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Admissible congruences on type B semigroups

  • Chunhua Li EMAIL logo , Lingxiang Meng , Jieying Fang and Baogen Xu
Published/Copyright: December 31, 2022
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Open Mathematics
From the journal Open Mathematics Volume 20 Issue 1

Abstract

The main aim of this article is to study admissible congruences on a type B semigroup. First, we give characterizations of the minimum admissible congruence whose trace is a normal congruence on a type B semigroup. After obtaining some properties of admissible congruences with the same trace on a type B semigroup, we introduce the notion of a normal subsemigroup and give characterizations of the minimum and maximum admissible congruences whose kernels are normal subsemigroups. Finally, the concept of a congruence pair of a type B semigroup is given, and two congruences associated with the congruence pair are obtained.

Keywords: admissible congruences; trace; type B semigroups; congruence pairs
MSC 2010: 20M10; 06F05

1 Introduction

Since the establishment of the algebraic theory of semigroups, regular semigroups have been an important research topic. In recent years, as a class of generalized regular semigroups, abundant semigroups have attracted more and more attention from semigroup scholars (see [1,2,3, 4,5,6, 7,8,9, 10,11,12, 13,14]). In 1979, the concept of a type B semigroup was introduced by Fountain in [15]. Recently, many semigroup scholars investigated type B semigroups and obtained some results (see [6,7,8,10,12,13]). Congruences play an important role in characterizations of properties of semigroups. The study of congruences of semigroups has been deeply developed (see [1,2,3,7,16,17, 18,19,20]). The kernel trace method is very successful in the study of an inverse semigroup congruence. In 1954, the concept of a kernel normal system was proposed for the first time by Preston, and congruences on inverse semigroups are characterized by the kernel normal system. On the basis of a kernel normal system, Reilly and Scheiblich proposed concepts of kernels and traces of congruences on inverse semigroups and obtained a kernel trace method [21]. Petrich studied congruences on inverse semigroups by using the method of a kernel trace in [18]. El-Qalliali extends it to the study of congruences of ample semigroups in [1]. It is well known that ample semigroups and type B semigroups are generalizations of inverse semigroups in the range of abundant semigroups. Therefore, it is a natural thing to characterize congruences of a type B semigroup by using the above method. This article is committed to extending these results to type B semigroups.

We proceed as follows: Section 2 provides some basic notions and properties of abundant semigroups. In particular, some properties of type B semigroups are given. In Section 3, we consider normal congruences of a type B semigroup S . It is shown that the minimum admissible congruence on S whose restriction to the idempotent set E of S is a normal congruence π and a congruence containing any admissible congruence whose restriction to E is a normal congruence π . In Section 4, we characterize admissible congruences with the same trace. In Section 5, we investigate kernels of congruences on a type B semigroup and obtain equivalent sets of kernels of σ π and μ π . In Section 6, we give the notion of a normal subsemigroup N of a type B semigroup and obtain some interesting results. In Section 7, we introduce the definition of a congruence pair on a type B semigroup. In Section 8, a conclusion is given.

2 Preliminaries

Throughout this article, we use notions and notations of [15,22, 23,24]. For undefined concepts, the reader can refer to [24].

In the following, we first recall some basic properties of Green ∗ relations ℒ ∗ and ℛ ∗ . For convenience, ℒ a ∗ and ℛ a ∗ denote the ℒ ∗ -class and ℛ ∗ -class containing a , respectively; E ( S ) denotes the set of idempotents of S ; a + and a ∗ denote the typical idempotent of the ℒ ∗ -class and ℛ ∗ -class containing a , respectively.

Lemma 2.1

[23] Let S be a semigroup and a , b ∈ S . Then the following statements are equivalent:

  1. a ℒ ∗ b ( a ℛ ∗ b ) ;

  2. for all x , y ∈ S 1 , a x = a y ( x a = y a ) if and only if b x = b y ( x b = y b ) .

Corollary 2.2

[23] Let S be a semigroup and a , e = e 2 ∈ S . Then the following statements are equivalent:

  1. a ℒ ∗ e ( a ℛ ∗ e ) ;

  2. a e = a ( a = e a ) and for all x , y ∈ S 1 , a x = a y ( x a = y a ) implies e x = e y ( x e = y e ) .

Obviously, the relations ℒ ∗ and ℛ ∗ are right and left congruences on semigroup S , respectively.

Definition 2.1

[15] A semigroup S is rpp (resp., lpp) if each ℒ ∗ -class (resp., ℛ ∗ -class) of S contains an idempotent. A semigroup S is said to be abundant if it is both rpp and lpp.

Definition 2.2

[15] An rpp (resp., lpp) semigroup S is right adequate (resp., left adequate) if E ( S ) is a semilattice. A semigroup is said to be adequate if it is both left and right adequate.

As in [15], if S is a right adequate semigroup, by Proposition 1.3 [15], any ℒ ∗ -class of S contains a unique idempotent. Dually, any ℛ ∗ -class of a left adequate semigroup S contains a unique idempotent.

Definition 2.3

[15] A right adequate semigroup S is right type B, if it satisfies the following conditions:

  1. for all e , f ∈ E ( S 1 ) , a ∈ S , ( e f a ) ∗ = ( e a ) ∗ ( f a ) ∗ ;

  2. for all a ∈ S , e ∈ E ( S ) , if e ≤ a ∗ , then there is f ∈ E ( S 1 ) such that e = ( f a ) ∗ .

Dually, a left adequate semigroup S is left type B, if it satisfies the following conditions:

  1. for all e , f ∈ E ( S 1 ) , a ∈ S , ( a e f ) + = ( a e ) + ( a f ) + ;

  2. for all a ∈ S , e ∈ E ( S ) , if e ≤ a + , then there is f ∈ E ( S 1 ) such that e = ( a f ) + .

A semigroup is said to be type B if it is both left and right type B.

Lemma 2.3

[15] Let S be an adequate semigroup and a , b ∈ S . Then the following statements hold:

  1. a ℒ ∗ b if and only if a ∗ = b ∗ ;  a ℛ ∗ b if and only if a + = b + ;

  2. ( a b ) ∗ = ( a ∗ b ) ∗ ; ( a b ) + = ( a b + ) + ;

  3. ( a b ) ∗ b ∗ = ( a b ) ∗ ; a + ( a b ) + = ( a b ) + ;

  4. ( a e ) ∗ = a ∗ e ; ( e a ) + = e a + .

Recall from [22] that a congruence ρ on an adequate semigroup S is said to be admissible if for all a ∈ S , x , y ∈ S 1 ,

a x ρ a y ⇒ a ∗ x ρ a ∗ y , x a ρ y a ⇒ x a + ρ y a + .

Lemma 2.4

[22] Let ρ be an admissible congruence on an adequate semigroup S . If a , b are two elements of S such that a ρ b , then a ∗ ρ b ∗ and a + ρ b + .

Lemma 2.5

[22] Let ρ be a congruence on an abundant semigroup S. Then ρ is an admissible congruence on S if and only if a ρ ℒ ∗ a ∗ ρ and a ρ ℛ ∗ a + ρ for all a ∈ S .

Lemma 2.6

[14] Let ρ be an admissible congruence on an adequate semigroup S. Then

  1. S / ρ is a type B semigroup if S is type B;

  2. for all a ∈ S , ( a ρ ) ∗ = a ∗ ρ and ( a ρ ) + = a + ρ .

Let ρ be an admissible congruence on a type B semigroup S . We note that if x ρ is an idempotent in S / ρ , then there exists an idempotent e in S such that ( x , e ) ∈ ρ .

Definition 2.4

[22] A homomorphism θ from an adequate semigroup S onto T is said to be an admissible homomorphism if

a ℒ ∗ ( S ) b ⇒ a θ ℒ ∗ ( T ) b θ ; a ℛ ∗ ( S ) b ⇒ a θ ℛ ∗ ( T ) b θ .

Definition 2.5

[15] Let S be an adequate semigroup. The relation μ is defined as follows:

( a , b ) ∈ μ if and only if for all e ∈ E ( S ) , ( e a ) ∗ = ( e b ) ∗ and ( a e ) + = ( b e ) + .

As in [22], μ is the maximum congruence contained in ℋ ∗ . By [15], the congruence μ is an admissible congruence if any two elements of E ( S / μ ) commute. An adequate semigroup S is said to be fundamental if μ is the identity relation on S .

Lemma 2.7

[22] Let S be an adequate semigroup. If E ( S / μ ) is a semilattice, then E ( S / μ ) = { e μ ∣ e ∈ E ( S ) } .

Lemma 2.8

[14] Let S be a type B semigroup. If any two elements of E ( S / μ ) are commutative, then

  1. S / μ is a type B semigroup;

  2. S / μ is fundamental.

Lemma 2.9

[6] Let S be a type B semigroup. The relation σ is defined as follows:

( a , b ) ∈ σ ⇔ ( ∃ e ∈ E ( S ) ) e a e = e b e .

Then σ is the least cancellative monoid congruence on S.

Lemma 2.10

Let S be an adequate semigroup and E ( S / μ ) be a semilattice. Then S is a full subdirect product of a cancellative monoid and a fundamental type B semigroup if and only if S is a type B semigroup and σ ⋂ μ = ι S , where ι S is an identity relation on S.

