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BY 4.0 license Open Access Published by De Gruyter Open Access November 24, 2022

The transitivity of primary conjugacy in regular ω-semigroups

  • Xin Liu and Shoufeng Wang EMAIL logo
From the journal Open Mathematics

Abstract

The conjugacy relation plays an important role in group theory and the conjugacy relation of groups has been generalized to semigroups in various methods by several authors. If a and b are elements of a semigroup S , then a is called primarily conjugate to b if a = u v and b = v u for some u , v S 1 . In general, primary conjugacy is reflexive and symmetric, but not transitive. Finding the classes of semigroups in which the primary conjugacy is transitive is an open problem raised by Araújo et al. in the literature. In this article, among other things we prove that the primary conjugacy is transitive in regular ω -semigroups.

MSC 2010: 20M10; 20M18

1 Introduction and preliminaries

The conjugacy relation of groups has many applications in the theory of groups, such as defining normal subgroups and studying the normalizers of elements in groups. It is an interesting topic to extend the concept of conjugation in groups to semigroups. Many authors have generalized the conjugacy relation of groups to some kinds of semigroups by various methods and obtained a series of results (see [1–15]). To state some necessary results on conjugacy relations of semigroups, we need to recall the notions of various kinds of conjugacy relations of semigroups.

Let G be a group. Then the conjugacy relation on G is defined as follows: For all a , b G , a b if and only if there exists g G such that b = g 1 a g . It is easy to see that

a b ( g G ) a g = g b ( u , v G ) a = u v , b = v u .

Inspired by this fact and the text [16], Lallement [17] has considered the conjugacy relations l and p on semigroups. Let S be a semigroup and a , b S . Recall that

a l b ( g S 1 ) a g = g b

and

a p b ( u , v S 1 ) a = u v , b = v u .

From [1], the relation p is called the primary conjugacy on S . According to Lallement [17], if S is a free semigroup, then p = l , and in this case, p is an equivalence. In a general semigroup, p is reflexive and symmetric, but not transitive. For this reason, the texts [1] and [2] investigated the transitive closure p of p and obtained some useful results. On the other hand, the relation l is reflexive and transitive, but not symmetric. In 1984, Otto [3] defined the relation o on a semigroup S as follows:

For all a , b S ,

a o b ( g , h S 1 ) a g = g b , b h = h a .

It is easy to see that o is an equivalence on S and p o as ( u v ) u = u ( v u ) and ( v u ) v = v ( u v ) for all u , v S 1 . Moreover, if e and f are two idempotents in S , we have e ( e f ) = e f = ( e f ) f and f ( f e ) = f e = ( f e ) e , and so e o f . Thus,

(1.1) e o f for all idempotents e and f in S .

However, if S has a zero, the o becomes the universal relation on S . To overcome this defect, Araújo et al. [4] defined the relation c on a semigroup S as follows: For all a , b S ,

a c b ( g P ( a ) ) ( h P ( b ) ) a g = g b , b h = h a ,

where P ( 0 ) = { 0 } and

P ( x ) = { g S 1 m S 1 ( m x 0 ( m x ) g 0 ) }

for all x S with x 0 . In view of the text [4], if S contains no zero, then c = o , and in general, c is contained in o properly. More recently, some new conjugacy relations on semigroups have been introduced and investigated, such as trace conjugacy t r for epigroups [5], i-conjugacy i for inverse semigroups [6], n -conjugacy n for general semigroups [8] and so on. Observe that i = n in inverse semigroups by [8, Theorem 2.6].

As mentioned above, the primary conjugacy p is not transitive in general semigroups. In fact, the primary conjugacy is not transitive in many well-known classes of semigroups, such as E -unitary inverse semigroups [5], polycyclic monoids [9], free inverse monoids [10], left or right patience sorting monoids [11], symmetric inverse semigroups [12,18], etc. In view of this fact, the study of p is divided into two aspects: One is to consider the descriptions of p and the transitive closure p of p for some interesting classes of semigroups (in particular, for various classes of (partial) transformation semigroups), the other is to find natural classes of semigroups in which p is transitive. For instance, Ganyushkin and Kormysheva [12] obtained the characterizations of p for finite symmetric inverse semigroups, Kudryavtseva [7] studied p for the class of regular epigroups and its some special classes, and Kudryavtseva and Mazorchuk [2] gave the descriptions of p for the full transformation semigroup and partial transformation semigroup on a finite set, symmetric inverse semigroup on a countable set and Brauer-type semigroups.

