 ≤ i ≤ n. Certainly E = {e_{1}, ..., e_{n}} ∈ E(P(Q)) and H ⊆ EH; consider i, j ∈ {1, ..., n} and put α_{k} = α_{i}α_{j}. Then e_{i}h_{j} = e_{i}e_{k}h_{j} = e_{k}h_{j} = h_{k} as α_{k} ≤ α_{j} and H is balanced. So EH ⊆ H and EH = H. Putting K = {h_{1}^{∗}, ..., h_{n}^{∗}} we have K is balanced so that also EK = K.

Clearly E ⊆ HK. Let i, j ∈ {1, ..., n} and again put α_{k} = α_{i}α_{j}. Then h_{i}h_{j}^{∗} = h_{i}e_{k}(e_{k}h_{j}^{∗}) and as Q is an inverse semigroup this gives h_{i}h_{j}^{∗} = h_{i}e_{k}(e_{k}h_{j})^{∗} = h_{k}h_{k}^{∗} = e_{k} and it follows that E = HK. Thus H is in a group H-class of P(Q) with H^{∗} = {h_{1}^{∗}, ..., h_{n}^{∗}} and HH^{∗} = {e_{1}, ..., e_{n}} as required.

3 Power semigroups of orders in commutative groups

In this section we show that if S is an order in a commutative group G then P(S) is an order in P(G). For such a G, P(G) has identity {e}, where e^{2} = e, and group of units the singletons of G. If S is an order in G then as is well known the Common Denominator Theorem holds, that is, given any g_{1}, ..., g_{n} ∈ G there exist a, b_{1}, ..., b_{n} ∈ S with g_{i} = a^{−1}b_{i}, 1 ≤ i ≤ n. Thus we immediately have the following lemma.

Lemma 3.1 Let S be an order in a commutative group G. Then P(S) is a weak order in P(G).

The more difficult task is to show that if S is an order in a commutative group G, then every square-cancellable element H of P(S) lies in a subgroup of P(G). By Corollary 2.3 this is equivalent to showing that |H^{2}| = |H|.

Theorem 3.2 Let S be an order in a commutative group G. Then P(S) is an order in P(G).

Proof From Lemma 3.1, P(S) is a weak order in P(G). Clearly the singletons of S are square-cancellable and lie in a subgroup (the group of units) of P(G).

At this point it is useful to make the following notational convention. If x ∈ X and ȳ ∈ X^{n} where X ∈ P(S), then by saying that x is an X-factor of ȳ, we mean that there exist x_{1}, ..., x_{n−1} ∈ X such that xx_{1}...x_{n−1} = ȳ.

Suppose now that X = {x, y} ∈ S(P(S)) where x ≠ y. Every element of X^{5} contains x^{3} or y^{3} as an X-factor. Indeed, X^{5} = {x^{3}, y^{3}}X^{2}. Since X is square-cancellable, X^{4} = {x^{3}, y^{3}}X. Looking at the element x^{2}y^{2} of X^{4} we have x^{2}y^{2} = x^{3}x, x^{3}y, y^{3}x or y^{3}y which each imply that y^{2} = x^{2}. It follows that X^{3} = x^{2}X and so

|X| ≤ |X^{2}| ≤ |X^{3}| = |x^{2}X| = |X|

so that |X| = |X^{2}| and by Corollary 2.3, X lies in a subgroup of P(G).

Now let n ≥ 3 and suppose that X = {x_{1}, ..., x_{n}} ∈ S(P(S)) where |X| = n. Every element of X^{n^{2}} contains an X-factor x_{i}^{n} for some i ∈ {1, ..., n}. Hence

X^{n^{2}} = {x_{1}^{n}, ..., x_{n}^{n}}X^{n^{2}−n}

and using the fact that X is square-cancellable we obtain X^{n+1} = {x_{1}^{n}, ..., x_{n}^{n}}X.

Suppose that X^{n+1} = {x_{1}^{n}, ..., x_{m}^{n}}X for some m with 1 < m ≤ n. We show that (with some re-labelling), X^{n+1} = {x_{1}^{n}, ..., x_{m−1}^{n}}X. Since it is certainly true that X^{n+1} = {x_{1}^{n}, ..., x_{m}^{n}}X with m = n, we obtain in n−1 steps that X^{n+1} = x_{1}^{n}X. As above it follows that |X| = |X^{2}| so that X lies in a subgroup of P(G).

