Semigroup Forum Vol. 65 (2002) 405–427c© 2002 Springer-Verlag New York Inc.

DOI: 10.1007/s002330010135

RESEARCH ARTICLE

The Structure of 0-BisimpleR-unipotent Semigroups

Anthony Hayes

Communicated by John B. Fountain

Abstract

We generalize theory of Lawson for 0-bisimple inverse monoids to wider classesof 0 -bisimple regular semigroups. We give a necessary and sufficient conditionfor a 0-bisimple R -unipotent semigroup to admit a 0-restricted idempotent-pure prehomomorphism to a primitive inverse semigroup. Several illustrationsof the theory are obtained as an application of the results in this paper.

Introduction

In this paper we extend results of Lawson [9] for 0-bisimple inverse monoidsto other classes of 0-bisimple regular semigroups, notably orthodox and R-unipotent. We give a necessary and sufficient condition for a 0-bisimple R-unipotent semigroup to admit a suitable prehomomorphism to a primitive in-verse semigroup, and illustrate the theory with the particular cases of left groupswith an adjoined zero and completely 0-simple semigroups.

In Section 1 we fix terminology and notation and recall elementary def-initions and results concerning various classes of regular semigroups. We alsoconsider the natural partial order relation on a regular semigroup, extensive useof which will be made later. All results will be stated without proof, and thereader is referred to [5] and [8] for any further information on the subject. Wedescribe some properties of primitive R-unipotent semigroups and investigateconnections between 0-morphisms and prehomomorphisms. In addition, we il-lustrate how concepts in general semigroup theory can be modified to give moreuseful ideas in the context of semigroups with zero, and establish the notationand terminology which will be used throughout the paper.

Section 2 deals with 0-bisimple regular semigroups and we begin by re-calling and proving some elementary properties of prehomomorphisms betweenregular semigroups. Next we consider 0-bisimple orthodox monoids and 0-bisimple R-unipotent monoids and show, among other things, that R1 , the R-class containing the identity, is a right cancellative submonoid. We prove that

The author would like to thank Professor J. B. Fountain, Dr. V. A. R. Gould, and ProfessorP. M. Higgins for much helpful discussion during the preparation of this paper.

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a 0-bisimple R-unipotent monoid is 0-E -unitary if and only if R1 is a can-cellative monoid. We then show that a 0-bisimple R-unipotent monoid admitsa 0-restricted idempotent-pure prehomomorphism to a group with zero if andonly if R1 is embeddable in a group. The particular cases of left groups with anadjoined zero and completely 0-simple R-unipotent semigroups are also consid-ered. We conclude the paper by considering a 0-bisimple R-unipotent monoidin which R1 is embedded in its fundamental group, and obtain a characterisationof the suitable prehomomorphism derived from the embedding homomorphism.

1. Definitions and preliminaries

If S is a regular semigroup then the relations ωl and ωr are defined as follows.If e, f ∈ E(S), then

e ωl f if and only if ef = e,

e ωr f if and only if fe = e.

Note that, on E(S), we have ≤= ωl ∩ωr , where ≤ is the natural partial orderon S .

A semigroup is called orthodox if it is regular and an E -semigroup,meaning that the idempotents form a subsemigroup. It is clear that, in anorthodox semigroup, any element below an idempotent in the natural partialorder is itself idempotent. In fact this property holds in all regular semigroups,as shown by the following lemma, which is a special case of [7, Section 1.4,Exercise 1(a)(ii)].

Lemma 1.1. In a regular semigroup, any element below an idempotent inthe natural partial order is itself idempotent.

It is well known that a semigroup is regular if and only if each R-class(L-class) contains an idempotent, and that a semigroup is inverse if and onlyif each R-class and each L-class contains exactly one idempotent. We say thata semigroup S with set of idempotents E is R-unipotent if it is regular andefe = ef for all e, f ∈ E .

Theorem 1.2. ([8, Exercise 6.4.6]) Let S be a regular semigroup with set ofidempotents E . Then the following statements are equivalent:

(1) S is R-unipotent;

(2) each R-class of S contains exactly one idempotent;

(3) for all e ∈ E , a ∈ S and a′ ∈ V (a) , aea′a = ae ;

(4) for all a ∈ S and a′, a′′ ∈ V (a) , aa′ = aa′′ .

Of course L-unipotent semigroups are defined dually, and there is thedual of Theorem 1.2. Warne [17] defines a semigroup to be L-unipotent if

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each L-class contains exactly one idempotent. Clifford [4], Venkatesan [16] andBailes [1] refer to semigroups which are L-unipotent as left unipotent, rightinverse and left inverse, respectively. Warne, Venkatesan and Bailes all showthat a semigroup S is L-unipotent if and only if S is orthodox and D = Rin the band E(S). Dually, a semigroup S is R-unipotent if and only if S isorthodox and D = L in E(S).

The two simple results below will be useful in the sequel.

Lemma 1.3. Let S be an R-unipotent (respectively L-unipotent) semi-group. Then for all e ∈ E(S) and e′ ∈ V (e) , we have that ee′ = e (respectivelye′e = e).

Proof. Since e ∈ E(S) it follows that e ∈ V (e), whence ee = ee′ , that is,e = ee′ . The L-unipotent case follows as a dual.

Corollary 1.4. Let S be an R-unipotent (respectively L-unipotent) semi-group. Then ωr ⊆ ωl (respectively ωl ⊆ ωr ).

Proof. Let e, f ∈ E(S) be such that e ωr f . Then fe = e . Since S isorthodox it follows that ef ∈ V (fe), whence (fe)(ef) = fe = e by Lemma 1.3.Thus e = fef = ef and e ωl f , as required. The L-unipotent case follows asa dual.

Observe that in an R-unipotent (respectively L-unipotent) semigroupS , we have ωr =≤ (respectively ωl =≤) on E(S).

We adopt the usual convention that a semigroup with zero must containat least two elements. If S is a semigroup then by S0 we mean the semigroupwith zero obtained from S be adjoining a zero element if necessary. Let S bea semigroup with zero and T a subset of S such that 0 ∈ T . We define T ∗ tobe the set of non-zero elements of T ; in particular, E∗(S) or simply E∗ is theset of non-zero idempotents of S .

A semigroup is said to be bisimple if it is a single D -class. Semigroupswith zero cannot be bisimple, because the zero is in its own D -class. For thisreason we define a semigroup with zero S to be 0-bisimple if the D -classes ofS are {0} and S∗ .

A semigroup S is said to be E -unitary if the idempotents form a unitarysubset of S . Because an E -unitary semigroup with zero is trivially a band,Szendrei [15] introduced the idea of an E∗ -unitary regular semigroup with zerounder the name (E\0)-unitary. The definition readily extends to arbitrarysemigroups and so a semigroup with zero S is E∗ -unitary if for all s ∈ S ande ∈ E(S), if se or es is in E∗(S), then s ∈ E∗(S). The next result can bededuced from [6, Lemma 1.1].

