A new characterization of a proper type B semigroup

Chunhua Li; Zhi Pei; Baogen Xu

doi:10.1515/math-2020-0104

Article Open Access

A new characterization of a proper type B semigroup

Chunhua Li , Zhi Pei and Baogen Xu

Published/Copyright: December 31, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

Abstract

In this paper, we develop the elementary theory of inverse semigroups to the cases of type B semigroups. The main aim of this paper is to study proper type B semigroups. We introduce first the concept of a left admissible triple. After obtaining some basic properties of left admissible triple, we give the definition of a Q-semigroup and get a structure theorem of Q-semigroup. In particular, we introduce the notion of an admissible triple and give some characterization of proper type B semigroups. It is proved that an arbitrary Q-semigroup with an admissible triple is an E-unitary type B semigroup.

Keywords: type B semigroup; Q-semigroup; proper; E-unitary

MSC 2010: 20M10; 06F05

1 Introduction

Let S be a semigroup and denote the set of idempotents of S by $E ( S )$ . As in [1], the relations $ℛ ⁎$ and $ℒ ⁎$ are generalizations of Green’s relations $ℛ$ and $ℒ$ , respectively. We define $a ℛ ⁎ b$ for the elements $a , b$ of a semigroup S if and only if $a , b$ are related by Green’s relation $ℛ$ in T containing S as a subsemigroup. $ℒ ⁎$ is defined dually. In the study by Fountain [2], 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. Fountain and others investigated some classes of abundant semigroups and got many interesting results (see [3,4,5,6,7,8,9,10,11]). From [2], a rpp semigroup S is said to be right adequate if $E ( S )$ is a semilattice (i.e., any two elements of $E ( S )$ commute). A left adequate semigroup is defined dually. A semigroup is adequate if it is both left and right adequate. Usually, we denote by $a ⁎$ (resp., $a +$ ) an idempotent $ℒ ∗$ - (resp., $ℛ ∗$ -) related to a. Following [2], a right adequate semigroup S is right type B, if it satisfies the following conditions (RB1) and (RB2):

(RB1) for all $e , f ∈ E ( S 1 ) , a ∈ S ,$ $( e f a ) ⁎ = ( e a ) ⁎ ( f a ) ⁎ ;$

(RB2) if for all $a ∈ S , e ∈ E ( S ) , e ≤ a ⁎ ,$ then there is an element $f ∈ E ( S 1 )$ such that $e = ( f a ) ⁎ ,$ where “ $≤$ ” is a natural partial order on $E ( S )$ (i.e., $( ∀ g , h ∈ E ( S ) ) g ≤ h ⇔ g = g h = h g$ ).

Dually, a left adequate semigroup S is left type B, if it satisfies the following conditions (LB1) and (LB2):

(LB1) for all $e , f ∈ E ( S 1 ) , a ∈ S ,$ $( a e f ) + = ( a e ) + ( a f ) +$ ;

(LB2) if for all $a ∈ S , e ∈ E ( S ) , e ≤ a + ,$ then there is an element $f ∈ E ( S 1 )$ such that $e = ( a f ) + .$

A semigroup is said to be type B if it is both right and left type B. Type B semigroups originally introduced by Fountain in [2] are generalizations of inverse semigroups in the range of abundant semigroups. Recently, Li and others investigated some classes of type B semigroups (see [12,13,14,15,16,17,18]).

Recall from Petrich [19] and Lawson [20] that a homomorphism $ϕ$ of semigroups S and T is idempotent-separated if for all $e , f ∈ E ( S ) ,$ $e ϕ f$ implies $e = f ,$ and that a semigroup S is said to be $E -$ unitary if $a , a b ∈ E ( S )$ implies $b ∈ E ( S )$ and $a , b a ∈ E ( S )$ implies $b ∈ E ( S )$ for all $a , b ∈ S .$ An inverse semigroup S is proper if $ℒ ∩ γ = ℛ ∩ γ = ι S ,$ where $γ$ and $ι S$ are the least group congruence on S and the identity relation on $S ,$ respectively. It is well known that the class of $E -$ unitary semigroups is one of the most important in inverse semigroup theory. There are two main reasons for this: first, McAlister theorem states that every inverse semigroup admits an $E -$ unitary cover (i.e., every inverse semigroup is an idempotent-separated homomorphic image of an $E -$ unitary semigroup); second, many naturally occurring inverse semigroups are $E -$ unitary. Generally, a proper inverse semigroup is $E -$ unitary, but the converse is not true. Therefore, it is an interesting thing to characterize a proper generalization inverse semigroup. Motivated by studying proper inverse semigroups and $E -$ unitary inverse semigroups in Petrich (Inverse Semigroups, Wiley, New York, 1984), and as a continuation of Petrich and Lawson’s work in inverse semigroups, in this paper, we shall characterize some proper type B semigroups.

This article is organized as follows. Section 2 provides some basic notions and some known results used in the sequel. In Section 3, we introduce the definitions of a left admissible triple and a $Q -$ semigroup and give some basic properties. In Section 4, we give the notion of an admissible triple and obtain a new characterization of a proper type B semigroup. In particular, we prove that an arbitrary $Q -$ semigroup with an admissible triple is an $E -$ unitary type B semigroup.

2 Preliminaries

The following basic terminologies and notations are from [1,2,8,16,17,20].

Lemma 2.1

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

$a ℒ ⁎ b$ $( a ℛ ⁎ b )$ ;
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

[1] Let S be a semigroup and $a ∈ S , e ∈ E ( S )$ . Then the following statements hold:

$a ℒ ⁎ e ( a ℛ ⁎ e )$ if and only if $a e = a$ $( e a = a )$ and for all $x , y ∈ S 1 ,$ $a x = a y$ $( x a = y a )$ implies $e x = e y$ $( x e = y e )$ ;
$ℒ ⁎$ (resp., $ℛ ⁎$ ) is a right (resp., left) congruence on S.

For convenience, $ℒ a ⁎$ and $ℛ a ⁎$ denote the $ℒ ⁎ -$ class and $ℛ ⁎ -$ class containing $a ,$ respectively. Obviously, every left (resp., right) adequate semigroup S is $ℛ ⁎ -$ unipotent (i.e., $| ℛ a ⁎ ∩ E ( S ) | = 1$ ) (resp., $ℒ ⁎ -$ unipotent) (i.e., $| ℒ a ⁎ ∩ E ( S ) | = 1$ ).

