CONDITIONS FOR THE PREFIX EXPANSION OF A MONOID TO BE (WEAKLY) LEFT AMPLE CHRISTOPHER HOLLINGS Abstract. We derive necessary and sufficient conditions for the BirgetRhodes prefix expansion [BR1] of a monoid to be (weakly) left ample, thereby proving analogues of the results already obtained for the related Szendrei expansion by Fountain, Gomes and Gould [FGG] and Fountain and Gomes [FG]. As a corollary, we obtain conditions for the prefix expansion to be inverse.

Introduction The concept of an expansion of a monoid was introduced by Birget and Rhodes in their 1984 paper [BR1] as a particular type of functor from one category of semigroups to a larger one. In [BR1], a number of different expansions were defined, including the so-called prefix expansion. After defining this expansion1, Birget and Rhodes went on to determine its major properties by, for example, investigating the nature of its subgroups and its R-classes. The ultimate goal of [BR1] was to apply techniques from finite semigroup theory to the study of infinite semigroups, something which Birget and Rhodes continued in [BR2], in which the usefulness of the constructions of [BR1] was demonstrated. In [BR2], we find a more comprehensive study of the prefix expansion, including the result that the prefix expansion of a group is an inverse monoid. Birget and Rhodes ultimately applied their “expansion techniques” to the Burnside problem for groups. Some time after the publication of [BR1], it was observed by Szendrei [Sz] that the prefix expansion of a group has a particularly simple form in terms of Date: March 28, 2007. 2000 Mathematics Subject Classification. 20 M 99. Key words and phrases. prefix expansion, (weakly) left ample monoid, inverse monoid. ´ Some of the results in this paper were obtained during a visit to the Centro de Algebra da Universidade de Lisboa in July 2005, funded by Treaty of Windsor grant LIS/992/2(205/06)U09. I wish to express my gratitude both to the British Council and to my hosts in Lisbon. I would also like to thank Victoria Gould for her extensive advice and encouragement. 1 In [BR1], Birget and Rhodes defined two prefix expansions: a left-hand and a right-hand version. In the present paper, we confine our study to the right-hand version. 1

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finite subsets of the group, rather than the R-chains which appear in the original definition. Szendrei’s observations were used to define a new expansion, which, for an arbitrary monoid, differs from the prefix expansion. This new expansion, appropriately enough, came to be known as the Szendrei expansion. Of the two expansions, it is the Szendrei expansion which has proved extremely useful in the study of non-regular semigroups. The Szendrei expansion of a right cancellative monoid, for example, is left ample. Left ample monoids are a special case of weakly left ample monoids, which are, in turn, a special case of weakly left E-ample monoids.2 These monoids occur very naturally as the (2,1,0)-subalgebras of semigroups of partial transformations of a set, where the unary operation is α 7→ Idom α . In [FGG] and [FG] can be found necessary and sufficient conditions for the Szendrei expansion of a monoid to be weakly left ample and left ample, respectively. In the present paper, we return to the prefix expansion and similarly obtain necessary and sufficient conditions for it to be (weakly) left ample. Since every inverse monoid is left ample, it will then be a simple matter to obtain as a corollary conditions for the prefix expansion to be inverse. In Section 1, we give a brief overview of weakly left ample and left ample monoids. In Section 2, we define the prefix expansion of a monoid and investigate some of its properties. In particular, because of their importance later on, we determine the idempotents of the prefix expansion when the monoid in question takes on certain special forms. By the end of Section 3, we will have determined the conditions for the prefix expansion of a monoid to be weakly left ample, and in Section 4, we specialise these conditions to the case when the prefix expansion is left ample. The concrete examples given in Section 5 will demonstrate that the conditions obtained in Sections 3 and 4 are not so unnatural as they might first appear. We conclude the paper with Section 6, in which we consider two special cases: the case when the monoid in question is finite, and the case when the prefix expansion is inverse. 1. An overview of (weakly) left ample monoids A summary of weakly left E-ample monoids can be found in [G]. In the present overview, we jump straight to the special case of weakly left ample monoids, since 2Weakly

left E-ample semigroups are precisely the (left) twisted C-semigroups of Jackson and Stokes [JS] and also the guarded semigroups of Manes [Ma]; weakly left ample semigroups are full (left) twisted C-semigroups and replete guarded semigroups.

PREFIX EXPANSION - (WEAKLY) LEFT AMPLE

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it is these which we deal with in this paper. Most of the following details apply equally well to semigroups, but we will only consider monoids, since a semigroup must be a monoid in order for us to be able to take its prefix expansion. Let S be a monoid with subset E(S) of idempotents. We define the (equivae on S by the rule that lence) relation R e b ⇐⇒ ∀e ∈ E(S)[ea = a ⇔ eb = b], aR

e for a, b ∈ S. Thus, two elements a, b are R-related if, and only if, they have the same idempotent left identities in S. It is easy to see that if e ∈ E(S), then e e ⇐⇒ ea = a and ∀f ∈ E(S)[f a = a ⇒ f e = e]. aR

(1.1)

