Transposition Regular AG-Groupoids and Their Decomposition Theorems

Abstract

In this paper, we introduce transposition regularity into AG-groupoids, and a variety of transposition regular AG-groupoids (L1/R1/LR, L2/R2/L3/R3-groupoids) are obtained. Their properties and structures are discussed by their decomposition theorems: (1) L1/R1-transposition regular AG-groupoids are equivalent to each other, and they can be decomposed into the union of disjoint Abelian subgroups; (2) L1/R1-transposition regular AG-groupoids are LR-transposition regular AG-groupoids, and an example is given to illustrate that not every LR-transposition regular AG-groupoid is an L1/R1-transposition regular AG-groupoid; (3) an AG-groupoid is an L1/R1-transposition regular AG-groupoid if it is an LR-transposition regular AG-groupoid satisfying a certain condition; (4) strong L2/R3-transposition regular AG-groupoids are equivalent to each other, and they are union of disjoint Abelian subgroups; (5) strong L3/R2-transposition regular AG-groupoids are equivalent to each other and they can be decomposed into union of disjoint AG subgroups. Their relations are discussed. Finally, we introduce various transposition regular AG-groupoid semigroups and discuss the relationships among them and the commutative Clifford semigroup as well as the Abelian group.

1. Introduction

The AG-groupoid (also known as the LA-semigroup [1]), first proposed by Kazim and Naseeruddin in 1972 (see [2]), is a groupoid that is a set together with the left invertive law. For example, let $G=\{a,b,c,d,e,f,g\}$ and define the operation on G as shown in

Table 1. This is an AG-groupoid.

*	a	b	c	d	e	f	g
a	a	a	a	a	a	a	a
b	a	a	c	c	a	c	c
c	a	b	a	b	a	b	b
d	a	b	c	d	e	f	g
e	a	a	a	e	e	a	a
f	a	b	c	f	a	f	f
g	a	b	c	g	a	f	g

:

In Table 1, for any elements $x,y,z\in G$, there is $\left(xy\right)z=\left(zy\right)x$; that is, G satisfies the left invertive law. So, G is an AG-groupoid.

The concept of a regular AG-groupoid is proposed in [3], the ideals of the AG-groupoid and regular AG-groupoid are studied, the definitions of completely regular AG-groupoid and fully regular AG-groupoid are given in [4], and the necessary and sufficient conditions for an AG-groupoid to be a fully regular AG-groupoid are explored. AG-groups and right AG-groups are proposed in [5], and AG-groups are studied more deeply (see [6,7,8]). At present, through the continuous research of many scholars, the nature and structure of the AG-groupoid have been deeply revealed, and many research results have been accumulated (see [9,10,11,12,13,14,15,16]).

The AG-groupoid is a kind of non-commutative and non-associative algebraic system midway a groupoid and commutative semigroup, but it retains the good properties in the structure of commutative and associative algebra. If an AG-groupoid contains a right identity, then it becomes a commutative monoid (see [17]). An AG-groupoid is the generalization of a semigroup theory and has vast applications in collaboration with semigroups (see [18]). Based on the close connections between semigroups and AG-groupoids, we introduce the transposition regularity proposed by Xiaohong Zhang and Yudan Du in [19] into AG-groupoids.

In this paper, we mainly introduce various transposition regular AG-groupoids, and their structures are characterized. The arrangement of this paper is as follows. In Section 2, we introduce the L1-transposition regular AG-groupoid and R1-transposition regular AG-groupoid, and their structures are described by decomposition theorems. In Section 3, we introduce the LR-transposition regular AG-groupoid and strong LR-transposition regular AG-groupoid, and their properties and structures are charaterzied. In Section 4, the concepts of the L2-transposition regular AG-groupoid and R2-transposition regular AG-groupoid are given, and their decompositon theorems are proved. In Section 5, we introduce the L3-transposition regular AG-groupoid and R3-transposition regular AG-groupoid and discuss the relationships among various transposition regular AG-groupoids. In Section 6, we introduce various transposition regular AG-groupoid semigroups, and their relationships among the commutative Clifford semigroup and Abelian group are given.

2. L1-Transposition Regular AG-Groupoid and R1-Transposition Regular AG-Groupoid

In this section, we give the definitions of transposition regular AG-groupoid and R1-transposition regular AG-groupoid.

Definition 1.

Let G be a groupoid. $a\in G$.

(1) If there exists $e\in G$ such that $ea=a(ae=a)$, e is said to be the local left (right) identity element of a. e is said to be the local identity element if e is both the local left identity element and local right identity element.

(2) Let e be a local left identity element/right identity element/identity elemnt of a. If there exists b such that $ba=e(ab=e)$, b is said to be a local left(right) inverse element of a relative to e. b is the local inverse element of a relative to e if b is both the local left inverse element of a relative to e and the local right inverse element of a relative to e.

Definition 2

([2]). A groupoid S satisfying the left invertive law is called as an AG-groupoid.

$$\left(ab\right)c=\left(cb\right)a,\forall a,b,c\in S.$$

In an AG-groupoid S, it satisfies the medial law:

$$\left(ab\right)\left(cd\right)=\left(ac\right)\left(bd\right),\forall a,b,c,d\in S.$$

Definition 3.

Let G be an AG-groupoid, $a\in G$. An element a is L1-transposition regular in G if $\exists p\in G$ s.t. $\left(pa\right)a=a=a\left(pa\right)$. The AG-groupoid G is said to be L1-transposition regular if all its elements are L1-transposition regular.

Example 1.

Let $G=\{a,b,c,d\}$. Define the operation on G as shown in

Table 2. L1-transposition regular AG-groupoid.

*	a	b	c	d
a	a	d	b	c
b	c	b	d	a
c	d	a	c	b
d	b	c	a	d

. Clearly, $(G,\ast )$ is an L1-transposition regular AG-groupoid. That is,

$$(a\ast a)\ast a=a=a\ast (a\ast a),(b\ast b)\ast b=b=b\ast (b\ast b),$$

$$(c\ast c)\ast c=c=c\ast (c\ast c),(d\ast d)\ast d=d=d\ast (d\ast d).$$

Clearly, G is not commutative.

Proposition 1.

Let G be an L1-transposition regular AG-groupoid. For any $a\in G$, let $pa=e,f=ee,m=ff$. Then, $\exists q\in G$, such that $ma=a=am,qm=q=mq,aq=m=qa$.

Proof. 

For any a in G, let $pa=e$. Then, $ea=a=ae$. So, e is a local identity element of a, and p is a local inverse element of a relative to e. According to the left invertive law,

$$\left(ee\right)a=\left(ae\right)e=ae=a,a\left(ee\right)=\left(ea\right)\left(ee\right)=\left(ee\right)\left(ae\right)=\left(ee\right)a=ae=a.$$

Furthermore,

$$ap=\left(ea\right)p=\left(pa\right)e=ee.$$

$$\left(pe\right)a=\left(ae\right)p=ap=ee.$$

So

$$a\left(pe\right)=\left(ea\right)\left(pe\right)=\left(ep\right)\left(ae\right)=\left(ep\right)a=\left(ap\right)e=\left(ee\right)e.$$

Furthermore,

$$a\left(pe\right)=\left(ae\right)\left(pe\right)=\left(ap\right)\left(ee\right)=\left(ee\right)\left(ee\right).$$

Then, $\left(ee\right)e=\left(ee\right)\left(ee\right).$

Let $ee=f$. Then, $fa=a=af$, $ap=f=\left(pe\right)a$, $fe=ff$.

