Transitive Compatible Pair Congruence

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Vol. 9, No. 1, 2015, 119-124

ISSN: 2279-087X (P), 2279-0888(online) Published on 30 January 2015

www.researchmathsci.org

119

Transitive Compatible Pair Congruence

Pei Yang1, Zhenlin Gao2 Hai-zhen Liu3 and Jing Zhang4 1

College of Science, University of Shanghai for Science and Technology Shanghai-200093, China. Email: [email protected]

2

3

4

College of Science, Dalian University of Technology Dalian-116024, China. Email: [email protected]

Corresponding author: Pei Yang

Received 19 January 2015; accepted 28 January 2015

Abstract. In this paper, we introduce the concepts of transitive compatible pair congruence

ρ

₍_{H K}_, ₎. Prove that any congruence on a semigroup S can be expressed as a transitive compatible pair congruence.

Keywords: Semigroup ; Transitive compatible pair; congruence AMS Mathematics Subject Classification (2010): 20M10 1. Introduction and preliminaries

The Green's star relations L∗，，，，R∗ D∗ H,∗ J ∗on semigroup S are defined as follows.

( , )a b ∈L∗⇔ 1

(∀x y, ∈S ) ax=ay⇔bx=by;

( , )a b ∈R∗⇔ (∀x y, ∈S1) xa= ya⇔ xb =yb;

D∗=L ∗∨ R∗ ，H ∗=L∗∧ R∗，a J ∗b⇔ J∗( )a =J b∗( ).

whereJ∗( )(x x∈S)represents the minimalJ ∗-ideal generated by x.

Definition 1.1. Assume that P represents a kind property about semigroups, if the semigroup S has P property, then S is called a P semigroup. A congruence ρ on the semigroup S is called P congruence, if S ρ is a P semigroup.

We shall call a semigroup S a rpp semigroup, if every L∗-class of S contains idempotents of S and for any a∈S, *

( a )

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The following basic conclusions are needed throughout this paper.

Lemma 1.1. Let S be a semigroup, for anya∈S,e∈E S( )the following statements are equivalent:

(1) aL ∗e(a R∗e)；

(2) ae=a ea( =a), and for anys t, ∈S1,as=at sa( =ta)implieses=et se( =te).

Lemma 1.2. (1)LandL∗both are right congruences on semigroupS;

(2)R⊆R∗，L⊆L∗，D⊆D H∗, ⊆H J∗, ⊆J∗； (3)R |Re ( )g S R L, |Re ( )g S L

∗ ₌ ∗ ₌

.

Lemma 1.3. (1) Band B has a semilattice decomposition Y

B J_α

α∈

= ∪ which rest with the

semilattice Y , whereJ_α

(

∀ ∈α Y

)

areJ-classes of B.

(2) Left regular bandIis a semilattice of left zero bandsL_α, whereL_αareL-classes of I.

2. Transitive compatible pair congruences

From this section, we always assume that semigroup S mentioned below contains zero element, unless otherwise stated.

Definition 2.1. Let H and K be two subsemigroups of semigroup S ,such that

S=H∪K,ρH andρK

be two relations on H and K respectively. We define a relation on S by

ρ

Hand

ρ

Kas follows:

( , ) { ( , ) | ( , ) or ( , ) }

H K

H K

H K _{a b} _S _S _{a b} _{a b}

ρ =ρ ∪ρ = ∈ × ∈ρ ∈ρ . (2.1)

If

ρ

Hand

ρ

Ksatisfy the following conditions:

( , )a b ∈ρH, ( , )b c ∈ρK ⇒( , )a c ∈ρ₍H K_, ₎ ; (2.2)

( )( , ) ( , ( , ) )

( , ) or ( , ) ,

(2.3) ( )( , ) , ( , ) ;

H K

x S a b resp a b

xa xb xa xb

or c H K xa c c xb

ρ ρ

∀ ∈ ∈ ∈

 _∈ _∈

 ⇒ 

∃ ∈ ∩ ∈ ∈



( , ) or ( , ) ,

( )( , ) , ( , )

_,

(2.4)

H K

ax bx ax bx

or c H K ax c c bx

ρ ρ

 _∈ _∈

 

∃ ∈ ∩ ∈ ∈



then,(H K, )is called a transitive compatible pair of

ρ

₍_{H K}_, ₎.

