# The Regular Part of a Semigroup of Full Transformations with Restricted Range: Maximal Inverse Subsemigroups and Maximal Regular Subsemigroups of Its Ideals

Authors: Worachead Sommanee
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

LetTX be the full transformation semigroup on a set X. For a fixed nonempty subset Y of a set X, let TX,Y be the semigroup consisting of all full transformations from X into Y. In a paper published in 2008, Sanwong and Sommanee proved that the set FX,Y=α∈TX,Y:Xα=Yα is the largest regular subsemigroup of TX,Y. In this paper, we describe the maximal inverse subsemigroups of FX,Y and completely determine all the maximal regular subsemigroups of its ideals.


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## Body

## 1. Introduction

LetT(X) be the set of all full transformations from a nonempty set X into itself. It is well-known that T(X) is a regular semigroup under composition of functions; see [1, p. 63]. Moreover, every semigroup can be embedded in T(X) for some nonempty set X; see [1, p. 7]. In [2, 3], Schein posed the problem to determine all maximal inverse subsemigroups of T(X); this problem is still unsolved. In 1976, Nichols [4] characterized one class of maximal inverse subsemigroup of T(X). Later in 1978, Reilly [5] has generalized Nichols construction and obtained a much wider class of maximal inverse subsemigroups of T(X). In 1999, Yang [6] described all of the maximal inverse subsemigroups of the finite symmetric inverse semigroup. Later in 2001, Yang [7] obtained the maximal subsemigroups of the finite singular transformation semigroups. In 2002, You [8] determined all the maximal regular subsemigroups of all ideals of the finite full transformation semigroup. In 2004, H. B. Yang and X. L. Yang [9] completely described the maximal subsemigroups of ideals of the finite full transformation semigroup. In 2014, Zhao et al. [10] showed that any maximal regular subsemigroup of ideals of the finite full transformation semigroup is idempotent generated. After that in 2015, East et al. [11] classified the maximal subsemigroups of T(X) when X is an infinite set containing certain subgroups of the symmetric group on X.LetY be a fixed nonempty subset of X and let T(X,Y) be the subsemigroup of T(X) of all elements with ranges contained in Y. In 1975, Symons [12] introduced and studied the semigroup T(X,Y). He described all the automorphisms of T(X,Y) and also determined the isomorphism theorem for two semigroups of type T(X,Y). In 2005, Nenthein et al. [13] characterized the regular elements of T(X,Y). In general, T(X,Y) is not a regular semigroup. In 2008, Sanwong and Sommanee [14] defined(1)FX,Y=α∈TX,Y:Xα=Yαand showed that FX,Y is the largest regular subsemigroup of T(X,Y). Obviously, F(X,Y)=T(X) when X=Y. Hence, we may regard F(X,Y) as a generalization of T(X). The authors also characterized Green’s relations on T(X,Y) and gave one class of maximal inverse subsemigroups of T(X,Y). Later in 2009, Sanwong et al. [15] described all maximal and minimal congruences on T(X,Y). In 2011, Mendes-Gonçalves and Sullivan [16] obtained all the ideals of T(X,Y). In the same year, Sanwong [17] described Green’s relations and ideals and all maximal regular subsemigroups of F(X,Y). The author also proved that every regular semigroup S can be embedded in F(S1,S). After that, in 2013, Sommanee and Sanwong [18] computed the rank of F(X,Y) when X is a finite set. They also obtained the rank and the idempotent rank of its ideals when X is finite. In 2014, Fernandes and Sanwong [19] calculated the rank of T(X,Y) when X is finite. In 2016, L. Sun and J. Sun [20] characterised the natural partial order on T(X,Y). In the same year, Tinpun and Koppitz [21] determined the relative rank of T(X,Y) modulo the semigroup S(X,Y) of all extensions of the bijections on Y when X is finite.LetX be a chain and OT(X) the full order-preserving transformation semigroup on X. In 2011, Dimitrova and Koppitz [22] characterized the maximal regular subsemigroups of ideals of OT(X) when X is a finite chain. In 2015, Sommanee and Sanwong [23] investigated the regularity and Green’s relations of the order-preserving transformation semigroup(2)OFX,Y=FX,Y∩OTX=α∈FX,Y:αis order-preserving.They also proved that OF(X,Y) is idempotent generated when Y is finite.LetV be a vector space and let T(V) denote the linear transformation semigroup from V into V. For a fixed subspace W of V, let T(V,W) be the semigroup consisting of all linear transformations from V into W. Recently in 2017, Sommanee and Sangkhanan [24] determined all the maximal regular subsemigroups of the semigroup(3)Q=α∈TV,W:Vα=Wα,when W is a finite dimensional subspace of V over a finite field.In this paper, we describe the maximal inverse subsemigroups ofF(X,Y) and completely determine all the maximal regular subsemigroups of its ideals.