According to left invertive law and medial law,

$$\left(ff\right)a=\left(af\right)f=af=a,a\left(ff\right)=\left(fa\right)\left(ff\right)=\left(ff\right)\left(af\right)=\left(ff\right)a=a.$$

Then $\left(ff\right)a=a=a\left(ff\right).$

However,

$$a\left(ep\right)=\left(ea\right)\left(ep\right)=\left(ee\right)\left(ap\right)=\left(ee\right)\left(ee\right)=ff.$$

$$\left(ep\right)a=\left(ep\right)\left(ea\right)=\left(ee\right)\left(pa\right)=\left(ee\right)e=fe=ff.$$

That is, $a\left(ep\right)=ff=\left(ep\right)a$. Furthermore, $ff$ is the local identity element of a, and $ep$ is local inverse element of a relative to $ff$. Let $m=ff$. Then, $ma=a=am$, $\left(ep\right)a=m=a\left(ep\right)$.

So, $mm=\left(\right(ep\left)a\right)m=\left(ma\right)\left(ep\right)=a\left(ep\right)=m$, and m is idempotent.

So,

$$m\left(m\right(ep\left)\right)=\left(\right(ep\left)a\right)\left(m\right(ep\left)\right)=\left(\right(ep\left)m\right)\left(a\right(ep\left)\right)=\left(\right(ep\left)m\right)m=\left(mm\right)\left(ep\right)=m\left(ep\right).$$

$$\begin{array}{ccc}\hfill \left(m\right(ep\left)\right)m& =& \left(m\right(ep\left)\right(mm)=(mm\left)\right(\left(ep\right)m)=m(\left(ep\right)m)\hfill \\ & =& \left(\right(ep\left)a\right)\left(\right(ep\left)m\right)=\left(\right(ep\left)\right(ep\left)\right)\left(am\right)\hfill \\ & =& \left(\right(ep\left)\right(ep\left)\right)a=\left(a\right(ep\left)\right)\left(ep\right)=m\left(ep\right).\hfill \end{array}$$

$$a\left(m\right(ep\left)\right)=\left(ma\right)\left(m\right(ep\left)\right)=\left(mm\right)\left(a\right(ep\left)\right)=mm=m.$$

$$\left(m\right(ep\left)\right)a=\left(a\right(ep\left)\right)m=mm=m.$$

Then, m is an identity element of $m\left(ep\right)$, and a is the local inverse element of $m\left(ep\right)$ relative to m. □

Theorem 1.

Let G be an L1-transposition regular AG-groupoid. Define the binary relation ≈ on G as follows:

$$a\approx b\iff e_a=e_b,\forall a,b\in G$$

where $e_a$ is a local identity element of a. Then

(1) 

The binary relation ≈ on G is equivalence relation, and we denote the equivalence class containing x by ${\left[x\right]}_{\approx }$;

(2) 

$\forall x\in G$, ${\left[x\right]}_{\approx }$ is an Abelian subgroup;

(3) 

$G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$, that is, every L1-transposition regular AG-groupoid is the disjoint union of Abelian subgroups.

Proof. 

(1) Clearly, $\forall x\in G$, $e_x=e_x$. That is, $x\approx x$.

Assume that $x\approx y$, then $e_x=e_y$, and $e_y=e_x$. So $y\approx x$.

If $x\approx y$ and $y\approx z$, then $e_x=e_y$, and $e_x=e_y$. Clearly, $e_x=e_z$. That is, $x\approx z$. So, ≈ is a equivalence relation on G.

(2) $\forall a,b\in {\left[x\right]}_{\approx }$, assume that $e_a=e_b=e$. $ab=\left(ae\right)b=\left(be\right)a=ba$. That is, ${\left[x\right]}_{\approx }$ satisfies the commutative law. Clearly, ${\left[x\right]}_{\approx }$ satisfies the left invertive law. Then, for any $a,b,c\in G$, $\left(ab\right)c=\left(cb\right)a=a\left(cb\right)=a\left(bc\right)$. So ${\left[x\right]}_{\approx }$ satisfies the associative law.

According to Proposition 1, there exists $p\in G$ s.t. $pa=e=ap,ea=a=ae,ep=p=pe.$ $\exists q\in G$ s.t. $qb=e=bq,eb=b=be,eq=q=qe.$ Then,

$$\left(ab\right)e=\left(eb\right)a=ba=ab, \mathrm{and} e\left(ab\right)=\left(ab\right)e=ab.$$

That is, $ab\in {\left[x\right]}_{\approx }$.

For any $a\in {\left[x\right]}_{\approx }$, $ee=e\left(ap\right)=\left(ea\right)p=ap=e$, then $e\in {\left[x\right]}_{\approx }$.

Because $ee=\left(pa\right)e=\left(ea\right)p=ap=e$, there is $e\in {\left[x\right]}_{\approx }$.

Because $ep=p=pe$, then $p\in {\left[x\right]}_{\approx }$. According to the definition of group, ${\left[x\right]}_{\approx }$ is an Abelian subgroup of G.

(3) For any $a\in {\left[x\right]}_{\approx }$, the local identity element of a is unique, then $G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$ and $\bigcap _{x\in G}{\left[x\right]}_{\approx }=⌀$. That is, every L1-transposition regular AG-groupoid is the disjoint union of subgroups. □

Definition 4.

Let G be an AG-groupoid. An element a is R1-transposition regular in G if $\exists p\in G$ s.t. $\left(ap\right)a=a=a\left(ap\right)$. The AG-groupoid G is said to be R1-transposition regular if all its elements are R1-transposition regular.

Example 2.

Let $G=\{a,b,c,d\}$. Define the operation on G as Table 2; then, we can get that $(G,\ast )$ is a R1-transposition regular AG-groupoid. There is,

$$(a\ast a)\ast a=a\ast a=a=a\ast (a\ast a),(b\ast b)\ast b=a\ast b=b=b\ast a=b\ast (b\ast b),$$

$$(c\ast d)\ast c=a\ast c=c=c\ast a=c\ast (c\ast d),(d\ast c)\ast d=a\ast d=d=d\ast a=d\ast (d\ast c).$$

Theorem 2.

G is an L1-transposition regular AG-groupoid if and only if it is an R1-transposition regular AG-groupoid.

Proof. 