Given X^{n+1} = {x_{1}^{n}, ..., x_{m}^{n}}X with 1 < m ≤ n we show that x_{i} is an X-factor of x_{j}^{n} for some i, j ∈ {1, ..., m} with i ≠ j. Consider first the element x_{1}x_{2}...x_{m−1}x_{m}^{2}x_{m+1}..x_{n} of X^{n+1}. This element has the following possible forms:

x_{1}...x_{m−1}x_{m}^{2}x_{m+1}...x_{n} = \{ \begin{matrix}& x_{i}^{n+1} & where & 1 ≤ i < m \\ or & x_{i}^{n}x_{j} & & 1 ≤ i < m, 1 ≤ j ≤ m, i ≠ j \\ or & x_{i}^{n}x_{m} & & 1 ≤ i < m \\ or & x_{i}^{n}x_{j} & & 1 ≤ i < m, m < j ≤ n \\ or & x_{m}^{n}x_{i} & & 1 ≤ i < m \\ or & x_{m}^{n+1} \\ or & x_{m}^{n}x_{j} & & m < j ≤ n. \end{matrix}

In the first four cases we have x_{m} is an X-factor of x_{i}^{n} for some i ∈ {1, ..., m − 1}. In the last two, x_{1} is an X-factor of x_{m}^{n}.

In the fifth case,

x_{1}...x_{m−1}x_{m}^{2}x_{m+1}...x_{n} = x_{m}^{n}x_{i},

where 1 ≤ i < m, giving

x_{1}...x_{i−1}x_{i+1}...x_{m−1}x_{m}^{2}x_{m+1}...x_{n} = x_{m}^{n}

so that, unless m = 2, we have x_{j} is an X-factor of x_{m}^{n} for some j ∈ {1, ..., m − 1}. In the case where m = 2, x_{2}^{2}x_{3}...x_{n} = x_{2}^{n} so that x_{3}...x_{n} = x_{2}^{n−2}. At this point we look at the element x_{1}^{2}x_{2}...x_{n}. Again using the hypothesis that X^{n+1} = {x_{1}^{n}, ..., x_{m}^{n}}X we have that

x_{1}^{2}x_{2}...x_{n} = \{ \begin{matrix}& x_{1}^{n+1} \\ or & x_{1}^{n}x_{2} \\ or & x_{1}^{n}x_{j} & where & 2 < j ≤ n \\ or & x_{2}^{n}x_{1} \\ or & x_{2}^{n+1} \\ or & x_{2}^{n}x_{j} & & 2 < j ≤ n. \end{matrix}

Except for the second case, this yields that x_{1} is an X-factor of x_{2}^{n} or x_{2} is an X-factor of x_{1}^{n}. In the second case we have x_{3}...x_{n} = x_{1}^{n−2}. Since also we know that x_{3}...x_{n} = x_{2}^{n−2} we have x_{1}^{n−2} = x_{2}^{n−2} and so x_{1} is certainly an X-factor of x_{2}^{n}.

We have now verified that there exist i, j ∈ {1, ..., m} with i ≠ j such that x_{i} is an X-factor of x_{j}^{n}. For convenience we re-label x_{1}, ..., x_{m} so that i = 1 and j = m, that is, x_{1} is an X-factor of x_{m}^{n}. Since every product of n + 1 elements of X contains an X-factor x_{i}^{n} for some i ∈ {1, ..., m}, by raising X to a high enough power t we have that every element of X^{t} contains an X-factor x_{i}^{n^{2}} for some i ∈ {1, ..., m}. But if i = m then as x_{1} is an X-factor of x_{m}^{n} we have that x_{1}^{n} is an X-factor of x_{m}^{n^{2}}. Thus every element of X^{t} contains an X-factor x_{i}^{n} for some i ∈ {1, ..., m − 1}. Then X^{t} = {x_{1}^{n}, ..., x_{m−1}^{n}}X^{t−n} and as X ∈ S(P(S)), X^{n+1} = {x_{1}^{n}, ..., x_{m−1}^{n}}X. This completes the proof that X lies in a subgroup of P(G).