Proof

Let C be a cancellative monoid and T be a fundamental type B semigroup. Put S = { ( a , t ) ∣ a ∈ C , t ∈ T } . Define a multiplication on S as follows:

( a , t ) ( b , x ) = ( a b , t x ) .

Clearly, S is a semigroup and E ( S ) = { ( 1 , e ) ∈ C × T ∣ e ∈ E ( T ) } is a semilattice. Now, we prove that S is a type B semigroup. Let ( c , t ) ∈ S such that ( c , t ) ( a , x ) = ( c , t ) ( b , y ) for all ( a , x ) , ( b , y ) ∈ S 1 . Then ( c a , t x ) = ( c b , t y ) . Hence, c a = c b and t x = t y . Note that C is a cancellative monoid and T is a fundamental type B semigroup. We have that a = b and t ∗ x = t ∗ y . Furthermore, ( 1 , t ∗ ) ( a , x ) = ( 1 , t ∗ ) ( b , y ) . But ( c , t ) = ( c , t ) ( 1 , t ∗ ) . Therefore, ( c , t ) ℒ ∗ ( 1 , t ∗ ) . Dually, ( c , t ) ℛ ∗ ( 1 , t + ) . Hence, S is an adequate semigroup. In other words, for all ( c , t ) ∈ S , we have that ( c , t ) ∗ = ( 1 , t ∗ ) and ( c , t ) + = ( 1 , t + ) . Let ( 1 , e ) , ( 1 , f ) ∈ E ( S ) and ( c , t ) ∈ S . Then

[ ( 1 , e ) ( 1 , f ) ( c , t ) ] ∗ = ( c , e f t ) ∗ = ( 1 , ( e f t ) ∗ ) = ( 1 , ( e t ) ∗ ( f t ) ∗ ) = ( 1 , ( e t ) ∗ ) ( 1 , ( f t ) ∗ ) = ( c , e t ) ∗ ( c , f t ) ∗ = [ ( 1 , e ) ( c , t ) ] ∗ [ ( 1 , f ) ( c , t ) ] ∗ .

Hence, S satisfies Condition (B1). Let ( 1 , e ) ∈ E ( S ) , ( c , t ) ∈ S . Then

( 1 , e ) ≤ ( c , t ) ∗ ⇒ ( 1 , e ) ≤ ( 1 , t ∗ ) ⇒ ( 1 , e ) = ( 1 , e ) ( 1 , t ∗ ) = ( 1 , t ∗ ) ( 1 , e ) ⇒ e = e t ∗ = t ∗ e ⇒ e ≤ t ∗ = ( ∃ f ∈ E ( T 1 ) ) e = ( f t ) ∗ ( by Condition ( B2 ) ) ⇒ ( 1 , e ) = ( 1 , ( f t ) ∗ ) = ( c , ( f t ) ∗ ) = [ ( 1 , f ) ( c , t ) ] ∗ ,

where ( 1 , f ) ∈ E ( S ) . Hence, S satisfies Condition (B2) and S is a right type B semigroup. Dually, we can prove that S is a left type B semigroup. This shows that S is a type B semigroup.

Let ( c , t ) , ( n , s ) ∈ S such that ( ( c , t ) , ( n , s ) ) ∈ σ ⋂ μ . Then there exists ( 1 , f ) ∈ E ( S ) such that

( 1 , f ) ( c , t ) ( 1 , f ) = ( 1 , f ) ( n , s ) ( 1 , f ) ,

where f ∈ E ( T ) . For any ( 1 , e ) ∈ E ( S ) , we have

[ ( 1 , e ) ( c , t ) ] ∗ = [ ( 1 , e ) ( n , s ) ] ∗ and [ ( c , t ) ( 1 , e ) ] + = [ ( n , s ) ( 1 , e ) ] + ,

where e ∈ E ( T ) . This shows that c = n , ( 1 , ( e t ) ∗ ) = ( 1 , ( e s ) ∗ ) and ( 1 , ( t e ) + ) = ( 1 , ( s e ) + ) . Therefore, c = n and ( t , s ) ∈ μ ( T ) . As T is fundamental, we have t = s . Thus, σ ⋂ μ = ι S .

Conversely, suppose that S is a type B semigroup. Then S / σ is a cancellative monoid and S / μ is a fundamental type B semigroup. Define a mapping as follows:

ψ : S → S / σ × S / μ , x ψ = ( x σ , x μ ) .

It is easy to see that ψ is a homomorphism. Let a μ be an idempotent of S / μ . Then there exists e ∈ E ( S ) such that ( e , a ) ∈ μ . But e σ is an identity of S / σ . Thus, e ψ = ( e σ , e μ ) and I m ψ is a full subdirect product of S / σ × S / μ . By the hypothesis, σ ⋂ μ = ι S . Therefore, ψ is one-to-one and S ≅ I m ψ .□

3 Normal congruences of a type B semigroup

In this section, we shall consider a normal congruence on a type B semigroup. For convenience, we replace E ( S ) by E in the remaining. As usual, for an arbitrary congruence ρ on a semigroup S , the restriction ρ ∣ E of ρ on the idempotent set E is called a trace of ρ , denoted by tr ρ . Obviously, tr ρ is a congruence on the idempotent set E and if e , f ∈ E with e ρ f and a ∈ S , we have that ( e a , f a ) ∈ ρ and ( a e , f e ) ∈ ρ . In particular, if ρ is an admissible, we have that ( e a ) ∗ ρ ( f a ) ∗ and ( a e ) + ρ ( a f ) + .

Definition 3.1

A congruence π on an idempotent set E of a type B semigroup S is said to be normal if for all e , f ∈ E and a ∈ S ,

e π f ⇒ ( e a ) ∗ π ( f a ) ∗ and ( a e ) + π ( a f ) + .

Definition 3.2

Let π be a normal congruence on E of a type B semigroup S . Define a relation on S as follows:

σ π = { ( a , b ) ∈ S × S ∣ ( ∃ e , f ∈ E ) e π a ∗ π b ∗ , f π a + π b + , f a e = f b e } .

We now give the main theorem in this section.

Theorem 3.1

Let S be a type B semigroup. Then σ π is the minimum congruence on S whose restriction to E is π . Furthermore, σ π is an admissible congruence on S.

Proof

First, we prove that σ π is an equivalence relation on S . Clearly, σ π is reflexive and symmetric. Now, we show that σ π is transitive. To see it, let a , b , c ∈ S such that ( a , b ) ∈ σ π and ( b , c ) ∈ σ π . Then there exist e , f ∈ E such that e π b ∗ π c ∗ , f π b + π c + and f a e = f b e ; and there exist g , h ∈ E such that g π b ∗ π c ∗ , h π b + π c + and h b g = h c g . Note that E is a semilattice. We have

f h a e g = h f a e g = h f b e g = f h b g e = f h c g e = f h c e g .

Since e π a ∗ π b ∗ and g π b ∗ π c ∗ , we have that e g π a ∗ b ∗ π a ∗ a ∗ = a ∗ and e g π b ∗ c ∗ π c ∗ c ∗ = c ∗ . That is, there exists e g ∈ E such that e g π a ∗ π c ∗ . Similarly, there exists f h ∈ E such that f h π a + π c + . Therefore, σ π is an equivalence relation on S .

Next, we prove that σ π is a congruence on S . Let a , b ∈ S such that ( a , b ) ∈ σ π . Then f a e = f b e for some e , f ∈ E , e π a ∗ π b ∗ and f π a + π b + . Hence, for all c ∈ S , c f a e = c f b e and c ∗ f a e = c ∗ f b e . Note that c ∗ f ≤ c ∗ . By Condition (B2), we have c ∗ f = ( g c ) ∗ for some g ∈ E 1 , and so ( g c ) ∗ a e = ( g c ) ∗ b e . Therefore, g c a e = g c b e . That is, g c + c a ( c a ) ∗ e = g c + c b ( c b ) ∗ e . Multiplying it on the left by c ∗ , we obtain that c ∗ g c + c a ( c a ) ∗ e = c ∗ g c + c b ( c b ) ∗ e . By the normality of π and Lemma 2.3(2), we have that

f π a + ⇒ ( c f ) + π ( c a + ) + = ( c a ) + .

Note that ℛ ∗ is a left congruence and S is ℛ ∗ -unipotent. We have that g c + = ( g c ) + . Again since c ∗ ℒ ∗ c and ℒ ∗ is a right congruence, we have that c ∗ ( g c ) + ℒ ∗ c ( g c ) + ℒ ∗ ( c ( g c ) + ) ∗ . Note that S is ℒ ∗ -unipotent. We have c ∗ ( g c + ) = ( c ( g c ) + ) ∗ . That is, c ∗ g c + π ( c ( g c ) + ) ∗ . Thus,

c ∗ g c + π ( c ( g c ) ∗ ) + π ( c c ∗ f ) + = ( c f ) + .