Definition 1.1. A monoid S is called weakly left ample if e (1) every element a is R-related to a (unique) idempotent, denoted a+ ; (2) idempotents commute; e is a left congruence; (3) R (4) for all a ∈ S and all e ∈ E(S), ae = (ae)+ a. e b if, and only if, a+ = b+ . The idempotent a+ is a left identity for a. Thus a R It is also clear that if e is idempotent, then e+ = e. Since it will also feature in the definition of a left ample monoid, the identity found in condition (4) will be referred to throughout as the “left ample identity.” The following result is worth noting, since it can be (indeed, will be!) very useful when showing that a monoid is weakly left ample. Lemma 1.2. Let M be a monoid whose subset E(M ) of idempotents forms a e complete ∧-semilattice. Then every element of M is R-related to an idempotent. In particular, this result holds if E(M ) is a finite semilattice. Proof. Let a ∈ M . In the notation of [G], we define the set E(M ) a

E(M ) a

⊆ E(M ) by

= {e ∈ E(M ) : ea = a},

i.e., E(M ) a is the set of idempotent left identities of a. It is clear that nonempty, since 1 ∈ E(M ) a. Notice that if e, f ∈ E(M ) a, then

E(M ) a

is

(ef )a = e(f a) = ea = a, so ef ∈ E(M ) a, hence E(M ) a is a subsemilattice of E(M ). V We now put u = E(M ) a. Clearly, ua = a. Let g ∈ E(M ) with ga = a, so e u. that g ∈ E(M ) a. Then, since u ≤ g, we have gu = u, hence a R

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A finite semilattice is necessarily complete, so, in particular, this result holds for the case when E(M ) is a finite semilattice. ¤ Weakly left ample monoids generalise inverse monoids, since every inverse e coinmonoid is weakly left ample with a+ = aa−1 (in an inverse monoid, R cides with Green’s relation R). A unipotent monoid is weakly left ample, with a+ = 1, for all elements a. We note a useful identity involving + , which follows immediately from the fact e is a left congruence: that R Lemma 1.3. Let S be a weakly left ample monoid, and let s, t ∈ S. Then (st)+ = (st+ )+ .

Weakly left ample monoids also generalise the left ample (formerly, left type-A) monoids of Fountain [F1, F2]. To see this, let us define the (equivalence) relation R∗ on a monoid3 S: a R∗ b ⇐⇒ ∀x, y ∈ S[xa = ya ⇔ xb = yb].

(1.2)

Equivalently, a R∗ b in S if, and only if, a R b in some oversemigroup T [F2, Lemma 1.1]. Hence R ⊆ R∗ . Notice also that if a R∗ b in S, then we can set y = 1 and x = e in (1.2), for any e ∈ E(S), to obtain ea = a ⇔ eb = b. e Hence R ⊆ R∗ ⊆ R. We now define a left ample monoid: Definition 1.4. A monoid S is called left ample if (1) every element a is R∗ -related to a (unique) idempotent, denoted a+ ; (2) idempotents commute; (3) for all a ∈ S and all e ∈ E(S), ae = (ae)+ a. Note that R∗ is always a left congruence, so we do not need to demand this e it is easy to see that every left ample explicitly. By our observation that R∗ ⊆ R, e so there monoid is weakly left ample, indeed, in a left ample monoid, R∗ = R, is no ambiguity in our use of + to denote the idempotent in an R∗ -class. In an e inverse monoid, we have R = R∗ = R. The following result involving R∗ will be useful in a later section: 3If

S were a semigroup but not a monoid, we would need x, y ∈ S 1 here.

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Lemma 1.5. Let M be an arbitrary monoid and let m ∈ M . Then m R∗ 1 in M if, and only if, m is right cancellable. Proof. Let m ∈ M . Then m R∗ 1 if, and only if, for all x, y ∈ M , xm = ym ⇔ x1 = y1, that is, m is right cancellable.

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2. The prefix expansion As noted in the Introduction, the concept of an expansion was introduced by Birget and Rhodes in [BR1]. The formal definition of an expansion is as follows: Definition 2.1. [BR1, p. 241] An expansion is a functor F from one category of semigroups to a larger one such that there exists a natural transformation η from F to the identity functor, with each arrow ηS surjective. Birget and Rhodes defined a number of different expansions; among them was the prefix expansion: Definition 2.2. [BR1, p. 266]4 Let M be a monoid and let s1 , s2 , · · · , sn ∈ M . We define the prefix set of the string (s1 , s2 , . . . , sn ) to be the set P (s1 , s2 , . . . , sn ) = {1, s1 , s1 s2 , s1 s2 s3 , . . . , s1 s2 s3 · · · sn−1 , s1 s2 s3 · · · sn−1 sn }, i.e., the set of all “prefixes” of s1 s2 · · · sn . The prefix expansion of M is the set Pr(M ) = {(P (s1 , s2 , . . . , sn ), s1 s2 · · · sn ) : s1 , s2 , . . . , sn ∈ M }, together with the operation (P (s1 , . . . , sn ), s1 · · · sn )(P (t1 , . . . , tm ), t1 · · · tm ) = (P (s1 , . . . , sn ) ∪ s1 · · · sn P (t1 , . . . , tm ), s1 · · · sn t1 · · · tm ). The prefix expansion, like any expansion, is a functor, so we should also specify its effect on the arrows of the category of all monoids. However, the ‘objects’ part is the only aspect of the expansion which we will consider. We will therefore omit all reference to the prefix expansion’s effect on arrows. Proposition 2.3. [BR1, p. 266] If M is a monoid then Pr(M ) is also a monoid, with identity ({1}, 1). 4Our

“prefix expansion” is Birget and Rhodes’ SeR .