Lemma 1.5. Let S be a semigroup with zero. Then the following conditionsare equivalent.

(1) S is E∗ -unitary;

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(2) for all elements a of S and e of E(S) , if ae ∈ E∗(S) , then a ∈ E∗(S) ;

(3) for all elements a of S and e of E(S) , if ea ∈ E∗(S) , then a ∈ E∗(S) .

The E∗ -unitary inverse semigroups are precisely the 0-E -unitary inversesemigroups studied by Lawson in [9]. A regular semigroup with zero is 0-E -unitary if any element above a non-zero idempotent in the natural partial orderis itself an idempotent. An E∗ -unitary regular semigroup S is certainly 0-E -unitary. For, if e ∈ E∗(S) and a ∈ S are such that e ≤ a , then by [8,Theorem 6.1.2(3)], e = fa for some f ∈ E∗(S). Thus f, fa ∈ E∗(S), whencea ∈ E∗(S) because S is E∗ -unitary. However, the converse is not true.

Example. Let S be the semigroup of [5, Exercise 1.2.2]:

e f g a 0e e a e a 0f 0 f g 0 0g g f g f 0a 0 a e 0 00 0 0 0 0 0

Then S is regular, for e, f, g, 0 ∈ E(S) and aga = (ag)a = ea = a . It is also0-E -unitary, because a , the only non-idempotent, is not above any of the non-zero idempotents in the natural partial order. However, it is not E∗ -unitary,since ag = e , so that g, ag ∈ E∗(S) but a �∈ E∗(S).

If S is a semigroup with zero, then 0 is immediately seen to be the uniqueminimum idempotent. By a primitive idempotent we mean an idempotent whichis minimal within the set of non-zero idempotents. Hence, if e is a primitiveidempotent and f ∈ E(S), then

ef = fe = f �= 0 implies that e = f.

We define a primitive regular (inverse) semigroup to be a regular (inverse) semi-group with zero in which every non-zero idempotent is primitive.

Lemma 1.6. ([6, Lemma 1.7]) Let S be a primitive inverse semigroup.

(i) If e, f ∈ E∗(S) , then ef �= 0 implies e = f .

(ii) If e ∈ E∗ and s ∈ S∗ then

es �= 0 implies es = s, se �= 0 implies se = s.

(iii) If a, s ∈ S∗ and as = a , then s = a−1a . Dually, if sa = a , thens = aa−1 .

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We now consider primitive R-unipotent (L-unipotent) semigroups. Cor-responding to Lemma 1.6 we have the following.

Lemma 1.7. Let S be a primitive R-unipotent (L-unipotent) semigroup.

(i) If e, f ∈ E∗(S) , then ef �= 0 implies ef = e (ef = f ).

(ii) If e ∈ E∗(S) and s ∈ S∗ then

se �= 0 implies se = s (es �= 0 implies es = s).

(iii) If a, s ∈ S∗ and sa = a , then s = aa′ for all a′ ∈ V (a) (if as = a , thens = a′a for all a′ ∈ V (a)).

Proof. (i) If ef �= 0, then since ef = efe we have ef ≤ e . Hence ef = eas S is primitive.

(ii) If se �= 0 then s′se �= 0 where s′ ∈ V (s). Therefore s′se = s′s by (i)and so

se = ss′se = s(s′se) = ss′s = s.

(iii) If sa = a , then saa′ = aa′ �= 0 where a′ ∈ V (a). Thus saa′ = sby (ii) and so s = aa′ . But aa′ = aa′′ for all a′, a′′ ∈ V (a) because S isR-unipotent, as required.

Dual arguments give the L-unipotent case.

Note that, for a primitive R-unipotent (respectively L-unipotent) semi-group S , if e ∈ E∗(S) and s, t ∈ Re (respectively Le ), then st ∈ Re (respec-tively Le ) whenever st �= 0. For, if s′ ∈ V (s) and t′ ∈ V (t), then t′s′ ∈ V (st)as S is orthodox. Now ss′ = ee′ and tt′ = ee′′ for some e′, e′′ ∈ V (e). Butee′ = ee′′ = e by Lemma 1.3. Thus

(st)(t′s′) = s(tt′)s′ = ses′ = (se)s′ = ss′ = e,

since se �= 0, by Lemma 1.7(ii). It follows that R0e = Re∪{0} is a subsemigroup

of S . The L-unipotent case is proved dually.

We see from Lemma 1.7(ii) that a primitive R-unipotent (L-unipotent)semigroup S is E∗ -unitary (and therefore 0-E -unitary). For, if e, s ∈ S aresuch that e, se ∈ E∗(S), then se �= 0, whence s = se ∈ E∗(S), and S isE∗ -unitary by Lemma 1.5. A dual argument gives the L-unipotent case.

We say that a semigroup with zero S is 0-simple if {0} and S are itsonly ideals and S2 �= {0} . It is clear that S is 0-simple if and only if S2 �= {0}and {0} and S∗ are the only J -classes.

A completely 0-simple semigroup is a 0-simple semigroup with a primitiveidempotent. It can be shown that every non-zero idempotent in a completely0-simple semigroup is primitive. Completely 0-simple semigroups are describedby the well known Rees Theorem, see [8, Theorem 3.2.3], for example.

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We define a function θ from a semigroup S to a regular semigroup Tto be a prehomomorphism if (xy)θ ≤ (xθ)(yθ) for all x, y ∈ S . If S andT have zeros, then θ is 0-restricted if 0θ−1 = {0} . Also, θ is said to beidempotent-pure if x ∈ E(S) whenever xθ ∈ E(T ). A 0-restricted idempotent-pure prehomomorphism will often be referred to as “suitable.”

Let S, T be semigroups with zero. Following [2], we say that a functionφ: S → T is a 0-morphism if (xy)φ = (xφ)(yφ) whenever x, y ∈ S and xy �= 0,and φ is 0-restricted.

The connection between 0-morphisms and prehomomorphisms is givenin the following results.

Lemma 1.8. Let S, T be semigroups with zero and suppose also that T isregular. If θ: S → T is a 0-morphism, then it is a 0-restricted prehomomor-phism.

Proof. First, θ is 0-restricted by definition. Let x, y be in S . If eitherof x, y is zero, then clearly, (xy)θ = 0θ = 0 = (xθ)(yθ). If x, y are non-zero and xy = 0, then (xy)θ = 0θ = 0 ≤ (xθ)(yθ). Finally, if xy �= 0 then(xy)θ = (xθ)(yθ), and hence θ is a prehomomorphism.

Lemma 1.9. Let S and T be semigroups with zero and suppose further thatT is primitive R-unipotent (L-unipotent). If θ: S → T is a 0-restrictedprehomomorphism, then it is a 0-morphism.