(⇒) Let G be an L1-transposition regular AG-groupoid. For any $a\in G$, there exists $p\in G$ such that $\left(pa\right)a=a=a\left(pa\right)$. Let $pa=e$. There is, $ea=a=ae$. That is, e is the local identity element of a, and p is the local left inverse element of a relative to e. According to the left invertive law,

$$\left(ee\right)a=\left(ae\right)e=ae=a,a\left(ee\right)=\left(ea\right)\left(ee\right)=\left(ee\right)\left(ae\right)=\left(ee\right)a=ae=a.$$

Furthermore,

$$ap=\left(ea\right)p=\left(pa\right)e=ee.$$

$$\left(pe\right)a=\left(ae\right)p=ap=ee.$$

At this time,

$$\left(ee\right)e=\left(ap\right)e=\left(ep\right)a=\left(ep\right)\left(ae\right)=\left(ea\right)\left(pe\right)=a\left(pe\right)=\left(ae\right)\left(pe\right)=\left(ap\right)\left(ee\right)=\left(ee\right)\left(ee\right).$$

Let $ee=f$. There is $fa=a=af$, and $ap=f=\left(pe\right)a$. Then, $\left(ap\right)a=a=a\left(ap\right)$. So, a is R1-transposition regular, and G is an R1-transposition regular AG-groupoid.

(⇐) Let G be an R1-transposition regular AG-groupoid. For any $a\in G$, there exists $p\in G$ such that $a\left(ap\right)=a=\left(ap\right)a$. Let $ap=e$, there is $ea=a=ae$. Then, e is a local identity element of a, p is the local right inverse element of a relative to e. According to the left invertive law, there is, $\left(pe\right)a=\left(ae\right)p=ap=e.$

Furthermore, $\left(ee\right)a=\left(ae\right)e=ae=a,a\left(ee\right)=\left(ea\right)\left(ee\right)=\left(ee\right)\left(ae\right)=\left(ee\right)a=a.$ $\left(ep\right)a=\left(ap\right)e=ee.$

Then,

$$\left(\right(ep\left)a\right)a=a=a\left(\right(ep\left)a\right).$$

That is, a is L1-transposition regular, and G is an L1-transposition regular AG-groupoid. □

3. LR-Transposition Regular AG-Groupoid and Strong LR-Transposition Regular AG-Groupoid

In [19], for a semigroup S, for any element $a\in S$, there exists $p,q\in S$ such that $\left(pa\right)a=a=a\left(aq\right)$. However, in an AG-groupoid, we can get the whole equation through the right equation. So, we give the following definition:

Definition 5.

Let G be an AG-groupoid, $a\in G$. An element a is LR-transposition regular in G $\exists q\in G$ s.t. $a=a\left(aq\right)$. G is an LR-transposition regular AG-groupoid if all elements of G are LR-transposition regular.

Proposition 2.

Let G be an LR-transposition regular AG-groupoid. $\forall a\in G$, there exists $q\in G$ s.t. $a=a\left(aq\right)$. Then, there exists $p\in G$ s.t. $\left(pa\right)a=a$.

Proof. 

Assume that $p=q^{2}a$. Then,

$$\left(pa\right)a=\left(\left(q^{2}a\right)a\right)a=\left(\left(aa\right)q^2\right)a=\left(\left(aa\right)\left(qq\right)\right)a=\left(\left(aq\right)\left(aq\right)\right)a=\left(a\left(aq\right)\right)\left(aq\right)=a\left(aq\right)=a.$$

At this time, $\left(pa\right)a=a$. □

Example 3.

Let $G=\{a,b,c,d,e\}$. Define the operation $\ast$ on G in

Table 3. This is an LR-transposition regular AG-groupoid.

*	a	b	c	d	e
a	a	b	c	d	e
b	b	a	d	c	b
c	d	c	b	a	d
d	c	d	a	b	c
e	a	b	c	d	e

. Clearly, G is an LR-transposition regular AG-groupoid. That is,

$$(a\ast a)\ast a=a\ast a=a=a\ast a=a\ast (a\ast a),$$

$$(b\ast b)\ast b=a\ast b=b=b\ast a=b\ast (b\ast b),$$

$$(d\ast c)\ast c=a\ast c=c=c\ast b=c\ast (c\ast c),$$

$$(c\ast d)\ast d=a\ast d=d=d\ast b=d\ast (d\ast d),$$

$$(e\ast e)\ast e=e\ast e=e=e\ast e=e\ast (e\ast e).$$

Theorem 3.

Let G be an LR-transposition regular AG-groupoid. $\forall a\in G$, there exists $q\in G$ s.t. $a=a\left(aq\right)$. $\left(q^{2}a\right)a=aq$ if and only if it is an L1-transposition regular AG-groupoid.

Proof. 

(⇒) For any element a in G, let $p=q^{2}a$, and $pa=e=aq$. That is, $\left(pa\right)a=a=a\left(pa\right)$. Then, G is an L1-transposition regular AG-groupoid.

(⇐) Let G be an L1-transposition regular AG-groupoid. For any $a\in G$, there exists $p\in G$ s.t. $\left(pa\right)a=a=a\left(pa\right)$. According to Proposition 1, let $e=pa,f=ee,ffff=ff$, and $a\left(a\right(ep\left)\right)=a$, $a\left(ep\right)=ff$.

Then, let $q=ep$, there is $\left(\left(q^{2}a\right)a\right)a=a,aq=ff$. So,

$$\left(q^{2}a\right)a=\left(\left(qq\right)a\right)a=\left(aa\right)\left(qq\right)=\left(aq\right)\left(aq\right)=\left(ff\right)\left(ff\right)=ff=aq.$$

That is, $\left(q^{2}a\right)a=aq$. □

Example 4 shows that not every LR-transposition regular AG-groupoid is an L1-trans-position regular AG-groupoid.

Example 4.

Let $G=\{a,b,c,d,e\}$. Define the operation on G as Table 3. Then, $(G,\ast )$ is an LR-transposition regular AG-groupoid. Where $(a\ast c)\ast c=b\ne c,(b\ast c)\ast c=a\ne c,(c\ast c)\ast c=d\ne c,(e\ast c)\ast c=b\ne c\ast (e\ast c)$, and $(d\ast c)\ast c=c\ne d=c\ast (d\ast c)$, that is, there don’t exist $p\in G$ s.t. $(p\ast c)\ast c=c=c\ast (p\ast c)$. So, G is not an L1-transposition regular AG-groupoid.

Definition 6.

Let G be an LR-transposition regular AG-groupoid, $a\in G$. An element a is strong LR-transposition regular in G if $q^{2}a=q$. G is an LR-transposition regular AG-groupoid if every element of G is strong LR-transposition regular.

Example 5 shows that not every LR-transposition regular AG-groupoid is a strong LR-transposition regular AG-groupoid.

Example 5.

Let $G=\{a,b,c,d,e\}$. Define the operation on G in Table 3. Then, $(G,\ast )$ is an LR-transposition regular AG-groupoid. However, $c\ast (c\ast c)=c\ast b=c$, and $c^2\ast c=b\ast c=d\ne c$. That is, $(G,\ast )$ is not a strong LR-transposition regular AG-groupoid.

Example 6.

Let $G=\{a,b,c,d,e,f,g,h\}$. Define the operation on G in

Table 4. This is a strong LR-transposition regular AG-groupoid.

