On transformation semigroups which are ℬ

ARE

Ꮾᏽ

-SEMIGROUPS

Received 23 December 2005; Revised 6 June 2006; Accepted 22 June 2006

A semigroup whose bi-ideals and quasi-ideals coincide is called aᏮᏽ-semigroup. The full transformation semigroup on a setXand the semigroup of all linear transformations of a vector spaceVover a fieldFinto itself are denoted, respectively, byT(X) andLF(V). It is known that every regular semigroup is aᏮᏽ-semigroup. Then bothT(X) andLF(V) areᏮᏽ-semigroups. In 1966, Magill introduced and studied the subsemigroupT(X,Y) ofT(X), where∅ =Y ⊆XandT(X,Y)= {α∈T(X)|Yα⊆Y}. IfW is a subspace of V, the subsemigroupLF(V,W) ofLF(V) will be defined analogously. In this paper, it is shown thatT(X,Y) is aᏮᏽ-semigroup if and only ifY=X,|Y| =1, or |X| ≤3, and LF(V,W) is aᏮᏽ-semigroup if and only if (i)W=V, (ii)W= {0}, or (iii)F=Z2, dimFV=2, and dimFW=1.

Copyright © 2006 Hindawi Publishing Corporation. All rights reserved. 1. Introduction

The cardinality of a setAis denoted by|A|. The image of a mapαatxin the domain of αwill be written byxα.

An elementaof a semigroupSis said to be regular ifa=abafor someb∈S, andSis called a regular semigroup if every element ofSis regular. The set of all regular elements ofSis denoted by Reg(S).

The full transformation semigroup on a nonempty setXis denoted byT(X), that is, T(X) is the semigroup of all mappings α:X→X under composition. The semigroup T(X) is known to be regular [4, page 4]. Magill [9] introduced and studied the subsemi-group

T(X,Y)=α∈T(X)|Yα⊆Y (1.1)

ofT(X), where∅ =Y⊆X. Note that 1X, the identity map onX, belongs toT(X,Y) and T(X,Y) containsT(X,Y) as a subsemigroup, whereT(X,Y)= {α∈T(X)|ranα⊆Y}

and ranαdenotes the range ofα. The semigroupT(X,Y) was introduced and studied by Symons [13].

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 12757, Pages1–10

For a vector spaceV over a fieldF, letLF(V) be the semigroup of all linear transfor-mationsα:V→V under composition. It is known thatLF(V) is a regular semigroup [5, page 63]. For a subspaceW ofV, we define the subsemigroupLF(V,W) ofLF(V) analogously, that is,

LF(V,W)=α∈LF(V)|Wα⊆W. (1.2)

Clearly, 1V∈LF(V,W) and 0, the zero map onV, also belongs toLF(V,W). In addition, LF(V,W) containsLF(V,W)= {α∈LF(V)|ranα⊆W}as a subsemigroup.

A subsemigroupQof a semigroupSis called a quasi-ideal ofSifSQ∩QS⊆Q, and a bi-ideal ofSis a subsemigroupBofSsuch thatBSB⊆B. The notions of quasi-ideal and bi-ideal for semigroups were introduced by Steinfeld [11] and Good and Hughes [3], respectively. Both quasi-ideals and bi-ideals are generalizations of one-sided ideals, and bi-ideals also generalize quasi-ideals. For a nonempty subsetAofS, let (A)qand (A)bbe the quasi-ideal and the bi-ideal ofSgenerated byA, respectively, that is, (A)q[(A)b] is the intersection of all quasi-ideals (bi-ideals) ofScontainingA[12, pages 10, 12]. Observe that (A)b⊆(A)q.

Proposition 1.1 [2, pages 84, 85]. For a nonempty subsetAof a semigroupS, (i) (A)q=S1A∩AS1,

(ii) (A)b=AS1A∪A.

Kapp [6] usedᏮᏽto denote the class of all semigroups whose bi-ideals and quasi-ideals coincide and Mielke [10] called a semigroup inᏮᏽaᏮᏽ-semigroup. Important Ꮾᏽ-semigroups are the following ones.