Proof. Let x, y ∈ S , with xy �= 0. Then (xy)θ ∈ T ∗ because θ is 0-restricted, and (xy)θ ≤ (xθ)(yθ). Hence there exists e ∈ E∗(T ) with (xy)θ =(xθ)(yθ)e �= 0. But e ∈ E∗(T ) and ((xθ)(yθ)) ∈ T ∗ , so that ((xθ)(yθ))e =(xθ)(yθ), by Lemma 1.7(ii). Thus (xy)θ = (xθ)(yθ), and θ is a 0-morphism.

The L-unipotent case is proved dually.

Combining Lemma 1.8 and Lemma 1.9 gives the following proposition.

Proposition 1.10. Let S, T be semigroups with zero and suppose also thatT is primitive R-unipotent (L-unipotent). Then θ: S → T is a 0-morphismif and only if it is a 0-restricted prehomomorphism.

2. 0-bisimple R-unipotent semigroups

In this section we give necessary and sufficient conditions for the existence ofa suitable prehomomorphism from a 0-bisimple R-unipotent semigroup to aprimitive inverse semigroup. We begin by recalling the following definitionfrom [9].

For an inverse semigroup, the restricted product or trace product is apartial binary operation defined as follows: x · y exists and is equal to xyprecisely when x−1x = yy−1 . We extend this notion to arbitrary regular semi-groups in the obvious way. If x and y are elements of a regular semigroup, thenx · y is defined and equals xy if and only if x′x = yy′ for some x′ ∈ V (x) andy′ ∈ V (y). It is easy to see that x · y L y and x · y R x .

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The following result is proved in [13].

Proposition 2.1. If θ is a prehomomorphism between regular semigroups Sand T , then the following hold.

(1) For all s ∈ S and s′ ∈ V (s) we have that s′θ ∈ V (sθ) .

(2) If x′x ωl yy′ or yy′ ωr x′x for some x′ ∈ V (x) and y′ ∈ V (y) , then(xy)θ = (xθ)(yθ) .

(3) θ preserves the natural partial order.

(4) θ maps L-related (R-related) elements to L-related (R-related) elements.

It follows from (2) that if e ∈ E(S), then eθ ∈ E(T ).

From [12] we know that, in the inverse case, a prehomomorphism θpreserves the restricted product and that θ is idempotent-pure if and onlyif all restrictions of θ to R-classes (L-classes) are injective. The next twopropositions generalize these facts.

Proposition 2.2. Let S and T be regular semigroups with zero. If θ is a0-restricted prehomomorphism from S to T and T is primitive R-unipotent(L-unipotent), then the following hold:

(1) θ preserves the restricted product;

(2) if all restrictions of θ to R-classes (L-classes) are injective, then θ isidempotent-pure.

Proof. (1) Let x, y ∈ S and suppose that x′x = yy′ for some x′ ∈ V (x) andy′ ∈ V (y). By Lemma 1.9 we have that θ is a 0-morphism. If x′x = 0, thenx = xx′x = 0, and similarly y = 0. Hence xθ = yθ = 0. Thus (x′θ)(xθ) =(yθ)(y′θ) = 0 and x′θ ∈ V (xθ) and y′θ ∈ V (yθ) by Proposition 2.1(1), so that(xθ) · (yθ) exists. On the other hand, if x′x �= 0, then

(x′θ)(xθ) = (x′x)θ = (yy′)θ = (yθ)(y′θ).

Since x′θ ∈ V (xθ) and y′θ ∈ V (yθ), we have that (xθ) · (yθ) is defined andequal to (xθ)(yθ).

(2) Let s ∈ S with sθ ∈ E(T ). If sθ = 0, then s = 0 as θ is 0-restricted,and so s ∈ E(S). Suppose that sθ �= 0. Then s �= 0, and if s′ ∈ V (s) we have

sθ = (ss′s)θ = ((ss′)θ)(sθ),

since θ is a 0-morphism. Hence sθ ωr (s′s)θ as sθ ∈ E(T ). Because T isR-unipotent it follows that sθ ≤ (ss′)θ . Thus sθ = (ss′)θ as T is primitive.Since s R ss′ and θ is injective when restricted to R-classes, we have thats = ss′ and s ∈ E(S), as required. A dual argument proves the L-unipotentcase.

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Proposition 2.3. Let θ be a prehomomorphism from an R-unipotent (L-unipotent) semigroup S to a regular semigroup T . Then if θ is idempotent-pure,all restrictions of θ to R-classes (L-classes) are injective.

Proof. Suppose that x, y ∈ S are such that x R y and xθ = yθ . Thenxx′ = yy′ for some x′ ∈ V (x) and y′ ∈ V (y). Since xθ = yθ , we have that y′θis an inverse of yθ , that is, of xθ . Also (y′x)θ ≤ (y′θ)(xθ) where (y′θ)(xθ) isidempotent, so that (y′x)θ ∈ E(T ) by Lemma 1.1. Hence y′x ∈ E(S) because θis idempotent-pure. Now x R y so that y′x R y′y , and since y′x is idempotentand S is R-unipotent, y′x = y′y . Thus

x = xx′x = yy′x = yy′y = y,

as required. A dual argument gives the L-unipotent case.

Combining Proposition 2.2(2) and Proposition 2.3 gives the followingtheorem.

Theorem 2.4. Let θ be a 0-restricted prehomomorphism from an R-unipotent (L-unipotent) semigroup S to a primitive R-unipotent (L-unipotent)semigroup T . Then θ is idempotent-pure if and only if all restrictions of θ toR-classes (L-classes) are injective.

We know from the classical structure theory of Clifford [3], Reilly [14]and McAlister [11] that the properties of a 0-bisimple inverse monoid S arecompletely determined by the properties of the R-class containing the identity1. Some aspects of this connection are described in [9, Theorem 3]. There it isshown, among other things, that R1 is a “CRM-monoid” (named after Clifford,Reilly and McAlister). This means that R1 is right cancellative and has theproperty that any two principal left ideals are either disjoint or intersect in aprincipal left ideal. The dual definition could obviously be given for L1 . Thefollowing theory extends part of the cited result to orthodox monoids. We firsthave a straightforward lemma.

Lemma 2.5. Let S be a regular semigroup. Suppose that e, s ∈ S are suchthat e ∈ E(S) and s R e .

(1) ss′ = e for some s′ ∈ V (s) .

(2) If S is R-unipotent, then ss′ = e for all s′ ∈ V (s) .

(3) If S is a monoid and e = 1 , then ss′ = 1 for all s′ ∈ V (s) .

Proof. (1) As e ∈ V (e), there exists s′ ∈ V (s) with ss′ = ee = e .

(2) This follows immediately from (1) and the fact that ss′ = ss′′ for alls′, s′′ ∈ V (s).

(3) If s′ ∈ V (s), then since ss′ R s R 1 and idempotents are leftidentities for their R-classes we have 1 = ss′1 = ss′ .