*	a	b	c	d	e	f	g	h
a	b	d	c	a	g	e	f	h
b	c	a	b	d	f	h	g	e
c	a	c	d	b	h	f	e	g
d	d	b	a	c	e	g	h	f
e	h	f	e	g	f	h	g	e
f	e	g	h	f	g	e	f	h
g	g	e	f	h	e	g	h	f
h	f	h	g	e	h	f	e	g

. Then, $(G,\ast )$ is a strong LR-transposition regular AG-groupoid. Let

$$p_a=\{p\in G|a=a\left(ap\right) and p^{2}a=p,a\in G\}.$$

Then

$$p_a=b,p_b=a,p_c=d,p_d=c,p_e=f,p_f=e,p_g=h,p_h=d.$$

Theorem 4.

Let G be a strong LR-transposition regular AG-groupoid. $\forall a\in G$, there exists $q\in G$ s.t. $a=a\left(aq\right)$ and $q^{2}a=q$. $qa=aq$ if and only if it is an L1-transposition regular AG-groupoid.

Proof. 

(⇒) For any a in G, there exists $q\in G$ s.t. $a=a\left(aq\right)$ and $q^{2}a=q$. According to Proposition 2, $\left(qa\right)a=a=a\left(aq\right)$. Because $qa=aq$, then $\left(qa\right)a=a=a\left(qa\right)$. So, G is an L1-transposition regular AG-groupoid.

(⇐) Let G be an L1-transposition regular AG-groupoid. For any $a\in G$, there exists $p\in G$ s.t. $\left(pa\right)a=a=a\left(pa\right)$. According to Proposition 1, let $e=pa,f=ee,m=ff$, and $\left(m\right(ep\left)\right)a=m=a\left(\right(m\left(ep\right)),m(m\left(ep\right))=m(ep)$. Let $q=m\left(ep\right)$. Then, $qa=m=aq,mq=q$. Then,

$$a\left(aq\right)=am=a,q^{2}a=\left(qq\right)a=\left(aq\right)q=mq=q.$$

Then, G is a strong LR-transposition regular AG-groupoid, and $qa=aq$. □

Example 7 shows that not every strong LR-transposition regular AG-groupoid is an L1-transposition regular AG-groupoid.

Example 7.

Let $G=\{a,b,c,d,e,f,g,h\}$. Define the operation on G in Table 4. Then, $(G,\ast )$ is a strong LR-transposition regular AG-groupoid. However, $a\ast (a\ast p_a)=a\ast (a\ast b)=a\ast d=a,p_a^2\ast a=(b\ast b)\ast a=a\ast a=b=p_a,p_a\ast a=b\ast a=c\ne a=a\ast d=a\ast p_a.$ So, $(G,\ast )$ is not an L1-transposition regular AG-groupoid.

4. L2-Transposition Regular AG-Groupoid and R2-Transposition Regular AG-Groupoid

In this section, we define the L2-transposition regular AG-groupoid and R2-transposition regular AG-groupoid.

Definition 7.

Let G be an AG-groupoid, $a\in G$. An element a is L2-transposition regular in G if there exists unique $x\in G$ s.t. $a\left(xa\right)=a$ and $x\left(xa\right)a=xa$. G is an L2-transposition regular AG-groupoid if all elements of G are L2-transposition regular.

Proposition 3.

Let G be L2-transposition regular AG-groupoid. $\forall a\in G$, there exists unique $x\in G$ s.t. $a\left(xa\right)=a$ and $x\left(xa\right)a=xa$. Let $xa=e$. Then, $xa=e=ax$, and $xe=x$.

Proof. 

For any a in G, let $xa=e$. Then, $ae=a\left(xa\right)=a$. According to left invertive law, there is, $x\left(xa\right)a=\left(xe\right)a=\left(ae\right)x=ax=xa=e$. Then, $a\left(ax\right)=a$. There is, $\left(ee\right)a=\left(ae\right)e=ae=a$.

Then,

$$a\left(\right(xe\left)a\right)=ae=a,$$

$$\left(\right(xe\left)\right(\left(xe\right)a\left)\right)a=\left(\right(xe\left)e\right)a=\left(ae\right)\left(xe\right)=\left(ax\right)\left(ee\right)=\left(xa\right)\left(ee\right)=\left(xe\right)\left(ae\right)=\left(xe\right)a=e.$$

Because x is unique, then $xe=x$. □

Example 8.

Let $G=\{a,b,c,d\}$. Define the operation on G in

Table 5. This is an L2-transposition regular AG-groupoid.

*	a	b	c	d
a	a	b	c	d
b	b	a	d	c
c	d	c	b	a
d	c	d	a	b

. Then, $(G,\ast )$ is an L2-transposition regular AG-groupoid. Let

$x_a$={$x\in G|$there exists unique $x\in G$ s.t. $a\left(xa\right)=a$ and $x\left(xa\right)a=xa$, $a\in G$}

There is,

$$x_a=a,x_b=b,x_c=c,x_d=d.$$

Definition 8.

Let G be an L2-transposition regular AG-groupoid. G is a strong L2-transposition regular AG-groupoid if ${\left(xa\right)}^2=xa$.

Example 9 shows that not every L2-transposition regular AG-groupoid is a strong L2-transposition regular AG-groupoid.

Example 9.

Let $G=\{a,b,c,d\}$. Define the operation on G in Table 5. That is, $x_c\ast c=c\ast c=b$, but $b^2\ne b$. So, $(G,\ast )$ is not a strong L2-transposition regular AG-groupoid.

Theorem 5.

Let G be a strong L2-transposition regular AG-groupoid. Then, for any $a\in G$,

(1) 

The local identity element of a is idempotent;

(2) 

The local identity element of a is unique;

(3) 

The local inverse element of a is unique;

(4) 

The local inverse element of local inverse element of a is a.

Proof. 

(1) For any $a\in G$, let $xa=e$. Then, $ae=a$, $\left(xe\right)a=\left(ae\right)x=ax=xa=e.$ Because ${\left(xa\right)}^2=xa$, and $e^2=e$. According to left invertive law, there is, $ea=\left(ee\right)a=\left(ae\right)e=ae=a.$ Then, $ea=a=ae,ax=e=xa.$ That is, e is the local identity element of a, x is the local inverse element of a relative to e. So, e is idempotent. That is to say, for any $a\in G$, the local identity element of a is idempotent.

(2) Assume that the local identity of a is not unique. Then, there exists $e,x,f,y\in G$ such that $ea=a=ae$, $xa=e=ax.$ Furthermore, $fa=a=af,ya=f=ay.$ According to the left invertive law,

$$ef=\left(xa\right)f=\left(fa\right)x=ax=e,$$

$$fe=\left(ya\right)e=\left(ea\right)y=ay=f,$$

$$e=ef=\left(ee\right)f=\left(fe\right)e=fe=f.$$

Then, for any $a\in G$, the local identity element of a is unique.

(3) According to (1), x is the local inverse element of a relative to e. By Definition 7, x is unique. That is, for $a\in G$, the local inverse element of a is unique.