Proposition 1.2 [8]. Every regular semigroup is aᏮᏽ-semigroup.

Proposition 1.3 [6]. Every left (right) simple semigroup or every left (right) 0-simple semi-group is aᏮᏽ-semigroup.

Recall that a semigroup Sis left (right) simple if S has no proper left (right) ideal, and a semigroupSwith 0 is called left (right) 0-simple ifS2_{= {0}andShas no proper} nonzero left (right) ideal. Kemprasit showed in [7] that ifXis an infinite set, then the subsemigroup{α∈T(X)|Xranαis infinite}ofT(X) is aᏮᏽ-semigroup but it is nei-ther regular nor left (right) simple. In fact,Ꮾᏽ-semigroups have been characterized by Calais [1] as follows.

Proposition 1.4 [1]. A semigroupSis aᏮᏽ-semigroup if and only if (x,y)b=(x,y)qfor allx,y∈S.

that ifX is infinite, then the semigroup{α∈T(X)|Xranαis infinite}is a left ideal ofT(X). Similarly, ifV has infinite dimension overF, then the semigroup{α∈LF(V)|

dimF(V/ranα) is infinite}is a left ideal ofLF(V).

In Section 2, we give a necessary and sufficient condition forT(X,Y) to be aᏮᏽ -semigroup in terms of|X|and|Y|. InSection 3, a necessary and sufficient condition for LF(V,W) to be aᏮᏽ-semigroup is given in terms of|F|, dimFV, and dimFW.

In the remainder, letXbe a nonempty set,∅ =Y⊆X,Va vector space over a fieldF, andWa subspace ofV.

2. The semigroupT(X,Y)

We begin this section by characterizing regular elements of the semigroupT(X,Y). Then it is shown thatT(X,Y) is a regular semigroup if and only ifY=XorY contains only one element.

Proposition 2.1. The following statements hold for the semigroupT(X,Y). (i) Forα∈T(X,Y),α∈Reg(T(X,Y)) if and only if ranα∩Y=Yα. (ii) The semigroupT(X,Y) is regular if and only if eitherY=Xor|Y| =1.

Proof. (i) SinceYα⊆Y, we haveYα⊆ranα∩Y. Assume thatα=αβα for someβ∈ T(X,Y). If x∈ranα ∩ Y, thenx∈Y andx=aαfor somea∈X which imply that x=aα=aαβα=xβα∈Yβα⊆Yα. Hence we have ranα∩Y=Yα.

Conversely, assume that ranα∩Y=Yα. Then for eachx∈ranα∩Y, we havexα−1_∩

Y= ∅. We choose an elementx∈xα−1_{∩Yfor eachx∈ranα∩Y. Also, forx∈ranα}

Y, choose an element x∈xα−1_{. Thenxα=x for allx∈ranα∩Y andxα=xfor all} x∈ranαY. Letabe a fixed element inY and defineβ:X→Xby a bracket notation as follows:

β=

x t Xranα

x _{t a}

x∈ranα∩Y t∈ranαY.

(2.1)

ThenYβ⊆ {x_{|x∈ranα∩Y} ∪ {a} ⊆Y, and forx∈X,}

xαβα=(xα)βα=

⎧ ⎨ ⎩

(xα)α=xα ifxα∈ranα∩Y,

(xα)α=xα ifxα∈ranαY. (2.2)

Henceβ∈T(X,Y) andα=αβα.

(ii) Suppose thatYXand|Y|>1. Letaandbbe two distinct elements ofY. Define α:X→Xby

α=

Y XY

a b

. (2.3)

IfY =X, thenT(X,Y)=T(X) which is regular. Next, assume thatY= {c}. In this case,T(X,Y) is isomorphic to the semigroupP(XY) consisting of all partial transfor-mations ofXY, via the mapP(XY)→T(X,Y),α→α, where

α=

x Xdomα

xα c

x∈domα.

(2.4)

It is well known thatP(XY) is regular [4, page 4]. HenceT(X,Y) is a regular

semi-group, as required.

To characterize whenT(X,Y) is aᏮᏽ-semigroup, Propositions1.1,1.2,1.4, and2.1 and the following three lemmas are needed.