## 2. Preliminaries and Notations

For all undefined notions, the reader is referred to [1].For any setA, A means the cardinality of the set A. If A is a subset of a semigroup S, then A denotes the subsemigroup of S generated by A. An element e of a semigroup S is calledidempotent if e2=e. If U is a subset of a semigroup S, then E(U) denotes the set of all idempotents in the set U. An element a of a semigroup S is calledregular if there exists x∈S such that a=axa. The semigroup S is called regular if all its elements are regular. If a is an element of a semigroup S, we say that a′ is aninverse of a if a=aa′a and a′=a′aa′. We denote the set of inverses of an element a by V(a). A semigroup S is called an inverse semigroup if every element a in S has a unique inverse a-1 in S; it is equivalent to Sbeing regular and idempotent elements commute. A proper(regular,inverse) subsemigroup M of a semigroup S is amaximal (regular,inverse) subsemigroup of S if whenever M⊆T⊆S for some a (regular, inverse) subsemigroup T of S, then M=T or T=S.The Green’s relationsL, R, H, D, and J on a semigroup S are defined as follows. For a,b∈S,(1)
aLb if and only if S1a=S1b;(2)
aRb if and only if aS1=bS1;(3)
aJb if and only if S1aS1=S1bS1;(4)
H=L∩R and D=L∘R.For eacha∈S, we denote L-class, R-class, H-class, D-class, and J-class containing a by La,Ra,Ha,Da, and Ja, respectively.Lemma 1 (see [5, Lemma 1]).
LetM be a maximal inverse subsemigroup of a semigroup S. If M contains a minimum idempotent e, then He⊆M.Lemma 2.
LetS be any inverse semigroup. If e is a minimum idempotent in S, then ea=ae for all a∈S.Proof.
Assume thate is a minimum idempotent in S. Then e≤f for all f∈E(S); that is, e=ef=fe. Let a∈S. Then aea-1,a-1ea∈E(S) and ea=e(aea-1)a=(aea-1)ea=a(ea-1ea)=ae.Throughout the paper we assume thatY is a nonempty subset of a set X. Let T(X,Y)=α∈TX:Xα⊆Y, where Xα denotes the image of α. Define(4)FX,Y=α∈TX,Y:Xα=Yα.It is known that FX,Y is the largest regular subsemigroup of T(X,Y).Ifα∈T(X) and x∈X, then the image of x under α is written as xα. The set of all inverse images of x under α is denoted by xα-1.Lemma 3 (see [17, Lemma 1]).
Letα∈T(X,Y). Then α∈F(X,Y) if and only if aα-1∩Y≠∅ for all a∈Xα.In [17], the author gave a complete description of Green’s relations on F(X,Y) as the following lemma.Lemma 4.
Letα,β∈F(X,Y). Then(1)
αLβ if and only if Xα=Xβ;(2)
αRβ if and only if πα=πβ, where πα={xα-1:x∈Xα};(3)
αDβ if and only if Xα=Xβ;(4)
D=J.
In addition,αHβ if and only if Xα=Xβ and πα=πβ.Lemma 5.
F ( X , Y ) is an inverse semigroup if and only if |Y|=1.Proof.
If|Y|=1, then FX,Y=1 and so it is inverse. Conversely, assume that F(X,Y) is an inverse semigroup. Let a and b be elements in Y. Then there exist constant maps Xa and Xb with ranges {a} and {b}, respectively. Thus they are idempotents in F(X,Y) such that Xa=XbXa=XaXb=Xb; it follows that a=b. Hence, |Y|=1.Proposition 6.
M is a maximal inverse subsemigroup of T(X,Y) if and only if M is a maximal inverse subsemigroup of F(X,Y).Proof.
One direction is clear. Indeed, each inverse subsemigroup ofT(X,Y) is contained in F(X,Y), the largest regular subsemigroup of T(X,Y). Conversely, assume that M is a maximal inverse subsemigroup of F(X,Y). Then M is an inverse subsemigroup of T(X,Y). Let V be an inverse subsemigroup of T(X,Y) such that M⊆VT(X,Y). Since F(X,Y) is the largest regular subsemigroup of T(X,Y), it follows that M⊆V⊆F(X,Y). We get that M=V or V=F(X,Y) by the maximality of M in F(X,Y). If F(X,Y)=V, then F(X,Y) is an inverse semigroup and so |Y|=1 by Lemma 5; this implies that T(X,Y)=F(X,Y). Therefore, M is a maximal inverse subsemigroup of T(X,Y).For eacha∈Y, let(5)Fa=α∈FX,Y:aα=a,αis injective onX∖aα-1.The authors [14] showed that Fa is a maximal inverse subsemigroup of T(X,Y). Thus by Proposition 6, we have that Fa is also a maximal inverse subsemigroup of F(X,Y).


## 3. Maximal Inverse Subsemigroups

In this section, we writeG(A) for the permutation group on a set A.Lete be any idempotent in T(X). In 1978, Reilly defined Ie to be the set of those elements α∈T(X) which satisfy the following three conditions:(1)
The restriction ofα to Xe is a permutation of Xe; that is, α|Xe∈G(Xe).(2)
Forx∈X, if xα∉Xe, then xαα-1=1.(3)
αe=eα.Lemma 7 (see [5, Theorem 2]).
Lete be an idempotent in T(X) and let Ie be defined as above. Then Ie is a maximal inverse subsemigroup of T(X) with minimum idempotent e.Notice that ife is a constant map with range {a}⊆X, then Ie=Ia, where(6)Ia=α∈TX:aα=a,αis injective onX∖aα-1;see [4, 5] for more details.Now, lete be any idempotent in FX,Y and define(7)Fe=Ie∩FX,Y.Since e∈Ie∩FX,Y=Fe, we obtain Fe≠∅ and so Fe is a subsemigroup of F(X,Y). We note that if X=Y, then F(X,Y)=T(X) and Fe=Ie∩F(X,Y)=Ie∩T(X)=Ie.Here, we aim to prove thatFe is the maximal inverse subsemigroups of F(X,Y) with minimum idempotent e.To prove our main result we need the following five lemmas.Lemma 8.