(4) By (1), for any $a\in G$, let $xa=e$. Then, $ae=a=ea$, and $ax=xa=e$. By Proposition 3, $x\left(ax\right)=xe=x,ex=\left(ee\right)x=\left(xe\right)e=xe=x$. So, the local inverse element of a is a. □

Theorem 6.

Let G be a strong L2-transposition regular AG-groupoid. Define the binary relation ≈ on G as follows:

$$a\approx b\iff e_a=e_b,\forall a,b\in G$$

where $e_a$ is the local identity element of a. Then,

(1) 

The binary relation ≈ on G is an equivalence relation, and we denote the equivalence class containing x by ${\left[x\right]}_{\approx }$;

(2) 

$\forall x\in G$, ${\left[x\right]}_{\approx }$ is a subgroup;

(3) 

$G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$, that is, every strong L2-transposition regular AG-groupoid is the disjoint union of subgroups.

Proof. 

(1) Clearly, $\forall x\in G$, $e_x=e_x$. That is, $x\approx x$.

Assume that $x\approx y$, then $e_x=e_y$, and $e_y=e_x$. So, $y\approx x$.

If $x\approx y$ and $y\approx z$, then $e_x=e_y$, and $e_x=e_y$. Clearly, $e_x=e_z$. That is, $x\approx z$. So, ≈ is an equivalence relation on G.

(2) $\forall a,b\in {\left[x\right]}_{\approx }$, assume that $e_a=e_b=e$. $ab=\left(ae\right)b=\left(be\right)a=ba$. That is, ${\left[x\right]}_{\approx }$ satisfies the commutative law. Clearly, ${\left[x\right]}_{\approx }$ satisfies the left invertive law. Then, for any $a,b,c\in G$, $\left(ab\right)c=\left(cb\right)a=a\left(cb\right)=a\left(bc\right)$. So, ${\left[x\right]}_{\approx }$ satisfies the associative law.

According to Proposition 1, there exists $p\in G$ s.t. $pa=e=ap,ea=a=ae,ep=p=pe.$ $\exists q\in G$ s.t. $qb=e=bq,eb=b=be,eq=q=qe.$ Then,

$$\left(ab\right)e=\left(eb\right)a=ba=ab, \mathrm{and} e\left(ab\right)=\left(ab\right)e=ab.$$

That is, $ab\in {\left[x\right]}_{\approx }$.

For any $a\in {\left[x\right]}_{\approx }$, $ee=e\left(ap\right)=\left(ea\right)p=ap=e$, then $e\in {\left[x\right]}_{\approx }$.

Because $ee=\left(pa\right)e=\left(ea\right)p=ap=e$, there is $e\in {\left[x\right]}_{\approx }$.

Because $ep=p=pe$, then $p\in {\left[x\right]}_{\approx }$. According to the definition of the group, ${\left[x\right]}_{\approx }$ is an Abelian subgroup of G.

(3) For any $a\in {\left[x\right]}_{\approx }$, the local identity element of a is unique; then, $G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$ and $\bigcap _{x\in G}{\left[x\right]}_{\approx }=⌀$. That is, every strong L2-transposition regular AG-groupoid is the disjoint union of subgroups. □

Example 10.

Let $G=\{a,b,c,d\}$. Define the operation on G as

Table 6. This is a strong L2-transpostion regular AG-groupoid.

*	a	b	c	d
a	a	c	d	b
b	b	b	a	c
c	c	d	c	a
d	d	a	b	d

. Then, $(G,\ast )$ is a strong L2-transpostion regular AG-groupoid. Let

$$x_a=\{x\in G|. There exists x\in G s.t. a\left(xa\right)=a,x\left(xa\right)a=xa and {\left(xa\right)}^2=xa, a\in G\}.$$

And

$$x_a=a,x_b=b,x_c=c,x_d=d.$$

Let $A,B,C,D$ be a set composed of elements whose local identity is $a,b,c,d$ respectively. Then, $A=\left\{a\right\},B=\left\{b\right\},C=\left\{c\right\},D=\left\{d\right\}$.

Clearly, $G=A\cup B\cup C\cup D$, where $A,B,C,D$ are subgroups of G.

According to Theorem 5 (2), every strong L2-transpostion regular AG-groupoid is an L1/R1-transpostion regular AG-groupoid. Example 11 shows that not every L1/R1-transpostion regular AG-groupoid is a strong L2-transpostion regular AG-groupoid.

Example 11.

Let $G=\{a,b,c,d,e\}$. Define the operation on G as

Table 7. This is an L1/R1-transposition regular AG-groupoid.

*	a	b	c	d	e
a	a	a	a	d	e
b	a	b	c	d	e
c	a	c	b	d	e
d	d	d	d	e	a
e	e	e	e	a	d

. Then, $(G,\ast )$ is an L1/R1-transpostion regular AG-groupoid.

Because $a\ast (a\ast a)=a,(a\ast (a\ast a\left)\right)\ast a=a\ast a,a\ast (b\ast a)=a\ast a=a,(b\ast (b\ast a\left)\right)\ast a=(b\ast a)\ast a=a\ast a=b\ast a$, and $a\ne b$. So, $(G,\ast )$ is not a strong L2-transpostion regular AG-groupoid.

Definition 9.

Let G be an AG-groupoid, $a\in G$. An element a is an R2-transposition regular in G if there exists a unique $x\in G$ s.t. $\left(ax\right)a=a$ and $a\left(\right(ax\left)x\right)=ax$. G is R2-transposition regular AG-groupoid if all its elements are R2-transposition regular.

Proposition 4.

Let G be an R2-transposition regular AG-groupoid. $\forall a\in G$, there exists unique $x\in G$ s.t. $\left(ax\right)a=a$ and $a\left(\right(ax\left)x\right)=ax$, let $ax=e$. Then, $x=ex$.

Proof. 

Let $ax=e$. Then, $ea=a,a\left(ex\right)=ax=e.ee=\left(ax\right)e=\left(ex\right)a$. There is,

$$\left(a\right(ex\left)\right)a=ea=a,$$

$$a\left(\right(a\left(ex\right)\left(ex\right))=a(e\left(ex\right))=(ea\left)\right(e\left(ex\right))=(ee\left)\right(a\left(ex\right))=(ee)e=(\left(ex\right)a)e=(ea\left)\right(ex)=a(ex).$$

Because x is unique, then $ex=x$. □

Example 12.

Let $X=\left\{\right(a,b\left)\right|a\in R$, and $a\ne 0,b\in \{1,-1\}\}$, define $(a,b)\ast (c,d)=(ac,-b/d).$ Then

$$\left[\right(a,b)\ast (c,d\left)\right]\ast (e,f)=(ac,-b/d)\ast (e,f)=(ace,b/df)$$

$$\left[\right(e,f)\ast (c,d\left)\right]\ast (a,b)=(ec,-f/d)\ast (a,b)=(ace,f/bd)$$

Because $b,f\in \{-1,1\}$, $b^2=f^2$, and $b/f=f/b$. We can verify $b/df=f/bd$. Thus

$$\left[\right(a,b)\ast (c,d\left)\right]\ast (e,f)=\left[\right(e,f)\ast (c,d\left)\right]\ast (a,b)$$

satisfies the left invertive law. So, $(X,\ast )$ is an AG-groupoid.