Similarly, ( c b ) + π c ∗ g c + . Therefore, c ∗ g c + π ( c a ) + π ( c b ) + . Clearly, c ∗ g c + c a ℒ ∗ ( c ∗ g c + c a ) ∗ . Again, since S is a right congruence, we have that

( c ∗ g c + c a ( c a ) ∗ e ) ∗ ℒ ∗ c ∗ g c + c a ( c a ) ∗ e ℒ ∗ ( c ∗ g c + c a ) ∗ ( c a ) ∗ e .

Hence, ( c ∗ g c + c a ( c a ) ∗ e ) ∗ = ( c ∗ g c + c a ) ∗ ( c a ) ∗ e since S is ℒ ∗ -unipotent. Again since e π a ∗ π b ∗ , we obtain

( c a ) ∗ e π ( c a ) ∗ a ∗ π ( c a ) ∗ and ( c b ) ∗ e π ( c b ) ∗ b ∗ π ( c b ) ∗ .

Thus, ( c ∗ g c + c a ) ∗ ( c a ) ∗ e π ( c ∗ g c + c a ) ∗ ( c a ) ∗ . But, c ∗ g c + π ( c a ) + . By the normality of π , ( c ∗ g c + c a ) ∗ ( c a ) ∗ π ( ( c a ) + c a ) ∗ ( c a ) ∗ . Therefore,

( c ∗ g c + c a ( c a ) ∗ e ) ∗ π ( ( c a ) + c a ) ∗ ( c a ) ∗ = ( c a ) ∗ .

Similarly, we can prove that ( c ∗ g c + c b ( c b ) ∗ e ) ∗ π ( c b ) ∗ . Since c ∗ g c + c a ( c a ) ∗ e = c ∗ g c + c b ( c b ) ∗ e . We obtain that ( c a ) ∗ π ( c b ) ∗ π ( c a ) ∗ e = e ( c a ) ∗ . But, g c a e = g c b e . Multiplying it on the left by c ∗ and on the right by ( c a ) + , we have that c ∗ g c + c a e ( c a ) ∗ = c ∗ g c + c b e ( c a ) ∗ . Thus, ( c a , c b ) ∈ σ π .

On the other hand, let a , b ∈ S such that ( a , b ) ∈ σ π . Then f a e = f b e for some e , f ∈ E , e π a ∗ π b ∗ , f π a + π b + and f a e c = f b e c . Hence, f a e c + = f b e c + . Note that e c + ≤ c + . By Condition ( B2 ′ ), there exists h ∈ E 1 such that e c + = ( c h ) + , and so f a ( c h ) + = f b ( c h ) + . Multiplying it on the right by c h , we obtain that f a c h = f b c h . That is, f ( a c ) + a c h c ∗ c + = f ( b c ) + b c h c ∗ c + . By the normality of π and Lemma 2.3(2),

e π a ∗ ⇒ ( e c ) ∗ π ( a ∗ c ) ∗ = ( a c ) ∗ .

Again since h c ∗ = c ∗ h ℒ ∗ c h ℒ ∗ ( c h ) ∗ and S is ℒ ∗ -unipotent, we have that h c ∗ = ( c h ) ∗ and so h c ∗ c + ℛ ∗ ( c h ) ∗ c ℛ ∗ ( ( c h ) ∗ c ) + . This gives that h c ∗ c + = ( ( c h ) ∗ c ) + since S is ℛ ∗ -unipotent. Thus,

h c ∗ c + π ( ( c h ) + c ) ∗ π ( e c + c ) ∗ = ( e c ) ∗ .

Therefore, h c ∗ c + π ( a c ) ∗ . Similarly, we can prove that h c ∗ c + π ( b c ) ∗ . Hence, h c ∗ c + π ( a c ) ∗ π ( b c ) ∗ . Obviously, a c h c ∗ c + ℛ ∗ ( a c h c ∗ c + ) + . Since ℛ ∗ is a left congruence, we have that

( f ( a c ) + a c h c ∗ c + ) + ℛ ∗ f ( a c ) + a c h c ∗ c + ℛ ∗ f ( a c ) + ( a c h c ∗ c + ) + .

Note that S is ℛ ∗ -unipotent. We have ( f ( a c ) + a c h c ∗ c + ) + = f ( a c ) + ( a c h c ∗ c + ) + . But f π a + π b + , we have that

f ( a c ) + π a + ( a c ) + π ( a c ) + and f ( b c ) + π b + ( b c ) + π ( b c ) + .

Thus, f ( a c ) + ( a c h c ∗ c + ) + π ( a c ) + ( a c h c ∗ c + ) + . Again h c ∗ c + π ( a c ) ∗ , by the normality of π , ( a c ) + ( a c h c ∗ c + ) + π ( a c ) + ( a c ( a c ) ∗ ) + . Therefore,

( f ( a c ) + a c h c ∗ c + ) + π ( a c ) + ( a c ( a c ) ∗ ) + = ( a c ) + .

Similarly, ( f ( b c ) + b c h c ∗ c + ) + π ( b c ) + . But, f ( a c ) + a c h c ∗ c + = f ( b c ) + b c h c ∗ c + . We have

( a c ) + π ( b c ) + π f ( a c ) + .

As f a c h = f b c h , so f ( a c ) + a c h c ∗ c + = f ( a c ) + b c h c ∗ c + . Thus, ( a c , b c ) ∈ σ π .

Summing up the above arguments, we conclude that σ π is a congruence on S .

Now, we prove that tr σ π = π . Clearly, tr σ π ⊆ π . Let e , f ∈ E such that e π f . Then f e e = f e = f f e . By the definition of σ π , we have e σ π f . Thus, tr σ π = π .

Next, we show that σ π is the minimum congruence whose restriction to E is π . Let ρ be a congruence whose restriction to E is π . If ( a , b ) ∈ σ π , then f a e = f b e for some e , f ∈ E , e π a ∗ π b ∗ and f π a + π b + . Since tr ρ = π , we have that e ρ a ∗ ρ b ∗ and f ρ a + ρ b + . Hence,

a ρ = a + a a ∗ ρ = a + ρ a ρ a ∗ ρ = f ρ a ρ e ρ = ( f a e ) ρ = ( f b e ) ρ = f ρ b ρ e ρ = b + ρ b ρ b ∗ ρ = b + b b ∗ ρ = b ρ .

Thus, ( a , b ) ∈ ρ . That is, σ π ⊆ ρ . Therefore, σ π is the minimum congruence whose restriction to E is π .

Finally, we prove that σ π is admissible. Let a ∈ S , s , t ∈ S 1 such that ( a s , a t ) ∈ σ π . Then f a s e = f a t e for some e , f ∈ E , e π ( a s ) ∗ π ( a t ) ∗ and f π ( a s ) + π ( a t ) + . By Lemma 2.3, ( a s ) ∗ = ( a ∗ s ) ∗ and ( a t ) ∗ = ( a ∗ t ) ∗ , and so e π ( a ∗ s ) ∗ π ( a ∗ t ) ∗ . But

f a s e = f a t e ⇒ f a a ∗ s e = f a a ∗ t e ⇒ ( f a ) ∗ a ∗ s e = ( f a ) ∗ a ∗ t e .

Thus, ( a ∗ s , a ∗ t ) ∈ σ π . Similarly, ( s a , t a ) ∈ σ π implies ( s a + , t a + ) ∈ σ π . Therefore, σ π is an admissible congruence. This completes the proof.□

Let π be an admissible congruence on S . By Theorem 3.1, we have that σ π is the minimum admissible congruence on S whose restriction to E is π . As usual, a congruence ρ on a semigroup S is called idempotent-separating if e ρ = f ρ implies e = f for all e , f ∈ E . Similarly, we can define an idempotent-separating homomorphism.

Proposition 3.2

Let ρ be an admissible congruence on a type B semigroup S whose restriction to E is π . Then S / ρ is an idempotent-separating homomorphic image of S / σ π .

Proof

Define a mapping ϕ as follows:

ϕ : S / σ π → S / ρ , ( s σ π ) ϕ = s ρ .

Then, it is easy to see that ϕ is a homomorphism of S / σ π onto S / ρ . By Lemma 2.7, we have that

E ( S / σ π ) = { e σ π ∣ e ∈ E } .

Let e σ π , f σ π be two idempotents of S / σ π , where e , f ∈ E . Then

( e σ π ) ϕ = ( f σ π ) ϕ ⇒ e ρ = f ρ ⇒ ( e , f ) ∈ ρ ⇒ ( e , f ) ∈ π ( tr ρ = π ) ⇒ e σ π = f σ π .

This means that ϕ is idempotent-separating. This completes the proof.□

Definition 3.3

Let S be a type B semigroup and π be a normal congruence on E . Define a relation on S as follows:

μ π = { ( a , b ) ∈ S × S ∣ ( ∀ e ∈ E ) ( e a ) ∗ π ( e b ) ∗ , ( a e ) + π ( b e ) + } .

Lemma 3.3

Let S be a type B semigroup and π be a normal congruence on E. Then the following statements are equivalent:

  1. ( a , b ) ∈ μ π ;

  2. for all e , f ∈ E , e π f implies ( e a ) ∗ π ( f b ) ∗ and ( a e ) + π ( b f ) + ;

  3. ( a σ π , b σ π ) ∈ μ ( S / σ π ) , where μ ( S / σ π ) denotes the relation μ on S / σ π .