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The next result gives two circumstances where Re is a right cancellativesubmonoid. A semigroup S is a left group if L = S × S and S is rightcancellative. By the dual of [8, Exercise 2.6.6(a)], we know that if S is aleft group then E(S) �= ∅ . It follows that a left group is regular.

Lemma 2.6. Let S be a regular semigroup and e ∈ E(S) .

(1) If S is a left group, then Re is a subgroup of S .

(2) If S is a monoid and e = 1 , then C = R1 is a right cancellativesubmonoid of S .

Proof. (1) First note that Re �= ∅ as e ∈ Re . Since S is a left group wehave Re = He . It now follows immediately that Re is a subgroup of S and istherefore cancellative.

(2) Let a, b ∈ C . Then b R 1, so that ab R a1 = a and a R 1. Henceab R 1 and C is a submonoid of S . If a, b, c ∈ C are such that ac = bc , thenfor c′ ∈ V (c) we have acc′ = bcc′ . But cc′ = 1 by Lemma 2.5(3), so that a = b ,and C is a right cancellative monoid.

Theorem 2.7. Let S be a 0-bisimple orthodox semigroup with e ∈ E∗(S)and put C = Re . Let a, b, c, d ∈ C .

(1) a L b if and only if there is an invertible element u ∈ He such thata = ub .

(2) S∗ = C ′C = {b′a | a, b ∈ C, b′ ∈ V (b)} and Le ⊆ C ′ . If S is a monoidand e = 1 , then C ′ = L1 .

(3) Suppose that there is an invertible element u ∈ He such that a = uc andb = ud . Then for all d′ ∈ V (d) , there exists b′ ∈ V (b) such that b′a = d′c ,and for all b′ ∈ V (b) , there exists d′ ∈ V (d) such that b′a = d′c .

(4) Let b′a = d′c where b′ ∈ V (b) and d′ ∈ V (b) are such that bb′ = dd′ = e .Suppose further that V (d′) = {d} or V (b′) = {b} . Then there is aninvertible element u ∈ He such that a = uc and b = ud .

(5) If S is a monoid and e = 1 , let x = b′a and y = d′c where b′ ∈ V (b)and d′ ∈ V (d) . Suppose that b = pd and a = pc for some p ∈ C . Then,if b′ = d′p′ where p′ ∈ V (p) , we have that x ≤ y .

Proof. (1) If a = ub for some u ∈ He , let u−1 be the inverse of u in He .Then u−1a = u−1ub = eb = b as b R e . Hence a L b .

Conversely, suppose that a L b . Let b′ ∈ V (b) be such that bb′ = e .Then a′a = b′b for some a′ ∈ V (a). It follows that a = aa′a = a(b′b) = (ab′)b .But ab′ is a restricted product (since a′a = b′b) and bb′ = e so that b′ L e .Thus ab′ H e .

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(2) Let x ∈ C ′C . Then x = t′s for some s, t ∈ C and t′ ∈ V (t). Ift′s = 0, then tt′s = 0. But s R t R tt′ , so that tt′s = s = 0, a contradiction.Thus t′s �= 0 and x ∈ S∗ .

Conversely, suppose that x ∈ S∗ . Then x D e and so x L s R e forsome s ∈ S∗ . Let s′ ∈ V (s) be such that ss′ = e . Since x L s there existsx′ ∈ V (x) such that x′x = s′s . In particular, s ∈ C . Also

x = xx′x = x(s′s) = (xs′)s.

Because S is orthodox, xs′ ∈ V (sx′). Put t = sx′ and t′ = xs′ . Then

tt′ = sx′xs′ = s(s′s)s′ = ss′ = e,

so that t ∈ C and we have x = t′s with s, t ∈ C .

To see that Le ⊆ C ′ , let x ∈ Le . Then there exists x′ ∈ V (x) withx′x = e . Now x ∈ V (x′) and x′ R e so that x ∈ C ′ .

Finally, suppose that S is a monoid and e = 1. Then L1 ⊆ C ′ by theabove. If x ∈ C ′ , then x = y′ for some y′ ∈ V (y) and y ∈ C . Now yy′ = 1 byLemma 2.5(3), whence yx = 1 and x ∈ L1 .

(3) Let d′ ∈ V (d). Then, if u−1 is the inverse of u in He , we have thatb′ = d′u−1 ∈ V (b). Hence b′a = d′u−1uc = d′ec = d′c . A dual argument givesthe second part of the statement.

(4) Let a′ ∈ V (a) be such that aa′ = e . Then b′aa′ = d′ca′ . Thereforeb′e = d′ca′ , and since bb′ = e we have b′ L e and b′e = b′ = d′ca′ . Similarly,if c′ ∈ V (c) is such that cc′ = e , then d′ = b′ac′ . Hence b′ R d′ . Supposethat V (d′) = {d} . Now, as b ∈ V (b′), there exists (d′)′ ∈ V (d′) such thatb′b = d′(d′)′ . But V (d′) = {d} by assumption, so that b′b = d′d and b L d .Thus, by (1), there is an invertible element u ∈ He such that b = ud . Hence,as aa′ = bb′ = e ,

a = aa′a = bb′a = bd′c.

But b = ud , whence bd′ = udd′ = ue = u as u L e , so that

a = bd′c = uc,

as required. A dual argument gives the case where V (b′) = {b} .(5) Note first that if c′ ∈ V (c), then x = b′a = d′p′pc = (d′c)c′p′pc as

cc′ = 1. Hence x = yc′p′pc ∈ yS . Now put x′ = c′p′pd ∈ V (x). Then

xx′y = d′p′pcc′p′pdd′c = d′p′pp′pc = d′p′pc = x,

so that by [8, Theorem 6.1.2], x ≤ y .

Note that in (4) we have V (d′) = {d} for all d′ ∈ V (d) or V (b′) = {b} for allb′ ∈ V (b), by [8, Theorem 6.2.4].

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When the semigroup is R-unipotent we have the following.

Proposition 2.8. Let S be a 0-bisimple R-unipotent semigroup with e ∈E∗(S) and put C = Re . Let a, b, c, d ∈ C .

(1) C ′ = Le .

(2) If b′a = d′c where b′ ∈ V (b) and d′ ∈ V (d) , then there is an invertibleelement u ∈ He such that a = uc and b = ud .

(3) There is an invertible element u ∈ He such that a = uc and b = ud ifand only if b′a = d′c for some b′ ∈ V (b) and d′ ∈ V (d) .

(4) Let b′ ∈ V (b) . Then b′a is idempotent if and only if a = b .

(5) Let x = b′a and y = d′c where b′ ∈ V (b) and d′ ∈ V (d) . Suppose thatb = pd and a = pc for some p ∈ C . Then, if b′ = d′p′ where p′ ∈ V (p) ,we have that x ≤ y .