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Proposition 2.4. The prefix expansion Pr(M ) of a monoid M is generated by the elements of the form ({1, s}, s), with s ∈ M . Proof. Let (P (s1 , . . . , sn ), s1 · · · sn ) ∈ Pr(M ). We claim that this element can be rewritten as (P (s1 , . . . , sn ), s1 · · · sn ) = ({1, s1 }, s1 )({1, s2 }, s2 ) · · · ({1, sn }, sn ). This result certainly holds when n = 1. Suppose, for induction, that it holds for n = k and note that (P (s1 , . . . , sk , sk+1 ), s1 · · · sk sk+1 ) = (P (s1 , . . . , sk ), s1 · · · sk )({1, sk+1 }, sk+1 ). The result then follows by induction.

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At this point, it is worth noting a couple of useful facts about elements of Pr(M ): Lemma 2.5. Let M be a monoid. Then (i) if (B, b), (C, c) ∈ Pr(M ) with B = C, then b R c; (ii) if u, v ∈ M with u R v, then ({1, u, v}, v), ({1, v, u}, u) ∈ Pr(M ). Proof. (i) We have b ∈ C and c ∈ B, so there exist b0 , c0 ∈ M with bb0 = c and cc0 = b, hence b R c. (ii) We have uw = v and vw0 = u, for some w, w0 ∈ M , so ({1, u, uw}, uw) = ({1, u, v}, v) and ({1, v, vw0 }, vw0 ) = ({1, v, u}, u) are elements of Pr(M ). ¤ Birget and Rhodes proved the following about the prefix expansion: Proposition 2.6. [BR2, Proposition II.6] If M is a group, then Pr(M ) is an inverse monoid with (S, s)−1 = (s−1 S, s−1 ). It was further observed by Szendrei [Sz] that if M is a group, then Pr(M ) has a particularly simple form: Proposition 2.7. [Sz, Proposition 1] If M is a group, then Pr(M ) = {(A, a) ∈ P1f (M ) × M : a ∈ A},

(2.1)

(A, a)(B, b) = (A ∪ aB, ab),

(2.2)

with multiplication where P1f (M ) denotes the collection of finite subsets of M which contain 1.

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We can use (2.1) and (2.2) to define a new expansion of a monoid M : the Szendrei expansion Sz(M ). Note that in general, for an arbitrary monoid M , Pr(M ) and Sz(M ) will be different. We now wish to characterise the idempotents of Pr(M ). For reasons which will become clear in later sections, we are interested in determining what forms the idempotents of Pr(M ) take when M satisfies certain conditions. First of all, we are interested in the case when M is unipotent, so we now make some comments on unipotent monoids. A unipotent monoid has a particular decomposition which we will make extensive use of in later sections. Let M be a unipotent monoid and suppose that s ∈ M has a right inverse in M , i.e., there exists an s−1 ∈ M such that ss−1 = 1. Then s−1 s is idempotent. However, there is only one idempotent which it can be, namely 1. Thus s−1 is a two-sided inverse for s, hence the group of units of M (H1 , the H-class of 1) coincides with R1 , the R-class of 1, and dually with L1 . In this case, M = H1 ∪ I, where I is an ideal of M consisting of all non-invertible elements, so that H1 ∩ I = ∅.5 We return now to the idempotents of Pr(M ). The following lemma follows immediately from the definition of multiplication in Pr(M ): Lemma 2.8. If M is an arbitrary monoid, then E(Pr(M )) = {(F, f ) ∈ Pr(M ) : f F ⊆ F and f ∈ E(M )}. In particular, Pr(M ) has idempotents of the form (F, 1); let E denote the collection of all such idempotents. Lemma 2.9. If M is an arbitrary monoid, then E is a subsemilattice of Pr(M ) and E = {(F, 1) ∈ Pr(M ) : F ⊆ R1 }. Proof. First of all, it is easy to see that E is a semilattice: (E, 1)(F, 1) = (E ∪ F, 1) = (F ∪ E, 1) = (F, 1)(E, 1). Now suppose that (F, 1) ∈ E. Then, by definition of Pr(M ), for each a ∈ F , there is a b ∈ M such that ab = 1, so F ⊆ R1 . ¤ Corollary 2.10. If M is a unipotent monoid, then E(Pr(M )) = E and E = {(F, 1) ∈ Pr(M ) : F ⊆ H1 } = E(Sz(H1 )). 5This decomposition was first demonstrated for cancellative monoids by Anton Suschkewitsch

in 1934 [Su].

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Returning to the general case, the next result follows immediately from the second part of Lemma 2.9: Lemma 2.11. If M is a monoid with R1 = {1}, then E = {({1}, 1)}. Corollary 2.12. If M is a unipotent monoid with R1 = {1}, then ({1}, 1) is the only idempotent in Pr(M ). Now suppose that M has precisely two idempotents. We introduce the term bipotent to refer to such a monoid. Unless otherwise stated, the two idempotents of a bipotent monoid will be denoted by 1 and e. Extending the above observations, if M is bipotent, then any idempotent in Pr(M ) has one of two possible forms: either (E, e), where eE ⊆ E, or (F, 1), where F ⊆ R1 . For reasons which, again, will become clear in later sections, we now impose extra conditions on the bipotent monoid M : Lemma 2.13. Let M be a bipotent monoid which satisfies the following conditions: • R1 = {1} and Re = {e}; • e
PREFIX EXPANSION - (WEAKLY) LEFT AMPLE