Theorem 2.1. Let ( ,H K be a transitive compatible pair of)

ρ

₍_{H K}_, ₎defined as Definition 2.1. Then the following conclusions hold :

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121 ( , )|

H

H K H

ρ =ρ or ρK =ρ₍_{H K}_, ₎|_K； (2.5) (2)

ρ

₍_{H K}_, ₎ is a congruence on S if and only if

ρ

Hand

ρ

Kare congruences on H and

K respectively.

Proof: (1)⇒. Assume that the relation

ρ

₍_{H K}_, ₎ is an equivalent relation on S, for any a∈H , ( , )a a ∈ρ₍_{H K}_, ₎ implies ( , ) H

a a ∈ρ , thus the relation ρH

is reflexive.

Let ( , )a b ∈ρH, by formula (2.1) we have( , )a b ∈

ρ

₍H K_, ₎, thus( , )b a ∈ρ(H K, ), namely

( , )b a ∈ρHby ( , )b a ∈ ×H Hand formula (2.1), this shows that the relation ρH is symmetric. Similarly, we have

( , ) ( , )

( , ), ( , )a b b c ∈ρH ⇒( , ),( , )a b b c ∈ρH K ⇒( , )a c ∈ρH K ⇒( , )a c ∈ρH. So we conclude that ρH

is an equivalent relation on H . Also, using the same argument as above, we obtain that ρK

is an equivalent relation on K . ⇐. Let ρ ρH, K

be equivalent relations on H and K respectively, for anya∈S,

then a∈H or a∈K . Hence ( , )a a ∈ρH or ( , )a a ∈ρK , this implies ( , )a a ∈ρ₍_{H K}_, ₎by

formula (2.1), thus the relation ρ₍_{H K}_, ₎ is reflexive. Let( , )a b ∈ρ₍_{H K}_, ₎, then ( , )a b ∈ H

ρ or ( , ) K

a b ∈ρ by formula (2.1). Hence, we can infer that ( , ) H

b a ∈ρ or ( , ) K b a ∈ρ , this fact shows( , )b a ∈ρ₍_{H K}_, ₎.Namely,ρ₍_{H K}_, ₎is symmetric. Next, we shall illustrate that

(H K, )

ρ

istransitive. Let( , ), ( , )a b b c ∈ρ(H K, ), according to formula (2.1), there exist four

cases as follows:

(i) ( , ),( , )a b b c ∈ρH; (ii)( , )a b ,( , )b c ∈ρK; (iii) ( , )a b ∈ρH,( , )b c ∈ρK; (iv) ( , )a b ∈ρK, ( , )b c ∈ρH.

For case (i) (resp, case(ii)),we know ( , )a c ∈ρH(resp, ( , )a c ∈ρK), it allows that ( , )a c

(H K, )

ρ

∈ . Now we considers the last two cases. Since ( , )H K is a transitive compatible pair of

ρ

₍_{H K}_, ₎, by formula (2.2) we have( , )a c ∈ρ(H K, )by formula(2.1). Summing up the

above cases, we know that the relation

ρ

(H K, )is transitive. Consequently, the relation

(H K, )

ρ

is an equivalent relation onS. Finally, we shall prove formula (2.5). Clearly,

( , )|

H

H K H

ρ ⊆ρ , ( , )|

K

H K K

ρ ⊆ρ . We assume that ( , )|

K

H K K

ρ ⊂ρ , then we will illustrate ( , )|

H

H K H

ρ =ρ . Since

( , )| ( ( , )|₍ ₎)

K

H K K H K H K

ρ =ρ ∪ ρ ∩ , thus according to Definition 2.1 we have that

( , ) ( ( , )| ) ( ( , )| ) ( ( , )| ) ( , )

K H K

H K H K H H K K H K H H K

ρ = ρ ∪ ρ ⊇ ρ ∪ρ ⊇ρ ∪ρ =ρ . So, we can include that ( , )|

H

H K H

ρ =ρ , and ( , )|( ) H

H K H K

ρ ⊇ρ ∩ .

(2)⇐.