Lemma 2.2. LetSbe a semigroup. If∅ =A⊆Reg(S), then (A)b=(A)q.

Proof. We know that (A)b⊆(A)q. Let x∈(A)q. ByProposition 1.1(i), x=sa=bt for somes,t∈S1_{anda,b∈A. Sincea∈Reg(S),a=aaafor somea∈S. Then}

x=sa=saa_{a=btaa∈ASA⊆(A)}

b (2.5)

byProposition 1.1(ii). Hence we have (A)b=(A)q, as desired. Lemma 2.3. LetSbe a semigroup, let∅ =A⊆S, and letB⊆Reg(S). If (A)b=(A)q, then (A∪B)b=(A∪B)q.

Proof. We first show thatS1_A∩BS1_andS1_B∩AS1_{are subsets of (A∪B)}

b. Letx∈S1_A∩

BS1_{. Thenx=sa=btfor somes,t∈S}1_{,a∈A, andb∈B. Sinceb∈Reg(S),b=bbbfor} someb∈S. It follows that

x=bt=bb_{bt=bbsa∈BSA⊆(A∪B)S(A∪B)⊆(A∪B)b. (2.6)}

This shows thatS1_A∩BS1_⊆(A∪B)

b. It can be shown similarly thatS1_B∩AS1_⊆(A∪

B)b. Consequently,

(A∪B)q=S1(A∪B)∩(A∪B)S1

= S1_A∪S1_{B∩ AS}1_∪BS1

= S1_A∩AS1_{∪ S}1_A∩BS1_{∪ S}1_B∩AS1_{∪ S}1_B∩BS1

=(A)q∪ S1A∩BS1∪ S1B∩AS1∪(B)q

=(A)b∪ S1A∩BS1∪ S1B∩AS1∪(B)b, from the assumption andLemma 2.2,

⊆(A)b∪(A∪B)b∪(A∪B)b∪(B)b=(A∪B)b.

(2.7)

Lemma 2.4. If|X| =3 and|Y| =2, then for allα,β∈T(X,Y), (α,β)b=(α,β)qinT(X,Y). Proof. For convenience, letXadenote the constant map whose domain and range areX and{a}, respectively.

Assume thatX= {a,b,c}andY= {a,b}. Clearly,

T(X,Y)=

1X,Xa,Xb,

a b c a a b

,

a b c a a c

,

a b c b b a

,

a b c b b c

,

a b c a b a

,

a b c a b b

,

a b c b a a

,

a b c b a b

,

a b c b a c

.

(2.8)

ByProposition 2.1(i),T(X,Y)Reg(T(X,Y))= {[a b c_{a a b}], [a b c_{b b a}]}. Letλ=[a b c_{a a b}] andη=

[a b c_{b b a}]. Note that λ2_{=Xa=ηλandη}2_{=Xb=λη. To show that (α,β)}

b=(α,β)q for all α,β∈T(X,Y), byLemma 2.3, it suffices to show that (λ)b=(λ)q, (η)b=(η)q, and (λ, η)b=(λ,η)q. By direct multiplication, we have

T(X,Y)λ=λ,Xa, λT(X,Y)=λ,Xa,Xb,η, λT(X,Y)λ=Xa, T(X,Y)η=η,Xb, ηT(X,Y)=η,Xa,Xb,λ, ηT(X,Y)η=Xb,

λT(X,Y)η=Xb, ηT(X,Y)λ=Xa.

(2.9)

Hence

(λ)b=λT(X,Y)λ∪ {λ} =Xa,λ=T(X,Y)λ∩λT(X,Y)=(λ)q, (η)b=ηT(X,Y)η∪ {η} =Xb,η=T(X,Y)η∩ηT(X,Y)=(η)q, (λ,η)b= {λ,η}T(X,Y){λ,η} ∪ {λ,η}

=λT(X,Y)λ∪λT(X,Y)η∪ηT(X,Y)λ∪ηT(X,Y)η∪ {λ,η} =Xa,Xb,λ,η,

(λ,η)q=T(X,Y){λ,η} ∩ {λ,η}T(X,Y)

= T(X,Y)λ∪T(X,Y)η∩ λT(X,Y)∪ηT(X,Y)

=λ,Xa,η,Xb=(λ,η)b.