(6) Let b′ ∈ V (b) and d′ ∈ V (d) . If b′a ≤ d′c , then b = pd and a = pc forsome p ∈ C .

Proof. (1) By Theorem 2.7(2), Le ⊆ C ′ . The opposite inclusion followsimmediately from the fact that xx′ = e for all x ∈ C and x′ ∈ V (x).

(2) Exactly as in Theorem 2.7(4), noting that bb′ = dd′ = e for allb′ ∈ V (b) and d′ ∈ V (d), and to see that b L d we observe that d′(d′)′ = d′dsince S is R-unipotent.

(3) This follows immediately from (2) and Theorem 2.7(3).

(4) Suppose that b′a ∈ E∗(S). Since a R b we have that b′a R b′b . Butb′a is idempotent and S is R-unipotent, whence b′a = b′b . Thus bb′a = bb′b ,that is, ea = b and so a = b . The converse is immediate.

(5) Follow the proof of Theorem 2.7(5), observing that x = b′a = d′p′pc =(d′c)c′p′pc because d′cc′ = d′e = d′ as d′ L e by (1). Also

xx′y = d′(p′pcc′p′p)dd′c = d′p′p(cc′dd′)c = d′p′peec = d′p′pc = x,

since S is R-unipotent.

(6) Since b′a ≤ d′c , we have that b′a ∈ d′cS and b′a = b′a(b′a)′d′c forsome (b′a)′ ∈ V (b′a). But, as S is R-unipotent, b′a(b′a)′ = b′a(b′a)′′ for all(b′a)′, (b′a)′′ in V (b′a) and so b′a = b′a(b′a)′d′c for all (b′a)′ ∈ V (b′a). Hence,if we put (b′a)′ = a′b where a′ ∈ V (a), we have (b′a)′ ∈ V (b′a) and therefore

b′a = b′aa′bd′c = b′ebd′c = b′bd′c

since aa′ = e . Thus bb′a = bb′bd′c , that is, ea = bd′c and a = bd′c . Putp = bd′ . Since aa′′ = e for all a′′ ∈ V (a) we have that

e = bd′cc′db′ = bd′edb′ = bd′db′

416 Hayes

where c′ ∈ V (c). Then p ∈ C and

b′b = b′eb = b′bd′db′b = b′bd′d

as S is R-unipotent. Thus b = bd′d and

pd = bd′d = b,

as required.

It is shown in [9] that a 0-bisimple inverse monoid is 0-E -unitary if andonly if C = R1 is a cancellative monoid. We show below that this equivalenceholds when S is R-unipotent.

Proposition 2.9. Let S be a 0-bisimple R-unipotent semigroup with e ∈E∗(S) and put C = Re . Then S is 0-E -unitary if and only if, for alla, b, c ∈ C ,

ca = cb �= 0 implies a = b.

If S is a monoid and e = 1 , then S is 0-E -unitary if and only if C is acancellative monoid.

Proof. Let S be 0-E -unitary and suppose that a, b, c ∈ C are such thatca = cb �= 0. Let a′ ∈ V (a) and c′ ∈ V (c). Since S is orthodox we have thata′c′ ∈ V (ca) and a′c′ca ∈ E(S). Now

(a′c′ca)(a′b) = a′c′c(aa′)b = a′c′ceb = a′c′cb = a′c′ca,

as a ∈ C . Further, if b′ ∈ V (b) it follows by Proposition 2.8(1) that a′c′ca =a′ec′ca = a′bb′c′ca . Thus a′c′ca ∈ a′bS , whence a′c′ca ≤ a′b , by [8, Theo-rem 6.1.2]. As ca �= 0 it follows that a′c′ca ∈ E∗(S), so that a′b ∈ E∗(S),because S is 0-E -unitary. Thus b = a by Proposition 2.8(4), as required.

Conversely, suppose that the cancellation law holds. Let e ∈ E∗(S) andx ∈ S be such that e ≤ x . Then e = a′a and x = b′c for some a, b, c ∈ C ,by Theorem 2.7(2) and Proposition 2.8(4). Now by Proposition 2.8(6), we havethat (a, a) = p(b, c) for some p ∈ C . Hence a = pb = pc �= 0. But by theassumed cancellation law b = c . Thus x = b′c = b′b is idempotent.

The second statement of the proposition follows immediately fromLemma 2.6(2) and the above.

The following corollary is immediate.

Corollary 2.10. Let S be a 0-bisimple R-unipotent semigroup. Considerthe condition

(1) for e ∈ E∗(S) and a, b, c ∈ Re ,

ca = cb �= 0 implies a = b.

Hayes 417

Then if condition (1) holds for SOME e ∈ E∗(S) , it holds for EVERY e ∈E∗(S) .

We now show that a left group with an adjoined zero is E∗ -unitary.

Lemma 2.11. If S is a left group then S0 is E∗ -unitary.

Proof. Let e, s ∈ S be such that e, se ∈ E∗(S0). Then sese = se , whenceby right cancellation ses = s . Also e L se , so that e = tse for some t ∈ S1 .If t = 1 then e = ee = se and by cancellativity s = e ∈ E∗(S0). If t �= 1 thene = ee = tsee = tse , whence e = ts and

s = ses = s(tse)s = st(ses) = sts = se ∈ E∗(S0).

Thus by Lemma 1.5, S0 is E∗ -unitary.

Note in particular that a left group S with an adjoined zero is 0-E -unitary. Thiscould also be proved by showing that S0 is 0-bisimple R-unipotent, and thenusing the fact that a left group is isomorphic to the direct product of a groupand a left zero semigroup to verify the cancellation property of Proposition 2.9.

We now turn our attention to completely 0-simple semigroups. If S =M0(G; I,Λ;P ) is a completely 0-simple semigroup, then it is routine to verifythat

E(S) = {(i, p−1λi , λ) | pλi �= 0} ∪ {0},

V ((i, a, λ)) = {(j, b, µ) | pλj , pµi �= 0, b = p−1λj a

−1p−1µi },

(i, a, λ)R(j, b, µ) if and only if i = j,

(i, a, λ)L(j, b, µ) if and only if λ = µ,

S is an E -semigroup if and only if pλi, pλj , pµj �= 0 implies pµi �= 0 and

p−1λi pλjp

−1µj = p−1

µi .

Lemma 2.12. Let S = M0(G; I,Λ;P ) be a completely 0-simple semigroup.Suppose that e ∈ E∗(S) . Then for all a, b, c ∈ Re ,

ca = cb �= 0 implies a = b.

Proof. Put a = (i, g, λ), b = (i, h, µ) and c = (i, k, γ). If ca = cb �= 0, then

(i, k, γ)(i, g, λ) = (i, k, γ)(i, h, µ) �= 0,

that is,(i, kpγig, λ) = (i, kpγih, µ) �= 0.