Lemma 2.3

[8] Let S be an abundant semigroup. Define the natural partial order relation “ $≤$ ” on S as follows:

$( ∀ a , b ∈ S ) a ≤ b ⇔ ( ∃ e , f ∈ E ( S ) ) a = e b = b f .$

Lemma 2.4

[16] Let S be a left type B semigroup. Define a relation “ $σ$ ” on S as follows:

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

Then $σ$ is the least right cancellative monoid congruence on $S .$

Lemma 2.5

[17] Let S be a type B semigroup. Define a relation “ $ξ$ ” on S as follows:

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

Then $ξ$ is the least cancellative monoid congruence on $S .$

In this paper, we call a left type B semigroup S proper if $σ ∩ ℛ ⁎ = ι S$ , and call a type B semigroup S proper if $ξ ∩ ℒ ⁎ = ι S$ and $ξ ∩ ℛ ⁎ = ι S$ , where $ι S$ is the identity relation on $S .$

Lemma 2.6

[18] Let S be a proper type B semigroup. Then S is $E -$ unitary.

Definition 2.1

[20] Let T be a monoid with identity 1 and A be a nonempty set. Define a mapping μ as follows:

$μ : T × A → A , ( t , a ) ↦ t a$

such that $1 ⋅ a = a$ and $( s t ) a = s ( t a )$ . Then A is a left $T -$ act or T acts on A on the left.

3 Definitions and basic results

First, we recall some terminologies and notations of partial order set. Suppose that X is a partial order set. If $a , b ∈ X$ have a greatest lower bound in S, it will be denoted by $a ∧ b$ . Let Y be a subset of X with $a , b ∈ Y$ . Then $a ∧ b ∈ Y$ means that a and b have a greatest lower bound in X that it belongs to Y. In particular, if for all $a , b ∈ Y$ , $a ∧ b ∈ Y$ , then the nonempty subset Y of X is said to be a subsemilattice of X.

Let X be a partial order set and let Y be a subsemilattice of X. Suppose that $i ∈ X$ and for all $a ∈ Y$ , $a ≤ i$ . Let T be a right cancellative monoid with identify 1. Let T act on X on the left such that for all $a , b ∈ X$ , $t ∈ T$ $a ≤ b$ implies $t a ≤ t b$ . We call $( T , X , Y )$ with the above properties a triple.

Definition 3.1

Let $( T , X , Y )$ be a triple satisfying the following conditions (Q1) to (Q6):

(Q1) $T Y i = X$ , where $Y i = Y ∪ { i }$ ;

(Q2) For all $t ∈ T$ , there exists $b ∈ Y$ such that $b ≤ t i$ ;

(Q3) If $a , b ∈ Y , t ∈ T$ and $a ≤ t i$ , then $a ∧ t b ∈ Y$ ;

(Q4) If $a , b , c ∈ Y , t , u ∈ T$ and $a ≤ t i , b ≤ u i$ , then $( a ∧ t b ) ∧ t u c = a ∧ t ( b ∧ u c )$ ;

(Q5) If $a , b ∈ Y , t ∈ T$ with $a ≤ t i$ and $b ≤ t i$ , then $a ∧ t ( a ∧ b ) = b ∧ t ( a ∧ b ) ;$

(Q6) If $a , b ∈ Y$ and $b = a ∧ b$ , then for all $t ∈ T$ there exists $c ∈ Y i$ such that $b = a ∧ t c$ .

Then we call $( T , X , Y )$ with the aforementioned conditions (Q1) to (Q6) a left admissible triple.

Now, we give an example which satisfies conditions (Q1) to (Q6) listed in Definition 3.1.

Example 3.1

Let $X = { I , A , A ′ , O }$ be the four element Boolean algebra with top element I, bottom element O and $A ′$ the complement of $A .$ Put $Y = { A , O }$ . Let $T = { 1 , g }$ with $g 2 = 1 ,$ where 1 is the identity automorphism of X and $g I = I ,$ $g O = O ,$ $g A = A ′ ,$ $g A ′ = A .$ Obviously, X is a partial order set. It is easy to see that Y is a subsemilattice of X satisfying $T Y I = X ,$ where I is the element i listed in Definition 3.1. That is, $( T , X , Y )$ satisfies condition (Q1) of Definition 3.1. Note that $1 I = I = g I .$ We have that $b ≤ t I = I$ for all $t ∈ T$ and $b ∈ Y ,$ which implies that condition (Q2) holds. Similarly, it is easily seen that conditions (Q3) and (Q4) hold. On the other hand, for $A , O ∈ Y ,$ we have that $A ≤ t I = I$ and $O ≤ t I = I$ for all $t ∈ T .$ Again, for all $t ∈ T ,$

$A ∧ t ( A ∧ O ) = A ∧ t O = A ∧ O = O ,$

$O ∧ t ( A ∧ O ) = O ∧ t O = O ∧ O = O .$

Thus, $( T , X , Y )$ satisfies condition (Q5). Finally, it is easily seen that $O = O ∧ O = O ∧ t O ,$ $A = A ∧ A = A ∧ I = A ∧ t I$ and $O = A ∧ O = A ∧ t O .$ Therefore, condition (Q6) is true.

Summarizing the aforementioned arguments, we conclude that $( T , X , Y )$ is a left admissible triple.

Definition 3.2

Let $( T , X , Y )$ be a left admissible triple and we define:

$Q ( T , X , Y ) = { ( a , t ) ∈ Y × T | a ≤ t i }$

with a multiplication

$( a , t ) ( b , u ) = ( a ∧ t b , t u ) .$

Obviously, $Q ( T , X , Y )$ is a semigroup with respect to the above multiplication.

Example 3.2

In our Example 3.1, put $Q = { ( A , 1 ) , ( O , 1 ) , ( A , g ) , ( O , g ) }$ . Obviously, $A ≤ 1 I = I , O ≤ 1 I = I , A ≤ g I = I$ and $O ≤ g I = I .$ This means that Q satisfies the property $Q ( T , X , Y ) = { ( a , t ) ∈ Y × T | a ≤ t i }$ listed in Definition 3.2. Furthermore, it is easy to check that Q is a semigroup with respect to the multiplication of Definition 3.2.

Theorem 3.1

Let $( T , X , Y )$ be a left admissible triple. Then the following statements are true:

$E ( Q ( T , X , Y ) ) = Y × { 1 } ;$
$Q ( T , X , Y )$ is a left type B semigroup;
Let $( a , t ) , ( b , u ) ∈ Q ( T , X , Y )$ . Then $( a , t ) ℛ ⁎ ( b , u )$ if and only if $a = b$ .