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Proposition 3.2. Let M be a monoid whose prefix expansion Pr(M ) is weakly left ample. Then either E(M ) = {1} or E(M ) = {1, e}, where e 6= 1. Proof. Let e, f ∈ E(M ). Then ({1, e}, e), ({1, f }, f ) ∈ E(Pr(M )). Suppose that e, f 6= 1. Since idempotents commute in Pr(M ), we have ({1, e}, e)({1, f }, f ) = ({1, f }, f )({1, e}, e), hence ({1, e, ef }, ef ) = ({1, f, f e}, f e). Thus ef = f e and either e = f or e = f e, since e ∈ {1, f, f e}. Also, either f = e or f = ef , since f ∈ {1, e, ef }. In any case, e = f , since ef = f e. ¤ We begin by considering the case when M = H1 ∪ I is unipotent. For (S, s) = (P (s1 , . . . , sn ), s1 · · · sn ), we put S = S ∩ H1 , i.e., S is the set of all units in S. Lemma 3.3. Let (s1 , . . . , sk ) be the longest initial substring (s1 , . . . , si ) of (s1 , . . . , sn ) such that s1 · · · sk is a unit. Then S = P (s1 , . . . , sk ) and (S, 1) ∈ Pr(M ). Proof. By assumption, (s1 , . . . , sk ) is the longest initial substring of (s1 , . . . , sn ) for which there exists a t ∈ M with s1 · · · sk t = 1. Then, for any h ≤ k, s1 · · · sh is a unit, with (s1 · · · sh )−1 = sh+1 · · · sk t. Moreover, if s1 · · · sh were a unit (with inverse u, say) for any k + 1 ≤ h ≤ n, then s1 · · · sk sk+1 would also be a unit with (s1 · · · sk sk+1 )−1 = sk+2 · · · sh u, which would contradict s1 · · · sk being the ‘longest’ unit. Hence s1 · · · sh is a non-unit for k + 1 ≤ h ≤ n. Therefore, P (s1 , . . . , sk ) is precisely the set of units in S, as required. Since every element ¤ of S is a unit, (S, 1) = (P (s1 , . . . , sk ), 1) is certainly an element of Pr(M ). Lemma 3.4. Let M = H1 ∪ I be a unipotent monoid and let (S, s) ∈ Pr(M ). e (S, 1). Then (S, s) R Proof. We see that (S, 1) is certainly a left identity for (S, s): (S, 1)(S, s) = (S ∪ S, s) = (S, s).

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Let (F, 1) be an idempotent in Pr(M ): (F, 1)(S, s) = (S, s) ⇒ (F ∪ S, s) = (S, s) ⇒F ⊆S ⇒ F ⊆ S, since F ⊆ H1 ⇒ (F, 1)(S, 1) = (S, 1) e (S, 1). Therefore (S, s) R

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e ({1}, 1). Corollary 3.5. If s ∈ I, then ({1, s}, s) R

Proposition 3.6. Let M be a unipotent monoid. If Pr(M ) is weakly left ample, then M satisfies the following condition: (U1) su = s, for all s ∈ I and all u ∈ H1 . Proof. Let s ∈ I and u ∈ H1 . Since I is an ideal, su ∈ I. Then, by Lemma 3.4 and the fact that Pr(M ) is weakly left ample, we have ({1, s, su}, s)+ = ({1}, 1). From the left ample identity ({1, s}, s)({1, u}, 1) = [({1, s}, s)({1, u}, 1)]+ ({1, s}, s), we deduce that {1, s, su} = {1, s}. If su = 1, then su is a unit — a contradiction. Therefore, su = s.

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Proposition 3.6 has the following converse: Proposition 3.7. Let M be a unipotent monoid which satisfies the condition: (U1) su = s, for all s ∈ I and all u ∈ H1 . Then Pr(M ) is weakly left ample. Proof. By Lemma 2.9, E(Pr(M )) = E is a semilattice. Let (S, s) ∈ Pr(M ). By e (S, s). Lemma 3.4, (S, 1) R We need to show that the left ample identity holds. The two sides of the identity are: (A, a)(E, 1) = (A ∪ aE, a) and

[(A, a)(E, 1)]+ (A, a) = (A ∪ aE ∪ A, a) = (A ∪ aE, a).

For the identity to hold, we need A ∪ aE = A ∪ aE. There are two cases to consider:

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(i) a ∈ H1 . In this case, aE = aE, so A ∪ aE = A ∪ aE. (ii) a ∈ I. Then aE = ∅ and aE = {a}, by property (U1), so A∪aE = A∪aE. e to be a left congruence. First note that: Finally, we need R e (B, b) ⇐⇒ (A, a)+ = (B, b)+ ⇐⇒ A = B. (A, a) R

e (C, c)(B, b) is a consequence of (A, a) R e (B, b), We wish to show that (C, c)(A, a) R for any (C, c) ∈ Pr(M ). To put this another way, we want to show that C ∪ cA = C ∪ cB, or C ∪ cA = C ∪ cB. Suppose that A = B. There are two cases to consider: (i) c ∈ I. In this case, cA = ∅ = cB, so it is immediate that C ∪cA = C ∪cB. (ii) c ∈ H1 . In this case, cA = cA and cB = cB. Then, since A = B, we have C ∪ cA = C ∪ cB. e is a left congruence, and Pr(M ) is weakly left ample. Therefore R ¤ We must next deal with the bipotent case which arises from Proposition 3.2.