(2.10)

Theorem 2.5. The semigroupT(X,Y) is aᏮᏽ-semigroup if and only if one of the following statements holds.

Proof. Assume that (i), (ii), and (iii) are false. ThenXY= ∅,|Y|>1, and|X|>3. Case 1 ((|Y| =2)). LetY= {a,b}. Since|X|>3,|XY|>1. Letc∈XY. ThenX

{a,b,c} = ∅. Defineα,β,γ∈T(X,Y) by

α=

a b c X{a,b,c}

b b a c

, β=

c x a x

x∈X{c}

, γ=

a b X{a,b}

b b c

.

(2.11)

Thenaαβ=b=aγα,bαβ=b=bγα,cαβ=a=cγα, and (X{a,b,c})αβ= {a} =(X

{a,b,c})γα=(X{a,b,c})α, soα=αβ=γα∈(α)q byProposition 1.1(i). Ifαβ∈(α)b, then byProposition 1.1(ii),αβ=αηαfor someη∈T(X,Y). Hence we havea=cαβ=

cαηα=(aη)α. This implies thataη=cwhich is contrary toa∈Y andc∈XY. Thus (α)b=(α)q, so byProposition 1.4,T(X,Y) is not aᏮᏽ-semigroup.

Case 2 ((|Y|>2)). Leta,b,cbe distinct elements ofY. Letα,β,γ∈T(X,Y) be defined by

α=

a Y{a} XY

b a c

, β=

a b x b a x

x∈X{a,b},

γ=

a Y{a} x

c a x

x∈XY.

(2.12)

Thenaαβ=a=aγα=aα, (Y{a})αβ= {b} =(Y{a})γα, and (XY)αβ= {c} =

(XY)γα. Thusα=αβ=γα∈(α)q. Ifαβ∈(α)b, thenαβ=αηαfor someη∈T(X,Y). Therefore we have for everyx∈XY,c=xαβ=xαηα=(cη)αwhich implies thatcη∈

XY. This is a contradiction sincec∈Y. Hence (α)b=(α)q, and so byProposition 1.4, T(X,Y) is not aᏮᏽ-semigroup.

If Y =X or|Y| =1, thenT(X,Y) is regular byProposition 2.1(ii) which implies by Proposition 1.2 that T(X,Y) is a Ꮾᏽ-semigroup. If |X| =3 and |Y| =2, then by Lemma 2.4andProposition 1.4,T(X,Y) is aᏮᏽ-semigroup.

Hence the theorem is completely proved.

Two direct consequences of Propositions1.2,2.1(ii),Theorem 2.5, and the proof of Lemma 2.4are as follows.

Corollary 2.6. If|X| =3, then the following statements are equivalent. (i)T(X,Y) is aᏮᏽ-semigroup.

(ii)Y=Xor|Y| =1.

(iii)T(X,Y) is a regular semigroup.

Corollary 2.7. The semigroupT(X,Y) is a nonregularᏮᏽ-semigroup if and only if|X| =

Therefore we deduce fromCorollary 2.7that if|X| =3 and|Y| =2, thenT(X,Y) is an example ofᏮᏽ-semigroup which is neither regular nor left (right) simple (see Proposi-tions1.2and1.3).

3. The semigroupLF(V,W)

In this section, we give a necessary and sufficient condition forLF(V,W) to be aᏮᏽ -semigroup. We first provide the conditions of the regularity of elements ofLF(V,W) and of the semigroupLF(V,W). The following facts about vector spaces and linear trans-formations will be used. IfU1 andU2are subspaces ofV,B1 is a basis of the subspace

U1∩U2,B2⊆U1andB3⊆U2are such thatB1∪B2andB1∪B3are bases ofU1andU2, respectively, thenB1∪B2∪B3is a basis of the subspaceU1+U2ofV. Ifα∈LF(V),B1is a basis of kerα,B2is a basis of ranα, and choose an elementu∈uα−1for everyu∈B2, thenB1∪ {u|u∈B2}is a basis ofV.