ρ

₍_{H K}_, ₎is an equivalent relation onS. Let( , )a b ∈ρ(H K, ), then ( , )a b ∈

H

ρ

or

( , )a b ∈

ρ

K. Now we consider the situation ( , ) H

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( , )a b ∈ρH and

ρ

His a congruence on H , thus we have (xa xb, )∈ρH, (ax bx, )∈

ρ

H. If x∈K, by formula (2.3) given in Definition 2.1, we have

( , ) H or ( , ) K, or ( )( , ) H,( , ) K xa xb ∈ρ xa xb ∈ρ ∃ ∈ ∩c H K xa c ∈ρ c xb ∈ρ . Hence, we obtain (xa xb, )∈

ρ

₍_{H K}_, ₎. Similarly, we have(ax bx, )∈

ρ

₍_{H K}_, ₎.

For the situation( , )a b ∈

ρ

K, similarly, we also have(xa xb, ), (ax bx, )∈

ρ

(H K, ).

Thus

ρ

₍_{H K}_, ₎ is a congruence onS.

⇒. Let ( , ) H

a b ∈ρ , then( , )a b ∈ρ(H K, ). Forh∈H, since H is a subsemigroup of S , thus (ha hb, ), (ah bh, )∈ ×H H.By the given condition,

ρ

₍_{H K}_, ₎ is a congruence onS, we

immediately have (ha hb, ), (ah bh, )∈

ρ

(H K, ) , therefore ( , ), ( , ) H ha hb ah bh ∈ρ by

formula(2.1) and (ha hb, ), (ah bh, )∈ ×H H. This shows that ρHis a congruence onH .

In a similar way, we also can obtain that ρK

is a congruence onK.

Definition 2.2. If the relation ( , )

H K

H K

ρ =ρ ∪ρ defined by formula (2.1) is a congruence on semigroupS,then we call

ρ

₍_{H K}_, ₎a transitive compatible pair congruence

onS.For the transitive compatible pair congruence ( , )

H K

H K

ρ =ρ ∪ρ onS, if at least one

of H and K is a proper subsemigroup ofSand H 1 K 1

Hor K

ρ ≠ ρ ≠ ,then we call

ρ

₍_{H K}_, ₎a

proper transitive compatible pair congruence onS,and (H K, )a proper transitive compatible pair of

ρ

₍_{H K}_, ₎.Otherwise, we call

ρ

₍_{H K}_, ₎trivial.

Obviously, by Definition 2.2, the Rees congruence =(ρ H×H)∪1_Sdetermined by a proper ideal is the proper transitive compatible pair congruence

ρ

(H S, )on S and ( , )H S is a proper transitive compatible pair ofρ.In order to study a nontrivial congruenceρ, we need to obtain some proper transitive compatible pair congruence representation of

ρ. So, we need to prove the following theorem.

Theorem 2.2. For any nontrivial congruenceρon semigroup S can be expressed as a

proper transitive compatible pair congruence

ρ

₍_{H K}_, ₎onS.

Proof: Let 1S ≠ ⊂ ×ρ S S.Now, we shall prove that ρcan be expressed as a proper transitive compatible pair congruence

ρ

₍_{H K}_, ₎onS.

Denote the zero element of S

ρ

by

__

0 , let

__

0 ={_ρ a∈S a| ∈0 },

__ __

={ | 0 } { 0 } K x∈S ≠ ∈x S ρ ∪ , __

0 { | is not a zero divisor of }

K = x∈K x S ρ , H=0_ρ∪(K K\ ₀).

IfK0≠∅, then for anyx y, ∈K0,

_ _ ___ __

0

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123

0

K is a proper subsemigroup ofSwithout zero divisors. Since for anya∈0_ρ,x∈S,

__ _ _ ___

0= ⋅ =a x ax, therefore we have ax∈0_ρ. Similarly, we obtainxa∈0_ρ. Hence we can deduce that 0_ρis a proper ideal of S . Next, we shall prove that H is a subsemigroup

ofS.For anya b, ∈H, we distinguish two cases as follows. Case 1. At least one of elements a and b belong to 0_ρ;

Case 2. Elements a and bboth are inK K\ 0, i.e. a b, ∈K K\ 0.