Proof

Let $( a , t ) ∈ E ( Q ( T , X , Y ) )$ . Then $( a , t ) = ( a , t ) 2 = ( a ∧ t a , t 2 )$ from Definition 3.2. Hence, $t = t 2$ . Note that T is a right cancellative monoid with identify 1 and so $t = 1$ . That is, $( a , t ) = ( a , 1 ) ∈ Y × { 1 }$ . Conversely, for all $a ∈ Y$ we have $a ≤ i$ . So $( a , 1) ∈ Q ( T , X , Y )$ and $( a , 1)$ is an idempotent of $Q ( T , X , Y )$ . Hence, the elements in $Y × {1}$ are the idempotents of $Q ( T , X , Y )$ . Therefore, $(1)$ holds.
Obviously, $Q ( T , X , Y )$ is a semigroup from Definition 3.2. Now, we prove that $Q ( T , X , Y )$ is left abundant. To see it, let $( a , t ) ∈ Q ( T , X , Y )$ , $( b , s ) , ( c , k ) ∈ Q ( T , X , Y ) 1$ such that $( b , s ) ( a , t ) = ( c , k ) ( a , t )$ . Then $( b ∧ s a , s t ) = ( c ∧ k a , k t )$ , and so $b ∧ s a = c ∧ k a$ , $s t = k t$ . Note that T is a right cancellative monoid with identify 1. We have $s = k$ . Hence,
$( b , s ) ( a , 1 ) = ( b ∧ s a , s ) = ( c ∧ k a , k ) = ( c , k ) ( a , 1 ) .$

But $( a , 1 ) ( a , t ) = ( a ∧ a , t ) = ( a , t )$ . We have $( a , t ) ℛ ⁎ ( a , 1 ) ∈ E ( Q ( T , X , Y ) )$ from Corollary 2.2. This gives that $Q ( T , X , Y )$ is left abundant. Clearly, $E ( Q ( T , X , Y ) )$ is a semilattice from (1). Therefore, $Q ( T , X , Y )$ is a left adequate semigroup. That is, $( a , t ) + = ( a , 1 )$ for all $( a , t ) ∈ Q ( T , X , Y )$ .

Now, we prove that $Q ( T , X , Y )$ is a left type B semigroup. First, we verify that $Q ( T , X , Y )$ satisfies condition (LB1). To see it, let $( a , t ) ∈ Q ( T , X , Y )$ , $( b , 1 ) , ( c , 1 ) ∈ E ( Q ( T , X , Y ) 1 )$ . Then
$( a , t ) ( b , 1 ) ( c , 1 ) = ( a ∧ t b , t ) ( c , 1 ) = ( a ∧ t b ∧ t c , t ) .$
Note that
$[ ( a , t ) ( b , 1 ) ( c , 1 ) ] + = ( a ∧ t b ∧ t c , t ) + = ( a ∧ t b ∧ t c , 1 ) ,$

$[ ( a , t ) ( b , 1 ) ] + = [ ( a ∧ t b , t ) ] + = ( a ∧ t b , 1 ) ,$

$[ ( a , t ) ( c , 1 ) ] + = [ ( a ∧ t c , t ) ] + = ( a ∧ t c , 1 ) .$
But $( a ∧ t b , 1 ) ( a ∧ t c , 1 ) = ( a ∧ t b ∧ t c , 1 )$ . Hence, $[ ( a , t ) ( b , 1 ) ( c , 1 ) ] + = [ ( a , t ) ( b , 1 ) ] + [ ( a , t ) ( c , 1 ) ] +$ . That is, $Q ( T , X , Y )$ satisfies condition (LB1).

Next, we prove that $Q ( T , X , Y )$ satisfies condition (LB2). To see it, let $( a , t ) ∈ Q ( T , X , Y )$ , $( b , 1) ∈ E ( Q ( T , X , Y ) )$ and $( b , 1) ≤ ( a , t ) +$ . That is, $( b , 1) ≤ ( a , 1)$ . Then $( b , 1 ) = ( b , 1 ) ( a , 1 ) = ( a , 1 ) ( b , 1 )$ and so $b = a ∧ b = b ∧ a$ . By condition (Q6), we know there exists $c ∈ Y$ such that $b = a ∧ t c$ . Hence, $( b , 1 ) = ( a ∧ t c , 1 ) = ( a ∧ t c , t ) + = [ ( a , t ) ( c , 1 ) ] +$ and $( c , 1 ) ∈ E ( Q ( T , X , Y ) )$ . This shows that $Q ( T , X , Y )$ satisfies condition (LB2). Therefore, $Q ( T , X , Y )$ is a left type B semigroup.
Let $( a , t ) , ( b , u ) ∈ Q ( T , X , Y )$ and $( a , t ) ℛ ⁎ ( b , u )$ . Then, by the proof of (2), we obtain that $( a , 1 ) ℛ ⁎ ( b , 1 )$ . By (2), each $ℛ ⁎$ class of $Q ( T , X , Y )$ is unipotent. Hence, $( a , 1 ) = ( b , 1 )$ . That is, $a = b$ . The converse is true from the proof of (2).□

Remark 3.1

From Lemma 2.4, it can be seen that every left type B semigroup has a least right cancellative monoid congruence. Therefore, there exists a least right cancellative monoid congruence $σ$ (refer to Lemma 2.4) on $Q ( T , X , Y )$ from Theorem 3.1(2).

Theorem 3.2

Let $( T , X , Y )$ be a left admissible triple and let $σ$ be the least right cancellative monoid congruence on $Q ( T , X , Y )$ . Then the following statements are true:

If $( a , t ) , ( b , u ) ∈ Q ( T , X , Y )$ , then $( a , t ) σ ( b , u )$ if and only if $t = u$ ;
$σ ∩ ℛ ⁎ = ι Q$ , where $ι Q$ is the identity relation on $Q ( T , X , Y )$ ;
$Q ( T , X , Y ) / σ ≅ T$ .

Proof

Let $( a , t ) , ( b , u ) ∈ Q ( T , X , Y )$ such that $( a , t ) σ ( b , u )$ . Then, by Lemma 2.4, there exists $( c , 1) ∈ E ( Q ( T , X , Y ) )$ such that $( a , t ) ( c , 1 ) = ( b , u ) ( c , 1 )$ . That is, $( a ∧ t c , t ) = ( b ∧ u c , u ) .$ Thus, $t = u$ .