Lete be any idempotent in FX,Y and α∈FX,Y. Then α∈Fe if and only if(1)
α|Xe∈G(Xe),(2)
fory∈Y, if yα∉Xe, then yαα-1=1,(3)
αe=eα.Proof.
One direction is clear. Conversely, assume that conditions (1), (2), and (3) hold. We prove thatα∈Ie. Suppose that x∈X and xα∉Xe. Since α∈F(X,Y), it follows that xα=yα for some y∈Y. By condition (2), we get that yαα-1=1. Thus |(xα)α-1|=1 and so α∈Ie.Lemma 9.
Lete be any idempotent in FX,Y and α∈FX,Y. Then condition (2) in Lemma 8 is equivalent to the statement(8)X∖Yα⊆Xe,αis injective onY∖Xeα-1.Proof.
Assume that, fory∈Y, if yα∉Xe, then yαα-1=1. We prove (X∖Y)α⊆Xe. Let xα∈(X∖Y)α for some x∈X∖Y. Since (X∖Y)α⊆Xα=Yα, there exists y∈Y such that xα=yα. Thus x,y∈(yα)α-1. If yα∉Xe, then |(yα)α-1|=1 by the assumption. Whence x=y, this is a contradiction. Hence yα∈Xe; that is, xα∈Xe. Next, we prove α is injective on Y∖(Xe)α-1. Let w,z∈Y∖(Xe)α-1 be such that wα=zα. Then w∈Y and wα∉Xe. Since w,z∈(wα)α-1, we obtain w=z by our assumption.
Conversely, suppose that(X∖Y)α⊆Xe and α is injective on Y∖(Xe)α-1. Let y∈Y and yα∉Xe. To show that |(yα)α-1|=1, we let u,v∈(yα)α-1. Then uα=vα=yα∉Xe; this implies that u,v∉(Xe)α-1. Since (X∖Y)α⊆Xe, we obtain u,v∈Y. Thus u,v∈Y∖(Xe)α-1 and so u=v since α is injective on Y∖(Xe)α-1.Lemma 10.
Letα be any element in TX and let e be an idempotent in T(X). If there is α′∈V(α) such that e=eα′α=αα′e,eα=αe and eα′=α′e, then α|Xe∈G(Xe).Proof.
Assume that the conditions hold. Since(9)Xe=Xeα′α=Xeα′α=Xα′eα⊆Xeα=Xeα=Xαe⊆Xe,it follows that (Xe)α=Xe; that is, α maps Xe onto Xe. We prove that α|Xe is injective. Let x,y∈Xe be such that xα=yα. Then(10)x=xe=xαα′e=xαα′e=yαα′e=yαα′e=ye=y.Hence αXe:Xe→Xe is bijective.Lemma 11.
Lete be an idempotent in FX,Y. Then Fe is an inverse subsemigroup of FX,Y with minimum idempotent e.Proof.
Letα be any element in Fe. Then α∈Ie. Since Ie is an inverse semigroup, there exists β∈Ie such that α=αβα and β=βαβ. We prove β∈F(X,Y). Since β=βαβ, we obtain Xβ=Xβαβ⊆Xαβ⊆Yβ⊆Xβ; that is, Xβ=Yβ. To show Xβ⊆Y, we let xβ∈Xβ for some x∈X. Since α∈Fe, (X∖Y)α⊆Xe by Lemma 9. If xβ∈X∖Y, then (xβ)α∈(X∖Y)α⊆Xe. It follows that xβ=(xβα)β∈(Xe)β=Xe⊆Y; this is a contradiction. Hence, xβ∈Y and so Xβ⊆Y. Thus, β∈Ie∩F(X,Y)=Fe. Whence Fe is a regular subsemigroup of Ie, it follows from Lemma 7 that Fe is an inverse subsemigroup of F(X,Y) and e is the minimum idempotent in Fe.Lemma 12.
Lete be an idempotent in F(X,Y). If M is any inverse subsemigroup of F(X,Y) such that Fe⊆M, then E(M)⊆Fe.Proof.
Assume thatM is an inverse subsemigroup of F(X,Y) containing Fe. Let f be any idempotent in M. Since e,f are idempotents in M, we get ef=fe. We prove that Xe⊆Xf. Suppose that Xe⊈Xf. Then Xef⊆Xe∩XfXe, so there exists z∈Xe∖Xef. We see that zf∈Xef and zf≠z. Now, we have z,zef∈Xe such that z≠zef. Let σ∈G(Xe) be such that zσ=zef and (zef)σ=z. We define α∈T(X) by(11)xα=xeσ∀x∈X.We see that Xα=(Xe)σ=(Ye)σ=Yα and Xα=(Xe)σ=Xe⊆Y; we obtain α∈F(X,Y). We prove α|Xe∈G(Xe). Let x,y∈X be such that (xe)α=(ye)α. Then ((xe)e)σ=((ye)e)σ; that is, (xe)σ=(ye)σ. Thus xe=ye since σ is injective on Xe. Since (Xe)α=((Xe)e)σ=(Xe)σ=Xe, α is surjective on Xe. Hence, α|Xe∈G(Xe). Since yα=(ye)σ∈Xe for all y∈Y, whence condition (2) in Lemma 8 holds. To prove αe=eα. Let x∈X. Since (xe)σ∈Xe and e is an idempotent, we get that xαe=((xe)σ)e=(xe)σ=((xe)e)σ=xeα. So, α∈Fe. By Lemma 11, we have Fe as an inverse subsemigroup of F(X,Y). Then there exists α-1∈V(α)∩Fe and σ-1=α-1|Xe∈G(Xe). Since zef=zfe∈Xe, we obtain (zef)α-1=(zef)σ-1=z. It is clear that α-1fα is an idempotent in M. We see that(12)zeffα-1fα=zefα-1fα=zfα=zfeσ=zefσ=z,zefα-1fαf=zfαf=zfeσf=zefσf=zf.Since z≠zf, this implies that f(α-1fα)≠(α-1fα)f which is a contradiction. Therefore, Xe⊆Xf. Hence ef=e and so e=ef=fe. It is easy to see that f|Xe=idXe, the identity map on Xe, which implies that f|Xe∈G(Xe). Thus, conditions (1) and (3) in Lemma 8 are satisfied for f.