Proof

( 1 ) ⇒ ( 2 ) is clear. Now, we show that ( 2 ) ⇒ ( 1 ) . For all b ∈ S , e , f ∈ E with e π f , we have ( e b ) ∗ π ( f b ) ∗ . If ( a , b ) ∈ μ π , then ( e a ) ∗ π ( e b ) ∗ . Hence, ( e a ) ∗ π ( f b ) ∗ . On the other hand, for all b ∈ S , e , f ∈ E with e π f , it follows that ( b e ) + π ( b f ) + . If ( a , b ) ∈ μ π , then ( a e ) + π ( b e ) + . Hence, ( a e ) + π ( b f ) + . That is, (2) holds.

( 1 ) ⇔ ( 3 ) For all a , b ∈ S , we have

□ ( a , b ) ∈ μ π ⇔ ( ∀ e ∈ E ) ( e a ) ∗ π ( e b ) ∗ , ( a e ) + π ( b e ) + ⇔ ( ∀ e ∈ E ) ( e a ) ∗ σ π = ( e b ) ∗ σ π , ( a e ) + σ π = ( b e ) + σ π ( since tr σ π = π ) ⇔ ( ∀ e ∈ E ) ( e σ π a σ π ) ∗ = ( e σ π b σ π ) ∗ ( a σ π e σ π ) + = ( b σ π e σ π ) + ( since σ π is admissible ) ⇔ ( a σ π , b σ π ) ∈ μ ( S / σ π ) .

Theorem 3.4

Let S be a type B semigroup. Then tr μ π = π . In particular, if ρ is an admissible congruence on S with tr ρ = π , then ρ ⊆ μ π .

Proof

Obviously, μ π is an equivalence relation on S . Now, we prove that μ π is a congruence on S . For any a , b , c ∈ S , e ∈ E , if ( a , b ) ∈ μ π , then ( e a ) ∗ π ( e b ) ∗ . By the normality of π and Lemma 2.3(2), we have that

( e a ) ∗ π ( e b ) ∗ ⇒ ( ( e a ) ∗ c ) ∗ π ( ( e b ) ∗ c ) ∗ ⇒ ( e a c ) ∗ = ( e a ) ∗ c ∗ π ( ( e b ) ∗ c ) ∗ = ( e b c ) ∗ .

Hence, ( e a c ) ∗ π ( e b c ) ∗ . Since ( c e ) + ∈ E , we have ( a ( c e ) + ) + π ( b ( c e ) + ) + from the definition of μ π . By Lemma 2.3(2), ( a ( c e ) + ) + = ( a c e ) + and ( b ( c e ) + ) + = ( b c e ) + . That is, ( a c e ) + π ( b c e ) + . Thus, ( a c , b c ) ∈ μ π . On the other hand, for all a , b , c ∈ S , e ∈ E , if ( a , b ) ∈ μ π , then ( a e ) + π ( b e ) + . By the normality of π and Lemma 2.3(2), we have

( a e ) + π ( b e ) + ⇒ ( c ( a e ) + ) + π ( c ( b e ) + ) + ⇒ ( c a e ) + π ( c b e ) + .

Since ( e c ) ∗ ∈ E , we obtain ( ( e c ) ∗ a ) ∗ π ( ( e c ) ∗ b ) ∗ . But ( ( e c ) ∗ a ) ∗ = ( e c a ) ∗ and ( ( e c ) ∗ b ) ∗ = ( e c b ) ∗ . This shows that ( e c a ) ∗ = ( e c b ) ∗ . Thus, ( c a , c b ) ∈ μ π . To sum up, μ π is a congruence.

Next, we prove that tr μ π = π . It is easy to see that π ⊆ μ π . For all f , g ∈ E with f μ π g , then e f π e g for all e ∈ E . Let e = f and e = g , respectively. This gives that f π f g and g f π g . As f g = g f . So f π g . Thus, tr μ π = π .

Finally, we show that μ π is the maximum admissible congruence whose trace is π . To see it, let ρ be any admissible congruence on S such that tr ρ = π . If ( a , b ) ∈ ρ , then ( e a , e b ) ∈ ρ and ( a e , b e ) ∈ ρ for all e ∈ E . By Lemma 2.4,

( ( e a ) ∗ , ( e b ) ∗ ) ∈ ρ and ( ( a e ) + , ( b e ) + ) ∈ ρ .

Hence, ( e a ) ∗ π ( e b ) ∗ , ( a e ) + π ( b e ) + . That is, ( a , b ) ∈ μ π . This completes the proof.□

4 Congruences with same trace on a type B semigroup

In this section, we mainly describe the relationship between two admissible congruences ρ and τ on a type B semigroup S , which have the same trace. Clearly, if we restrict ρ to E , then tr ρ is normal, and σ tr ρ , μ tr ρ are the minimum and the maximum admissible congruences on S , respectively. In particular, tr σ tr ρ = tr ρ = tr μ tr ρ , where

σ tr ρ = { ( a , b ) ∈ S × S ∣ a ∗ ρ b ∗ , a + ρ b + , ( ∃ e ∈ a ∗ ρ ∩ E , ∃ f ∈ a + ρ ∩ E ) f a e = f b e }

μ tr ρ = { ( a , b ) ∈ S × S ∣ ( ∀ e ∈ E ) ( e a ) ∗ π ( e b ) ∗ , ( a e ) + π ( b e ) + } .

For convenience, we denote σ tr ρ by σ ρ and denote μ tr ρ by μ ρ .

Theorem 4.1

Let ρ be an arbitrary admissible congruence on a type B semigroup S. Then σ ρ ⊆ ρ ⊆ μ ρ and tr σ ρ = tr ρ = tr μ ρ .

Proof

It follows directly from Theorems 3.1 and 3.4.□

Definition 4.1

Let ρ and τ be two congruences on a type B semigroup S with τ ⊆ ρ . Define a congruence ρ / τ on S / τ as follows:

( a , b ∈ S ) a τ ( ρ / τ ) b τ ⇔ a ρ b .

Theorem 4.2

Let ρ and τ be two admissible congruences on a type B semigroup S. Then the following statements are equivalent:

  1. tr ρ = tr τ ;

  2. ρ ⊆ μ τ and μ τ / ρ = μ ( S / ρ ) ;

  3. ( ∀ a , b ∈ S ) a ρ μ ( S / ρ ) b ρ ⇔ a τ μ ( S / τ ) b τ ;

  4. ( ∀ a , b ∈ S ) a ρ ℋ ∗ ( S / ρ ) b ρ ⇔ a τ ℋ ∗ ( S / τ ) b τ ;

  5. ρ ∩ τ ∣ e ρ and ρ ∩ τ ∣ e τ are cancellative congruences, where e ∈ E ;

  6. ρ / ρ ∩ τ and τ / ρ ∩ τ are congruences contained in ℋ ∗ ( S / ρ ∩ τ ) .

Proof

(1) ⇒ (2) Since μ tr ρ = μ ρ , μ tr τ = μ τ and tr ρ = tr τ , we have μ ρ = μ τ . Thus, ρ ⊆ μ τ . For all a , b ∈ S , it follows that

a ρ ( μ τ / ρ ) b ρ ⇔ a ρ ( μ ρ / ρ ) b ρ ⇔ a μ ρ b ⇔ ( ∀ e ∈ E ) ( e a ) ∗ ρ ( e b ) ∗ , ( a e ) + ρ ( b e ) + ⇔ ( a ρ , b ρ ) ∈ μ ( S / ρ ) .

(2) ⇒ (1) It is easy to see that tr ρ ⊆ tr μ τ ⊆ tr τ . For all e , f ∈ E , we have

e τ f ⇒ e μ τ f ⇒ e ρ ( μ τ / ρ ) f ρ ⇒ e ρ = f ρ ⇒ e ρ f .

That is, tr τ ⊆ tr ρ . Thus, tr τ = tr ρ .

(1) ⇒ (3) For all a , b ∈ S , we have

a ρ μ ( S / ρ ) b ρ ⇔ ( ∀ e ∈ E ) ( e a ) ∗ ρ = ( e b ) ∗ ρ , ( a e ) + ρ = ( b e ) + ρ ⇔ ( ∀ e ∈ E ) ( e a ) ∗ τ = ( e b ) ∗ τ , ( a e ) + τ = ( b e ) + τ ⇔ a τ μ ( S / τ ) b τ .

(3) ⇒ (1) For all e , f ∈ E , it follows that

e ρ f ⇔ e ρ = f ρ ⇔ e ρ μ ( S / ρ ) f ρ ⇔ e τ μ ( S / τ ) f τ ⇔ e τ = f τ ⇔ e τ f .

(1) ⇒ (4) For all a , b ∈ S with a ρ ℋ ∗ ( S / ρ ) b ρ . We have a ρ ℒ ∗ ( S / ρ ) b ρ and a ρ ℛ ∗ ( S / ρ ) b ρ . By the hypothesis, ρ is admissible. Thus, by Lemma 2.5, we have a ∗ ρ ℒ ∗ ( S / ρ ) b ∗ ρ and a + ρ ℛ ∗ ( S / ρ ) b + ρ . By Lemma 2.6(1), S / ρ is a type B semigroup, it shows that S / ρ is ℒ ∗ -unipotent and ℛ ∗ -unipotent. Thus, a ∗ ρ = b ∗ ρ and a + ρ = b + ρ . But, tr ρ = tr τ , this shows that a ∗ τ = b ∗ τ and a + τ = b + τ . Therefore, a τ ℋ ∗ ( S / τ ) b τ . Similarly, a τ ℋ ∗ ( S / τ ) b τ implies a ρ ℋ ∗ ( S / ρ ) b ρ .