Conversely, let $( a , t ) , ( b , t ) ∈ Q ( T , X , Y )$ . Then $a ≤ t i$ , $b ≤ t i$ . By condition (Q5) of Definition 3.1, we have that $a ∧ t ( a ∧ b ) = b ∧ t ( a ∧ b )$ and so $( a , t ) ( a ∧ b , 1 ) = ( b , t ) ( a ∧ b , 1 )$ . In other words, $( a , t ) σ ( b , t )$ .
It follows directly from Theorems 3.1(3) and 3.2(1).
Define the mapping $ϕ$ as follows:

$ϕ : Q ( T , X , Y ) / σ → T , ( a , t ) σ ↦ t .$

Then $ϕ$ is well-defined. In fact, let $( a , t ) σ , ( b , u ) σ ∈ Q ( T , X , Y ) / σ$ and $( a , t ) σ = ( b , u ) σ$ . It follows that $t = u$ from $(1)$ . This shows that $ϕ$ is well-defined. Clearly, $ϕ$ is a homomorphism from $Q ( T , X , Y ) / σ$ onto T. Next, we only show that $ϕ$ is injective. To see it, let $( a , t ) σ , ( b , u ) σ ∈ Q ( T , X , Y ) / σ$ such that $ϕ [ ( a , t ) σ ] = ϕ [ ( b , u ) σ ]$ . Then $t = u$ from the definition of $ϕ$ . It follows that $( a , t ) σ = ( b , u ) σ$ from Theorem 3.2(1). Therefore, $ϕ$ is injective.

Summarizing the aforementioned arguments, we conclude that $Q ( T , X , Y ) / σ ≅ T$ . This completes the proof.□

For convenience, we denote $Q ( T , X , Y )$ by Q and call $Q ( T , X , Y )$ a $Q -$ semigroup.

Theorem 3.3

Every proper left type B semigroup is isomorphic to a $Q -$ semigroup.

Proof

Let S be a proper left type B semigroup. Then $E ( S )$ is a semilattice. Put $S / σ = T$ , where $σ$ is the least right cancellative monoid congruence on S. Then T is a right cancellative monoid and $E ( S )$ is the identify of T. Define a relation “ $⊲$ ” on $E ( S ) × T$ as follows:

$( ∀ e , f ∈ E ( S ) , a , b ∈ S ) ( e , a σ ) ⊲ ( f , b σ ) ⇔ ( ∃ c ∈ S ) c + = e , c = c f , and b σ = a σ c σ .$

Obviously, “ $⊲$ ” is reflexive. Next, we prove that “ $⊲$ ” is transitive. To see it, let $( e , a σ ) , ( b σ ) , ( g , c σ ) ∈ E ( S ) × T$ with $( e , a σ ) ⊲ ( f , b σ ) , ( f , b σ ) ⊲ ( g , c σ )$ . Then there exist $x , y ∈ S$ such that $b σ = a σ x σ$ and $c σ = b σ y σ$ and that $x + = e , x = x f , y + = f$ and $y = y g$ . Hence, $c σ = a σ x σ y σ = a σ ( x y ) σ$ . Since $y ℛ ⁎ f$ and $ℛ ⁎$ is a left congruence. We have $x y ℛ ⁎ x f = x$ . Note that each $ℛ ⁎$ class of S contains only one idempotent. We get $( x y ) + = x + = e$ . But $x y g = x ( y g ) = x y$ . That is, $( e , a σ ) ⊲ ( g , c σ )$ . We deduce that “ $⊲$ ” is transitive. However, the relation “ $⊲$ ” is not a partial order. Now, we define a relation “ $π$ ” on $E ( S ) × T$ as follows:

$( e , a σ ) π ( f , b σ ) ⇔ ( e , a σ ) ⊲ ( f , b σ ) and ( f , b σ ) ⊲ ( e , a σ ) .$

Obviously, “ $π$ ” is an equivalence relation. For $A , B ∈ ( E ( S ) × T ) / π$ , we give a definition as follows:

$A ≤ B ⇔ ( ∃ ( e , a σ ) ∈ A , ( f , b σ ) ∈ B ) ( e , a σ ) ⊲ ( f , b σ ) .$

It is clear that $≤$ is a partial order of $( E ( S ) × T ) / π$ . We write $X = ( E ( S ) × T ) / π$ , which is a partial order set by $≤$ . Note that for all $A , B ∈ X$ . We have that:

$A ≤ B ⇔ ( ∀ ( e , a σ ) ∈ A , ( f , b σ ) ∈ B ) ( e , a σ ) ⊲ ( f , b σ ) .$

For $( e , a σ ) π ∈ X$ and $b σ ∈ T$ , we define $b σ ( e , a σ ) π = ( e , b a σ ) π$ . Obviously, it is well-defined. In fact, let $( e , a σ ) π = ( f , c σ ) π$ and $b σ = d σ$ . Then there exist $x , y ∈ S$ such that $c σ = a σ x σ$ and $a σ = c σ y σ$ , where $x + = e , x = x f , y + = f$ and $y = y e$ . Hence, we have

$( d c ) σ = d σ a σ x σ = ( b a ) σ x σ , ( b a ) σ = b σ a σ = b σ c σ y σ = ( d c ) σ y σ .$

Therefore, $( e , b a σ ) π = ( f , d c σ ) π .$ It means that for all $A , B ∈ X , t ∈ T$ , $A ≤ B$ implies $t A ≤ t B$ .

Now, we put $Y = { ( e , E ( S ) ) π | e ∈ E ( S ) }$ . Then, by the definition of “ $π$ ,” we obtain that $( e , E ( S ) ) ⊲ ( f , E ( S ) ) ⇔ e ≤ f$ , where $≤$ is the partial order of $E ( S )$ . Hence,

(3.1)

$( e , E ( S ) ) π ≤ ( f , E ( S ) ) π ⇔ e ≤ f$

Therefore, the largest element of Y is $( 1 , E ( S ) ) π$ .