In order to verify condition (2) in Lemma8 for f to be an element of Fe, let y∈Y be such that yf∉Xe. We show that |(yf)f-1|=1. Let u,v∈(yf)f-1. Then uf=vf=yf∉Xe. Suppose that u≠v. Assume that v∈Y. We define d∈T(X) by(13)xd=xifx∈Y∖u,xeifx∈u∪X∖Y.Obviously, Xd⊆Y and it is easy to verify that (xe)d=xe=x(de) for all x∈X. Hence, (Xe)d=Xe and ed=de. It is easy to show that d|Xe is injective, whence d|Xe∈G(Xe). We prove Xd⊆Yd. Let xd∈Xd for some x∈X. If x∈Y∖{u}, then xd∈Yd. But if x∈{u}∪(X∖Y), then xd=xe=(xe)d∈Yd. Thus, Xd⊆Yd and so d∈F(X,Y). We prove d satisfies the condition in Lemma 9. By the definition of d, we have (X∖Y)d⊆Xe. Let a,b∈Y∖(Xe)d-1. Then a,b∈Y and ad,bd∉Xe, which implies that a,b∈Y∖{u}. So, if ad=bd, then a=b. Hence, d is injective on Y∖(Xe)d-1 and d∈Fe. Next, we prove d is an idempotent. For x∈Y∖{u}, (xd)d=xd. If x∈{u}∪(X∖Y), then (xd)d=(xe)d=xe=xd. Thus, d2=d. Since d and f are idempotents in M, df=fd. However,(14)udf=uef=ue∈Xe,ufd=vfd=vdf=vf∉Xe,a contradiction. Hence, v∈X∖Y. Define g∈TX by(15)xg=xifx∈Y,xeifx∈X∖Y.Similarly, by using the same argument as in the proof above for d, we can show that g is an idempotent in Fe. Since f,g∈E(M), we obtain fg=gf. We see that(16)vfg=vf∉Xebutvgf=vef=ve∈Xe,and this is a contradiction. Thus, u=v and so condition (2) in Lemma 8 is also satisfied. Whence f∈Fe, therefore, E(M)⊆Fe as required.Now, we are ready to prove our main result (Theorem13).Theorem 13.
Lete be an idempotent in F(X,Y). Then Fe is a maximal inverse subsemigroup of F(X,Y) with minimum idempotent e.Proof.
From Lemma11, Fe is an inverse subsemigroup of F(X,Y) with minimum idempotent e. To prove the maximality, let M be any inverse semigroup of F(X,Y) such that Fe⊆MF(X,Y). Then by Lemma 12, we have E(M)⊆Fe, which implies that e is the minimum idempotent in M. We prove M=Fe by letting α∈M. Since α,α-1∈M, we obtain eα=αe and eα-1=α-1e by Lemma 2. And since αα-1 and α-1α are idempotents in M, e=αα-1e and e=eα-1α. Then by Lemma 10, we get that α|Xe∈G(Xe). Thus, conditions (1) and (3) in Lemma 8 are satisfied for α to be an element of Fe. Next, we prove that α satisfies condition (2) in Lemma 8. Let y∈Y be such that yα∉Xe. Let u,v∈(yα)α-1. Then uα=vα=yα∉Xe; this implies that uαα-1=vαα-1=yαα-1∉Xe. Now, we have αα-1∈E(Fe),y∈Y and y(αα-1)∉Xe. Since u,v∈[y(αα-1)](αα-1)-1, we obtain that u=v by using condition (2) in Lemma 8 for αα-1∈Fe. Hence, α∈Fe and so M⊆Fe. Therefore, M=Fe and Fe is a maximal inverse subsemigroup of F(X,Y).Corollary 14.
Lete be an idempotent in F(X,Y). Then Fe is a maximal inverse subsemigroup of T(X,Y).Proof.
It follows directly from Theorem13 and Proposition 6.Remark 15.
F e ≠ F f for all distinct idempotents e,f∈F(X,Y). In fact, if there exist idempotents e and f in F(X,Y) such that Fe=Ff, then f∈Fe and e∈Ff such that e and f are minimum idempotents of Fe and Ff, respectively. Thus, we get e≤f and f≤e; it follows that e=f.Corollary 16.
Lete be an idempotent in F(X,Y) such that Xe=Y. Then Fe=He is a maximal inverse subsemigroup of F(X,Y).Proof.