For any $a\in R$,

$$(a,1)\ast (\frac{1}{a},1)\ast (a,1)=(1,-1)\ast (a,1)=(a,1),$$

$$(a,1)\ast ((1,-1)\ast (\frac{1}{a},1))=(a,1)\ast (\frac{1}{a},1);$$

$$(a,-1)\ast (\frac{1}{a},-1)\ast (a,-1)=(1,-1)\ast (a,-1)=(a,-1),$$

$$(a,-1)\ast ((1,-1)\ast (\frac{1}{a},-1))=(a,-1)\ast (\frac{1}{a},-1).$$

We can get that $(X,\ast )$ is an R2-transposition regular AG-groupoid.

Definition 10.

Let G be an R2-transposition regular AG-groupoid. G is a strong R2-transposition regular AG-groupoid if ${\left(ax\right)}^2=ax$.

Example 13 shows that not every R2-transposition regular AG-groupoid is a strong R2-transposition regular AG-groupoid.

Example 13.

In Example 12, for any $a\in R$, $(1,-1)$ is the local left identity element of $(a,1)$, and $(1,-1)$ is the local left identity element of $(a,-1)$. However, $(1,1)\ast (1,1)=(1,-1)$; that is, $(1,1)$ is not idempotent. Then, G is not a strong R2-transposition regular AG-groupoid.

Theorem 7.

Let G be a strong R2-transposition regular AG-groupoid. Then, for any $a\in G$,

(1) The local left identity element of a is idempotent;

(2) The local left identity element of a is unique;

(3) The local inverse element of a is unique;

(4) The local inverse element of the local inverse element of a is a.

Proof. 

(1) For any element a in G, let $ax=e$. Then, $ea=a$. According to the definition, $e^2=e.$

(2) For any element a in G, assume that the local left identity element of a is not unique. Then, there exist $e,x,f,y\in G$ such that $ea=a,ax=e.fa=a,ay=f$, and $ex=x,fy=y$. So,

$$xa=\left(ex\right)a=\left(ax\right)e=ee=e,ya=\left(fy\right)a=\left(ay\right)f=ff=f.$$

Then,

$$ef=\left(xa\right)f=\left(fa\right)x=ax=e,$$

$$fe=\left(ya\right)e=\left(ea\right)y=ay=f.$$

$$e=ef=\left(ee\right)f=\left(fe\right)e=fe=f.$$

So, for any $a\in G$, the local left identity element of a is unique.

(3) According to Definition 9 and (1), $xa=e=ax$, and the local inverse element of a exists, because the local right inverse element of a is unique. So, the local inverse element of a is unique.

(4) According to Proposition 4, let $ax=e$, then $ex=x$. So, the local inverse element of a is a. □

Theorem 8.

Let G be a strong R2-transposition regular AG-groupoid. Define the binary relation ≈ on G as follows:

$$a\approx b\iff e_a=e_b, \forall a,b\in G$$

where $e_a$ is the local left identity element of a. Then

(1) 

The binary relation ≈ on G is the equivalence relation, and we denote the equivalence class containing x by ${\left[x\right]}_{\approx }$;

(2) 

$\forall x\in G$, ${\left[x\right]}_{\approx }$ is an AG-subgroup;

(3) 

$G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$, that is, every strong R2-transposition regular AG-groupoid is the disjoint union of AG subgroups.

Proof. 

(1) Clearly, $\forall x\in G$, $e_x=e_x$. That is, $x\approx x$.

Assume that $x\approx y$, then $e_x=e_y$, and $e_y=e_x$. So $y\approx x$.

If $x\approx y$ and $y\approx z$, then $e_x=e_y$, and $e_x=e_y$. Clearly, $e_x=e_z$. That is, $x\approx z$. So, ≈ is a equivalence relation on G.

(2) For any $a\in {\left[x\right]}_{\approx }$, let $e_a=e$. Then, $ea=a,am=e$.

$\forall a,b\in {\left[x\right]}_{\approx }$, assume that $e_a=e_b=e$. $ea=a,am=e.eb=b,by=e$. Then,

$$e\left(ab\right)=\left(ee\right)\left(ab\right)=\left(ea\right)\left(eb\right)=ab.$$

That is, $ab\in {\left[x\right]}_{\approx }$.

Clearly, $e\in {\left[x\right]}_{\approx }$.

According to Proposition 4, there is $em=m$. Furthermore, $ma=\left(em\right)a=\left(am\right)e=ee=e$. Then, $m\in {\left[x\right]}_{\approx }$. So ${\left[x\right]}_{\approx }$ is an AG-subgroup of G.

(3) For any $a\in {\left[x\right]}_{\approx }$, the local left identity element of a is unique; then, $G=\bigcup _{x\in G}{\left[x\right]}_{\approx }$ and $\bigcap _{x\in G}{\left[x\right]}_{\approx }=⌀$. That is, every strong R2-transposition regular AG-groupoid is the disjoint union of AG subgroups. □

Theorem 9.

Let G be a strong R2-transposition regular AG-groupoid. Let $G=\bigcup _{n\in I}{G}_n$, and $i,j\in I$, $G_i\bigcap _{i\ne j}{G}_j=⌀$, where $G_n$ is an AG-subgroup of G, and I is the index set. Then, $\left|n\right|\ne 2$ and $\left|n\right|\ne 3$.

Proof. 

Assume that $\left|n\right|=2$; then, $G=G_1\bigcup G_2$. Let $a_1$ be the left identity element of $G_1$, and $a_1{a}_1=a_1$. Let $b_1$ be the identity element of $G_2$, and $b_1{b}_1=b_1$. Because the subset composed of idempotents in the AG-groupoid is the AG-subgroupoid, $a_1{b}_1=a_1$ or $a_1{b}_1=b_1$, it is divided into the following two conditions:

Case 1: Assume that $a_1{b}_1=a_1$; at that time, $\left(a_1{b}_1\right)a_1=a_1{a}_1=a_1,a_1\left(\left(a_1{b}_1\right)b_1\right)=a_1\left(a_1{b}_1\right)=a_1{a}_1=a_1{b}_1.$ Because $\left(a_1{a}_1\right)a_1=a_1{a}_1=a_1,a_1\left(\left(a_1{a}_1\right)a_1\right)=a_1\left(a_1{a}_1\right)=a_1{a}_1$. Then, $a_1=b_1.$

Case 2: Assume that $a_1{b}_1=b_1$; in a similar way, $b_1=a_1.$

Above all, $\left|n\right|\ne 2$. Then, $\left|n\right|=1.$

Then, we consider $\left|n\right|=3$, $G=G_1\bigcup G_2\bigcup G_3$. Let $a_1$ be the left identity element of $G_1$, and $a_1{a}_1=a_1$. Let $b_1$ be the identity element of $G_2$, and $b_1{b}_1=b_1$. Let $c_1$ be the identity element of $G_3$, and $c_1{c}_1=c_1$. According to Case 1 and Case 2, if in any case of $a_1{b}_1=a_1$, $a_1{b}_1=b_1$, $b_1{a}_1=a_1$, $b_1{a}_1=b_1$, $\left|n\right|=2$, then following Case 1 and Case 2. So, we should consider $a_1{b}_1=c_1$ and $b_1{a}_1=c_1$:

We can get that $\left(a_1{b}_1\right)b_1=\left(b_1{b}_1\right)a_1=b_1{a}_1=c_1$, because $a_1{b}_1=c_1$, $c_1{b}_1=c_1$. Then, $\left|n\right|=2$, according to Case 1 and Case 2, $\left|n\right|=1$. □

Similarly, if G is a strong L2-transposition regular AG-groupoid, let $G=\bigcup _{n\in I}{G}_n$ and $\forall i,j\in I$, $G_i\bigcap _{i\ne j}{G}_j=⌀$, where $G_n$ is the subgroup of G, and I is the index set. Then, $\left|n\right|\ne 2$ and $\left|n\right|\ne 3$.