Next, we verify that for all $f ∈ E ( S ) , a ∈ S$ we have

(3.2)

$( f , a σ ) π ∧ ( a + , E ( S ) ) π = ( ( a f ) + , E ( S ) ) π .$

In fact, from $( a f ) + a + = ( a f ) +$ , we have that $( a f ) + ≤ a +$ . By (3.1), $( ( a f ) + , E ( S ) ) π ≤ ( a + , E ( S ) ) π$ . Again since $a σ = ( a f ) σ = E ( S ) ⋅ ( a f ) σ$ and $a f = a f f$ . We have that $( ( a f ) + , E ( S ) ) ⊲ ( f , a σ )$ from the definition of “ $π$ .” Hence, $( ( a f ) + , E ( S ) ) π ≤ ( f , a σ ) π .$ This shows that $( ( a f ) + , E ( S ) ) π$ is a lower bound of $( f , a σ ) π$ and $( a + , E ( S ) ) π$ . Next, we show that $( ( a f ) + , E ( S ) ) π$ is the greatest lower bound of $( f , a σ ) π$ and $( a + , E ( S ) ) π$ . To see it, suppose that $( k , b σ ) π$ is another lower bound of $( f , a σ ) π$ and $( a + , E ( S ) ) π$ . Then $( k , b σ ) π ≤ ( a + , E ( S ) ) π$ . Hence, there exists $c ∈ S$ such that $E ( S ) = b σ ⋅ c σ$ , $c + = k$ and $c a + = c$ . Note that T is a right cancellative monoid. We have that $b σ ⋅ c σ$ is the identify of T and that the inverse of $b σ$ is $c σ$ . Hence, there exists $d ∈ S$ such that $( c a ) σ = E ( S ) ⋅ d σ$ , $d + = k$ and $d f = d .$ Again since $c a + = c$ and $ℛ ⁎$ is a left congruence, $c a ℛ ⁎ c a + = c$ . However, $c ℛ ⁎ k ℛ ⁎ d$ . Hence, $( c a , d ) ∈ ℛ ⁎ ∩ σ .$ Since S is a proper type B semigroup, we have that $c a = d$ . Furthermore, $c a f = d f = d = c a$ . That is, $c a = c ( a f ) ℛ ⁎ c ( a f ) +$ . And since $c a ℛ ⁎ k$ , we have that $c ( a f ) + ℛ ⁎ c$ . But $c ( a f ) + ℛ ⁎ σ c$ and S is proper. We have that $c = c ( a f ) +$ . These together with $c ℛ ⁎ k$ , we obtain that $( k , E ( S ) ) π ≤ ( ( a f ) + , c σ ) π$ . Therefore, $( k , b σ ) π ≤ ( ( a f ) + , E ( S ) ) π$ . That is, $( ( a f ) + , E ( S ) ) π$ is the greatest lower bound of $( f , a σ ) π$ and $( a + , E ( S ) ) π$ . Thus, (3.2) holds.

In particular, if we choose $a = e$ in (3.2), then

$( ∀ e , f ∈ E ( S ) ) ( e , E ( S ) ) π ∧ ( f , E ( S ) ) π = ( e f , E ( S ) ) π .$

Hence, Y is a subsemilattice of X.

Define a mapping as follows:

$θ : E ( S ) → Y , e ↦ ( e , E ( S ) ) π .$

It is easy to see that $θ$ is an order automorphism mapping. Now, we define another mapping as follows:

$ψ : S → Y × T , x ↦ ( ( x + , E ( S ) ) π , x σ ) .$

Then we can obtain that $ψ$ is an injective mapping. Define a multiplication on $Im ψ$ as follows:

$( ( x + , E ( S ) ) π , x σ ) ( ( y + , E ( S ) ) π , y σ ) = ( ( x + , E ( S ) ) π ∧ x σ ( y + , E ( S ) ) π , x y σ ) .$

Since $ℛ ⁎$ is a left congruence, we have that $x y ℛ ⁎ x y +$ and each $ℛ ⁎$ class of S contains only one idempotent. We have that $( x y ) + = ( x y + ) + .$ Hence,

$( ( x + , E ( S ) ) π , x σ ) ( ( y + , E ( S ) ) π , y σ ) = ( ( x + , E ( S ) ) π ∧ x σ ( y + , E ( S ) ) π , x y σ ) = ( ( x + , E ( S ) ) π ∧ ( y + , x σ ) π , x y σ ) = ( ( ( x y + ) + , E ( S ) ) π , x y σ ) = ( ( ( x y ) + , E ( S ) ) π , x y σ ) .$

Therefore, $Im ψ$ is closed with respect to the above multiplication. This means that $Im ψ$ is a monoid with identify $( ( 1 , E ( S ) ) π , E ( S ) )$ .

Finally, we consider the Q-semigroup $Q = Q ( T , X , Y )$ , where

$T = S / σ , X = ( E ( S ) × S / σ ) / π , Y = { ( e , E ( S ) ) π | e ∈ E ( S ) } .$

It is clear that Q is a subsemigroup of $Y × T$ and that $Im ψ$ is a subsemigroup of Q. We get $Im ψ = Q$ from (3.2). On the other hand, it is easily seen that $ψ : S → Im ψ$ is an isomorphism. Therefore, $S ≅ Q$ . This completes the proof.□

Remark 3.2

From Theorem 3.3, S is an arbitrary proper left type B semigroup with the least right cancellative monoid congruence $σ$ on it. Put
$T = S / σ , X = ( E ( S ) × S / σ ) / π , Y = { ( e , E ( S ) ) π | e ∈ E ( S ) } ,$
where $π$ is an equivalence relation on $E ( S ) × T .$ In our Theorem 3.3, we have proved that Y is a subsemilattice of a partial set $X$ . It is easy to see that $Y = Y i ,$ where $i = ( 1 , E ( S ) ) π .$
In fact, in the first half proof of Theorem 3.3, we can see that $( T , X , Y )$ satisfies conditions (Q1–Q6) listed in Definition 3.1. For example, let $a σ ∈ T , ( e , E ( S ) ) π ∈ Y i$ . Then $a σ ⋅ ( e , E ( S ) ) π = ( e , a σ E ( S ) ) π = ( e , a σ ) π ∈ X$ since $E ( S )$ is the identity of $T ,$ which implies that $T Y i ⊆ X$ . Conversely, for all $( e , a σ ) π ∈ X$ , we have $( e , a σ ) π = ( e , a σ E ( S ) ) π = a σ ⋅ ( e , E ( S ) ) π ∈ T Y i .$ That is, condition (Q1) holds. Similarly, let $a σ ∈ T$ then $a σ ⋅ i = a σ ⋅ ( 1 , E ( S ) ) π = ( 1 , a σ ) π$ . From the proof of Theorem 3.3, it is easy to see that there is $( a + , E ( S ) ) π ∈ Y$ such that $( a + , E ( S ) ) π ≤ ( 1 , a σ ) π$ since $( a + , E ( S ) ) π$ is the greatest lower bound of $( 1 , a σ ) π$ and $( 1 , E ( S ) ) π$ . This means that condition (Q2) holds. Again, let $( e , E ( S ) ) π , ( f , E ( S ) ) π ∈ Y$ and $a σ ∈ T$ such that $( e , E ( S ) ) π ≤ a σ ⋅ i = a σ ( 1 , E ( S ) ) π = ( 1 , a σ ) π .$ By the proof of Theorem 3.3, $( ( a f ) + , E ( S ) ) π$ is the greatest lower bound of $( f , a σ ) π$ and $( a + , E ( S ) ) π$ . Again, since $( a + , E ( S ) ) π$ is the greatest lower bound of $( 1 , a σ ) π$ and $( a + , E ( S ) ) π$ . We have that $( e , E ( S ) ) π ≤ ( a + , E ( S ) ) π$ , and so that