4 Power semigroups of orders in semilattices of torsion-free commutative groups

Let Q be a semilattice Y of torsion-free commutative groups G_{α}, α ∈ Y. We recall from Section 2 that E ∈ P(Q) is idempotent if and only if E is a finite subsemilattice of the idempotents of Q. Further, H ∈ P(Q) is in a subgroup of P(Q) if and only if supp H is a subsemilattice of Y and H is balanced.

Suppose now that S is an order in Q. As shown in Theorem 3.1 of [3], S is a semilattice Y of commutative cancellative semigroups S_{α} = S ∩ G_{α}, and S_{α} is an order in G_{α}, α ∈ Y. It is not true that for any such S and Q, P(S) is an order in P(Q). In fact, as we show below, P(S) need not be a weak order in P(Q).

At this stage it is useful to make an elementary remark about weak orders, which we will use without further comment. Given a commutative semigroup Q we write ≤_{H} for the preorder associated with Green’s relation H. Now if S is a weak order in Q and q ∈ Q, then q = a^{∗}b for some a, b ∈ S. But then q = (a^{2})^{∗}ab and ab≤_{H}a^{2}. Thus given q ∈ Q, we may write q as q = c^{∗}d for some c, d ∈ S with d≤_{H}c in Q. Further, if h, k ∈ Q and k lies in a subgroup of Q then from h≤_{H}k we deduce that hkHh. So if q = c^{∗}d with d≤_{H}c we have qHd; it follows that S intersects every H-class of Q.

Consider the three element semilattice Q = {α, β, γ} where α ≤ β and α ≤ γ. Clearly Q is an order in Q; if P(Q) were a weak order in P(Q) then {β, γ} could be written as {β, γ} = U^{∗}V for some U, V ∈ P(Q) with V ≤_{H}U. As commented above, V H{β, γ} so that by Lemma 2.1, V = {β, γ} and {β, γ} = U{β, γ}. But no such U ∈ P(Q) exists. Thus P(Q) is not a weak order in itself.

For the remainder of this section suppose that S is an order in Q, where Q is a semilattice Y of torsion-free commutative groups G_{α}, α ∈ Y. Our aim is to give necessary and sufficient conditions on S such that P(S) is a weak order in P(Q). We then show that in this case, square-cancellable elements of P(S) lie in subgroups of P(Q) so that P(S) is an order in P(Q).

Proposition 4.1 The semigroup P(S) is a weak order in P(Q) if and only if (*) holds: (*) given any distinct α_{1}, ..., α_{n} ∈ Y with s_{i} ∈ S_{α_{i}}, 1 ≤ i ≤ n, there exists α ∈ Y with α ≥ α_{i}, 1 ≤ i ≤ n and elements t_{i} ∈ S_{α_{i}}, a ∈ S_{α} such that ae_{α_{i}} = s_{i}t_{i}, 1 ≤ i ≤ n.

Proof Suppose that (*) holds and X ∈ P(Q); say X = {x_{ij} : 1 ≤ i ≤ n, 1 ≤ j ≤ m_{i}} where x_{i1}, ..., x_{im_{i}} ∈ G_{α_{i}}, 1 ≤ i ≤ n and α_{1}, ..., α_{n} are distinct elements of Y. Since S_{α_{i}} is an order in G_{α_{i}}, 1 ≤ i ≤ n, there are elements s_{i}, y_{i1}, ..., y_{im_{i}} ∈ S_{α_{i}} with x_{ij} = s_{i}^{∗}y_{ij}, 1 ≤ j ≤ m_{i}.

Let α ∈ Y and a, t_{1}, ..., t_{n} ∈ S be chosen as in (*). For i ∈ {1, ..., n} and j ∈ {1, ..., m_{i}} we have

x_{ij} = s_{i}^{∗}y_{ij} = s_{i}^{∗}t_{i}^{∗}t_{i}y_{ij} = (s_{i}t_{i})^{∗}t_{i}y_{ij} = (ae_{α_{i}})^{∗}t_{i}y_{ij} = a^{∗}e_{α_{i}}t_{i}y_{ij} = a^{∗}t_{i}y_{ij}.