(4) ⇒ (1) For all e , f ∈ E with e ρ f , we have e ρ ℋ ∗ ( S / ρ ) f ρ , and so e τ ℋ ∗ ( S / τ ) f τ . Hence, e τ = f τ . That is, e τ f . Similarly, e τ f implies e ρ f .

(1) ⇒ (5) Clearly, e ρ is an adequate semigroup, where e ∈ E . Let a , b , c ∈ e ρ and ( a b , a c ) ∈ ρ ∩ τ . By the hypothesis, ρ and τ are admissible congruences. Hence, ( a ∗ b , a ∗ c ) ∈ ρ ∩ τ . Since e ρ is an adequate semigroup, we have a ∗ , b + ∈ e ρ . Thus, ( a ∗ , b + ) ∈ ρ ∩ τ and ( b , a ∗ b ) ∈ ρ ∩ τ . Similarly, we can show ( a ∗ , c + ) ∈ ρ ∩ τ such that ( a ∗ c , c ) ∈ ρ ∩ τ . According to transitivity, ( b , c ) ∈ ρ ∩ τ . Therefore, ρ ∩ τ is left cancellative. Similarly, ρ ∩ τ is right cancellative. Thus, ρ ∩ τ ∣ e ρ is a cancellative congruence. Dually, we can prove that ρ ∩ τ ∣ e τ is a cancellative congruence.

(5) ⇒ (1) Let g , h ∈ e ρ ∩ E . Clearly, g g h = g h , where g , h , g h ∈ e ρ ∩ E . It follows that ( g g h , g h ) ∈ ρ ∩ τ . By the hypothesis, ρ ∩ τ is a cancellative congruence. Hence, ( g h , h ) ∈ ρ ∩ τ . Similarly, we have ( h g , g ) ∈ ρ ∩ τ . As g h = h g , we have ( g , h ) ∈ ρ ∩ τ . In particular, for all e , f ∈ E ,

e ρ f ⇒ f ∈ e ρ ⇒ ( e , f ) ∈ ρ ∩ τ ⇒ ( e , f ) ∈ τ ⇒ e τ f .

Similarly, e τ f ⇒ e ρ f is also true.

(1) ⇒ (6) For any a , b ∈ S , we obtain

a ( ρ ∩ τ ) ρ / ρ ∩ τ b ( ρ ∩ τ ) ⇒ a ρ b ⇒ a ∗ ρ b ∗ , a + ρ b + ⇒ a ∗ ( ρ ∩ τ ) ρ / ρ ∩ τ b ∗ ( ρ ∩ τ ) a + ( ρ ∩ τ ) ρ / ρ ∩ τ b + ( ρ ∩ τ ) ⇒ a ( ρ ∩ τ ) ℒ ∗ ( S / ρ ∩ τ ) b ( ρ ∩ τ ) a ( ρ ∩ τ ) ℛ ∗ ( S / ρ ∩ τ ) b ( ρ ∩ τ ) ⇒ a ( ρ ∩ τ ) ℋ ∗ ( S / ρ ∩ τ ) b ( ρ ∩ τ ) .

Hence, ρ / ρ ∩ τ ⊆ ℋ ∗ ( S / ρ ∩ τ ) . Similarly, τ / ρ ∩ τ ⊆ ℋ ∗ ( S / ρ ∩ τ ) .

(6) ⇒ (1) For all e , f ∈ E , we have

e ρ f ⇒ e ( ρ ∩ τ ) ρ / ρ ∩ τ f ( ρ ∩ τ ) ⇒ e ( ρ ∩ τ ) ℋ ∗ ( S / ρ ∩ τ ) f ( ρ ∩ τ ) ⇒ e ( ρ ∩ τ ) = f ( ρ ∩ τ ) ⇒ e τ f .

Similarly, e τ f implies e ρ f . This completes the proof.□

Corollary 4.3

Let ρ be an admissible congruence on a type B semigroup S. Then ρ = μ ρ if and only if S / ρ is fundamental.

Proof

By Theorem 4.2,

ρ = μ ρ ⇔ μ ρ / ρ = ι s ⇔ μ ( S / ρ ) = ι S ⇔ S / ρ is fundamental ,

where ι S is an identity relation on S . This completes the proof.□

5 Kernels of congruences on a type B semigroup

In this section, we investigate kernels of admissible congruences on a type B semigroup. Recall from the study of Petrich [19] that for any congruence ρ on a semigroup S , the kernel ker ρ of ρ is defined as follows:

ker ρ = { a ∈ S ∣ ( ∃ e ∈ E ) ( e , a ) ∈ ρ } .

Proposition 5.1

Let ρ be an admissible congruence on a type B semigroup S . If a ∈ ker ρ , then ( a + , a ∗ ) ∈ tr ρ .

Proof

Let a ∈ ker ρ . Then there exists e ∈ E such that ( e , a ) ∈ ρ . Since ρ is admissible, by Lemma 2.4, ( e , a ∗ ) ∈ ρ and ( e , a + ) ∈ ρ . Thus, ( a ∗ , a + ) ∈ ρ . Clearly, a ∗ , a + ∈ E . Therefore, ( a + , a ∗ ) ∈ tr ρ .□

Proposition 5.2

Let ρ be an admissible congruence on a type B semigroup S . Then e ρ = e μ ρ ∩ ker ρ for all e ∈ E .

Proof

Let a ∈ S , e ∈ E such that a ∈ e μ ρ ∩ ker ρ . It follows that a μ ρ e and there is f ∈ E such that a ρ f . Since ρ is admissible, we have that a ∗ μ ρ e and a ∗ ρ f . Again since tr ρ = tr μ ρ , we have a ∗ μ ρ f and a ∗ ρ e . By transitivity, e ρ f and a ρ e . Thus, a ∈ e ρ . That is, e μ ρ ∩ ker ρ ⊆ e ρ .

Conversely, if a ∈ e ρ , then a ∈ ker ρ . Since ρ ⊆ μ ρ , we have ( a , e ) ∈ μ ρ . That is, a ∈ e μ ρ . Therefore, a ∈ e μ ρ ∩ ker ρ and e ρ ⊆ e μ ρ ∩ ker ρ .□

Proposition 5.3

Let ρ be an admissible congruence on a type B semigroup S . Then the following sets are the same:

  1. K 1 = ker σ ρ = { a ∈ S ∣ ( ∃ e ∈ E ) ( e , a ) ∈ σ ρ } ;

  2. K 2 = { a ∈ S ∣ ( ∃ e ∈ a ∗ ρ ∩ E , f ∈ a + ρ ∩ E ) f a e = f e } .

Proof

For all a ∈ S , we have

a ∈ K 1 ⇒ ( ∃ g ∈ E ) ( a , g ) ∈ σ ρ ⇒ a ∗ ρ g , a + ρ g , ( ∃ e ∈ a ∗ ρ ∩ E , f ∈ a + ρ ∩ E ) f a e = f g e ⇒ ( e ∈ a ∗ ρ ∩ E , f g ∈ a + ρ ∩ E ) f g a e = f g e ⇒ a ∈ K 2 .

On the other hand,

a ∈ K 2 ⇒ ( ∃ e ∈ a ∗ ρ ∩ E , f ∈ a + ρ ∩ E ) f a e = f e ⇒ f e = f a e ρ a + a a ∗ = a ⇒ f e ρ a + , f e ρ a ∗ ( since  ρ  is admissible ) .

Again since f e a f e = e f a e f = e f f e e f = f e = f e f e f e , f e ρ a + and f e ρ a ∗ , we have ( a , f e ) ∈ σ ρ . That is, a ∈ K 1 . This completes the proof.□

Corollaries directly from Proposition 5.3 are as follows.

Corollary 5.4

Let ρ be an admissible congruence on a type B semigroup S. Then the following sets are the same:

  1. K 1 = ker σ π = { a ∈ S ∣ ( ∃ e ∈ E ) ( e , a ) ∈ σ π } ;

  2. K 2 = { a ∈ S ∣ ( ∃ e ∈ a ∗ ρ ∩ E , f ∈ a + ρ ∩ E ) f a e = f e }

Corollary 5.5

Let ρ be an admissible congruence on a type B semigroup S. Then the following sets are the same:

  1. K 1 = ker μ ρ = { a ∈ S ∣ ( ∃ e ∈ E ) ( e , a ) ∈ μ ρ } ;

  2. K 2 = { a ∈ S ∣ ( ∀ e ∈ E ) e a π a e } .

Corollary 5.6

Let ρ be an admissible congruence on a type B semigroup S. Then the following sets are the same:

  1. K 1 = ker μ π = { a ∈ S ∣ ( ∃ e ∈ E ) ( e , a ) ∈ μ π } ;

  2. K 2 = { a ∈ S ∣ ( ∀ e ∈ E ) e a π a e } .

Proposition 5.7

Let ρ be an admissible congruence on a type B semigroup S. Then the following sets are the same:

  1. ker σ = { a ∈ S ∣ ( ∃ e ∈ E ) e a e = e } ;

  2. ker μ = { a ∈ S ∣ ( ∀ e ∈ E ) e a = a e } .