On the other hand, many authors have found some natural classes of semigroups in which p is transitive. For example, Otto [3] has proved that the primary conjugacy p is always transitive in monoids defined by special presentations (see also in Zhang [13]), and Kudryavtseva [7] has shown that p is transitive in completely regular semigroups. As a consequence, we have the following result which will be used several times in the next sections. Recall that Clifford semigroups are just semigroups that are both completely regular and inverse. It is well known that each -class is a subgroup in a Clifford semigroup and a Clifford semigroup is a strong semilattice of its -classes (see Howie [19]).

Lemma 1.1

The primary conjugacy p is transitive in Clifford semigroups.

Recently, Araújo et al. [5] have introduced the class W of semigroups which consists of semigroups S such that the subsemigroup S 2 = { a b a , b S } is completely regular, and proved that p is transitive in every epigroup in W . As a consequence, it is shown that p is transitive in the variants of completely regular semigroups. However, Example 4.19 in [5] indicates that there is an epigroup that has a variant in which p is not transitive. Based on this fact, the paper [5] has raised the following question (see Problem 6.18 in [5]): Is it true that p is transitive in the variants of the members in W ? In 2020, Borralho and Kinyon[14] provided an affirmative answer to this question (see Corollary 11 in [14]). In the same year, Borralho [15] has also shown that for semigroups satisfying x y { y x , ( x y ) n } for some fixed integer n > 1 , the primary conjugacy p is transitive.

From the aforementioned statements, we can see that it is an interesting research topic to find new classes of semigroups in which primary conjugacy is transitive. In fact, the four kinds of conjugacy relations of abstract semigroups including p , o , c , t r have already been studied systematically by Araújo et al. in [5] where the task to find new classes of semigroups in which primary conjugacy is transitive is also proposed (see Problem 6.3 in [5]). Specifically, they have raised the following question.

Problem 1.2

Find other classes of semigroups in which primary conjugacy is transitive. Describe the ( E -unitary) inverse semigroups in which primary conjugacy is transitive. Ultimately, classify the class of semigroups in which primary conjugacy is transitive.

The aim of this article is to continue the study of primary conjugacy relations of semigroups around the above problem. It is well known that the class of regular ω -semigroups is an important class of inverse monoids. Recall that a regular ω -semigroup is a regular monoid in which the idempotents form a chain that is isomorphic to the chain N 0 (the set of non-negative integers) with respect to the partial order

0 > 1 > 2 > 3 > 4 > .

By using the classification and construction of regular ω -semigroups given in Kočin [20], Munn [21] and Reilly [22] (see also in Howie [19], Petrich [23] and McAlister and Medeiros [24]), among other things we shall prove that the primary conjugacy p is transitive in any regular ω -semigroup.

In the remainder of this section, we shall recall the classification and construction of regular ω -semigroups which are given in Kočin [20], Munn [21] and Reilly [22]. We first recall the notion of the Bruck-Reilly extension of a monoid determined by a morphism θ . Let T be a monoid with identity 1 and let θ be a morphism from T into H 1 the “group of units” of T (i.e. the -class containing 1). Then we can make N 0 × T × N 0 into a monoid with identity ( 0 , 1 , 0 ) by defining

( m , a , n ) ( p , b , q ) = ( m n + t , ( a θ t n ) ( b θ t p ) , q p + t )

where t = max { n , p } and θ 0 is interpreted as the identity map of T . From Howie [19], this semigroup is denoted by BR ( T , θ ) and called the Bruck-Reilly extension of T determined by θ . Obviously, BR ( T , θ ) contains no zero whence c = o in BR ( T , θ ) . Now we can state the constructions of regular ω -semigroups.