So

X = a^{∗}{t_{i}y_{ij} : 1 ≤ i ≤ n, 1 ≤ j ≤ m_{i}},

giving that P(S) is a weak order in P(Q).

Conversely, suppose that P(S) is a weak order in P(Q). First we take mutually incomparable elements β_{1}, ..., β_{n} in Y. We show that {β_{1}, ..., β_{n}} has an upper bound in Y.

Pick x_{i} ∈ S_{β_{i}}, 1 ≤ i ≤ n. Then {x_{1}, ..., x_{n}} = H^{∗}K for some H, K ∈ P(S) with K≤_{H}H in P(Q). Thus H^{∗}K H K so that by Lemma 2.1, K = {y_{1}, ..., y_{n}} where y_{i} ∈ S_{β_{i}}, 1 ≤ i ≤ n. Since x_{1} ∈ H^{∗}K we have that x_{1} = pk for some p ∈ H^{∗} and k ∈ K; if p ∈ G_{γ} then β_{1} ≤ γ. If i ∈ {2, ..., n} then py_{i} ∈ H^{∗}K = {x_{1}, ..., x_{n}} and py_{i} ∈ G_{δ} where δ = γβ_{i}. If β_{i} ≰ γ then δ < β_{i}. But this is impossible since δ = β_{j} for some j and β_{1}, ..., β_{n} are mutually incomparable. Thus β_{i} ≤ γ for all i ∈ {1, ..., n}.

Now suppose that α_{1}, ..., α_{n} ∈ Y are distinct and s_{i} ∈ S_{α_{i}}, 1 ≤ i ≤ n. By the above there exists γ ∈ Y with γ ≥ α_{i}, 1 ≤ i ≤ n. Choose x ∈ S_{γ}, where if γ = α_{i} we take x = s_{i}. Then {x^{∗}, s_{1}^{∗}, ..., s_{n}^{∗}} = X^{∗}Y for some X, Y ∈ P(S) with Y ≤_{H}X in P(Q), so that Y H {x^{∗}, s_{1}^{∗}, ..., s_{n}^{∗}}. From Lemma 2.1, Y = {y, t_{1}, ..., t_{n}} where y ∈ S_{γ} and t_{i} ∈ S_{α_{i}}, 1 ≤ i ≤ n; if γ = α_{i} then also y = t_{i}.

Since x^{∗} ∈ X^{∗}Y we have that x^{∗} = zw for some z ∈ X^{∗} and w ∈ Y. If z ∈ G_{α} then γ ≤ α so that α_{i} ≤ α for all i ∈ {1, ..., n}. By Lemma 2.5, z = a^{∗} for some a ∈ S_{α}. Now for i ∈ {1, ..., n}, a^{∗}t_{i} ∈ X^{∗}Y = {x^{∗}, s_{1}^{∗}, ..., s_{n}^{∗}} and a^{∗}t_{i} ∈ G_{α_{i}} so that a^{∗}t_{i} = s_{i}^{∗}. The semigroup Q is inverse so that at_{i}^{∗} = s_{i} and ae_{α_{i}} = at_{i}^{∗}t_{i} = s_{i}t_{i}, 1 ≤ i ≤ n. Thus (*) holds.

We note that in the above proposition we showed that if P(S) is a weak order in P(Q) then any X ∈ P(Q) can be written as X = a^{∗}Y where a ∈ S_{α}, Y ∈ P(S) and β ≤ α for all β ∈ supp Y. Hence X H Y in P(Q). We now set out to prove that if P(S) is a weak order in P(Q) then it is necessarily an order.

Lemma 4.2 If P(S) is a weak order in P(Q) and X ∈ P(S) is square-cancellable in P(S), then X is square-cancellable in P(Q).

Proof Suppose that P(S) is a weak order in P(Q) and X ∈ S(P(S)). Let U, V ∈ P(Q) with UX^{2} = V X^{2}. We can express U ∪V as U ∪V = a^{∗}W for some a ∈ S_{α} and W ∈ P(S) such that β ≤ α for all β ∈ supp W. Clearly there exist subsets W_{1}, W_{2} of W such that U = a^{∗}W_{1} and V = a^{∗}W_{2}. Now a^{∗}W_{1}X^{2} = a^{∗}W_{2}X^{2} so that e_{α}W_{1}X^{2} = e_{α}W_{2}X^{2}; but e_{α} is an identity for all elements of W, so that W_{1}X^{2} = W_{2}X^{2}. Since X ∈ S(P(S)) we have W_{1}X = W_{2}X and so a^{∗}W_{1}X = a^{∗}W_{2}X and UX = V X.