Example 10 shows the example of the strong R2-transposition regular AG-groupoid where $\left|n\right|=4$.

Example 14.

Let $G=\{a,b,c,d\}$. Define the operation on G as shown in Table 6. Then, $(G,\ast )$ is a strong R2-transposition regular AG-groupoid. Let

$$x_a=\{x\in G| there exists unique x\in G s.t. \left(ax\right)a=a,a\left(\right(ax\left)x\right)=ax and {\left(ax\right)}^2=ax, a\in G\}.$$

There is,

$$x_a=a,x_b=b,x_c=c,x_d=d.$$

Let $A,B,C,D$ be a set composed of elements whose local left identity element is $a,b,c,d$ respectively. Then, $A=\left\{a\right\},B=\left\{b\right\},C=\left\{c\right\},D=\left\{d\right\}.$

Then, $G=A\cup B\cup C\cup D$, where $A,B,C,D$ is the AG-subgroup of G.

Theorem 10.

Let G be a strong L2-transposition regular AG-groupoid. Then, it is a strong R2-transposition regular AG-groupoid.

Proof. 

Let G be a strong L2-transposition regular AG-groupoid. Then, for any $a\in G$, let $xa=e$, according to Proposition 3 and Theorem 6, $ae=a=ea,xa=e=ax$. So, $\left(ax\right)a=ea=a,a\left(\right(ax\left)x\right)=a\left(ex\right)=\left(ea\right)\left(ex\right)=\left(ee\right)\left(ax\right)=e$. Uniqueness holds; G is a strong R2-transposition regular AG-groupoid. □

However, not every strong R2-regular AG-groupoid is a strong L2-transposition regular AG-groupoid in Example 15.

Example 15.

Let $G=\{0,1,2,3\}$. Define the operation on G as shown in

Table 8. This is a strong R2-transposition regular AG-groupoid.

*	0	1	2	3
0	0	1	2	3
1	3	0	1	2
2	2	3	0	1
3	1	2	3	0

. Clearly, $(G,\ast )$ is a strong R2-transposition regular AG-groupoid.

However, G is not a strong L2-transposition regular AG-groupoid, since $1\ast (3\ast 1)=1$, $(3\ast (3\ast 1\left)\right)\ast 1=3\ast 1$, but $3\ast 1=2$ and $2\ast 2\ne 2$.

Table 8 shows a distance matrix that contains information about distances (Figure 1).

5. L3-Transposition Regular AG-Groupoid and R3-Transposition Regular AG-Groupoid

In this section, we propose the notion of an L3-transposition regular AG-groupoid and R3-transposition regular AG-groupoid.

Definition 11.

Let G be an AG-groupoid, $a\in G$. An element a is L3-transposition regular in G if there exists unique $x\in G$ s.t.$\left(xa\right)a=a$. G is an L3-transposition regular AG-groupoid if all its elements are L3-transposition regular.

Example 16.

Let $G=\{a,b,c,d\}$; define the operation on G as shown in

Table 9. This is an L3-transposition regular AG-groupoid.

*	a	b	c	d
a	b	d	c	a
b	c	a	b	d
c	a	c	d	b
d	d	b	a	c

. Then, $(G,\ast )$ is an L3-transposition regular AG-groupoid. Let

$$x_a=\{x\in G| there exists unique x\in G s.t. \left(xa\right)a=a, a\in G\}.$$

Furthermore,

$$x_a=b,x_b=a,x_c=d,x_d=c.$$

Definition 12.

Let G be an L3-transposition regular AG-groupoid. G is a strong L3-transposition regular AG-groupoid if ${\left(xa\right)}^2=xa$.

Example 17 shows that not every L3-transposition regular AG-grooupoid is a strong L3-transposition regular AG-groupoid.

Example 17.

Let $G=\{a,b,c,d\}$, define the operation on G as shown in Table 9. $x_a\ast a=b\ast a=c$, and c is not idempotent. Then, G is not a strong L3-transposition regular AG-groupoid.

Theorem 11.

Let G be an AG-groupoid. G is a strong L3-transposition regular AG-groupoid if and only if it is a strong R2-transposition regular AG-groupoid.

Proof. 

(⇒) Let G be a strong L3-transposition regular AG-groupoid. For any element a in G, let $xa=e$, then $ea=a$, and $ee=e$. Then,

$$ax=\left(ea\right)x=\left(xa\right)e=ee=e.$$

Then, e is the local left identity element of a, and x is the local inverse element of a relative to e, because $\left(\right(ex\left)a\right)a=\left(\right(ax\left)e\right)a=\left(ee\right)a=ea=a.$ According to the definition of a strong R3-transposition regular AG-groupoid, $ex=x.$

According to the definition, x, that is, the local left inverse element of a is unique, x is the local inverse element of a; thus, the local inverse element of a is unique. Assume that the local right inverse element of a is not unique; that is, there exists $y\in G$ and $x\ne y$ such that $\left(ay\right)a=a$. Let $ay=f$, there is, $fa=a$, because $ef=\left(xa\right)f=\left(fa\right)x=ax=e$, and $\left(ee\right)e=ee=e$, but $\left(fe\right)e=\left(ee\right)f=ef=e$. According to the definition of a strong L3-transposition regular AG-groupoid, $e=f$. Then, $ay=e$. That is, $\left(\right(ey\left)a\right)a=\left(aa\right)\left(ey\right)=\left(ae\right)\left(ay\right)=\left(ae\right)e=\left(ee\right)a=ea=a$. According to the definition of a strong L3-transposition regular AG-groupoid, $ey=ex=x.$

At this time, $\left(ax\right)x=ex=x$, and $\left(\right(\left(ya\right)a\left)x\right)x=\left(\right(xa\left)\right(ya\left)\right)x=\left(e\right(ya\left)\right)x=\left(\right(ee\left)\right(ya\left)\right)x$ $=\left(\right(ey\left)\right(ea\left)\right)x=\left(xa\right)x=ex=x$. According to definition of a strong L3-transposition regular AG-groupoid, $\left(ya\right)a=a$. Then, $x=y$; that is, the local right inverse element of a is unique. So

$$\left(ax\right)a=ea=a,a\left(\right(ax\left)x\right)=a\left(ex\right)=\left(ea\right)\left(ex\right)=\left(ee\right)\left(ax\right)=ee=e=ax \mathrm{and} ee=e.$$

Furthermore, x is unique, so G is a strong R2-transposition regular AG-groupoid.