Proof

Obviously,

a ∈ ker σ ⇒ ( ∃ f ∈ E ) ( a , f ) ∈ σ ⇒ ( ∃ e ∈ E ) e a e = e f e ⇒ ( ∃ e ∈ E ) f e a f e = f e a e f = f e f e f = f e .

That is, ker σ ⊆ { a ∈ S ∣ ( ∃ e ∈ E ) e a e = e } . Conversely, let a ∈ S such that e a e = e for some e ∈ E . Then ( e , a ) ∈ σ , and so a ∈ ker σ . That is, { a ∈ S ∣ ( ∃ e ∈ E ) e a e = e } ⊆ ker σ . Thus, ker σ = { a ∈ S ∣ ( ∃ e ∈ E ) e a e = e } . This completes the proof.□

The aforementioned corollaries show that the kernel of μ on S is the centralizer E ξ of E (i.e., for all e ∈ E , s ∈ E ξ , e s = s e ).

Corollary 5.8

Let S be a type B semigroup and x ∈ S . If x ∈ E ξ , then

x ∗ = x + , x ℋ ∗ x + .

Proof

For all x ∈ S , e ∈ E , if x ∈ E ξ , then e x = x e . Let e = x ∗ . Then x = x x ∗ = x ∗ x . By Lemma 2.1, x + = x ∗ x + . Let e = x + . Then x = x x + = x + x . By Lemma 2.1, x ∗ = x ∗ x + . To sum up, x ∗ = x + . In other words, x ℒ ∗ x ∗ = x + . Clearly, x ℛ ∗ x + . Therefore, x ℋ ∗ x + .□

6 Congruences with the same kernel on a type B semigroup

In this section, we shall extend the notion of normal subsemigroups in the class of inverse semigroups to the class of type B semigroups. By using the concept of a normal subsemigroup of a type B semigroup, some characterizations of congruences with the same kernel on a type B semigroup are given.

Definition 6.1

Let S be a type B semigroup and N be a full subsemigroup of S (i.e., E ( N ) = E ( S ) ). Then N is said to be a normal subsemigroup of S if it satisfies the following conditions:

  1. ( ∀ x , y ∈ S ) ( ∀ n ∈ N ) x y ∈ N ⇒ x n y ∈ N ;

  2. ( ∀ x , y ∈ S ) ( ∀ n ∈ N ) x n y ∈ N ⇒ x n ∗ y ∈ N , x n + y ∈ N .

Remark 6.1

  1. Obviously, the idempotent set E of S is a full subsemigroup of S and it satisfies Condition (b). If S is commutative, then Condition (a) holds for any subsemigroup of S . Thus, if S is commutative, then E is normal.

  2. There are full subsemigroups of commutative type B semigroups which do not satisfy Condition (b). For example, under the general multiplication, Z , Q and R are commutative type B semigroups. Clearly, the set of non-negative integers is a full subsemigroup of Z and a normal subsemigroup of Z . While Z is a full subsemigroup of Q and Z is not normal. Therefore, Condition (a) is not sufficient for N to be normal in S .

  3. When N and S are groups, for all x , y ∈ S , n ∈ N , x y ∈ N , we have n − 1 ∈ N and x n n − 1 y ∈ N , x n − 1 n y ∈ N . Hence, for all x , y ∈ S , n ∈ N , x y ∈ N implies that x n y ∈ N ⇔ N is normal.

  4. There are non-normal subsemigroups which satisfy Condition (b). That is, Condition (b) is not sufficient for N to be normal in S .

Example 6.1

Let R be a real set. Put

S = x 0 0 y ∣ x , y ∈ R , x y ≠ 0 , T = 0 x y 0 ∣ x , y ∈ R , x y ≠ 0 .

Let G = S ∪ T . Then G is a group under the matrix multiplication. Clearly, G is a type B semigroup.

Let N = 1 0 0 x ∣ x ≠ 0 . Obviously, N is a subgroup of G . It is easy to see that N is a full subsemigroup of G . For all A = 0 a b 0 ∈ T , A − 1 = 0 b − 1 a − 1 0 and for all M = 1 0 0 x ∈ N , M ∗ = M + = 1 0 0 1 ∈ N , A A − 1 = 1 0 0 1 ∈ N . But A M A − 1 = x 0 0 1 ∉ N . This means that N is not a normal subsemigroup of G . Let A , B ∈ G , C ∈ N . Then C ∗ = C + = 1 0 0 1 ∈ N . Hence, A C ∗ B = A C + B = A B . Note that

If A , B ∈ S and A C B ∈ N , then A B ∈ N ;

If A ∈ S and B ∈ T , then A C B ∉ N ;

If A ∈ T and B ∈ S , then A C B ∉ N ;

If A , B ∈ T with B = 0 x y 0 , x y = 1 and C = 1 0 0 1 , then A C B ∈ N and so A B ∈ N .

From Example 6.1, Conditions (a) and (b) of normal semigroups are independent.

A non-empty subset A of a semigroup S is said to be unitary if for all a ∈ A , s ∈ S , a s , s a ∈ A imply that s ∈ A .

Theorem 6.1

Let ρ be an admissible congruence on a type B semigroup S. Then

  1. If ker ρ is unitary, then ker ρ is a normal subsemigroup of S;

  2. If ker ρ is cancellative, then ker ρ is a normal subsemigroup of S.

Proof

(1) By hypothesis, it is clear that E ⊆ ker ρ and ker ρ is a full subsemigroup of S . Let x , y ∈ S , n ∈ ker ρ and x y ∈ ker ρ . Then there is e ∈ E such that ( n , e ) ∈ ρ and ( x n y , x e y ) ∈ ρ . Note that e y + ≤ y + . We have that there exists f ∈ E 1 such that e y + = ( y f ) + from Condition ( B2 ′ ). Hence,

x e y f = x e y + y f = x ( y f ) + y f ( since e y + = ( y f ) + ) = x y f ∈ ker ρ ( since x y ∈ ker ρ , f ∈ ker ρ ) .

Note that f ∈ ker ρ and ker ρ is unitary. We have that x e y ∈ ker ρ , and so x n y ∈ ker ρ . Thus, Condition (a) holds.

Since ρ is an admissible congruence on S and ( n , e ) ∈ ρ , we have ( n ∗ , e ) ∈ ρ . By transitivity, ( n , n ∗ ) ∈ ρ . Hence, for all x , y ∈ S , ( x n y , x n ∗ y ) ∈ ρ . Thus, x n y ∈ ker ρ implies x n ∗ y ∈ k e r ρ . Similarly, x n y ∈ ker ρ implies x n + y ∈ k e r ρ . This means that Condition (b) holds. To sum up, ker ρ is a normal subsemigroup of S .

(2) By the proof of (1), for all x , y ∈ S , n ∈ ker ρ , there exist e , f ∈ E such that x e y f = x y f . Since ρ is a congruence on S , we have x e y f ρ x y f . By the hypothesis, x e y ρ x y . Note that x y ∈ ker ρ , we have x e y ∈ ker ρ . Hence, x n y ∈ ker ρ . Thus, Condition (a) holds. By the proof of (1), Condition (b) holds. To sum up, ker ρ is a normal subsemigroup of S .□

Definition 6.2

Let N be any subsemigroup of a type B semigroup S . Define a syntactic congruence η N of N as follows:

η N = { ( a , b ) ∈ S × S ∣ ( ∀ x , y ∈ S 1 ) x a y ∈ N ⇔ x b y ∈ N } .

Obviously, η N is a congruence and η N is the maximum congruence. If E ( S ) ⊆ N , then ker η N ⊆ N . If S is a type B semigroup and N is normal, then we have that the maximum congruence on S whose kernel is N on S .

Proposition 6.2

Let S be a type B semigroup and N be a normal subsemigroup of S. Then η N is the maximum admissible congruence on S whose kernel is N.

Proof

It is easy to see that η N is a congruence on S . Now, we prove that ker η N = N . To see it, let a ∈ S . Then

a ∈ ker η N ⇒ ( ∃ e ∈ E ) ( e , a ) ∈ η N ⇒ ( ∀ x , y ∈ S 1 ) x a y ∈ N if and only if x e y ∈ N .

Let x = a + and y = a ∗ . We have that a + e a ∗ ∈ N since N is a full subsemigroup. Hence, a + a a ∗ ∈ N . That is, a ∈ N . On the other hand, let n ∈ N . Then, for all x , y ∈ S 1 , we have

x n y ∈ N ⇒ x n ∗ y ∈ N ⇒ x n n ∗ y ∈ N ⇒ x n y ∈ N .

Hence, x n y ∈ N if and only if x n ∗ y ∈ N . Thus, ( n , n ∗ ) ∈ η N . This shows that n ∈ ker η N . Therefore, ker η N = N .