Proposition 3.1. The following statements hold for the semigroupLF(V,W). (i) Forα∈LF(V,W),α∈Reg(LF(V,W)) if and only if ranα∩W=Wα. (ii) The semigroupLF(V,W) is regular if and only if eitherW=V orW= {0}. Proof. (i) The proof thatα∈Reg(LF(V,W)) implies ranα∩W=Wαis analogous to the proof of the “only if ” part ofProposition 2.1(i).

Conversely, assume that ranα∩W=Wα. LetB1be a basis of ranα∩W,B2⊆ranα

B1, andB3⊆WB1such thatB1∪B2andB1∪B3are bases of ranαandW, respectively. ThenB1∪B2∪B3is a basis of ranα+W. LetB4⊆V(B1∪B2∪B3) be such thatB1∪

B2∪B3∪B4 is a basis ofV. SinceB1⊆ranα∩W=Wα, we haveuα−1∩W= ∅for everyu∈B1. For eachu∈B1, choose an elementu∈uα−1∩W. SinceB2⊆ranα, for eachv∈B2,vα−1= ∅, so choose an elementv∈vα−1. Defineβ∈LF(V) on the basis

B1∪B2∪B3∪B4by

β=

u v B3∪B4

u _{v 0}

u∈B1

v∈B2.

(3.1)

It follows thatWβ= B1∪B3β= {u|u∈B1} ⊆W, soβ∈LF(V,W). LetB0 be a basis of kerα. ThenB0∪ {u|u∈B1} ∪ {v|v∈B2}is a basis ofV. Since

B0αβα= {0} =B0α, uαβα=uβα=uα ∀u∈B1,

vαβα=vβα=vα ∀v∈B2,

(3.2)

we haveα=αβα, soαis a regular element ofLF(V,W).

(ii) Assume that{0} =WV. LetB1be a basis ofWandBa basis ofV containing

B1. ThenB1= ∅ =BB1. Letw∈B1andu∈BB1. Defineα∈LF(V) by

α=

u B{u}

w 0

ThenWα= B1α⊆ B{u}α= {0}, soα∈LF(V,W). Since ranα∩W= w = {0} =Wα, by (i), we deduce thatαis not a regular element ofLF(V,W). HenceLF(V,W) is not a regular semigroup.

SinceLF(V,V)=LF(V)=LF(V,{0}), the converse holds.

To prove the main theorem, the following lemma is also needed. Lemma 2.3 and Proposition 3.1(i) are useful to obtain this result.

Lemma 3.2. IfF=Z2, dimFV=2, and dimFW=1, then for allα,β∈LF(V,W), (α,β)b= (α,β)qinLF(V,W).

Proof. Let{w} be a basis of W and {w,u} a basis ofV. Since F=Z2, it follows that

W= {0,w}andV= {0,w,u,u+w}. Clearly, both{u,u+w}and{w,u+w}are also bases ofV. Thusw ∩ u = w ∩ u+w = u ∩ u+w = {0}. All elements ofLF(V,W) defined on the basis{w,u}ofVcan be given as follows:

LF(V,W)=

0, 1V,

w u

0 w

,

w u

0 u

,

w u

0 w+u

,

w u

w 0

,

w u w w

,

w u

w w+u

.

(3.4)

ByProposition 3.1(i), LF(V,W)Reg(LF(V,W))= {[w u0 w]}. Let λ=[w u0w]. Note that λ2_{=0. To prove the lemma, byLemma 2.3, it suffices to show that (λ)}

b=(λ)q. By direct multiplication, we have

LF(V,W)λ= {0,λ}, λLF(V,W)= {0,λ}, λLF(V,W)λ= {0}. (3.5)

Consequently, (λ)b=λLF(V,W)λ∪ {λ} = {0,λ} =LF(V,W)λ∩ λLF(V,W)=(λ)q. Theorem 3.3. The semigroupLF(V,W) is aᏮᏽ-semigroup if and only if one of the follow-ing statements holds.