If U ∈ P(Q) and UX^{2} = X^{2}, then in view of (*) UX^{2} = e_{α}X^{2} where α ≥ β for all β ∈ supp X. Thus UX = e_{α}X = X. Hence X is square-cancellable in P(Q).

For X ∈ P(Q) we put θ(X) = |supp X|.

Lemma 4.3 Suppose that P(S) is a weak order in P(Q) and X ∈ S(P(S)). If α ∈ max X then |X_{α}| = 1.

Proof If θ(X) = 1 then X ⊆ S_{α} for some α ∈ Y and certainly X ∈ S(P(S_{α})) so that by Theorem 3.2 X lies in a subgroup of P(G_{α}). Since {e_{α}} is the only idempotent of P(G_{α}) we have by Corollary 2.4 that |X| = 1.

Suppose now that θ(X) > 1. Take α ∈ max X; we aim to show that X_{α} ∈ S(P(S_{α})), so that as above |X_{α}| = 1. Put Y = X \ X_{α} and note that if y ∈ Y and x ∈ X then yx ∉ X_{α}. Let M = max Y X_{α}. Suppose now that U, V ∈ P(S_{α}) and UX_{α}^{2} = V X_{α}^{2}. Put

\overline{U} = U ∪ ⋃\{e_{γ}U : γ ∈ M\} ∪ ⋃\{e_{γ}V : γ ∈ M\}

and

\overline{V} = V ∪ ⋃\{e_{γ}U : γ ∈ M\} ∪ ⋃\{e_{γ}V : γ ∈ M\}.

Then

\overline{U}X^{2} = UX^{2} ∪ T = U(X_{α} ∪ Y)^{2} ∪ T = U(X_{α}^{2} ∪ X_{α}Y ∪ Y^{2}) ∪ T,

where

T = (⋃{e_{γ}U : γ ∈ M} ∪ ⋃{e_{γ}V : γ ∈ M})X^{2}.

If β ∈ supp UX_{α}Y then β ≤ γ for some γ ∈ M so that if x ∈ (UX_{α}Y)_{β} then x = e_{γ}x ∈ T. Similarly, UY^{2} ⊆ T. Thus

\overline{U}X^{2} = UX_{α}^{2} ∪ T;

dually,

\overline{V} X^{2} = V X_{α}^{2} ∪ T

so that \overline{U}X^{2} = \overline{V} X^{2} and as X ∈ S(P(S)), \overline{U}X = \overline{V} X. It follows easily that UX_{α} = V X_{α}.

Finally, if U ∈ P(S_{α}) and UX_{α}^{2} = X_{α}^{2}, then taking a ∈ S_{α} we have aUX_{α}^{2} = aX_{α}^{2} and the above shows that aUX_{α} = aX_{α}; hence UX_{α} = X_{α} and X_{α} ∈ S(P(S_{α}).

We are now in a position to prove our final result.

Theorem 4.4 The semigroup P(S) is an order in P(Q) if and only if S satisfies condition (*).

Proof In view of Proposition 4.1 it only remains to show that if P(S) is a weak order in P(Q) then if X ∈ S(P(S)), X lies in a subgroup of P(Q). We use induction on θ(X) to show that X is balanced. If θ(X) = 1 then X ⊆ S_{α} for some α ∈ Y and then from Theorem 3.2 X lies in a subgroup of P(Q) so that in particular, X is balanced by Lemma 2.5.

We make the inductive assumption that θ(X) > 1 and if Y ∈ S(P(S)) with θ(Y) < θ(X), then Y is balanced.