$( ( e , E ( S ) ) π ∧ a σ ( f , E ( S ) ) π ) = ( ( e , E ( S ) ) π ∧ ( f , a σ E ( S ) ) π ) = ( ( e , E ( S ) ) π ∧ ( f , a σ ) π ) = ( ( e , E ( S ) ) π ∧ ( a + , E ( S ) ) π ∧ ( f , a σ ) π ) = ( ( e , E ( S ) ) π ∧ ( ( a f ) + , E ( S ) ) π ) = ( e ( a f ) + , E ( S ) ) π ) ∈ Y .$

This means that (T, X, Y) satisfies condition (Q3) of Definition 3.1. By using a simple calculation similar to the proof of (Q3), it is easily seen that conditions (Q5) and (Q6) hold.

Finally, we explain (Q4) is also true. To see it, let $( e , E ( S ) ) π ,$ $( f , E ( S ) ) π ,$ $( g , E ( S ) ) π ∈ Y$ and $a σ , b σ ∈ T$ such that $( e , E ( S ) ) π ≤ a σ ( 1 , E ( S ) ) π$ and $( f , E ( S ) ) π ≤ b σ ( 1 , E ( S ) ) π$ . That is, $( e , E ( S ) ) π ≤ ( 1 , a σ ) π$ and $( f , E ( S ) ) π ≤ ( 1 , a σ ) π$ . Then $( e , E ( S ) ) π ≤ ( a + , E ( S ) ) π$ , $( f , E ( S ) ) π ≤ ( b + , E ( S ) ) π$ and $( f , a σ ) π ≤ ( 1 , ( a b ) σ ) π$ . Hence, $( f , a σ ) π ≤ ( ( a b ) + , E ( S ) ) π$ , and so

$[ ( e , E ( S ) ) π ∧ a σ ( f , E ( S ) ) π ] ∧ [ ( a σ ⋅ b σ ) ( g , E ( S ) ) π ] = ( e , E ( S ) ) π ∧ ( f , a σ ) π ∧ ( g , ( a b ) σ ) π = ( e , E ( S ) ) π ∧ ( a + , E ( S ) ) π ∧ ( f , a σ ) π ∧ ( ( a b ) + , E ( S ) ) π ∧ ( g , ( a b ) σ ) π = ( e , E ( S ) ) π ∧ ( ( a f ) + , E ( S ) ) π ∧ ( ( a b g ) + , E ( S ) ) π = ( e ( a f ) + ( a b g ) + , E ( S ) ) π = ( e ( a f ) + ( a ( b g ) + ) + , E ( S ) ) π ( since ( a b g ) + = ( a ( b g ) + ) + ) = ( e ( a f ( b g ) + ) + , E ( S ) ) π ( by ( L B 1 ) )$

and

$( e , E ( S ) ) π ∧ a σ [ ( f , E ( S ) ) π ∧ b σ ( g , E ( S ) ) π ] = ( e , E ( S ) ) π ∧ a σ [ ( f , E ( S ) ) π ∧ ( g , b σ ) π ] = ( e , E ( S ) ) π ∧ a σ [ ( f , E ( S ) ) π ∧ ( b + , E ( S ) ) π ∧ ( g , b σ ) π ] = ( e , E ( S ) ) π ∧ a σ [ ( f , E ( S ) ) π ∧ ( ( b g ) + , E ( S ) ) π ] = ( e , E ( S ) ) π ∧ a σ ( f ( b g ) + , E ( S ) ) π = ( e , E ( S ) ) π ∧ ( f ( b g ) + , a σ ) π = ( e , E ( S ) ) π ∧ ( a + , E ( S ) ) π ∧ ( f ( b g ) + , a σ ) π = ( e , E ( S ) ) π ∧ ( ( a f ( b g ) + ) + , E ( S ) ) π = ( e ( a f ( b g ) + ) + , E ( S ) ) π .$

Therefore, $( T , X , Y )$ listed in Theorem 3.3 satisfies condition (Q4) of Definition 3.1. In other words, $( T , X , Y )$ listed in Theorem 3.3 is a left admissible triple.

4 A characterization of a proper type B semigroup

In this section, we give a new characterization of a proper type B semigroup. For this purpose, we first give a characterization of Green $∗$ relation $ℒ ⁎$ on the Q-semigroup $Q = Q ( T , X , Y )$ and replace the condition “T is a right cancellative monoid” by “T is a cancellative monoid.”

Proposition 4.1

Let $Q = Q ( T , X , Y )$ be a $Q$ -semigroup and let T be a cancellative monoid with identify 1. If $( b , 1 ) , ( a , t ) ∈ Q$ , then $( b , 1 ) ℒ ⁎ ( a , t )$ if and only if it satisfies the following conditions:

$a ≤ t b$ ;
For all $c , d ∈ Y$ , $a ∧ t c = a ∧ t d$ implies $b ∧ c = b ∧ d$ .

Proof

(Necessity) Let $( b , 1 ) , ( a , t ) ∈ Q$ and $( b , 1 ) ℒ ⁎ ( a , t )$ . Then $( a , t ) ( b , 1 ) = ( a , t )$ from Corollary 2.2. That is, $( a ∧ t b , t ) = ( a , t )$ . Hence, $a ∧ t b = a$ and so $a ≤ t b$ . This shows that condition (1) is satisfied. Let $( c , u ) , ( d , v ) ∈ Q 1$ with $( a , t ) ( c , u ) = ( a , t ) ( d , v )$ . Then $a ∧ t c = a ∧ t d$ and $t u = t v$ . Note that T is a cancellative monoid. We have that $u = v$ . Again since $( b , 1 ) ℒ ⁎ ( a , t )$ , it follows that $( b , 1 ) ( c , u ) = ( b , 1 ) ( d , v )$ and so $b ∧ c = b ∧ d$ . Therefore, condition (2) is satisfied.