Hence λ = µ , and kpγig = kpγih . Thus by cancellation g = h and

a = (i, g, λ) = (i, h, µ) = b,

as required.

418 Hayes

Note that, by [5, Theorem 2.51], a completely 0-simple semigroup is 0-bisimple. Therefore Proposition 2.9 and Lemma 2.12 show that a completely0-simple R-unipotent semigroup is 0-E -unitary. In fact we can say more.

Lemma 2.13. Let S = M0(G; I,Λ;P ) be a completely 0-simple orthodoxsemigroup. Then S is E∗ -unitary.

Proof. Suppose that e, es ∈ E∗(S), and put e = (i, p−1λi , λ), s = (j, b, µ).

Thenes = (i, p−1

λi , λ)(j, b, µ) = (i, p−1λi pλjb, µ) = (i, p−1

µi , µ)

as es ∈ E∗(S). Therefore p−1λi pλjb = p−1

µi , that is, b = p−1λj pλip

−1µi . But S is an

E -semigroup, whence b = p−1µj , and

s = (j, b, µ) = (j, p−1µj , µ) ∈ E∗(S),

as required.

In particular, Lemma 2.13 shows that a completely 0-simple R-unipotentsemigroup is E∗ -unitary.

From [9, Theorem 8] we know that a 0-bisimple inverse monoid admitsa 0-restricted idempotent-pure prehomomorphism to a primitive inverse semi-group if and only if R1 can be embedded in a group. We shall give an analogueof this result for 0-bisimple R-unipotent semigroups. We first need two pre-liminary lemmas.

Lemma 2.14. Let S be an orthodox semigroup and e ∈ E(S) . If s, t ∈ Sare such that st R e , then stt′ R e for all t′ ∈ V (t) .

Proof. Since st R e and t′s′ ∈ V (st) where s′ ∈ V (s), it follows thatstt′s′ = ee′ for some e′ ∈ V (e). Now tt′s′ ∈ V (stt′) and

(stt′)(tt′s′) = s(tt′tt′)s′ = stt′s′ = ee′,

so that stt′ R e , as required.

Lemma 2.15. Let S be an R-unipotent semigroup and e ∈ E(S) . Supposethat a, b, c ∈ Re and b′ ∈ V (b) . Then

ab′ = c implies a = cb.

Proof. Since ab′ = c it follows that ab′ ∈ Re and ab′b = cb . Furthermore,if a′ ∈ V (a), then ba′ ∈ V (ab′) and (ab′)(ba′) = e by Lemma 2.5(2). Henceab′ba′a = ea = a as a R e . Now

a = ab′ba′a = a(a′ab′ba′a) = a(a′ab′b) = ab′b

because S is R-unipotent, whence a = ab′b = cb , as required.

Hayes 419

Proposition 2.16. Let θ: S → T be a suitable prehomomorphism from a0-bisimple regular semigroup S to a primtive regular semigroup T and putC = Re , where e ∈ E∗(S) .

(1) If S is a monoid, the images of every non-zero element of S lie within asingle (group) H -class G of T .

(2) If S is a monoid and e = 1 , then θ|C is a homomorphism from themonoid C to the group G .

(3) If S is R-unipotent and T is primitive R-unipotent (L-unipotent), thenthere exists an injective function φ from C to an R-class of T such thatfor all a, b ∈ C ,

ab ∈ C implies (ab)φ = (aφ)(bφ).

If S is a monoid and e = 1 , then the monoid C is embedded in the group G .

Proof. (1) Let f ∈ E∗(S). Then f ≤ 1. By Proposition 2.1(3), θ preservesthe natural partial order, so that fθ ≤ 1θ . Now fθ �= 0 as θ is 0-restricted,whence fθ = 1θ because T is primitive. Hence (E∗(S))θ = {1θ} . If x ∈ S∗

and x′ ∈ V (x) then x′x �= 0 and xx′ �= 0 so that (x′x)θ = 1θ = (xx′)θ . But,by Proposition 2.1(4), θ preserves the L- and R-relations, and so xθ H 1θ .

(2) If we denote H1θ by G , then by (1), θ restricts to a function from Cto G . If a, b ∈ C with a′ ∈ V (a) and b′ ∈ V (b), then certainly a′a ≤ bb′ = 1,whence the restriction of θ to C is a homomorphism, by Proposition 2.1(2).

(3) Because θ maps R-related elements to R-related elements, we candefine φ = θ|C : C → Reθ . Since θ is idempotent-pure, it is injective whenrestricted to R-classes by Proposition 2.3. If a, b ∈ C with ab ∈ C then wehave that ab �= 0, whence

(ab)φ = (ab)θ = (aθ)(bθ) = (aφ)(bφ)

as θ is a 0-morphism.

The final statement follows immediately from (2) and (3).

Remark. The condition

ab ∈ C implies (ab)φ = (aφ)(bφ)

for all a, b ∈ C is reminiscent of the definition of 0-morphism in Section 1.

In particular, we have found a necessary condition for the existence ofa suitable prehomomorphism from a 0-bisimple R-unipotent semigroup to aprimitive inverse semigroup. We now show that this condition is also sufficient.

420 Hayes

Proposition 2.17. Let S be a 0-bisimple R-unipotent semigroup with e ∈E∗(S) and put C = Re . Suppose that there exists an injective function φ fromC to a (non-zero) R-class of a primitive inverse semigroup such that for alla, b ∈ C ,

ab ∈ C implies (ab)φ = (aφ)(bφ).

Then there exists a 0-restricted idempotent-pure prehomomorphism from S toa primitive inverse semigroup.

If S is a monoid and e = 1 , we may assume that C can be embedded ina group and the primitive inverse semigroup can be chosen to be a group withzero adjoined (a 0-group).

Proof. Let T be the primitive inverse semigroup. Define a function θ: S →T by 0θ = 0 and if x ∈ S∗ and x = b′a where a, b ∈ C and b′ ∈ V (b), then

xθ = (bφ)−1(aφ).

We first show that xθ �= 0. If (bφ)−1(aφ) = 0, then (bφ)(bφ)−1(aφ) = 0. Butaφ R bφ R (bφ)(bφ)−1 , whence (bφ)(bφ)−1(aφ) = aφ = 0, a contradiction.

To see that θ is well defined, suppose that

x = b′a = d′c

where a, b, c, d ∈ C and b′ ∈ V (b) and d′ ∈ V (d). Then by Proposition 2.8(2)there is an invertible element u ∈ C such that b = ud and a = uc . Henceud, uc ∈ C and

(bφ)−1(aφ) = ((ud)φ)−1((uc)φ) = ((uφ)(dφ))−1(uφ)(cφ)

= (dφ)−1(uφ)−1(uφ)(cφ) = (dφ)−1(cφ)

by Lemma 1.6(ii), so that xθ is defined independently of its representation.