For the former, since 0_ρis an ideal ofS, thenab∈0_ρ ⊆H. For the latter, leta b, ∈K

0

\K , if

___ __

0 ba= (or

___ __

0

ab= ), thenba(orab)∈K K\ ₀. If

___ __ ___ __

0, 0 ba≠ ab≠ , since

__ __

0 a≠ is a

zero divisor in S ρ, then there existsx∈K K\ ₀such that

_ _ ___ __

0 a x=ax= (or

_ _ __

0 x a= ).

Therefore

___ __ __ ___ __

0 ba x =b ax= (or

__ ___ ___ __ __

0

x ab=xa b = ), and consequently

___ __

0 ba≠ (or

___ __

0 ab≠ ) is a zero divisor in S ρ. We immediately have ba (or ab )∈K K\ 0. Hence, we obtain

thatHis a subsemigroup of S . In particular, letK₀= ∅, then K=K K\ ₀ is a proper subsemigroup of S.

Now we distinguish the following two situations.

(i) IfK0= ∅, then H=0ρ∪K=S. Since 0ρ is a proper ideal of S , it is obvious that

(0_ρ,K) ( |0_ρ) ( |K)

ρ = ρ ∪ ρ _{is a proper transitive compatible pair congruence on S} and

(0 ,K_ρ )

ρ ρ= _.

(ii) If K₀≠ ∅ , we shall prove that

0

(H K_{, 0}) ( | )H ( |K )

ρ = ρ ∪ ρ is a proper transitive compatible pair congruence on S . Clearly,

ρ

|_Hand

0 |K

ρ

are congruences on H and

0

K respectively. SinceH∩K0= ∅, thus formula (2.2) naturally holds.

Let ( , )a b ∈ρ|_H, for any x∈S, then x∈H orx∈K0. If x∈H, we have (xa xb, ),(ax,

) |_H

bx ∈ρ . Ifx∈K0, since ( , )a b ∈ρ|H, then ( , )a b ∈ρ, and therefore (xa xb, )∈ρ, (ax bx, )∈ρ. For the case (xa xb, )∈ρ, we will illustrate that either (xa xb, )∈ ×H H or

0 0

(xa,xb)∈K ×K . Ifxa∈K₀, letϑ:S→S ρ ϑ,x =xρbe the natural homomorphism.

We have 1

0

(( ) )

xb∈ xa ρ ϑ− ⊆K by (xa)ρ=(xb)ρ, hencexb∈K₀. If xa∈H, we also can get xb∈Husing the same means as above. Thus, we deduce that either (xa xb, )∈ ×H H or(xa xb, )∈K0×K0. Since (xa xb, )∈ρ, then either (xa xb, )∈ρ|Hor (xa xb, )∈ρ|K₀

by (xa xb, )∈ ×H Hor(xa,xb)∈K₀×K₀. Consider the other case (ax bx, )∈ρ, x∈S,we also obtain that either (ax bx, )∈ρ|_H or

0

(ax bx, )∈ρ|_K using the similar arguments. Similarly, for the situation

0

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Finally, we will prove ₍

0

, )

H K

ρ ρ= .Suppose that( , )a b ∈ρ, if ( , )a b ∉ρ|_K₀, since

0

H∩K = ∅ , then ( , )a b ∈ ×H H and consequently ( , )a b ∈ρ|_H . If ( , )a b ∉ρ|_H , similarly we can get

0

( , )a b ∈ρ|_K . This fact implies ₍ ₎ 0 ,

H K

ρ ρ

⊆ . On the contrary, let

( , ₀)

( , )a b ∈

ρ

_{H K} , then either ( , )a b ∈ρ|_H or

0

( , )a b ∈ρ|_K , namely ( , )a b ∈ρ . To

sum up, we know ₍ ₎ 0 , = _{H K}

ρ ρ

.

By Theorem 2.2, we know that transitive compatible pair congruence on semigroup

S is a kind of universal representation method fitting to any congruence onS . Concretely speaking, let H and K be subsemigroups of semigroup S , for a congruence

ρon S , so long as (H K, ) is a transitive compatible pair of ρ, then ρ can be represented as ρ ρ= ₍_{H K}_, ₎=( | )ρ _H ∪( | )ρ _K .

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