Proposition 3.8. Let M be a bipotent monoid. If Pr(M ) is weakly left ample, then M satisfies the following conditions: (B1) R1 = {1} and Re = {e}; (B2) e
R1 = {1}; e ({1}, 1); e 6∈ A ⇒ (A, a) R se ∈ {e, s}, for all s ∈ M ; Re = {e}; e ≤R a ⇒ ea = e; e
Proof. (A) Suppose that s R 1. Now as ({1, s}, 1) and ({1, e}, e) are idempotents, we have ({1, s}, 1)({1, e}, e) = ({1, e}, e)({1, s}, 1).

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We deduce that {1, s, e} = {1, e, es}. Hence s ∈ {1, e, es} ∩ R1 . As e 6= 1 and s R 1 it follows that s = 1. (B) Suppose that e 6∈ A and let (F, f ) ∈ E(Pr(M )). Then either f = 1 or f = e. If f = 1, then F = {1}, by (A), since F ⊆ R1 by Lemma 2.9. If f = e, then e ∈ F and (F, e)(A, a) = (A, a) ⇒ F ⊆ A ⇒ e ∈ A, e ({1}, 1). a contradiction. So if e 6∈ A, then (A, a) R (C) If s = 1 or e, then the result certainly holds, so suppose that s 6∈ {1, e}. Then ({1, s}, s)({1, e}, e) = [({1, s}, s)({1, e}, e)]+ ({1, s}, s) = ({1, s, se}, se)+ ({1, s}, s). If e 6∈ {1, s, se}, then, by (B), ({1, s, se}, se)+ = ({1}, 1) and so ({1, s, se}, se) = ({1, s}, s). Hence se = s. e ({1}, 1), by (B). We multiply (D) Let w R e, with w 6= e. Then ({1, w}, w) R on the left by ({1, e}, e) to obtain e ({1, e}, e). ({1, e}, e)({1, w}, w) R

Every idempotent is a left identity for its R-class, so ew = w. Hence

Notice that

e ({1, e}, e). ({1, e, w}, w) R ({1, e, w}, e)({1, e, w}, e) = ({1, e, w}, e) ∈ E(Pr(M )).

Also, ({1, e, w}, e)({1, e, w}, w) = ({1, e, w}, w), so ({1, e, w}, e)({1, e}, e) = ({1, e}, e). Then w ∈ {1, e}, so w = 1 and 1 R e, a contradiction. Thus Re = {e}. (E) We have e ≤R a ⇒ e ≤R ea ≤R e ⇒ e R ea ⇒ e = ea, by (D). (B2) Suppose that e
PREFIX EXPANSION - (WEAKLY) LEFT AMPLE

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Therefore ({1, a, e}, e)({1, e}, e) = ({1, e}, e)({1, a, e}, e), from which we deduce that {1, a, e} = {1, e}, hence a = 1. (B3) Let a ∈ M \ {1}. Then ae ∈ {e, a}, by (C). If ae = e, then we get e ≤R a. If e
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which case e ∈ A. Conversely, if e ∈ A, then ({1, e}, e)(A, a) = (A, a), by Lemma 3.10. We have shown:  ({1}, 1) if e 6∈ A; + (A, a) = ({1, e}, e) if e ∈ A.

e is a left congruence. Suppose that (A, a) R e (B, b), We must next show that R for (A, a), (B, b) ∈ Pr(M ). Then e is either in both A and B, or it is in neither. Consider (C, c)(A, a) = (C ∪ cA, ca). Suppose that e ∈ C ∪ cA. If e ∈ C, then e ∈ C ∪ cB, so we assume that e ∈ cA but that e 6∈ C. There must then exist an x ∈ A such that e = cx, and so e ≤R c. If e R c, then e = c, by (B1), a contradiction, since e 6∈ C. We must therefore have e
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We can now gather together the results of this section into the following analogue of Theorem 3.1: Theorem 3.12. The prefix expansion Pr(M ) of a monoid M is weakly left ample if, and only if, either • M is unipotent and satisfies the condition: (U1) su = s, for all s ∈ I and all u ∈ H1 . or • M is bipotent and satisfies the conditions: (B1) R1 = {1} and Re = {e};

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(B2) e
su = s, for all s ∈ I and all u ∈ H1 ; R = ι in I; u R v and us = vs ⇒ u = v; u R us = vs ⇒ u = v,

where ι denotes the equality relation on M . Proof. Condition (U1) holds by Proposition 3.6. (U2) Let u ∈ I and suppose that u R v, for some v ∈ M . Then v = us, for some s ∈ M , and so v ∈ I, hence Ru ⊆ I. If s ∈ H1 , then v = us = u, by (U1). Now