(Sufficiency) Suppose that conditions (1) and (2) hold. We have that $( a , t ) ( b , 1 ) = ( a ∧ t b , t ) = ( a , t )$ from condition (1). On the other hand, if there are $( c , u ) , ( d , v ) ∈ Q 1$ such that $( a , t ) ( c , u ) = ( a , t ) ( d , v )$ , then $( a ∧ t c , t u ) = ( a ∧ t d , t v )$ , and so $a ∧ t c = a ∧ t d$ and $t u = t v$ . Hence, $b ∧ c = b ∧ d$ from condition (2) and $u = v$ since T is cancellative. Therefore, $( b ∧ c , u ) = ( b ∧ d , v ) .$ That is, $( b , 1 ) ( c , u ) = ( b , 1 ) ( d , v )$ . By Corollary 2.2, we have that $( b , 1 ) ℒ ⁎ ( a , t )$ .□

Remark 4.1

In our Proposition 4.1, if both $( b ,1)$ and $( b ′ ,1)$ satisfy the conditions of Proposition 4.1, that is, $( b , 1 ) ℒ ⁎ ( a , t ) ℒ ⁎ ( b ′ , 1 )$ , then $a ≤ t b$ , $a ≤ t b ′$ and $a ∧ t b = a ∧ t b ′ = a$ . By condition (2), we obtain that $b ∧ b = b ∧ b ′$ and $b ′ ∧ b = b ′ ∧ b ′$ . Hence, $b = b ′$ . Thus, the element $( b , 1)$ is uniquely determined by the properties in Proposition 4.1. For given $( a , t ) ∈ Q$ , we denote b by $[ a , t ]$ . That is, $b = [ a , t ]$ . In other words, $( a , t ) ⁎ = ( [ a , t ] , 1 )$ .

For convenience, we restate the conditions in Proposition 4.1 as follows:

(Q7) Let T be a cancellative monoid with identify 1. For all $t ∈ T , a ∈ Y$ with $a ≤ t i$ and there exists a unique $[ a , t ] ∈ Y$ such that:

$a ≤ t [ a , t ]$ ;
For all $c , d ∈ Y$ , $a ∧ t c = a ∧ t d$ implies that $[ a , t ] ∧ c = [ a , t ] ∧ d$ .

It is easily seen that every $Q$ -semigroup $Q ( T , X , Y )$ satisfying conditions (Q1) to (Q7) is adequate.

Theorem 4.2

Let $Q = Q ( T , X , Y )$ be a $Q$ -semigroup satisfying conditions (Q1) to (Q7). Then Q is a type B semigroup if and only if the following two conditions hold:

(Q8) For all $a , b ∈ Y , t ∈ T ,$ $[ a ∧ b ∧ c , t ] = [ a ∧ b , t ] ∧ [ c ∧ b , t ]$ ;

(Q9) If for all $c ∈ Y , t ∈ T$ , $c = c ∧ [ a , t ]$ , then there exists $b ∈ Y$ such that $c = [ b ∧ a , t ]$ .

Proof

(Necessity) Let $Q ( T , X , Y )$ be a type B semigroup. Then for all $( b , t ) ∈ Q$ , $( a , 1),( c , 1) ∈ E ( Q 1 )$ we have

$[ ( a , 1 ) ( c , 1 ) ( b , t ) ] ⁎ = ( a ∧ b ∧ c , t ) ⁎ = ( [ a ∧ b ∧ c , t ] , 1 ) ( by Proposition 4.1 ) = [ ( a , 1 ) ( b , t ) ] ⁎ [ ( c , 1 ) ( b , t ) ] ⁎ ( by condition ( R B 1 ) ) = ( a ∧ b , t ) ⁎ ( c ∧ b , t ) ⁎ = ( [ a ∧ b , t ] , 1 ) ( [ c ∧ b , t ] , 1 ) ) ( by Proposition 4.1 ) = ( [ a ∧ b , t ] ∧ [ c ∧ b , t ] , 1 ) .$

Hence, $[ a ∧ b ∧ c , t ] = [ a ∧ b , t ] ∧ [ c ∧ b , t ]$ . Therefore, condition (Q8) holds.

On the other hand, if for all $( a , t ) ∈ Q$ , $( c , 1) ∈ E ( Q 1 )$ such that $( c , 1) ≤ ( a , t ) ⁎$ , then $( c , 1 ) ≤ ( [ a , t ] , 1 )$ from Proposition 4.1. Hence, $( c , 1 ) = ( c , 1 ) ( [ a , t ] , 1 ) = ( c ∧ [ a , t ] , 1 )$ and so $c = c ∧ [ a , t ]$ . Since $Q ( T , X , Y )$ is type B, we have that $( c , 1 ) = [ ( b , 1 ) ( a , t ) ] ⁎$ for some $( b , 1) ∈ E ( Q 1 )$ . That is, $( c , 1 ) = ( b ∧ a , t ) ⁎ = ( [ b ∧ a , t ] , 1 )$ from Proposition 4.1. Hence, $c = [ b ∧ a , t ]$ . Therefore, condition (Q9) holds.

(Sufficiency) By conditions (Q1) to (Q6), Q is a left type B semigroup from Theorem 3.1(2). Furthermore, it is easy to see that Q is an adequate semigroup from condition (Q7). Finally, by conditions (Q8) and (Q9), Q is a right type B semigroup. Therefore, Q is a type B semigroup.□

Proposition 4.3

Let $Q = Q ( T , X , Y )$ be a $Q$ -semigroup satisfying conditions (Q1) to (Q9). Then $ξ = σ .$

Proof

Note that Q is a type B semigroup from Theorem 4.2. By Lemmas 2.4 and 2.5, it is easy to see that $σ ⊆ ξ .$ Now, we prove that $ξ ⊆ σ .$ To see it, let $( a , t ) , ( b , u ) ∈ Q$ such that $( a , t ) ξ ( b , u ) .$ Then, by Lemma 2.5, there exists $( c , 1) ∈ E ( Q 1 )$ satisfying $( c , 1 ) ( a , t ) ( c , 1 ) = ( c , 1 ) ( b , u ) ( c , 1 ) .$ That is, $( c ∧ a ∧ t c , t ) = ( c ∧ b ∧ u c , u ) .$ Hence, $t = u$ . By Theorem 3.2(1), $( a , t ) σ ( b , u ) .$ This gives that $ξ ⊆ σ .$ Therefore, $ξ = σ .$ This completes the proof.□

Remark 4.2

From Proposition 4.3, it is easily seen that the least right cancellative monoid congruence on a $Q -$ semigroup $Q ( T , X , Y )$ is the least cancellative monoid congruence on $Q ( T , X , Y )$ (and vice versa) if $Q ( T , X , Y )$ satisfies conditions (Q1) to (Q9). Therefore, in the remaining, we can write $ξ = σ$ on $Q ( T , X , Y )$ satisfying conditions (Q1) to (Q9).