Lemma 1.3

[20,21] Let S be a regular ω -semigroup. If S has no kernel, then S is a Clifford semigroup. If S has kernel K and S = K (or equivalently, S is simple), then S is a Bruck-Reilly extension BR ( T , θ ) of T determined by θ where T is a Clifford monoid whose idempotents form a chain.

Lemma 1.4

[20,21] Let G 0 , , G l 1 be a set of pairwise disjoint groups for some non-negative integer l and let K be a simple regular ω -semigroup, disjoint from each G i , with group of units G . Write G l = G . For each i such that 0 i l 1 , let γ i be a morphism of G i into G i + 1 . For all integers i , j with 0 i < j l , define α i , j to be γ i γ i + 1 γ j 1 and let α i , i be the identity automorphism of G i for all integers i with 0 i l 1 . Let

S = G 0 G 1 G l 1 K .

Define a multiplication on S, extending that of K and of each G i , as follows:

a i b j = ( a i α i , t ) ( b j α j , t ) , a i x = ( a i α i , l ) x , x a i = x ( a i α i , l ) , x y = x y ,

where a i G i , b j G j , t = max { i , j } , and x , y K , i , j = 0 , 1 , , l 1 . Then S is a regular ω -semigroup with kernel K S , and the identity of G 0 is the identity of the whole semigroup S. Conversely, every such semigroup can be constructed in this way.

2 Primary conjugacy in Bruck-Reilly extensions

In this section, we shall explore some technical results of the primary conjugacy in Bruck-Reilly extensions. We begin our discussions by giving the lemma below.

Lemma 2.1

Let T be a monoid and S = BR ( T , θ ) the Bruck-Reilly extension of T determined by θ . Then ( m , a , m ) p ( n , b , n ) in S if and only if a p b in T for all ( m , a , m ) , ( n , b , n ) S .

Proof

Suppose that ( m , a , m ) , ( n , b , n ) S and ( m , a , m ) p ( n , b , n ) . Then there are ( u , c , v ) , ( s , d , t ) S such that

( m , a , m ) = ( u , c , v ) ( s , d , t ) = ( u v + p , ( c θ p v ) ( d θ p s ) , t s + p ) , ( n , b , n ) = ( s , d , t ) ( u , c , v ) = ( s t + q , ( d θ q t ) ( c θ q u ) , v u + q ) ,

where p = max { v , s } and q = max { t , u } . This implies that

u v = t s , a = ( c θ p v ) ( d θ p s ) , b = ( d θ q t ) ( c θ q u ) ,

and so v s = u t . If v s 0 , then p = v and q = u . Otherwise, we have p = s and q = t . In either case, we have p v = q u and p s = q t , and hence a p b in T .

Conversely, let a p b in T . Then a = u v and b = v u with u , v T . This implies that

( m , a , n ) = ( m , u , n ) ( n , v , m ) and ( n , b , n ) = ( n , v , m ) ( m , u , n ) ,

which gives that ( m , a , m ) p ( n , b , n ) in S .□

Lemma 2.2

Let T be a monoid, S = BR ( T , θ ) the Bruck-Reilly extension of T determined by θ and ( m 1 , a , n 1 ) , ( m 2 , b , n 2 ) S such that ( m 1 , a , n 1 ) p ( m 2 , b , n 2 ) . Then m 1 n 1 = m 2 n 2 , say p, and at least one of the following conditions is satisfied:

  1. There exist q Z , c , d T such that p q 0 , a = c ( d θ q p ) , and b = d ( c θ q ) .

  2. There exist q Z , c , d T such that

    0 q min { m 1 , n 1 } , q m 1 m 2 , a = c ( d θ n 1 q ) , b = ( d θ m 1 q ) c .

  3. There exist q Z , c , d T such that

    0 q min { m 1 , n 1 } , q m 1 m 2 , a = ( c θ m 1 q ) d , b = d ( c θ n 1 q ) .