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suppose that s ∈ I. Since u R v, we have ({1, u, v}, v), ({1, v, u}, u) ∈ Pr(M ), by Lemma 2.5(ii), and ({1, v, u}, u)({1, s}, s) = ({1, u, us}, us), since v = us, = ({1, u}, u)({1, s}, s). By Lemma 4.2, ({1, s}, s) is right cancellable, so ({1, v, u}, u) = ({1, u}, u). Since v ∈ I, v 6= 1, so we conclude that v = u. (U3) Suppose that u R v and that us = vs. If s ∈ H1 , then u = v. If s ∈ I, then ({1, s}, s) is right cancellable, by Lemma 4.2. We therefore deduce from ({1, u, v}, v)({1, s}, s) = ({1, v, u}, u)({1, s}, s) that ({1, u, v}, v) = ({1, v, u}, u), hence u = v. (U4) Let u R us = vs. If s ∈ H1 , then it follows immediately that u = v. On the other hand, if s ∈ I, then us ∈ I, hence u = us = vs, by (U2). Note that ({1, v, u}, u) ∈ Pr(M ), since u = vs. Then ({1, v, u}, u)({1, s}, s) = ({1, v, u, us}, us) = ({1, v, vs}, vs), since u = us = vs = ({1, v}, v)({1, s}, s). Once again, ({1, s}, s) is right cancellable so we obtain ({1, v}, v) = ({1, v, u}, u), from which we deduce that u = v. ¤ Before we prove a converse for Proposition 4.3, we will prove a useful preliminary result: Lemma 4.4. Let M be a unipotent monoid with the properties listed in Proposition 4.3 and let ({1, s}, s) ∈ Pr(M ). Then ({1, s}, s) R∗ ({1, s}, 1). e ({1, s}, 1). We suppose that Proof. By Lemma 3.4, we already have ({1, s}, s) R (U, u)({1, s}, s) = (V, v)({1, s}, s),

for (U, u), (V, v) ∈ Pr(M ), that is, us = vs and U ∪ {us} = V ∪ {vs}. We want to prove that (U, u)({1, s}, 1) = (V, v)({1, s}, s),

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where {1, s} = {1, s} ∩ H1 . There are two cases to consider: (i) s ∈ H1 . In this case, u = v and {1, s} = {1, s}, so (U, u)({1, s}, 1) = (V, v)({1, s}, 1), as required. (ii) s ∈ I. If us ∈ U , then us = u0 , where u0 u00 = u, for some u0 ∈ U and some u00 ∈ M . So u = usu00 , hence u R us. Then, by property (U4), u = v. Note also that since s ∈ I, we have u, us ∈ I, so u = us, by property (U2). Similarly, v = vs. Thus us = u = v = vs, hence (U, u) = (V, v) and the desired equality follows immediately. Similarly, if vs ∈ V . Suppose now that us 6∈ U and vs 6∈ V . It is easy to deduce that U = V , hence u R v, by Lemma 2.5. Then u = v, by property (U3). Therefore, (U, u) = (V, v). Hence ({1, s}, s) R∗ ({1, s}, 1), for all generators ({1, s}, s).

¤

Proposition 4.5. Let M be a unipotent monoid which satisfies the following conditions: (U1) (U2) (U3) (U4)

su = s, for all s ∈ I and all u ∈ H1 ; R = ι in I; u R v and us = vs ⇒ u = v; u R us = vs ⇒ u = v.

Then Pr(M ) is left ample. Proof. By Proposition 3.7, we know that Pr(M ) is weakly left ample. We also know, from Lemma 4.4, that every generator of Pr(M ) is R∗ -related to an idempotent. It therefore only remains to show that a general element of Pr(M ) is R∗ -related to an idempotent. Once again, we claim that (S, s) R∗ (S, 1). There are two cases to consider: (i) s ∈ H1 . In this case, S = S, by Lemma 3.3, and s may be cancelled from us = vs to obtain u = v. It therefore easily follows from (U, u)(S, s) = (V, v)(S, s) that (U, u)(S, 1) = (V, v)(S, 1). (ii) s ∈ I. Put (S, s) = (P (s1 , . . . , sn ), s1 · · · sn ). We aim to show that (U, u)(S, s) = (V, v)(S, s) ⇒ (U, u)(S, 1) = (V, v)(S, 1),

(4.1)

for all n ∈ N. This is certainly true for n = 1, by Lemma 4.4. We put (T, t) = (P (s1 , . . . , sn−1 ), s1 · · · sn−1 ), so that (S, s) = (T, t)({1, sn }, sn ), and suppose for induction that (U, u)(T, t) = (V, v)(T, t) ⇒ (U, u)(T , 1) = (V, v)(T , 1).

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There are two subcases to consider: (a) sn ∈ I. By Lemma 4.2, ({1, sn }, sn ) is right cancellable. Therefore (U, u)(S, s) = (V, v)(S, s) ⇒ (U, u)(T, t) = (V, v)(T, t) ⇒ (U, u)(T , 1) = (V, v)(T , 1), by assumption ⇒ (U, u)(S, 1) = (V, v)(S, 1), since S = T , as sn 6∈ H1 . (b) sn ∈ H1 . By Lemma 4.4, ({1, sn }, sn ) R∗ ({1, sn }, 1), so (U, u)(T, t)({1, sn }, sn ) = (V, v)(T, t)({1, sn }, sn ) ⇒ (U, u)(T, t)({1, sn }, 1) = (V, v)(T, t)({1, sn }, 1) ⇒ (U, u)(T ∪ {tsn }, t) = (V, v)(T ∪ {tsn }, t).