Let α ∈ max X; by Lemma 4.3, |X_{α}| = 1. Suppose there exists β ∈ supp X with β < α. Write

X = {s, t_{1}, ..., t_{n}, u_{1}, ..., u_{m}}

where s ∈ X_{α}, t_{1}, ..., t_{n} are distinct elements of X_{β} and u_{1}, ..., u_{m} are distinct elements of X \ (X_{α} ∪ X_{β}). Pick any a ∈ S_{β} and put Z = aX; since βα = ββ we have that θ(Z) < θ(X). It is then easy to see that Z ∈ S(P(S)) so that by the inductive assumption, Z is balanced. In particular, |Z_{β}| = 1. Thus at_{1} = ... = at_{n} and so t_{1} = ... = t_{n}, that is, |X_{β}| = 1. Moreover as = at_{1} so that se_{β} = t_{1}.

We can carry out this procedure for any α ∈ max X and β ∈ supp X with β < α; it follows that |X_{γ}| = 1 for all γ ∈ supp X. Suppose now that β, γ ∈ supp X and β < γ. Then β < γ ≤ α for some α ∈ max X; from the above, if x ∈ X_{β}, y ∈ X_{γ} and z ∈ X_{α}, then x = ze_{β} and y = ze_{γ}. Thus ye_{β} = ze_{γ}e_{β} = ze_{β} = x and so X is balanced. By induction on θ(X), every element of S(P(S)) is balanced.

By Lemma 2.5, to show that X ∈ S(P(S)) lies in a subgroup of P(Q), it remains to show that supp X is a subsemilattice of Y. If θ(X) = n then supp X^{n} is a subsemilattice and also X^{n} ∈ S(P(S)). By the above, X^{n} is balanced so that by Lemma 2.5, X^{n} lies in a subgroup of P(Q). Further, if supp X^{n} = {α_{1}, ..., α_{m}} then EX^{n} = X^{n} where E = {e_{1}, ..., e_{m}}. We now call on Lemma 4.2 to give that EX = X. Let α, β ∈ supp X; then α = α_{i} ∈ supp X^{n} and if x ∈ X_{β} then e_{i}x ∈ EX = X so that αβ ∈ supp X. Thus supp X is a subsemilattice of Y and this completes the proof that X lies in a subgroup of P(Q). Hence P(S) is an order in P(Q).

Theorem 4.4 simplifies slightly in the case where Y is a chain.

Corollary 4.5 If Y is a chain then P(S) is an order in P(Q) if and only if S satisfies condition (**): (**) for α, β ∈ Y with α > β and x ∈ S_{α}, y ∈ S_{β}, there exists p ∈ S_{α} and q ∈ S_{β} with xpe_{β} = yq.

Proof If P(S) is an order in P(Q) then by Theorem 4.4, (*) holds. Let α, β ∈ Y with α > β and x ∈ S_{α}, y ∈ S_{β}. By (*) there exists γ ∈ Y with γ ≥ α > β and elements p ∈ S_{α}, q ∈ S_{β} and a ∈ S_{γ} with ae_{α} = xp, ae_{β} = yq. Then xpe_{β} = ae_{α}e_{β} = ae_{β} = yq and (**) holds.

Conversely, suppose that (**) holds. Let α_{1}, ..., α_{n} be distinct elements of Y and s_{i} ∈ S_{α_{i}}, 1 ≤ i ≤ n. Note that (*) holds trivially when n = 1. We assume inductively that n > 1 and (*) holds for all natural numbers less than n.

Without loss of generality we assume that α_{1} > α_{2} > ... > α_{n}. By the inductive assumption there exists α′ ∈ Y with α′ ≥ α_{2} and elements a′ ∈ S_{α′}, t′_{i} ∈ S_{α_{i}}, 2 ≤ i ≤ n with a′e_{α_{i}} = s_{i}t′_{i}, 2 ≤ i ≤ n. Let b ∈ S_{α_{2}}; by (**) with x = s_{1}, y = a′b there exist p ∈ S_{α_{1}} and q ∈ S_{α_{2}} with s_{1}pe_{α_{2}} = a′bq. Put α = α_{1}, a = s_{1}p, t_{1} = p and t_{i} = t′_{i}bq ∈ S_{α_{i}}, 2 ≤ i ≤ n. Then ae_{α_{1}} = s_{1}pe_{α_{1}} = s_{1}p = s_{1}t_{1} and for i ∈ {2, ..., n} ae_{α_{i}} = s_{1}pe_{α_{i}} = s_{1}pe_{α_{2}}e_{α_{i}} = a′bqe_{α_{i}} = s_{i}t′_{i}bq = s_{i}t_{i}. Thus (*) holds.