  4. There exist q Z , c , d T such that 0 q p , a = ( c θ p q ) d and b = ( d θ q ) c .

Proof

Suppose that ( m 1 , a , n 1 ) p ( m 2 , b , n 2 ) . Then there are ( m 3 , c , n 3 ) , ( m 4 , d , n 4 ) S such that

( m 1 , a , n 1 ) = ( m 3 , c , n 3 ) ( m 4 , d , n 4 ) = ( m 3 n 3 + t , ( c θ t n 3 ) ( d θ t m 4 ) , n 4 m 4 + t ) , ( m 2 , b , n 2 ) = ( m 4 , d , n 4 ) ( m 3 , c , n 3 ) = ( m 4 n 4 + s , ( d θ s n 4 ) ( c θ s m 3 ) , n 3 m 3 + s ) ,

where t = max { n 3 , m 4 } , s = max { n 4 , m 3 } . This implies that

(2.1) m 1 = m 3 n 3 + t , n 1 = n 4 m 4 + t , a = ( c θ t n 3 ) ( d θ t m 4 ) , m 2 = m 4 n 4 + s , n 2 = n 3 m 3 + s , b = ( d θ s n 4 ) ( c θ s m 3 ) ,

and so m 1 n 1 = m 3 n 3 + m 4 n 4 = m 2 n 2 . Denote

(2.2) m 1 n 1 = m 3 n 3 + m 4 n 4 = m 2 n 2 = p .

If m 4 n 3 and m 3 n 4 , then t = n 3 and s = n 4 , and so m 1 = m 3 , m 2 = m 4 and n 2 = n 3 m 3 + n 4 by (2.1). Let q = n 3 n 2 . Then

0 m 3 n 4 = q = n 3 n 2 m 4 n 2 = m 2 n 2 = p .

Observe that t m 4 = n 3 m 2 = n 3 n 2 p = q p by (2.2) and s m 3 = n 4 m 3 = q , it follows that a = c ( d θ q p ) and b = d ( c θ q ) by (2.1) again. In this case, (i) holds.

If m 4 n 3 and m 3 n 4 , then t = n 3 and s = m 3 , and so m 1 = m 3 , n 2 = n 3 and n 1 = n 4 m 4 + n 3 by (2.1).

This implies that

(2.3) n 4 = n 1 n 3 + m 4 = n 1 n 2 + m 4 = m 1 m 2 + m 4

by (2.2). Let q = n 4 . Then by the facts that 0 m 4 n 3 and m 3 n 4 0 ,

0 q = n 4 m 3 = m 1 , q = n 1 + m 4 n 3 n 1 , q = m 1 m 2 + m 4 m 1 m 2 .

Since t m 4 = n 3 m 4 = n 1 n 4 = n 1 q by (2.3) and s n 4 = m 3 n 4 = m 1 q , we have a = c ( d θ n 1 q ) and b = ( d θ m 1 q ) c by (2.1) again. In this case, (ii) holds.

If m 4 n 3 and m 3 n 4 , then t = m 4 and s = n 4 , and so m 2 = m 4 , n 1 = n 4 and m 1 = m 3 n 3 + m 4 by (2.1). This yields that

(2.4) m 3 = m 1 m 4 + n 3 = m 1 m 2 + n 3 .

Let q = m 3 . Then by the facts m 4 n 3 0 and 0 m 3 n 4 ,

q = m 3 n 4 = n 1 , q = m 1 + n 3 m 4 m 1 , q = m 1 m 2 + n 3 m 1 m 2 .

As t n 3 = m 4 n 3 = m 1 m 3 = m 1 q by (2.4) and s m 3 = n 4 m 3 = n 1 q , we obtain a = ( c θ m 1 q ) d and b = d ( c θ n 1 q ) by (2.1) again. In this case, (iii) is true.

If m 4 n 3 and m 3 n 4 , then t = m 4 and s = m 3 , and so n 2 = n 3 , n 1 = n 4 and m 1 = m 3 n 3 + m 4 by (2.1). This implies that

m 3 n 1 = m 1 + n 3 m 4 n 1 = m 1 n 1 + n 3 m 4 = n 3 m 4 + p

by (2.2). Let q = m 3 n 1 . Then 0 m 3 n 4 = q = m 3 n 1 = n 3 m 4 + p p by the facts m 4 n 3 and m 3 n 4 = n 1 . Observe that

t n 3 = m 4 n 3 = m 1 m 3 = n 1 + p m 3 = p q

by the fact that m 1 = m 3 n 3 + m 4 and (2.2), and s n 4 = m 3 n 4 = m 3 n 1 = q , it follows that a = ( c θ p q ) d and b = ( d θ q ) c by (2.1) again. In this case, (iv) is valid.□

Lemma 2.3

Let T be a Clifford monoid with the identity 1 in which the idempotents form a chain, and S = BR ( T , θ ) be the Bruck-Reilly extension of T determined by θ . Then p on S is transitive.