(4.2)

Recall that s ∈ I, in which case t ∈ I also, since s = tsn and sn ∈ H1 . Therefore, by (U1), tsn = t and (4.2) becomes (U, u)(T, t) = (V, v)(T, t), whence (U, u)(T , 1) = (V, v)(T , 1), by assumption. Since s ∈ I, we have T = S, hence (U, u)(S, 1) = (V, v)(S, 1). Thus, by induction in both cases, (4.1) holds for all n, hence (S, s) R∗ (S, 1). ¤ We turn at last to the bipotent case and specialise Proposition 3.9 (and hence Proposition 3.8) to the left ample case with the modification of one of the listed properties and the addition of three more: Proposition 4.6. Let M be a bipotent monoid. If Pr(M ) is left ample, then M satisfies the following conditions: (A) (B0 ) (C) (D) (E) (B2) (B3) (F) ∗ ( B1) (B4)

R1 = {1}; e 6∈ A ⇒ (A, a) R∗ ({1}, 1); se ∈ {e, s}, for all s ∈ M ; Re = {e}; e ≤R a ⇒ ea = e; e
Proof. Conditions (A)-(E), as well as (B2) and (B3), follow from Proposition 3.9; e in a left ample semigroup. in the case of (B0 ), we have in mind that R∗ = R

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(F) Since u R us, ({1, us, u}, u) ∈ Pr(M ) and ({1, us, u}, u)({1, s}, s) = ({1, u}, u)({1, s}, s). If s 6= e, then ({1, s}, s) R∗ ({1}, 1), by (B0 ), in which case, ({1, s}, s) is right cancellable, by Lemma 1.5. We deduce that either us = 1 or us = u. If us = 1, by (A), we get s = u = 1, so u = us. If s = e, then u 6= 1, since u R us and 1R 6 e, so ue = u, by (B3). (∗ B1) Let u, v ∈ S be such that u R v. Then v = us, for some s ∈ M . Thus u R us, from which we deduce that u = us = v, by (F). (B4) Assuming the hypothesis, notice that ({1, c}, c)({1, a}, a) = ({1, c, b}, b)({1, a}, a). Since a 6= e, we have ({1, a}, a) R∗ ({1}, 1), by (B0 ). Then ({1, a}, a) is right cancellable, by Lemma 1.5, so ({1, c}, c) = ({1, c, b}, b). We read off the second coordinates to obtain b = c. ¤ By extracting particular conditions from Proposition 4.6, we have proved the the first half of the following result: Proposition 4.7. Let M be a bipotent monoid. Then Pr(M ) is left ample if, and only if, M satisfies the following conditions: (∗ B1) (B2) (B3) (B4)

R = ι; e
Proof. It only remains to prove that if M satisfies the above conditions, then Pr(M ) is left ample. By Proposition 3.11, Pr(M ) is certainly weakly left ample. We must now show that every element is R∗ -related to an idempotent. We will show that (A, a) R∗ ({1}, 1) if e 6∈ A, and (A, a) R∗ ({1, e}, e) if e ∈ A. The first case is equivalent to (A, a) being right cancellable so this is the approach we will take. We first show that if a 6= e, then ({1, a}, a) is right cancellable. Let (B, b), (C, c) ∈ Pr(M ). Then (B, b)({1, a}, a) = (C, c)({1, a}, a) ⇒ B ∪ {ba} = C ∪ {ca} and ba = ca. There are three separate cases to consider: (i) ba ∈ B, ca ∈ C ⇒ B = C ⇒ b R c (by Lemma 2.5) ⇒ b = c, by (∗ B1). (ii) ba ∈ 6 B, ca ∈ 6 C: as (i).

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(iii) ba ∈ B, ca 6∈ C ⇒ b ≤R ba ≤R b, so b = ba = ca. Hence b = c, by (B4). However, we now have: ca = b = c ∈ C, a contradiction. Thus case (iii) cannot occur. Similarly, ba 6∈ B, ca ∈ C cannot occur. We see then that in the two permissible cases (B, b) = (C, c), i.e., ({1, a}, a) is right cancellable. We next extend this last result to a general (A, a) with e 6∈ A. Let A = {1, a1 , a1 a2 , . . . , a1 a2 · · · an = a}, with no ai equal to 1. Then (A, a) = ({1, a1 }, a1 )({1, a2 }, a2 ) · · · ({1, an }, an ), by Proposition 2.4. If ak = e, for some k ≥ 2, then ({1, ak−1 }, ak−1 )({1, ak }, ak ) = ({1, ak−1 , ak−1 ak }, ak−1 ak ) = ({1, ak−1 }, ak−1 ), using (B3). We can therefore assume that no ak is equal to e, for k ≥ 2. The only remaining ai which could equal e is a1 . However, a1 6= e, since e 6∈ A. Therefore, (A, a) is a product of right cancellable elements, so it is itself right cancellable. Hence (A, a) R∗ ({1}, 1) if e 6∈ A. We must now deal with those (A, a) for which e ∈ A. Let A be as above. We conclude from the proof of Lemma 3.10 that a1 = e. We can also assume that a2 6= e, or else a1 = a1 a2 . If aj 6= e and aj+1 = e, for some j ≤ n, then ({1, aj }, aj )({1, aj+1 }, aj+1 ) = ({1, aj }, aj ), by (B3). We can therefore assume that a2 , . . . , an 6= e. Thus ({1, ai }, ai ) is right cancellable for i > 2, from which we obtain (B, b)(A, a) = (C, c)(A, a) ⇒ (B, b)({1, e}, e) = (C, c)({1, e}, e). Hence (A, a) R∗ ({1, e}, e).

¤

We can finally gather together the results of this section into the following analogue of Theorem 4.1: Theorem 4.8. The prefix expansion Pr(M ) of M is left ample if, and only if, either • M is unipotent and satisfies the conditions: (U1) su = s, for all s ∈ I and all u ∈ H1 ; (U2) R = ι in I; (U3) u R v and us = vs ⇒ u = v; (U4) u R us = vs ⇒ u = v.