SinceFe is a maximal inverse subsemigroup of F(X,Y) with the minimum idempotent e, we obtain that He⊆Fe by Lemma 1. To show that Fe⊆He, let α∈Fe. Then α|Xe∈G(Xe) and αe=eα. Since (Xe)α=Xe, it follows that Xα=Yα=(Xe)α=Xe. We prove that πα=πe. Let x,y∈X such that xα=yα. Then (xe)α=(xα)e=(yα)e=(ye)α; this implies that xe=ye since α|Xe is injective. Hence πα⊆πe. On the other hand, let xe=ye. Then (xα)e=(xe)α=(ye)α=(yα)e. Since xα,yα∈Y=Xe and e is an idempotent, we obtain xα=yα. Hence πe⊆πα and πα=πe. Then by Lemma 4, we get that α∈He. Thus, Fe⊆He. Therefore, He=Fe is the maximal inverse subsemigroup of F(X,Y).Notice that ife is a constant map with range {a}⊆Y, then Ie=Ia and(17)Fe=FX,Y∩Ia=α∈FX,Y:aα=a,αis injective onX∖aα-1;that is, Fe=Fa. Thus, we get the following corollary which appeared in [14, Theorem 4.3].Corollary 17.
Ife∈FX,Y is a constant map with range {a}⊆Y, then Fe=Fa is a maximal inverse subsemigroup of FX,Y and TX,Y.Next, we count the number of elements inFe. Let e be any idempotent in F(X,Y). We know that He≅GXe; see [17, Section 4]. Thus, if Xe=Y, then Fe=He≅GXe=G(Y) by Corollary 16. Hence, |Fe|=|G(Y)| when Xe=Y. In what follows, we assume that XeY.Recall that the number of ways thatr objects can be chosen from n distinct objects written as nr is given by(18)nr=n!n-r!r!.Let A and B be (possibly empty) sets and let PI(A,B) be the set of all partial injective maps from A into B. We note that PI(A,B) contains the empty map. Moreover, if A=∅ or B=∅, then |PI(A,B)|=1. And if A and B are finite sets such that |A|=m and |B|=n, then we shall write PI(m,n) instead of PI(A,B). By a consideration of the cardinality of a domain of partial injective maps, we can verify that(19)PIm,n=∑j=0min⁡m,nmjnjj!form,n≥1.Since XeY, there exists x∈Y∖Xe such that xe∈Xe. We see that x∈((xe)e-1∖{xe})∩Y. For each y∈Xe, we define(20)Ay=ye-1∖y∩Y,By=ye-1∩X∖Y.Then ye-1=y∪Ay∪By is a disjoint union such that ye-1∩Xe=y and A(y)∩Xe=∅=B(y)∩Xe. Moreover, A(y)≠∅ for some y∈Xe since XeY. Let(21)A=yi∈Xe:Ayi≠∅,i∈I,B=yj∈Xe:Byj≠∅,j∈J.Then yie=yi and yie-1∩Xe={yi} for all i∈I. It is easy to see that(22)X∖Xe=⋃i∈IAyi∪⋃j∈JByjis adisjoint union.Letα be any element in Fe. Then αe=eα and α|Xe=σ for some σ∈G(Xe). For each x∈X∖Xe, x∈A(yi)∪B(yj) for some yi∈A and yj∈B. If x∈A(yi), then xe=yi∈Xe. Since αe=eα, we obtain (xα)e=(xe)α=yiα=yiσ; that is, xα∈(yiσ)e-1={yiσ}∪A(yiσ)∪B(yiσ). Since xα∈Y and B(yiσ)⊆X∖Y, we obtain that xα∈{yiσ}∪A(yiσ). If x∈B(yj)⊆X∖Y, then xα∈(X∖Y)α⊆Xe by Lemma 9. Since e is an idempotent, we get xα=(xα)e=(xe)α=yjα=yjσ. By condition (2) in Lemma 8, we see that α is injective on all elements of A(yi) which are not mapped into {yiσ} for all i∈I. Thus for each i∈I, α|A(yi) corresponds with an element θi of PI(A(yi),A(yiσ)); that is, for u∈A(yi),(23)
u
α
=
u
θ
i
if
u
∈
dom
θ
i
,
y
i
σ
if
u
∈
A
y
i
∖
dom
θ
i
.We have shown the following. Givenα∈Fe. Then α is a union of a permutation σ∈G(Xe), a union of partial injections θi:A(yi)→A(yiσ) for each yi∈A, and a union of functions from B(yj) into {yjσ} for each yj∈B.Theorem 18.
Lete be any idempotent in F(X,Y) with XeY. Let G(Xe)={σk:k∈Λ} and A=yi∈Xe:Ayi≠∅,i∈I. Then there is a one-to-one correspondence between Fe and ⋃k∈Λσk×∏i∈IPIAyi,Ayiσk.Proof.