We end the paper by illustrating Theorem 4.4 and Corollary 4.5 with a number of examples.

Example 4.6 .

We consider several cases of orders in a two-element chain of torsion-free commutative groups.

Let G be the infinite cyclic group generated by an element z and let T be the infinite monogenic semigroup generated by z, so that T is an order in G.

Put Q_{1} = G^{1} and S_{1} = T^{1}, so that S_{1} is an order in Q_{1}. For any z^{n} ∈ T,

1 1 z^{0} ≠ z z^{n}

so that (**) fails with x = 1 and y = z.

Now take Q_{2} = G^{0} and S_{2} = T^{0} so that again S_{2} is an order in Q_{2}. Then for any z^{n} ∈ T,

z^{n} z 0 = 0 0

so that (**) holds.

With G and T as above let H be the infinite cyclic group generated by t and let U be the infinite monogenic semigroup generated by t, so that U is an order in H. Moreover if we let ϕ : G → H be the trivial homomorphism then S_{3} = T ∪U is an order in Q_{3} = G∪H, where Q_{3} is the two element chain of groups G and H with structure homomorphism ϕ. Now for any r, s ≥ 1,

z z^{r} t^{0} = t^{0} ≠ t t^{s}

so that (**) does not hold with x = z and y = t.

On the other hand let ψ : G → H be the homomorphism given by ψ(z) = t^{n} where n ≥ 1. Put S_{4} = T ∪ U and Q_{4} = G ∪ H with structure homomorphism ψ. Again, S_{4} is an order in Q_{4}. Let r, s ≥ 1 and pick u with n(r + u) > s. Then

z^{r} z^{u} t^{0} = z^{r+u} t^{0} = ψ(z^{r+u}) t^{0} = (ψ(z))^{r+u} t^{0} = t^{n(r+u)} t^{0} = t^{n(r+u)} = t^{s} t^{n(r+u)−s}

which shows that (**) holds.

Moving now to the example of an order S in Q, where Q is an arbitrary semilattice Y of torsion-free commutative groups, we remark that by (*), if Y has a finite set of elements without an upper bound, then P(S) is not a weak order in P(Q). In particular, if Q is the three element semilattice with two incomparable elements then P(Q) is not a weak order in itself, as shown at the beginning of this section.

Example 4.7 .

Let Y be the four element semilattice {α_{1}, ..., α_{4}} where α_{1} ≥ α_{2}, α_{3} ≥ α_{4}. For i ∈ {1, ..., 4} let G_{α_{i}} be the infinite cyclic group generated by x_{i} and let T_{α_{i}} be the order in G_{α_{i}} consisting of the positive powers of x_{i}. Let Q be the semilattice Y of groups G_{α_{i}} with structure homomorphisms ϕ_{α_{i}α_{j}} : G_{α_{i}} → G_{α_{j}} given by ϕ_{α_{i}α_{j}}(x_{i}) = x_{j}. Putting S to be the subsemigroup of Q consisting of positive words we have that S is an order in Q.

Consider positive integers a_{1}, ..., a_{4} and choose n with n > max {a_{1}, ..., a_{4}}. Put α = α_{1}, a = x_{1}^{n} and t_{i} = x_{i}^{n−a_{i}}, 1 ≤ i ≤ 4. Then

ae_{i} = x_{1}^{n}e_{i} = x_{i}^{n} = x_{i}^{a_{i}}x_{i}^{n−a_{i}} = x_{i}^{a_{i}}t_{i},

thus showing that (*) holds.

References

[1] A.H. Clifford and G.B. Preston, The algebraic theory of semigroups, (Mathematical Surveys 7, Vol. 1 American Math. Soc. 1961).

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

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

[4] J.M Howie, An Introduction to Semigroup Theory, (Academic Press 1976).

[5] J.-E. Pin, ‘Power semigroups and related varieties of finite semigroups’, Semigroups and their applications (eds. S.M. Goberstein and P.M. Higgins 1987), 139-152.