Theorem 4.4

Every $Q -$ semigroup $Q ( T , X , Y )$ satisfying conditions (Q1) to (Q9) is proper type B if and only if it satisfies the following condition:

(Q10) For all $a , b ∈ Y$ , if $a ≤ t i$ and $b ≤ t i$ , then $[ a , t ] = [ b , t ]$ implies $a = b$ .

Proof

(Sufficiency) Clearly, $Q ( T , X , Y )$ is a type B semigroup from Theorem 4.2. Suppose that condition (Q10) is satisfied. Let $( a , t ) , ( b , s ) ∈ Q$ with $( a , t )$ $( σ ∩ ℒ ⁎ ) ( b , s )$ . Then, we get $t = s$ from Theorem 3.2(1). By Proposition 4.1, $[ a , t ] = [ b , s ]$ . Hence, we have that $a = b$ from the hypothesis. This means that $( a , t ) = ( b , s )$ . That is, $σ ∩ ℒ ⁎ = ι Q .$ On the other hand, we obtain that $σ ∩ ℛ ⁎ = ι Q$ from Theorem 3.2(2). Therefore, $Q ( T , X , Y )$ is a proper type B semigroup.

(Necessity) Let $a , b ∈ Y , t ∈ T$ such that $a ≤ t i$ and $b ≤ t i .$ Then $( a , t ) , ( b , t ) ∈ Q$ from the definition of $Q ( T , X , Y )$ . Hence, by Theorem 2.2(2), $( a , t ) σ ( b , t )$ . Suppose that $[ a , t ] = [ b , t ]$ . By the proof of Proposition 4.1, we have $( a , t ) ⁎ = ( b , t ) ⁎$ . Hence, $( a , t ) ℒ ⁎ ( b , t )$ and so $( a , t ) ( σ ∩ ℒ ⁎ ) ( b , t )$ . Note that Q is proper. We have that $( a , t ) = ( b , t )$ , which implies that $a = b .$ Therefore, condition (Q10) is satisfied.□

Definition 4.1

Let $( T , X , Y )$ be a triple. Then $( T , X , Y )$ is said to be an admissible triple if it satisfies conditions (Q1) to (Q10).

Example 4.1

In our Example 3.1, denote $Q = { ( A , 1 ) , ( O , 1 ) , ( O , g ) }$ . It is easy to see that Q satisfies the condition of Definition 3.2 with $E ( Q ) = { ( A , 1 ) , ( O , 1 ) }$ and that Q satisfies conditions (Q1) to (Q6). Obviously, T is a cancellative monoid and that $( O , g ) ℒ ⁎ ( O , 1 )$ . By a simple calculation, we have that $[ A , 1 ] = A , [ O , 1 ] = O = [ O , g ]$ . Clearly, Q satisfies condition (Q7). Note that $E ( Q )$ is a semilattice and that $( O , g ) ℛ ⁎ ( O , 1 )$ , $( A , 1 ) ℛ ⁎ ( A , 1 )$ . We have that Q is an adequate semigroup. By Theorem 3.1, Q is left type B. It is easily seen that Q is also right type B. In other words, Q is a type B semigroup.

Next, we prove that Q satisfies conditions (Q8) to (Q10). In fact, for $A , O ∈ Y ,$ we have

$[ A ∧ O ∧ O , g ] = [ O , g ] = O = [ A ∧ O , g ] ∧ [ O ∧ O , g ]$

and

$[ A ∧ O ∧ O , 1 ] = [ O , 1 ] = O = [ A ∧ O , 1 ] ∧ [ O ∧ O , 1 ] ,$

which implies that condition (Q8) holds. On the other hand, for all $A , O ∈ Y ,$ $1 , g ∈ T ,$ we have

$A = A ∧ [ A , 1 ] ⇒ A = [ A ∧ A , 1 ] , O = O ∧ [ O , g ] ⇒ O = [ A ∧ O , g ] , O = O ∧ [ O , 1 ] ⇒ O = [ A ∧ O , 1 ] , O = O ∧ [ A , 1 ] ⇒ O = [ O ∧ A , 1 ] .$

This means that condition (Q9) holds. Finally, by Theorems 3.1(3) and 3.2(1), it is easy to check that Q is proper. Furthermore, we have that

$[ A , 1 ] = [ A , 1 ] , A ≤ 1 I and A = A , [ O , 1 ] = [ O , 1 ] , O ≤ 1 I and O = O , [ O , g ] = [ O , g ] , O ≤ g I and O = O .$

Therefore, condition (Q10) holds.

Summarizing the aforementioned arguments, Q is an admissible triple satisfying conditions (Q1) to (Q10).

Theorem 4.5

Let S be a proper type B semigroup. Then for some admissible triple $( T , X , Y )$ , $S ≅ Q ( T , X , Y )$ . Conversely, every $Q -$ semigroup constructed by an admissible triple of Definition 4.1 is a proper type B semigroup.

Proof

Let S be a proper type B semigroup. Then, by Theorem 3.3, $S = Q ( T , X , Y )$ for some left admissible triple $( T , X , Y )$ . By Theorem 3.2(3), $Q ( T , X , Y ) / σ ≅ T$ . But S is a type B semigroup and so $Q ( T , X , Y ) / σ$ is a cancellative monoid. Thus, T is a cancellative monoid. Then, by Proposition 4.1, Theorems 4.2 and 4.4, $Q ( T , X , Y )$ is a proper type B semigroup, where $( T , X , Y )$ is an admissible triple. Conversely, it follows directly from Proposition 4.1, Theorems 4.2 and 4.4.□

Corollary 4.6

Let $Q = Q ( T , X , Y )$ be a $Q -$ semigroup satisfying conditions (Q1) to (Q10). Then Q is an $E -$ unitary type B semigroup. Conversely, every $E -$ unitary type B semigroup can be constructed so.

Proof

It follows directly from Lemma 2.6 and Theorem 4.5.□

Acknowledgments

The authors are very grateful to the referees for their valuable suggestions which lead to an improvement of this paper. This work was supported by the NSF(CN) (No. 11261018 and 11961026) and the NSF of Jiangxi Province (No. 20181BAB201002).

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Received: 2020-02-27

Revised: 2020-09-06

Accepted: 2020-10-17

Published Online: 2020-12-31

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

Articles in the same Issue

https://doi.org/10.1515/math-2020-0104

Keywords for this article

type B semigroup; Q-semigroup; proper; E-unitary

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