Proof

Let ( m 1 , a , n 1 ) , ( m 2 , b , n 2 ) , and ( m 3 , c , n 3 ) be in S such that ( m 1 , a , n 1 ) p ( m 2 , b , n 2 ) and ( m 2 , b , n 2 ) p ( m 3 , c , n 3 ) . Then by Lemma 2.2, we can denote

(2.5) m 1 n 1 = m 2 n 2 = m 3 n 3 = p ,

and have 16 cases including

( i , i ) , ( i , ii ) , ( i , iii ) , ( i , iv ) , ( ii , i ) , ( ii , ii ) , ( ii , iii ) , ( ii , iv ) , ( iii , i ) , ( iii , ii ) , ( iii , iii ) , ( iii , iv ) , ( iv , i ) , ( iv , ii ) , ( iv , iii ) , ( iv , iv )

to consider. In the following statements, we shall check all the cases one by one. We first observe that

( x θ k ) 1 ( x θ k ) = ( x 1 θ k ) ( x θ k ) = ( x 1 x ) θ k = 1 = ( x x 1 ) θ k = ( x θ k ) ( x 1 θ k ) = ( x θ k ) ( x θ k ) 1

for any positive integer k and x T .

Case 1–(i, i). There exist q 1 , q 2 Z and u , v , s , t T such that

p q 1 0 , a = u ( v θ q 1 p ) , b = v ( u θ q 1 ) , p q 2 0 , b = s ( t θ q 2 p ) , c = t ( s θ q 2 ) .

In this case, by (2.5) we have

(2.6) n 3 + q 2 n 3 + p = m 3 , m 1 q 2 m 1 , n 3 + q 1 n 3 + p = m 3 , m 1 q 1 m 1 .

(1.1) q 1 p , q 1 0 , q 2 p , q 2 0 . In this case, we have 2 p < q 1 + q 2 < 0 , and so p < q 1 + q 2 p < p . If p < q 1 + q 2 p < 0 , then

0 < p q 1 q 2 < p , a = u ( s θ q 1 p ) ( c θ q 1 + q 2 p p ) [ ( s 1 θ q 1 p ) u 1 ] θ p ,

and m 1 + p q 1 q 2 m 1 = p q 1 q 2 > 0 and

( n 3 + q 1 + q 2 p ) m 3 = n 3 m 3 + q 1 + q 2 p = q 1 + q 2 p p > p p = 0

by (2.5). This implies that

( m 1 , a , n 1 ) = ( m 1 , u ( s θ q 1 p ) , n 3 + q 1 + q 2 p ) ( m 3 , c [ ( s 1 θ q 1 p ) u 1 ] θ p q 1 q 2 , m 1 + p q 1 q 2 ) , ( m 3 , c , n 3 ) = ( m 3 , c [ ( s 1 θ q 1 p ) u 1 ] θ p q 1 q 2 , m 1 + p q 1 q 2 ) ( m 1 , u ( s θ q 1 p ) , n 3 + q 1 + q 2 p ) ,

which gives that ( m 1 , a , n 1 ) p ( m 3 , c , n 3 ) .

If 0 < q 1 + q 2 p < p , then we have

c = t ( v θ q 2 ) ( a θ ( q 1 + q 2 ) ) [ ( v 1 θ q 2 ) t 1 ] θ p ,

and m 1 q 1 q 2 m 1 = q 1 q 2 > 0 and

( n 3 + q 1 + q 2 ) m 3 = q 1 + q 2 + n 3 m 3 = q 1 + q 2 p > 0 ,

by (2.5). This implies that

( m 1 , a , n 1 ) = ( m 1 , a [ ( v 1 θ q 2 ) t 1 ] θ q 1 + q 2 p , n 3 + q 1 + q 2 ) ( m 3 , t ( v θ q 2 ) , m 1 q 1 q 2 ) , ( m 3 , c , n 3 ) = ( m 3 , t ( v θ q 2 ) , m 1 q 1 q 2 ) ( m 1 , a [ ( v 1 θ q 2 ) t 1 ] θ q 1 + q 2 p , n 3 + q 1 + q 2 ) .