If f ∈ E∗(S), then f = a′a for some a ∈ C and a′ ∈ V (a), byProposition 2.8(4). It follows that

fθ = (aφ)−1(aφ) ∈ E∗(T ).

Now suppose that x ∈ S with xθ ∈ E∗(T ). Let x = b′a where a, b ∈ C andb′ ∈ V (b). Then

xθ = (bφ)−1(aφ) = g ∈ E∗(T ),

whence (bφ)(bφ)−1(aφ) = (bφ)g , that is, aφ = bφ . Since φ is injective we havethat a = b and so x ∈ E(S). Thus θ is idempotent-pure.

To complete the proof we show that θ is a 0-morphism. This, byLemma 1.8, will imply that θ is a 0-restricted prehomomorphism. We haveseen that θ is 0-restricted. Let x, y ∈ S be such that xy �= 0. Put x = b′a andy = d′c . Now ad′ �= 0 and so ad′ = t′s for some s, t ∈ C and t′ ∈ V (t). Then

Hayes 421

tad′ = tt′s = es = s so that tad′ ∈ C , whence sd = tad′d ∈ C by Lemma 2.14.If s′ ∈ V (s), then from ad′ = t′s we also get ad′s′ = t′ss′ = t′e = t′ , sincet′ ∈ Le . If a′ ∈ V (a) it follows that

(ta)(a′t′) = t(aa′)t′ = tet′ = t(ea)d′s′ = (tad′)s′ = ss′ = e,

and ta ∈ C as a′t′ ∈ V (ta). Because (ta)d′ = s we have that ta = sd ∈ C byLemma 2.15. Hence (tφ)(aφ) = (sφ)(dφ) and so

(tφ)−1(tφ)(aφ)(dφ)−1 = (tφ)−1(sφ)(dφ)(dφ)−1 �= 0,

that is, (aφ)(dφ)−1 = (tφ)−1(sφ).

Now

xy = b′ad′c

= b′t′sc

and b′t′ is an inverse of tb , say (tb)′ = b′t′ . Thus xy = (tb)′sc . We can slightlymodify the argument that ta ∈ C to show that tb ∈ C . Also, if c′ ∈ V (c),

(sc)(c′s′) = s(cc′)s′ = ses′ = ta(d′e)s′ = (tad′)s′ = ss′ = e,

whence sc ∈ C . Therefore

(xy)θ = ((tb)φ)−1(sc)φ

= ((tφ)(bφ))−1(sφ)(cφ)

= (bφ)−1(tφ)−1(sφ)(cφ)

= (bφ)−1(aφ)(dφ)−1(cφ)

= (xθ)(yθ).

Thus θ is a 0-morphism and the proof is complete.

If S is a monoid and e = 1, let φ be an embedding of C into a group G .Since G is a single R-class, a group with zero is a primitive inverse semigroup,and C is a submonoid, the result follows by replacing T with G0 in the aboveproof.

Note that θ extends φ . For, if a ∈ C , then a = ea where e ∈ V (e) and e ∈ C .Hence

aθ = (ea)θ = (eφ)−1(aφ) = (eφ)(aφ) = (ea)φ = aφ,

since eφ ∈ E∗(T ).Combining Proposition 2.16(3) and Proposition 2.17 gives the following

theorem.

422 Hayes

Theorem 2.18. Let S be a 0-bisimple R-unipotent semigroup with e ∈E∗(S) and put C = Re . Then S admits a 0-restricted idempotent-pure pre-homomorphism to a primitive inverse semigroup T if and only if there existsan injective function φ from C to a (non-zero) R-class of T such that for alla, b ∈ C ,

ab ∈ C implies (ab)φ = (aφ)(bφ).

If S is a monoid and e = 1 , we can take T to be a group with zero and assumethat C can be embedded in a group.

The following corollary is immediate.

Corollary 2.19. Let S be a 0-bisimple R-unipotent semigroup. Considerthe condition

(1) for e ∈ E∗(S) , there exists an injective function φ from Re to a (non-zero) R-class of a primitive inverse semigroup such that for all a, b ∈ Re .

ab ∈ Re implies (ab)φ = (aφ)(bφ),

Then if condition (1) holds for SOME e ∈ E∗(S) , it holds for EVERY e ∈E∗(S) .

Of course, this poses the question, for which 0-bisimple R-unipotentmonoids is it the case that R1 is embeddable in a group? In the semigroupcase, the lemma below shows that left groups with an adjoined zero satisfy thecondition. Recall that, if S is a left group and e ∈ E(S), then Re = He .

Lemma 2.20. If S is a left group then S0 admits a 0-restricted idempotent-pure prehomomorphism to a group with zero.

Proof. It is clear that S0 is 0-bisimple. We know by the remark precedingLemma 2.6 that S is regular, whence every R-class of S contains an idempo-tent. If f and g are R-related idempotents of S then f H g as S is a leftgroup, so that f = g . Thus each R-class of S0 contains exactly one idempotentand S0 is R-unipotent. Let e ∈ E(S). Since He is a group, it is clear thatHe can be embedded in a group (take the identity mapping from He to itself).The result now follows from the proof of Proposition 2.17.

Remark. In fact, we know from the dual of [8, Exercise 2.6.6] that a semigroupis a left group if and only if it is isomorphic to the direct product of a groupand a left zero semigroup. Let G be a group and M a left zero semigroup, andconsider the direct product G×M . It is straightforward to verify that

E(G×M) = {(1,m) | m ∈ M},

and(g,m) R (h, n) if and only if m = n.

Hayes 423

Hence, for all m ∈ M , the set {(g,m) | g ∈ G} is an R-class (and indeed anH -class) of G × M . It is immediate that the function φ: H(1,m) → G givenby (g,m)φ = g is an embedding (and indeed an isomorphism). If n ∈ M ,then (g,m) = (1,m)(g, n), where (1,m) ∈ V ((1, n)) and (g, n), (1, n) ∈ R(1,n) .Therefore, we can follow the proof of Proposition 2.17 to define a functionθ: (G×M)0 → G0 by 0θ = 0 and

(g,m)θ = ((1, n)φ)−1((g, n)φ) = 1−1g = 1g = g.

Then θ is clearly a 0-restricted idempotent-pure prehomomorphism.

We now return to completely 0-simple semigroups.

Lemma 2.21. A completely 0-simple semigroup S = M0(G; I,Λ;P ) is R-unipotent (L-unipotent) if and only if every column (row) of P has exactly onenon-zero entry, that is, if and only if

for all i ∈ I, there exists a unique λ ∈ Λ such that pλi �= 0

(for all λ ∈ Λ, there exists a unique i ∈ I such that pλi �= 0).

Proof. Let S be R-unipotent and suppose that i ∈ I . If λ, µ ∈ Λ are suchthat pλi, pµi �= 0, then (i, p−1

λi , λ) and (i, p−1µi , µ) are R-related idempotents.