For eachα∈Fe, we have α|Xe=σk for some k∈Λ and partial injections θi:A(yi)→A(yiσk) for all yi∈A. We define(24)Ψ:Fe⟶⋃k∈Λσk×∏i∈IPIAyi,Ayiσk,by αΨ=(σk,(θi)i∈I) for all α∈Fe. We verify that Ψ is injective. Let α,β∈Fe be such that αΨ=βΨ. Then (σk,(θi)i∈I)=(σl,(τi)i∈I) for some σk,σl∈G(Xe),(θi)i∈I∈∏i∈IPI(A(yi),A(yiσk)) and (τi)i∈I∈∏i∈IPI(A(yi),A(yiσl)). Thus, σk=σl and θi=τi for all i∈I; it follows that xα=xβ for all x∈Xe∪⋃i∈IAyi=Y. For x∈X∖Y, x∈B(yj) for some yj∈B=yj∈Xe:Byj≠∅,j∈J. We get that xα=yjσk=yjσl=xβ. Whence α=β, next, we prove Ψ is surjective. Let σk∈G(Xe) and (θi)i∈I∈∏i∈IPI(A(yi),A(yiσk)). We define α∈T(X,Y) which is determined by σk and θi for all i∈I as follows. For x∈X,(25)xα=xσkifx∈Xe,xθiifx∈dom⁡θi≠∅,yiσkifx∈Ayi,x∉dom⁡θiorθiis the empty map,yjσkifx∈Byj.We see that, for each x∈X∖Y, x∈B(yj) for some yj∈B. Then yj∈Xe⊆Y and xα=yjσk=yjα. So, Xα⊆Yα, that is α∈F(X,Y). By the construction of α, conditions (1) and (2) in Lemma 8 are satisfied for α to be an element of Fe. To prove αe=eα, we let x∈X. If x∈Xe, then (xα)e=(xσk)e=xσk=xα=(xe)α, since x,xσk∈Xe and e is an idempotent. If x∈domθi≠∅, then x∈A(yi); it follows that xe=yi and xθi∈A(yi)θi⊆A(yiσk). Thus, (xθi)e=yiσk and (xα)e=(xθi)e=yiσk=yiα=(xe)α since yi∈Xe. If x∈Ayiandx∉domθiorθiis the empty map, then xe=yi and (xα)e=(yiσk)e=yiσk=yiα=(xe)α. And if x∈B(yj), then xe=yj and (xα)e=(yjσk)e=yjσk=yjα=(xe)α, since yj∈Xe. Hence, αe=eα. Whence α∈Fe and αΨ=(σk,(θi)i∈I), therefore, Ψ is a bijection.The following corollary is a straightforward consequence of Theorem18.Corollary 19.
Lete be an idempotent in F(X,Y) such that XeY and Xe=p∈Z+. Let A=yi∈Xe:Ayi≠∅,1≤i≤t≤p. For each yi∈A, we let |A(yi)|=mi≥1 and |A(yiσk)|=ni,k≥0(1≤i≤t,1≤k≤p!). Then(26)Fe=∑k=1p!∏i=1tPImi,ni,k.It is known that ifX=Y, then F(X,Y)=T(X) and Fe=Ie. Then by Corollary 19, we have the following corollary.Corollary 20.
Lete be an idempotent in TX such that XeX and Xe=p∈Z+. Let A=yi∈Xe:Ayi≠∅,1≤i≤t≤p. For each yi∈A, we let Ayi=mi≥1 and Ayiσk=ni,k≥01≤i≤t,1≤k≤p!. Then(27)Ie=∑k=1p!∏i=1tPImi,ni,k.Ife is a constant map with range {a}⊆Y, then we get the following corollary which appeared in [14, Theorem 4.5].Corollary 21.
LetY be a finite subset of X with Y=r≥2 and e a constant map in FX,Y with range {a}⊆Y; then Fa=∑j=0r-1j!r-1j2.Proof.
We note thatFe=Fa by Corollary 17. Since Xe=a, we obtain Xe=1 and there is a unique Aa=Y∖a; that is, Aa=r-1. By Corollary 19, we have(28)Fa=Fe=∑k=11!∏i=11PIAa,Aa=PIAa,Aa=PIr-1,r-1,where(29)PIr-1,r-1=∑j=0min⁡r-1,r-1r-1jr-1jj!=∑j=0r-1r-1j2j!.Therefore, Fa=∑j=0r-1j!r-1j2.


## 4. Maximal Regular of Ideals

Throughout this section,X is a finite set with n elements and Y a nonempty subset of X with r≥2 elements. For convenience, we write Fn,r for FX,Y.For eachα∈Fn,r with |Xα|=k∈Z+, it follows from Lemma 3 that we can write(30)α=A1A2⋯Aka1a2⋯ak,where a1,a2,…,ak∈Y and {A1,A2,…,Ak} is a partition of X such that Ai∩Y≠∅ for all 1≤i≤k.For1≤k≤r, we define J(F;k)={α∈Fn,r:|Xα|=k}. Then J(F;k) is a J-class of the semigroup Fn,r. Let Q(F;k)=J(F;1)∪J(F;2)∪⋯∪J(F;k)={α∈Fn,r:|Xα|≤k}, where 1≤k≤r. Then Q(F;k) is an ideal of Fn,r and it is a regular subsemigroup of Fn,r; see [17, Lemma 7].Recall that the principal factor ofFn,r is the Rees quotient(31)Pk=QF;k/QF;k-1,where 2≤k≤r. It is usually convenient to think of it as J(F;k)∪{0}, and the product of two elements of Pk is taken to be zero if it falls in Q(F;k-1). Since Pk is finite and it is not a zero semigroup, we obtain that Pk is a completely 0-simple semigroup; see [18, p. 233] for more details.We need the following lemmas for proof of our main result.Lemma 22 (see [1, p. 98]).
LetS be a completely 0-simple semigroup. If a,b∈S and ab≠0, then ab∈Ra∩Lb.Lemma 23 (see [1, Proposition 2.3.7]).
Leta,b be elements in D-class D. Then ab∈Ra∩Lb if and only if La∩Rb contains an idempotent.Lemma 24.
Leta be any element in a semigroup S and T a regular subsemigroup of S.(1)
IfLa∩T≠∅, then La∩T contains an idempotent.(2)
IfRa∩T≠∅, then Ra∩T contains an idempotent.Proof.