From now on, for each x T , we shall denote the idempotent in the -class containing x by 1 x . Thus for all x , y T , it follows that

1 x x = x = x 1 x , 1 x = x x 1 = x 1 x and 1 x y = y 1 x .

If q 1 + q 2 p = 0 , then we have

a = u ( s θ q 1 p ) ( c θ p ) [ ( s 1 θ q 1 p ) u 1 ] θ p , c = t ( v θ q 2 ) ( a θ p ) [ ( v 1 θ q 2 ) t 1 ] θ p

and so 1 u a = a and 1 t c = c . This implies that

a ( v 1 θ q 2 ) 1 u = a ( v 1 θ q 2 ) , c ( s 1 θ q 1 p ) 1 t = c ( s 1 θ q 1 p ) .

Since the idempotents of T form a chain, we have 1 u 1 t = 1 t or 1 u 1 t = 1 u . If 1 u 1 t = 1 u , then

a ( v 1 θ q 2 ) t 1 t ( v θ q 2 ) = a ( v 1 θ q 2 ) 1 u 1 t ( v θ q 2 ) = a ( v 1 θ q 2 ) 1 u ( v θ q 2 ) = a ( v 1 θ q 2 ) ( v θ q 2 ) = a .

This implies that

( m 1 , a , n 1 ) = ( m 1 , a ( v 1 θ q 2 ) t 1 , m 3 ) ( m 3 , t ( v θ q 2 ) , n 1 ) , ( m 3 , c , n 3 ) = ( m 3 , t ( v θ q 2 ) , n 1 ) ( m 1 , a ( v 1 θ q 2 ) t 1 , m 3 ) .

If 1 u 1 t = 1 t , then

c ( s 1 θ q 1 p ) u 1 u ( s θ q 1 p ) = c ( s 1 θ q 1 p ) 1 t 1 u ( s θ q 1 p ) = c ( s 1 θ q 1 p ) 1 t ( s θ q 1 p ) = c ( s 1 θ q 1 p ) ( s θ q 1 p ) = c .

This implies that

( m 1 , a , n 1 ) = ( m 1 , u ( s θ q 1 p ) , n 3 ) ( m 3 , c ( s 1 θ q 1 p ) u 1 , m 1 ) , ( m 3 , c , n 3 ) = ( m 3 , c ( s 1 θ q 1 p ) u 1 , m 1 ) ( m 1 , u ( s θ q 1 p ) , n 3 ) .

(1.2) q 1 = p , q 1 0 , q 2 p , q 2 0 . In this case, we have

a = u v , b = v ( u θ p ) , b = s ( t θ q 2 p ) , c = t ( s θ q 2 ) ,

and so a = u s ( c θ q 2 p ) ( s 1 u 1 ) θ p . This implies by (2.6) that

( m 1 , a , n 1 ) = ( m 1 , u s , n 3 + q 2 ) ( m 3 , c ( s 1 u 1 ) θ q 2 , m 1 q 2 ) , ( m 3 , c , n 3 ) = ( m 3 , c ( s 1 u 1 ) θ q 2 , m 1 q 2 ) ( m 1 , u s , n 3 + q 2 ) .

(1.3) q 1 p , q 1 = 0 , q 2 p , q 2 0 . In this case, we obtain

a = u ( v θ p ) , b = v u , b = s ( t θ q 2 p ) , c = t ( s θ q 2 ) ,

whence c = t ( v θ q 2 ) ( a θ q 2 ) [ ( v 1 θ q 2 ) t 1 ] θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , a [ ( v 1 θ q 2 ) t 1 ] θ q 2 p , n 3 + q 2 ) ( m 3 , t ( v θ q 2 ) , m 1 q 2 ) , ( m 3 , c , n 3 ) = ( m 3 , t ( v θ q 2 ) , m 1 q 2 ) ( m 1 , a [ ( v 1 θ q 2 ) t 1 ] θ q 2 p , n 3 + q 2 ) .