Since S is R-unipotent it follows that (i, p−1λi , λ) = (i, p−1

µi , µ), whence λ = µ ,as required.

Conversely, suppose that every column of P contains exactly one non-zero entry. Let (i, p−1

λi , λ) and (j, p−1µj , µ) be R-related idempotents. Then

i = j , and as pλi, pµi �= 0, we have that λ = µ by assumption. Thus(i, p−1

λi , λ) = (j, p−1µj , µ), so that each R-class contains exactly one idempotent,

and S is R-unipotent by Theorem 1.2(2).

A dual argument gives the result when S is L-unipotent.

It is clear that, if M0(G; I,Λ;P ) is a completely 0-simple R-unipotent (re-spectively L-unipotent) semigroup, then |I| ≥ |Λ| (respectively |Λ| ≥ |I|).

Lemma 2.22. Let S be a completely 0-simple R-unipotent semigroup withe ∈ E∗(S) and put C = Re . Then there exists an injective 0-morphism from C0

to a (non-zero) R-class of a primitive inverse semigroup with a zero adjoined.Moreover, the primitive inverse semigroup can be chosen to be a group with zero.

Proof. Suppose that S = M0(G; I,Λ;P ). Let K be a group in one-onecorrespondence with Λ which we associate with Λ. Then S = M0(G; I,K;P ).Put e = (i, p−1

mi,m), where m is the unique element of K such that pmi �= 0.For (i, g, k) ∈ C , define φ: C0 → (G×K)0 by 0φ = 0 and

(i, g, k)φ = (pmig,m−1k).

To see that φ is injective, let (i, g, k), (i, h, l) ∈ Re . Suppose that (i, g, k)φ =(i, h, l)φ . Then (pmig,m

−1k) = (pmih,m−1l). Hence k = l and g = h since

424 Hayes

G and H are groups and therefore cancellative. Thus (i, g, k) = (i, h, l), asrequired.

Now let a, b ∈ C be such that ab �= 0. Then a = (i, g, k), b = (i, h, l),and since

ab = (i, g, k)(i, h, l) �= 0

we have that pki �= 0, whence k = m . Clearly ab ∈ C whenever ab �= 0. Itfollows that

(ab)φ = ((i, g,m)(i, h, l))φ

= (i, gpmih, l)φ

= (pmi(gpmih),m−1l)

= ((pmig)(pmih), (m−1m)(m−1l))

= (pmig,m−1m)(pmih,m

−1l)

= (i, g,m)φ(i, h, l)φ

= (aφ)(bφ).

By definition, φ is 0-restricted. Therefore φ is a 0-morphism and the proof iscomplete.

Corollary 2.23. Let S be a completely 0-simple R-unipotent semigroup.Then S admits a 0-restricted idempotent-pure prehomomorphism to a groupwith zero.

Proof. Let e ∈ E∗(S) and put C = Re . If a, b ∈ C , then ab ∈ C if andonly if ab �= 0. The result now follows from Theorem 2.18 and Lemma 2.22.

Remark. If S = M0(G; I,K;P ) is a completely 0-simple R-unipotent semi-group where K is a group, we can use the construction in the proof of Propo-sition 2.17 to obtain a 0-restricted idempotent-pure prehomomorphism from Sto a group with zero. Let j ∈ I and n be the unique element of K such thatpnj �= 0. Suppose that (i, g, k) ∈ S∗ and m is the unique element of K suchthat pmi �= 0. Then

(i, g, k) = (i, p−1mi, n)(j, p

−1nj pmig, k),

where (i, p−1mi, n) ∈ V ((j, p−1

nj ,m)), and (j, p−1nj ,m), (j, p−1

nj pmig, k) ∈ R(j,p−1nj

,n) .

Then θ: S → (G×K)0 defined by 0θ = 0 and

(i, g, k)θ = ((j, p−1nj ,m)φ)−1((j, p−1

nj pmig, k)φ)

= (pnjp−1nj , n

−1m)−1(pnjp−1nj pmig, n

−1k)

= (1, n−1m)−1(pmig, n−1k)

= (1,m−1n)(pmig, n−1k)

= (pmig,m−1k),

is a 0-restricted idempotent-pure prehomomorphism.

Hayes 425

The result below is [10, Proposition 3.6].

Theorem 2.24. For every monoid M there exists a group U(M) and amonoid homomorphism η: M → U(M) such that Mη generates U(M) andfor every homomorphism θ: M → G to a group G such that Mθ generates Gthere exists a unique homomorphism θ∗: U(M) → G such that ηθ∗ = θ .

We call the group U(M) the fundamental group of the monoid M .The following proposition, [9, Proposition 10], shows that the homomorphismη: M → U(M) determines whether or not a monoid M can be embedded in agroup.

Proposition 2.25. A monoid M can be embedded in a group if and only ifit is embedded in U(M) by η .

Now let S be a 0-bisimple R-unipotent semigroup with e ∈ E∗(S),and suppose that Re is a submonoid of S . Assume that η embeds Re inG = U(Re). Let β: S → G0 be the suitable prehomomorphism constructedfrom η as in Proposition 2.17. We conclude this paper with the followingcharacterisation of β .

Theorem 2.26. Let θ: S → H0 be any suitable prehomomorphism to a0-group such that the image of θ generates H0 . Then there is a unique 0-restricted prehomomorphism θ∗: G0 → H0 such that βθ∗ = θ .

If S is a monoid and e = 1 , then H0 can be taken to be a primitiveregular semigroup T .

Proof. We can adapt the proof of Proposition 2.16(3) to show that θ restrictsto an embedding of Re in H . Now by Theorem 2.7(2) every element of S∗ canbe written as b′a where a, b ∈ Re and b′ ∈ V (b). It follows that H is generatedby the image of θ|Re

. Hence there exists a unique group homomorphismθ∗: G → H such that ηθ∗ = θ|Re

, by Theorem 2.24. Extend θ∗ by defining0θ∗ = 0. We now have a 0-restricted (pre)homomorphism from G0 to H0 .

To verify that βθ∗ = θ , let s ∈ S∗ . Then s = a′b where a, b ∈ Re anda′ ∈ V (a). By Proposition 2.17,

(sβ)θ∗ = ((aη)−1(bη))θ∗

which is equal to (aθ)−1(bθ). Now by Lemma 1.9 θ is a 0-morphism, and sincea′b �= 0 we have that (a′b)θ = (a′θ)(bθ) = (aθ)−1(bθ). Thus (sβ)θ∗ = sθ , asrequired. The uniqueness of θ∗ with the stated properties is clear.

If S is a monoid and e = 1, then we know from Proposition 2.16(1) thatT must be a group with zero, say H0 . Then following the above proof givesthe result.

426 Hayes

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Department of MathematicsUniversity of YorkHeslington, York YO10 5DD, UKtonydhayes@hotmail.com

Received January 8, 2002Online publication May 14, 2002