(1) Assume thatLa∩T≠∅. Then there exists b∈La∩T. So, bLa and b=bb′b for some b′∈T. We get that b′b is an idempotent such that b′b∈T and b′bLbLa. Therefore, b′b is an idempotent in La∩T.
(2) If there isb∈Ra∩T, then we can write b=bb′b for some b′∈T and show that bb′ is an idempotent in Ra∩T.Lemma 25 (see [18, Lemma 3.5]).
For1≤k≤r-1, QF;k=JF;k.Lemma 26 (see [18, Lemma 4.2]).
Ifα∈J(F;k), then α can be written as a finite product of idempotents in J(F;k).The following fact can be obtained from Lemmas25 and 26 immediately.Lemma 27.
For1≤k≤r-1, Q(F;k)=E(J(F;k)).Lemma 28.
Let1≤k≤r-1 and α∈J(F;k). If β∈Lα, then β=γα for some γ∈J(F;k) such that Lγ≠Lα.Proof.
Suppose thatβ∈Lα. Thus Xα=Xβ; we write(32)α=A1A2⋯Aka1a2⋯ak,β=B1B2⋯Bka1a2⋯ak,where a1,a2,…,ak∈Y and Ai∩Y≠∅≠Bi∩Y for all 1≤i≤k. Since Xα≠Y, there is y∈Y∖Xα such that y∈Aj for some j∈1,2,…,k. We choose dai∈Ai∩Y for all i∈1,2,…,j-1,j+1,…,k and define(33)γ=B1⋯Bj-1BjBj+1⋯Bkda1⋯daj-1ydaj+1⋯dak.Then γ∈J(F;k) such that Xγ≠Xα and β=γα.Lemma 29.
Let2≤k≤r-1 and α∈J(F;k). If β∈Rα, then β=αγ for some γ∈J(F;k) such that Rγ≠Rα.Proof.
Assume thatβ∈Rα. Thus πα=πβ; we write(34)α=A1A2⋯Aka1a2⋯ak,β=A1A2⋯Akb1b2⋯bk,where ai,bi∈Y and Ai∩Y≠∅ for all 1≤i≤k. Define(35)γ=a1a2⋯ak-1ak∪X∖Xαb1b2⋯bk-1bk.It is clear that γ∈JF;k. If γ∉Rα, then Rγ≠Rα and β=αγ. If such γ∈Rα, we define(36)γ′=a1,ya2⋯ak-1ak∪X∖Xα∖yb1b2⋯bk-1bk,where y∈Y∖Xα. So, γ′∈J(F;k) such that Rγ′≠Rγ=Rα and β=αγ′.For the casek=1, the ideal QF;1=JF;1 consisting of all constant maps Xa with range a⊆Y. It is easy to verify that J(F;1)∖Xa is a maximal regular subsemigroup of QF;1 for all a∈Y.Lemma 30.
For2≤k≤r-1, M=Q(F;k-1)∪(J(F;k)∖Lα) is a maximal regular subsemigroup of Q(F;k) for all α∈J(F;k).Proof.
Letα be any element in J(F;k). Let β,γ∈J(F;k)∖Lα. If βγ∈Q(F;k-1), then βγ∈M. If βγ∉Q(F;k-1), then βγ≠0 in Pk=Q(F;k)/Q(F;k-1); that is, βγ∈Rβ∩Lγ by Lemma 22. Thus, βγ∈Lγ≠Lα and so βγ∈M. Hence, M is a subsemigroup of Q(F;k). We prove M is regular. Let λ∈M. If λ∈Q(F;k-1), then λ is a regular element in Q(F;k-1). If λ∈J(F;k)∖Lα, we write(37)λ=A1A2⋯Aka1a2⋯ak,where a1,a2,…,ak∈Y and Ai∩Y≠∅ for all 1≤i≤k. Since Xα≠Y, there exists y∈Y∖Xα such that y∈Aj for some j∈{1,2,…,k}. We choose dai∈Ai∩Y for all i∈{1,2,…,j-1,j+1,…,k} and define(38)μ=a1⋯aj-1ajaj+1⋯ak-1ak∪X∖Xλda1⋯daj-1ydaj+1⋯dak-1dak.Then μ∈J(F;k), Xμ≠Xα, and λ=λμλ. Hence, M is a regular semigroup. Next, we prove M is maximal. Let S be a regular subsemigroup of Q(F;k) such that MS⊆Q(F;k). Then there exists ξ∈S∖M; that is, ξ∈Lα∩S. Let τ be any element in Lα=Lξ. Then there is γ∈J(F;k) such that γ∉Lξ and τ=γξ by Lemma 28; that is, γ∈M⊆S. Thus, τ=γξ∈S. Hence, Lα⊆S and therefore S=Q(F;k).Lemma 31.
For2≤k≤r-1, M=Q(F;k-1)∪(J(F;k)∖Rα) is a maximal regular subsemigroup of Q(F;k) for all α∈J(F;k).Proof.