(1.4) q 1 p , q 1 0 , q 2 = p , q 2 0 . In this case, we obtain

a = u ( v θ q 1 p ) , b = v ( u θ q 1 ) , b = s t , c = t ( s θ p ) ,

whence a = u ( s θ q 1 p ) ( c θ q 1 p ) [ ( s 1 θ q 1 p ) u 1 ] θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , u ( s θ q 1 p ) , n 3 + q 1 ) ( m 3 , c [ ( s 1 θ q 1 p ) u 1 ] θ q 1 , m 1 q 1 ) , ( m 3 , c , n 3 ) = ( m 3 , c [ ( s 1 θ q 1 p ) u 1 ] θ q 1 , m 1 q 1 ) ( m 1 , u ( s θ q 1 p ) , n 3 + q 1 ) .

(1.5) q 1 p , q 1 0 , q 2 p , q 2 = 0 . In this case, we obtain

a = u ( v θ q 1 p ) , b = v ( u θ q 1 ) , b = s ( t θ p ) , c = t s ,

whence c = ( t v ) ( a θ q 1 ) ( v 1 t 1 ) θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , a ( v 1 t 1 ) θ q 1 p , n 3 + q 1 ) ( m 3 , t v , m 1 q 1 ) , ( m 3 , c , n 3 ) = ( m 3 , t v , m 1 q 1 ) ( m 1 , a ( v 1 t 1 ) θ q 1 p , n 3 + q 1 ) .

(1.6) q 1 = p , q 1 = 0 . In this case, we obtain

q 2 = p = 0 , m 1 n 1 = m 2 n 2 = m 3 n 3 = p = 0 , a = u v , b = v u , b = s t , c = t s .

By Lemmas 1.1 and 2.1, ( m 1 , a , m 1 ) p ( m 3 , c , m 3 ) . It is easy to see that q 1 = p , q 1 = 0 if and only if q 2 = p , q 2 = 0 . Thus, all the cases that there exist at least three items are equal to zero in q 1 p , q 1 , q 2 p , q 2 reduce to this case.

(1.7)   q 1 = p , q 1 0 , q 2 = p , q 2 0 . In this case, we obtain

p = q 1 0 , a = u v , b = v ( u θ p ) , b = s t , c = t ( s θ p ) ,

whence a = ( u s ) c ( s 1 u 1 ) θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , u s , m 3 ) ( m 3 , c ( s 1 u 1 ) θ p , n 1 ) , ( m 3 , c , n 3 ) = ( m 3 , c ( s 1 u 1 ) θ p , n 1 ) ( m 1 , u s , m 3 ) .

(1.8)   q 1 = p , q 1 0 , q 2 p , q 2 = 0 . In this case, we obtain

p = q 1 0 , a = u v , b = v ( u θ p ) , b = s ( t θ p ) , c = t s ,

whence 1 t c = 1 t t s = t s = c , u 1 a = u 1 u v = 1 u v , t 1 c = t 1 t c = 1 t s and

1 u b = 1 u v ( u θ p ) = u 1 u v ( u θ p ) = u 1 a ( u θ p ) , 1 t b = 1 t s ( t θ p ) = t 1 c ( t θ p ) .

Since the idempotents of T form a chain, we have 1 u 1 t = 1 t or 1 u 1 t = 1 u . If 1 u 1 t = 1 t , then

1 t u 1 a ( u θ p ) = 1 t 1 u b = 1 u 1 t b = 1 u t 1 c ( t θ p ) , 1 t 1 t u 1 a ( u θ p ) = 1 t 1 u t 1 c ( t θ p ) = 1 t t 1 c ( t θ p ) .