(i)W=V. (ii)W= {0}.

(iii)F=Z2, dimFV=2, and dimFW=1.

Proof. Assume that (i), (ii), and (iii) are false. Then (1){0} =WV and (2)F=Z2, dimFV >2, or dimFW >1. LetB1be a basis ofWandBa basis ofVcontainingB1. Then

B1= ∅andBB1= ∅.

Case 1 ((F=Z2)). Leta∈F{0, 1},w∈B1, andu∈BB1. Defineα,β,γ∈LF(V,W) by

α=

u B{u}

w 0

, β=

w B{w}

aw 0

, γ=

u B{u}

au 0

. (3.6)

for someη∈LF(V,W). Thenaw=uαβ=uαηα=(wη)α. Butwη∈WandWα= B1α⊆ B{u}α= {0}, soaw=0 which is contrary toa=0. Thus (α)q=(α)b, soLF(V,W) is not aᏮᏽ-semigroup byProposition 1.4.

Case 2 ((dimFW >1)). Then |B1|>1. Letw1,w2∈B1 be such that w1=w2 andu∈

BB1. Defineα,β,γ∈LF(V,W) by

α=

w1 u B

w1,u

w2 w1 0

, β=

w1 B

w1

w1 0

, γ=

u B{u}

u 0

. (3.7)

Then αβ=[_wu B₁ 0{u}]=γα=α, so αβ∈(α)q. If αβ∈(α)b, then αβ=αηα for some

η∈LF(V,W). Thusw1=uαβ=uαηα=(w1η)α. Sincew1η∈W= B1, we havew1η=

aw1+vfor some a∈F andv∈ B1{w1}. ButB1{w1} ⊆B{w1,u}, sovα=0. Consequently,w1=(aw1+v)α=aw2which is contrary to the independence ofw1and

w2. ByProposition 1.4,LF(V,W) is not aᏮᏽ-semigroup.

Case 3 ((dimFV >2 and dimFW=1)). Then|B1| =1 and|BB1|>1. LetB1= {w} andu1,u2∈BB1be such thatu1=u2. Letα,β,γ∈LF(V,W) be defined by

α=

u1 u2 B

u1,u2

w u1 0

, β=

w B{w}

w 0

, γ=

u1 B

u1

u1 0

.

(3.8)

Then we haveαβ=[u1B{u1}

w 0 ]=γα=α, soαβ∈(α)q. Suppose thatαβ∈(α)b. It follows thatαβ=αηαfor someη∈LF(V,W). Thusw=u1αβ=u1αηα=(wη)α. Butwη∈W= wandwα=0, sow=(wη)α=0, a contradiction. Hence (α)q=(α)b, soLF(V,W) is not aᏮᏽ-semigroup, as before.

For the converse, if (i) or (ii) holds, thenLF(V,W)=LF(V) which is aᏮᏽ-semigroup byProposition 1.2. If (iii) holds, thenLF(V,W) is aᏮᏽ-semigroup byProposition 1.4

andLemma 3.2.

The following corollaries follow directly from Propositions1.2,3.1(ii),Theorem 3.3, and the proof ofLemma 3.2.

Corollary 3.4. IfF=Z2or dimFV=2, then the following statements are equivalent. (i)LF(V,W) is aᏮᏽ-semigroup.

(ii)W=VorW= {0}.

(iii)LF(V,W) is a regular semigroup.

Corollary 3.5. The semigroup LF(V,W) is a nonregularᏮᏽ-semigroup if and only if F=Z2, dimFV=2, and dimFW=1. Hence ifF=Z2and dimFV=2, there are exactly 3 semigroupsLF(V,W) which are nonregularᏮᏽ-semigroups, and each of suchLF(V,W) contains 8 elements.

Acknowledgments

We wish to express our appreciation to the referees for helpful comments and suggestions on the original manuscript.

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S. Nenthein: Department of Mathematics, Faculty of Science, Chulalongkorn University, Bangkok 10330, Thailand

E-mail address:[email protected]

Y. Kemprasit: Department of Mathematics, Faculty of Science, Chulalongkorn University, Bangkok 10330, Thailand