Next, we prove that η N is an admissible congruence on S . For all a ∈ S , m , n ∈ S 1 , if a m η N a n , then for all x , y ∈ S 1 , x a m y ∈ N if and only if x a n y ∈ N . Since N is a normal semigroup of S , we obtain that x a ∗ m y ∈ N if and only if x a ∗ n y ∈ N . Thus, a ∗ m η N a ∗ n . Similarly, for all a ∈ S , m , n ∈ S 1 , m a η N n a implies m a + η N n a + . Therefore, η N is an admissible congruence.

Finally, we show that η N is the maximum admissible congruence on S whose kernel is N . To see it, suppose ρ is an admissible congruence on S and ker ρ = N . Let ( a , b ) ∈ ρ . Then for all x , y ∈ S 1 , we have ( x a y , x b y ) ∈ ρ . Hence, x a y ∈ ker ρ if and only if x b y ∈ ker ρ . That is, x a y ∈ N if and only if x b y ∈ N . Therefore, ( a , b ) ∈ η N .

To sum up, η N is the maximum admissible congruence on S whose kernel is N .□

Let S be a type B semigroup and N be a normal subsemigroup of S . In order to give the minimum admissible congruence on S whose kernel is N , we first give the following lemma:

Lemma 6.3

Let S be a type B semigroup and N be a normal subsemigroup of S. Define a relation τ N on S as follows:

τ N = { ( x n 1 y , x n 2 y ) ∣ x , y ∈ S 1 ; n 1 , n 2 ∈ N ; n 1 + = n 2 + } .

Then the following statements hold:

  1. τ N is an equivalence relation on S ;

  2. N = { a ∈ S ∣ ( ∃ e ∈ E ) ( a , e ) ∈ τ N } ;

  3. τ N is contained in an arbitrary admissible congruence on S whose kernel is N.

Proof

(1) It is clear.

(2) Let a ∈ N . Clearly, a = a + a a ∗ and a + a ∗ = a + a + a ∗ . Since a , a + ∈ N and a + = ( a + ) + = a + , we have ( a , a + a ∗ ) ∈ τ N . That is, there exists e = a + a ∗ ∈ E such that ( a , e ) ∈ τ N .

Conversely, let a ∈ S and ( a , f ) ∈ τ N for some f ∈ E . By the definition of τ N , we have

a = x n 1 y , f = x n 2 y ; n 1 , n 2 ∈ N ; n 1 + = n 2 + .

Note that E ⊆ N . We have x n 2 y = f ∈ N . Since N is normal. We obtain x n 1 + y = x n 2 + y ∈ N . Hence, x n 1 y = x n 1 + n 1 y ∈ N . That is, a ∈ N and (2) holds.

(3) Let ρ be any admissible congruence on S whose kernel is N . Let ( a , b ) ∈ τ N . By the definition of τ N ,

a = x n 1 y , b = x n 2 y ; n 1 , n 2 ∈ N ; n 1 + = n 2 + .

Since ker ρ = N , there are e , f ∈ E such that ( n 1 , e ) ∈ ρ , ( n 2 , e ) ∈ ρ . Note that ρ is an admissible congruence on S . We have that ( n 1 + , e ) ∈ ρ and ( n 2 + , f ) ∈ ρ . But n 1 + = n 2 + , we have ( e , f ) ∈ ρ . Thus, ( x e y , x f y ) ∈ ρ . It follows that:

( n 1 , e ) ∈ ρ , ( n 2 , f ) ∈ ρ ⇒ ( x n 1 y , x e y ) ∈ ρ , ( x n 2 y , x f y ) ∈ ρ ⇒ ( x n 1 y , x n 2 y ) ∈ ρ ⇒ ( a , b ) ∈ ρ .

Therefore, (3) holds.□

As usual, let S be an arbitrary relation on X . Define S ∞ , the transitive closure of S , by S ∞ = ⋃ { S n ∣ n ≥ 1 } .

Proposition 6.4

Let S be a type B semigroup and λ N be a transitive closure of τ N ( λ N = τ N t ) . Then λ N is the minimum admissible congruence on S whose kernel is N .

Proof

Obviously, λ N is a congruence on S . Now, we prove that ker λ N = N . To see it, let n ∈ N . Then there exists e ∈ E such that ( n , e ) ∈ τ N . By Lemma 6.3, we have that ( n , e ) ∈ λ N for some e ∈ E and n ∈ ker λ N .

Conversely, let a ∈ ker λ N . Then ( a , f ) ∈ λ N for some f ∈ E . By hypothesis, λ N = τ N t . Hence, there exist a 1 , a 2 , … , a n ∈ S such that

( a , a 1 ) ∈ τ N , ( a 1 , a 2 ) ∈ τ N , … , ( a n − 1 , a n ) ∈ τ N , ( a n , f ) ∈ τ N .

Note that

( a n , f ) ∈ τ N ⇒ a n = x n 1 y , f = x n 2 y ; n 1 , n 2 , f ∈ N ; n 1 + = n 2 + ;

x n 2 y = f ∈ N ⇒ x n 2 + y ∈ N ⇒ x n 1 + y ∈ N ⇒ x n 1 y ∈ N ⇒ a n ∈ N .

Similarly, a n − 1 , a n − 2 , … , a 1 , a ∈ N . That is, a ∈ N . Therefore, ker λ N = N .

Finally, we show that λ N is the minimum admissible congruence on S whose kernel is N . Let ρ be any admissible congruence on S whose kernel is N . By Lemma 6.3, we have that τ N ⊆ ρ , and so λ N ⊆ ρ .□

The following corollary can be obtained from Propositions 6.2 and 6.4.

Corollary 6.5

Let S be a type B semigroup and ρ be an admissible congruence on S whose kernel is N. Then λ N ⊆ ρ ⊆ η N and ker λ N = ker ρ = ker η N .

A congruence on S is said to be idempotent-pure if for all a ∈ S , e ∈ E and ( a , e ) ∈ ρ imply a ∈ E .

Proposition 6.6

Let S be a type B semigroup and ρ be an admissible congruence. If λ N is an admissible congruence on S, then S / ρ is an idempotent-pure image of S / λ N .

Proof

Define a mapping ϕ as follows:

ϕ : S / λ N → S / ρ , ( a λ N ) ϕ = a ρ .

It is easy to see that ϕ is a homomorphism from S / λ N to S / ρ . Let a ∈ S , e ∈ E such that ( a λ N ) ϕ = ( e λ N ) ϕ . Then a ρ = e ρ . That is, a ρ e . By Proposition 5.7, ker λ N = N = ker ρ . Hence, a λ N e . That is, a λ N = e λ N . Therefore, S / ρ is an idempotent-pure image of S / λ N .□

7 Congruence pairs on a type B semigroup

In this section, we extend the concept of congruence pairs from inverse semigroup to type B semigroups.

Definition 7.1

Let N be a normal subsemigroup of a type B semigroup and π be a normal congruence on E . Then ( π , N ) is a congruence pair on S if the following statements hold:

  1. for all n ∈ N , n + π n ∗ ;

  2. for all x , y ∈ S , e , f ∈ E , x e y ∈ N and e π f imply x f y ∈ N .

Lemma 7.1

Let ρ be an admissible congruence on a type B semigroup and ker ρ is unitary. Then ( tr ρ , ker ρ ) is a congruence pair of S .

Proof

Obviously, tr ρ is a normal congruence. Since ker ρ is unitary, by Theorem 6.1, we have that ker ρ is a normal subsemigroup of S . Let n ∈ ker ρ . By Proposition 5.1, ( n + , n ∗ ) ∈ tr ρ . Let x , y ∈ S , e , f ∈ E such that x e y ∈ ker ρ and e tr ρ f . Then

e tr ρ f ⇒ e ρ f ⇒ x e y ρ x f y .

Since x e y ∈ ker ρ , we obtain x f y ∈ ker ρ . Therefore, ( tr ρ , ker ρ ) is a congruence pair of S .□

Let N be a normal subsemigroup of a type B semigroup and π be a normal congruence on E such that ( π , N ) is a congruence pair of S . A congruence ρ is said to be associated with the congruence pair ( π , N ) if ker ρ = N , tr ρ = π . The following theorem will show that a congruence associated with the congruence pair ( π , N ) can be constructed on S by using relations μ π and η N .

Theorem 7.2

The relation μ π ∩ η N is a congruence on a type B semigroup S associated with the congruence pair ( π , N ) .

Proof

Let ρ = μ π ∩ η N . Clearly, ρ is a congruence on S . Now, we prove that ker ρ = N . Let n ∈ N . Then for all e ∈ E ⊆ N , we have e n ∈ N . Since ( π , N ) is a congruence pair, we obtain that ( e n ) + π ( e n ) ∗ , n + π n ∗ and e n + π e n ∗ . Note that e n ∗ = ( e n ∗ ) ∗ and e n + = ( e n ) + . We have

( e n ) ∗ π ( e n ) + π e n + π e n ∗ π ( e n ∗ ) ∗ .

That is, ( e n ) ∗ π ( e n ∗ ) ∗ . Similarly, ( n e ) + π ( n ∗ e ) + . Thus, ( n , n ∗ ) ∈ μ π . By the proof of Proposition 6.2, ( n , n ∗ ) ∈ η N . Thus, ( n , n ∗ ) ∈ ρ , n ∈ ker ρ . That is, N ⊆ ker ρ . On the other hand, it is easy to see that ker ρ ⊆ ker η N = N . Therefore, ker ρ ⊆ N and ker ρ = N .