Multiplying by t from the left on both sides of the second equation above, we have

( t u 1 ) a ( u θ p ) = t 1 t t 1 c ( t θ p ) = t t 1 c ( t θ p ) = 1 t c ( t θ p ) = c ( t θ p ) ,

and hence c = t u 1 a ( u θ p ) ( t 1 θ p ) = t u 1 a ( u t 1 ) θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , a ( u t 1 ) θ p , n 3 ) ( m 3 , t u 1 , m 1 ) , ( m 3 , c , n 3 ) = ( m 3 , t u 1 , m 1 ) ( m 1 , a ( u t 1 ) θ p , n 3 ) .

Similarly, we can prove the case that 1 u 1 t = 1 u .

(1.9)   q 1 p , q 1 = 0 , q 2 = p , q 2 0 . In this case, we obtain

p 0 , a = u ( v θ p ) , b = v u , b = s t , c = t ( s θ p ) ,

whence 1 u a = a , t 1 c = 1 t ( s θ p ) , u 1 a = 1 u ( v θ p ) and

1 u ( b θ p ) = 1 u ( v θ p ) ( u θ p ) = u 1 a ( u θ p ) , 1 t ( b θ p ) = 1 t ( s θ p ) ( t θ p ) = t 1 c ( t θ p ) , 1 u t 1 c ( t θ p ) = 1 u 1 t ( b θ p ) = 1 t 1 u ( b θ p ) = 1 t u 1 a ( u θ p ) .

Since the idempotents of T form a chain, we have 1 u 1 t = 1 t or 1 u 1 t = 1 u . If 1 u 1 t = 1 u , then

1 u 1 u t 1 c ( t θ p ) = 1 u 1 t u 1 a ( u θ p ) = 1 u u 1 a ( u θ p ) = u 1 a ( u θ p ) ,

Multiplying by u from the left on both sides of the equation above, we have

( u t 1 ) c ( t θ p ) = u u 1 a ( u θ p ) = 1 u a ( u θ p ) = a ( u θ p ) ,

and hence a = ( u t 1 ) c ( t θ p ) ( u 1 θ p ) = ( u t 1 ) c ( t u 1 ) θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , u t 1 , m 3 ) ( m 3 , c ( t u 1 ) θ p , n 1 ) , ( m 3 , c , n 3 ) = ( m 3 , c ( t u 1 ) θ p , n 1 ) ( m 1 , u t 1 , m 3 ) .

Similarly, we can prove the case that 1 u 1 t = 1 t .

(1.10)   q 1 p , q 1 = 0 , q 2 p , q 2 = 0 . In this case, we obtain

p 0 , a = u ( v θ p ) , b = v u , b = s ( t θ p ) , c = t s ,

whence c = t v a ( v 1 t 1 ) θ p . This yields that

( m 1 , a , n 1 ) = ( m 1 , a ( v 1 t 1 ) θ p , n 3 ) ( m 3 , t v , m 1 ) , ( m 3 , c , n 3 ) = ( m 3 , t v , m 1 ) ( m 1 , a ( v 1 t 1 ) θ p , n 3 ) .

Case 2–(i, ii). There exist q 1 , q 2 Z and u , v , s , t T such that

m 2 n 2 = p q 1 0 , a = u ( v θ q 1 p ) , b = v ( u θ q 1 ) ,

0 q 2 m 2 n 2 , q 2 m 2 m 3 , b = s ( t θ n 2 q 2 ) , c = ( t θ m 2 q 2 ) s .

In this case,

n 1 m 1 = n 3 m 3 = p 0 , n 3 + q 1 n 3 + p = m 3 , m 1 q 1 m 1 .

(2.1)   q 1 p , q 1 0 , q 2 n 2 , q 2 m 2 . In this case, we obtain

a = u ( t 1 θ n 2 + q 1 q 2 ) ( c θ q 1 p ) [ ( t θ n 2 + q 1 q 2 ) u 1 ] θ p .

This yields that

( m 1 , a , n 1 ) = ( m 1 , u ( t 1 θ n 2 + q 1 q 2 ) , n 3 + q 1 ) ( m 3 , c [ ( t θ n 2 + q 1 q 2 ) u 1 ] θ q 1 , m 1 q 1 ) , ( m 3 , c , n 3 ) = ( m 3 , c [ ( t θ n 2 + q 1 q 2 ) u 1 ] θ q 1 , m 1 q 1 ) ( m 1 , u ( t 1 θ n 2 + q