Letα be any element in J(F;k). Let β,γ∈J(F;k)∖Rα. If βγ∈Q(F;k-1), then βγ∈M. If βγ∉Q(F;k-1), then βγ≠0 in Pk and so βγ∈Rβ∩Lγ by Lemma 22. Thus, βγ∈Rβ≠Rα. So, βγ∈M. Hence, M is a subsemigroup of Q(F;k). We prove M is regular. Let λ∈M. If λ∈Q(F;k-1), then λ is a regular element in Q(F;k-1). If λ∈J(F;k)∖Rα, we write(39)λ=A1A2⋯Aka1a2⋯ak,where a1,a2,…,ak∈Y and Ai∩Y≠∅ for all 1≤i≤k. We choose dai∈Ai∩Y for all 1≤i≤k and define(40)λ′=a1a2⋯ak-1ak∪X∖Xλda1da2⋯dak-1dak.Then λ′∈J(F;k) and λ=λλ′λ. If λ′∉Rα, then λ′ is a regular element in M. If λ′∈Rα, we define(41)λ″=a1,ya2⋯ak-1ak∪X∖Xα∖yda1da2⋯dak-1dak,where y∈Y∖Xλ since k<r. We obtain λ″∉Rλ′=Rα and λ=λλ″λ. Hence, M is a regular semigroup. For a maximality of M, we let S be a regular subsemigroup of Q(F;k) such that MS⊆Q(F;k). Then there exists ξ∈S∖M; that is, ξ∈Rα∩S. Let τ be any element in Rα=Rξ. Then there is γ∈J(F;k) such that γ∉Rξ=Rα and τ=ξγ by Lemma 29; that is, γ∈M⊆S. Thus, τ∈S and so Rα⊆S. Therefore, S=Q(F;k).Theorem 32.
Each maximal regular subsemigroup ofQ(F;k),2≤k≤r-1 must be one of the following forms:(1)
Q(F;k-1)∪(J(F;k)∖Lα);(2)
Q(F;k-1)∪(J(F;k)∖Rβ),whereα,β∈J(F;k), Lα is the L-class containing α in J(F;k), and Rβ is the R-class containing β in J(F;k).Proof.
By Lemmas30 and 31, both (1) and (2) are maximal regular subsemigroups of Q(F;k).
On the other hand, letM be an arbitrary maximal regular subsemigroup of Q(F;k). It is easy to see that M∪Q(F;k-1) is a regular subsemigroup of Q(F;k) such that M⊆M∪Q(F;k-1)⊆Q(F;k). Then M=M∪Q(F;k-1) or M∪Q(F;k-1)=Q(F;k) by the maximality of M. If M∪Q(F;k-1)=Q(F;k), then J(F;k)=M∩J(F;k)⊆M. We obtain M⊆Q(F;k)=J(F;k)⊆M by Lemma 25; this implies that M=Q(F;k), a contradiction. Whence M=M∪Q(F;k-1) and so Q(F;k-1)⊆M, assume that Lα∩M≠∅ and Rα∩M≠∅ for all α∈J(F;k). We first prove that E(J(F;k))⊆M. Let e be an idempotent in J(F;k). Thus, Le∩M≠∅≠Re∩M. By Lemma 24, Le∩M and Re∩M contain idempotents, say f and g respectively. So, fLe and gRe. Since Lf∩Rg=Le∩Re contains the idempotent e, fg∈Rf∩Lg⊆J(F;k) by Lemma 23. Since fg∈M and M is regular, fg=(fg)h(fg) for some h∈M. We see that f(ghf)g=fg∈J(F;k), which implies that ghf∈J(F;k); that is, (gh)f≠0 in Pk. Then by Lemma 22, we get (gh)f∈Rgh∩Lf. Also, we have g(hf)∈Rg∩Lhf. Whence ghf∈Lf∩Rg=Le∩Re=He, the group H-class contains e. So, there exists a positive integer t such that e=(ghf)t∈M since g,h,f∈M. Therefore, E(J(F;k))⊆M. It follows from Lemma 27 that Q(F;k)=E(J(F;k))⊆M⊆Q(F;k); hence M=Q(F;k), a contradiction. Therefore, Lα∩M=∅ for some α∈J(F;k) or Rβ∩M=∅ for some β∈J(F;k). If M∩Lα=∅ for some α∈J(F;k), then(42)M⊆QF;k-1∪JF;k∖Lα⊊QF;k.Thus, M=Q(F;k-1)∪(J(F;k)∖Lα) by the maximality of M. If M∩Rβ=∅ for some β∈J(F;k), then(43)M⊆QF;k-1∪JF;k∖Rβ⊊QF;k.Thus, M=QF;k-1∪JF;k∖Rβ by the maximality of M.Notice that the number ofL-classes in J(F;k) equals rk and the number of R-classes in J(F;k) equals S(r,k)kn-r where S(r,k) is the Stirling number of the second kind; see [18, Section 2] for details. Therefore, the number of maximal regular subsemigroups of Q(F;k) is equal to rk+S(r,k)kn-r when 2≤k≤r-1.We shall normally writeTn instead of T(X). For 1≤k≤n, define(44)Jk=α∈Tn:Xα=k,Kn,k=α∈Tn:Xα≤k.It is well-known that K(n,k)=J1∪J2∪⋯∪Jk and K(n,k) is an ideal of Tn for all 1≤k≤n. If n=r, then Fn,r=Fn,n=Tn, J(F;k)=Jk, and Q(F;k)=K(n,k). Therefore, we establish the following corollary which first appeared in [8, Theorem 2].Corollary 33.
Each maximal regular subsemigroup ofK(n,k),2≤k≤n-1 must be one of the following forms:(1)
K(n,k-1)∪(Jk∖Lα);(2)
K(n,k-1)∪(Jk∖Rβ),
whereα,β∈Jk, Lα is the L-class containing α in Jk of Tn, and Rβ is the R-class containing β in Jk of Tn.We note that ifn=r, then we immediately obtain that the number of maximal regular subsemigroups of K(n,k) is equal to nk+S(n,k) when 2≤k≤n-1.



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