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or • M is bipotent and satisfies the conditions: (∗ B1) R = ι; (B2) e
22

CHRISTOPHER HOLLINGS

(B2): If e
PREFIX EXPANSION - (WEAKLY) LEFT AMPLE

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Observe further that condition (B4) follows immediately from right cancellation in BL(X)e,1 . We see then that condition (∗ B1) is independent of conditions (B2)(B4). We know that the Szendrei expansion of the 2-element chain is left ample [FG, Proposition 4.4]. We now consider its prefix expansion: Example 5.6. Let C2 be the 2-element chain: 1 0 1 1 0 0 0 0 (∗ B1): Any chain is a semilattice, so R = ι. (B2): It is clear that if 0
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CHRISTOPHER HOLLINGS

Suppose now that M is a type-II monoid of (finite) order n. Let 1 and e 6= 1 be the idempotents of M , and suppose that a ∈ M . Then, since M is finite, there exists a k ∈ N with either ak = 1 or ak = e. If ak = 1, then a R 1, in which case a = 1, by (B1). Similarly, if ak = e, then e ≤R a and either a = 1 or a = e, by (B2) and (B1). Therefore n = 2 and M = C2 . ¤ From Proposition 6.1 and the observation preceding it, we can write down the following specialisation of Theorems 3.12 and 4.8 to the finite case: Theorem 6.2. The prefix expansion Pr(M ) of a finite monoid M is (weakly) left ample if, and only if, M is a group or the 2-element chain. In our second special case, we obtain, as a corollary to our earlier results, conditions for the prefix expansion of a monoid to be inverse:6 Theorem 6.3. The prefix expansion Pr(M ) of a monoid M is inverse if, and only if, M is a group or the 2-element chain. Proof. As already observed (Proposition 2.6), the prefix expansion of a group is inverse. The prefix expansion of the 2-element chain is inverse, by Example 5.6. Conversely, suppose that Pr(M ) is inverse. Then Pr(M ) is certainly left ample, so, by Theorem 4.8, M is type-Ia or type-IIa. Suppose first of all that M is typeIa and let s ∈ I. Then, since Pr(M ) is inverse, the element ({1, s}, s) ∈ Pr(M ) must have an inverse (T, t) ∈ Pr(M ). The products ({1, s}, s)(T, t) = ({1, s} ∪ sT, st) and (T, t)({1, s}, s) = (T ∪ {t, ts}, ts) are therefore idempotent in Pr(M ). This means that we must have both st and ts idempotent in M . But M is unipotent, so st = 1 = ts, which contradicts s being a non-unit in M . We therefore conclude that I = ∅, hence M is a group. Now suppose that M is type-IIa and let x be a non-idempotent element of M . Then, since Pr(M ) is inverse, the element ({1, x}, x) ∈ Pr(M ) must have an inverse (Y, y) ∈ Pr(M ). Recall Lemma 2.13, which told us that the prefix expansion of a type-IIa monoid has only two idempotents, namely ({1}, 1) and ({1, e}, e). These are therefore the only two possibilities for the product ({1, x}, x)(Y, y) = ({1, x} ∪ xY, xy). 6Conditions

for the prefix expansion to be regular can be found in [Me]. It is an easy exercise to obtain Theorem 6.3 from Meakin’s conditions.

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If xy = 1, then, by (∗ B1), we must have x = 1. If xy = e, then, as x 6∈ {1, e}, we have x R 6 e, by (∗ B1). Then, by (B2), x = 1, since e
J.-C. Birget and J. Rhodes, ‘Almost finite expansions of arbitrary semigroups’, J. Pure Appl. Algebra, 32 (1984), 239-287. [BR2] J.-C. Birget and J. Rhodes, ‘Group theory via global semigroup theory’, J. Algebra, 120 (1989), 284-300. [CP] A. H. Clifford and G. B. Preston, The Algebraic Theory of Semigroups, Volume 2, Mathematical Surveys of the American Mathematical Society, No. 7, Providence, R. I. (1967). [F1] J. Fountain, ‘A class of right PP monoids’, Quart. J. Math. Oxford (2), 28 (1977), 285-300. [F2] J. Fountain, ‘Adequate semigroups’, Proc. Edinb. Math. Soc. (2), 22 (1979), 113-125. [FG] J. Fountain and G. M. S. Gomes, ‘The Szendrei expansion of a semigroup’, Mathematika, 37 (1990), 251-260. [FGG] J. Fountain, G. Gomes and V. Gould, ‘Enlargements, semiabundancy and unipotent monoids’, Comm. Algebra, 27 (1999), 595-614. [G] V. Gould, ‘(Weakly) left E-ample semigroups’, http://www-users.york.ac.uk/∼varg1/finitela.ps [JS] M. Jackson and T. Stokes, ‘An invitation to C-semigroups’, Semigroup Forum, 62 (2001), 279-310. [Ma] E. Manes, ‘Guarded and banded semigroups’, Semigroup Forum, 72 (2006), 94-120. [Me] J. Meakin, ‘Regular expansions of semigroups’, Simon Stevin, 60 (1986), 241-255. ¨ [Su] A. Suschkewitsch, ‘Uber Semigruppen’, Zap. Inst. Mat. Meh. (4) (Comm. Soc. Math. Kharkow), 8 (1934), 25-28. [Sz] M. B. Szendrei, ‘A note on Birget-Rhodes expansion of groups’, J. Pure Appl. Algebra, 58 (1989), 93-99. Department of Mathematics, University of York, Heslington, York, YO10 5DD, UK E-mail address: [email protected]