Next, we show that tr ρ = π . Obviously, ρ ⊆ μ π . For all e , f ∈ E and ( e , f ) ∈ ρ , we have ( e , f ) ∈ μ π . Furthermore, e π f and tr ρ ⊆ π . On the other hand, let e , f ∈ E with e π f . Then, by the definition of ( π , N ) , we have that x e y ∈ N if and only if x f y ∈ N . Hence, ( e , f ) ∈ η N . Since ( e , f ) ∈ μ π , we have that ( e , f ) ∈ ρ and π ⊆ tr ρ . Therefore, tr ρ = π .□

Let ρ be an admissible congruence on a type B semigroup S . If ker ρ = N and tr ρ = π , then ρ ⊆ μ π from Theorem 3.4 and ρ ⊆ η N from Proposition 6.2. Thus, we have the following corollary:

Corollary 7.3

Let ρ be an admissible congruence on a type B semigroup S. If ker ρ = N and tr ρ = π , then ρ ⊆ μ π ∩ η N .

Theorem 7.4

The relation σ π ∨ λ N is a congruence on a type B semigroup S associated with the congruence pair ( π , N ) .

Proof

Let τ = σ π ∨ λ N . Then

τ = ⋂ { ρ ∣ ρ is a congruence , σ π ⊆ ρ and λ N ⊆ ρ } ,

and for all a , b ∈ S , ( a , b ) ∈ τ if and only if there exist a 1 , a 2 , … , a n ∈ S such that

( a , a 1 ) ∈ λ N , ( a 1 , a 2 ) ∈ σ π , ( a 2 , a 3 ) ∈ λ N , … , ( a n − 1 , a n ) ∈ σ π , ( a n , b ) ∈ λ N .

Therefore, τ is the minimum congruence containing both σ π and λ N .

Now, we prove ker τ = N . To see it, let a ∈ ker τ . Then there exists e ∈ E such that ( a , e ) ∈ τ . Hence, there exist a 1 , a 2 , … , a n ∈ S such that

( a , a 1 ) ∈ λ N , ( a 1 , a 2 ) ∈ σ π , ( a 2 , a 3 ) ∈ λ N , … , ( a n − 1 , a n ) ∈ σ π , ( a n , e ) ∈ λ N .

Since ( a n , e ) ∈ λ N , by Proposition 6.4, a n ∈ N , ( a n − 1 , a n ) ∈ σ π . By the definition of σ π , there are e , f ∈ E such that

e π a n ∗ π a n − 1 ∗ , f π a n + π a n − 1 + and f a n e = f a n − 1 e .

Since N is normal, a n ∈ N , f a n e ∈ N and f a n − 1 e ∈ N , we have

f a n − 1 e a n − 1 ∗ ∈ N , e π a n − 1 ∗ .

By the definition of congruence pairs, we obtain

f a n − 1 = f a n − 1 a n − 1 ∗ a n − 1 ∗ ∈ N .

Thus, a n − 1 + f a n − 1 ∈ N . Again, since f π a n − 1 + , we have

a n − 1 = a n − 1 + a n − 1 + a n − 1 ∈ N .

Hence, a n − 1 ∈ N and ( a n − 2 , a n − 1 ) ∈ λ N . By the proof of Theorem 6.4, a n − 2 ∈ N . Since ( a n − 3 , a n − 2 ) ∈ σ π , we have a n − 3 ∈ N from the above proof. The process will continue until we reach a ∈ N . Thus, ker τ ⊆ N . Conversely,

n ∈ N ⇒ n ∈ ker λ N ⇒ ( ∃ e ∈ E ) ( n , e ) ∈ λ N ⇒ ( ∃ e ∈ E ) ( n , e ) ∈ τ ⇒ n ∈ ker τ .

Hence, N ⊆ ker τ . Therefore, ker τ = N .

Next, we prove that tr τ = π . Let e , f ∈ E . Then

( e , f ) ∈ π ⇒ ( e , f ) ∈ σ π ⇒ ( e , f ) ∈ τ .

This means that π ⊆ tr τ .

On the other hand, let e , f ∈ E and ( e , f ) ∈ tr τ . That is, ( e , f ) ∈ τ . Then there exist a 1 , a 2 , … , a n ∈ S such that

( e , a 1 ) ∈ λ N , ( a 1 , a 2 ) ∈ σ π , ( a 2 , a 3 ) ∈ λ N , … , ( a n − 1 , a n ) ∈ σ π , ( a n , f ) ∈ λ N .

Note that ( e , a 1 ) ∈ λ N . There are b 1 , b 2 , … , b k ∈ S such that

( e , b 1 ) ∈ τ N , ( b 1 , b 2 ) ∈ τ N , … , ( b k − 1 , b k ) ∈ τ N , ( b k , a 1 ) ∈ τ N .

Again ( e , b 1 ) ∈ τ N , by the definition of τ N ,

e = x n 1 y , b 1 = x n 2 y ; n 1 , n 2 ∈ N ; n 1 + = n 2 + .

According to the definition of congruence pairs, for all n ∈ N , we have n ∗ π n + . Thus, for all x , y ∈ S , n y + ∈ N , which follows ( n y + ) + π ( n y + ) ∗ . Since ( n y ) + = ( n y + ) + , we have ( n y ) + π ( n y + ) ∗ . By the normality of π ,

( x ( n y ) + ) + π ( x ( n y + ) ∗ ) + .

That is, ( x n y ) + π ( x ( n y + ) ∗ ) + . In particular, since n 1 , n 2 ∈ N , we have

( x n 1 y ) + π ( x ( n 1 y + ) ∗ ) + , ( x n 2 y ) + π ( x ( n 2 y + ) ∗ ) + .

Again since n 1 ∗ π n 1 + , n 1 + = n 2 + and n 2 ∗ π n 2 + , we obtain n 1 ∗ π n 2 ∗ , n 1 ∗ y + π n 2 ∗ y + and ( n 1 y + ) ∗ π ( n 2 y + ) ∗ . By the normality of π , ( x ( n 1 y + ) ∗ ) + π ( x ( n 2 y + ) ∗ ) + . By transitivity of π , ( x n 1 y ) + π ( x n 2 y ) + . That is, e π b 1 + . Note that ( b 1 , b 2 ) ∈ τ N . We have

b 1 = x 1 n 3 y 1 , b 2 = x 1 n 4 y 1 ; n 3 , n 4 ∈ N ; n 3 + = n 4 + .

Similarly, we can prove b 1 + π b 2 + , b 2 + π b 3 + , … , b k + π a 1 + . By transitivity, we have e π a 1 + . Since ( a 1 , a 2 ) ∈ σ π , we obtain a 1 + π a 2 + . Note that ( a 2 , a 3 ) ∈ λ N . According to the above procedure, a 2 + π a 3 + , a 3 + π a 4 + , a 4 + π a 5 + , … , a n − 1 + π a n + and a n + π f . By transitivity, e π f . Therefore, tr τ ⊆ π .

To sum up, σ π ∨ λ N is a congruence on a type B semigroup associated with the congruence pair ( π , N ) . This completes the proof.□

Let ρ be an admissible congruence on a type B semigroup. If ker ρ = N , tr ρ = π . By Theorem 3.1, we have σ π ⊆ ρ . By Proposition 6.4, we obtain λ N ⊆ ρ . Therefore, we obtain the following corollary:

Corollary 7.5

Let ρ be an admissible congruence on a type B semigroup S. If ker ρ = N and tr ρ = π , then σ π ∨ λ N ⊆ ρ .

Theorem 7.6

Let ρ be an admissible congruence on a type B semigroup S. If ker ρ = N and tr ρ = π , then μ π ∩ η N and σ π ∨ λ N are congruences on S associated with the congruence pair ( π , N ) . Furthermore, σ π ∨ λ N ⊆ ρ ⊆ μ π ∩ η N .

Proof

It follows from Theorems 7.2 and 7.4 and Corollary 7.5.□

8 Conclusion

As we know, abundant semigroups are generalized regular semigroups and type B semigroups are generalized inverse semigroups in the range of abundant semigroups. The kernel-trace approach consists in splitting the analysis of a congruence on a regular semigroup into two parts: the kernel and the trace. Here we develop the kernel-trace approach of inverse semigroups to the cases of type B semigroups. As a concrete application of the above approach, we introduce admissible congruences of type B semigroups. Admissible congruences introduced here could improve studies of the inverse semigroup theory.

Acknowledgements

The authors are very grateful to the referees for their valuable suggestions which lead to an improvement of this article.

  1. Funding information: This work was supported by the NNSF (CN) (Nos 11261018 and 11961026), the NSF of Jiangxi Province (No. 20181BAB201002).

  2. Conflict of interest: The authors state no conflict of interest.

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Received: 2022-07-06
Revised: 2022-11-22
Accepted: 2022-12-15
Published Online: 2022-12-31

© 2022 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2022-0551
Keywords for this article
admissible congruences; trace; type B semigroups; congruence pairs
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