(⇐) Let G be a strong R2-transposition regular AG-groupoid. Then, for any $a\in G$, there exists a unique $x\in G$ such that $\left(ax\right)a=a,a\left(\right(ax\left)x\right)=ax$, and ${\left(ax\right)}^2=ax$. Let $ax=e$; that is, $ea=a,a\left(ex\right)=ax=e,ee=e$. According to Proposition 4, $ex=x$. That is, $xa=\left(ex\right)a=\left(ax\right)e=ee=e$. According to definition of a strong R2-transposition regular AG-groupoid, the local right inverse element of a is unique.

Assume that the local left inverse element of a relative to the local left identity element is not unique; that is, there exists $y\in G$ and $x\ne y$, such that $\left(ya\right)a=a$. Let $ya=f$, there is $fa=a$. However, $ef=\left(xa\right)f=\left(fa\right)x=ax=e$, and $\left(ee\right)e=ee=e,e\left(\right(ee\left)e\right)=e\left(ee\right)=ee$, $\left(ef\right)e=ee=e,e\left(\right(ef\left)f\right)=e\left(ef\right)=ee=e$. According to the definition of a strong R2-transposition regular AG-groupoid, $e=f$, that is, $ya=e$. At this time, $ay=\left(ea\right)y=\left(ya\right)e=ee=e$, and $\left(ay\right)a=ea=a,a\left(\right(ay\left)y\right)=a\left(ey\right)=\left(ea\right)\left(ey\right)=\left(ee\right)\left(ay\right)=ee=e=ay$. According to the definition of a strong R2-transposition regular AG-groupoid, $x=y$. That is, the local left inverse element of a is relative to the local left identity element.

Because

$$\left(xa\right)a=ea=a, \mathrm{and} ee=e.$$

Furthermore, x is unique, and G is strong L3-transposition regular AG-groupoid. □

Definition 13.

Let G be an AG-groupoid, $a\in G$. An element a is R3-transposition regular in G if there exists a unique $x\in G$ s.t. $a\left(ax\right)=a$. G is an R3-transposition regular AG-groupoid if all its elements are R3-transposition regular.

Example 18.

Let $G=\{a,b,c,d\}$; define the operation on G as shown in Table 9. Then, $(G,\ast )$ is an R3-transposition regular AG-groupoid. Let

$$x_a=\{x\in G| there exists a unique x\in G s.t. a\left(ax\right)=a, a\in G\}.$$

Furthermore,

$$x_a=b,x_b=a,x_c=d,x_d=c.$$

Definition 14.

Let G be an R3-transposition regular AG-groupoid. G is a strong R3-transposition regular AG-groupoid if ${\left(ax\right)}^2=ax$.

Example 19.

Let $G=\{a,b,c,d\}$; define the operation on G as shown in Table 9. Here, a, b, c, d are not idempotent. Then, $(G,\ast )$ is not a strong R3-transposition regular AG-groupoid.

Theorem 12.

Let G be an AG-groupoid. G is a strong L2-transposition regular AG-groupoid if and only if it is a strong R3-transposition regular AG-groupoid.

Proof. 

(⇒) Let G be a strong L2-transposition regular AG-groupoid. For any element a of G, let $xa=e$, then $ae=a$, $\left(x\right(xa\left)\right)a=\left(xe\right)a=\left(ae\right)x=ax=xa$. Then, $a\left(ax\right)=a\left(xa\right)=a$. If x is unique, G is a strong R3-transposition regular AG-groupoid.

(⇐) Let G be a strong R3-transposition regular AG-groupoid. Then, for any $a\in G$, there exists a unique $x\in G$ s.t. $a\left(ax\right)=a$, and ${\left(ax\right)}^2=ax$. Let $ax=e$, then $ae=a$, and $ee=e$. That is, $ea=\left(ee\right)a=\left(ae\right)e=ae=a$. Because $a\left(ex\right)=\left(ea\right)\left(ex\right)=\left(ee\right)\left(ax\right)=\left(ee\right)e=ee=e$, $a\left(xe\right)=\left(ae\right)\left(xe\right)=\left(ax\right)\left(ee\right)=e\left(ee\right)=ee=e$. According to the definition of a strong R3-transposition regular AG-groupoid, $ex=x=xe$. Then, $xa=\left(ex\right)a=\left(ax\right)e=ee=e$. That is, $ea=a=ae$, $xa=e=ax$, $xe=x=ex$. According to the definition of a strong R3-transposition regular AG-groupoid, the local right inverse element of a and the relative local identity element are unique.

Assume that the local left inverse element of a relative to the local identity element is not unique; that is, there exists $y\in G$ and $x\ne y$ such that $ya=f$, and $fa=a=af$. Then, $ef=\left(xa\right)f=\left(fa\right)x=ax=e.$ Then, $e\left(ee\right)=ee=e$, $e\left(ef\right)=ee=e$. So, $e=f$. That is, $ya=e$. Then, $ay=\left(ea\right)y=\left(ya\right)e=ee=e$. According to the definition of a strong R3-transposition regular AG-groupoid, $x=y$. That is, the local left inverse element of a relative to the local identity element is unique.

At this moment, $a\left(xa\right)=ae$, $\left(x\right(xa\left)\right)a=\left(xe\right)a=xa=e$, and ${\left(xa\right)}^2=xa$, and the local left inverse element is unique; then, G is a strong L2-transposition regular AG-groupoid. □

Theorem 13.

Let G be an AG-groupoid. If G is an L2-transposition regular AG-groupoid, G is an R3-transposition regular AG-groupoid.

Proof. 

Let G be an L2-transposition regular AG-groupoid. Then, for any element a in G, there exists a unique x such that $a\left(xa\right)=a,x\left(xa\right)a=xa.$ Let $xa=e$; then, $ae=a,\left(xe\right)a=\left(ae\right)x=ax=xa.$ Then, $a\left(ax\right)=a$, and x is unique. Then, G is an R3-transposition regular AG-groupoid. □

Example 20.

Let $G=\{a,b,c,d\}$ define the operation on G as shown in Table 9. Clearly, G is an R3-transposition regular AG-groupoid. However, G is not an L2-transposition regular AG-groupoid, since $a\ast (d\ast a)=a$, $(d\ast (d\ast a\left)\right)\ast a=(d\ast d)\ast a=c\ast a=a\ne d=d\ast a$.

Figure 2 shows the relation among various transposition regular AG-groupoids, where A represents a strong L2-transposition regular AG-groupoid and a strong R3-transposition regular AG-groupoid;

B represents an L1/R1-transposition regular AG-groupoid that is not a strong L2-transposition regular AG-groupoid; see Example 11;

C represents an L2-transposition regular AG-groupoid that is not a strong L2-transposition regular AG-groupoid; see Example 9;

D represents an R3-transposition regular AG-groupoid that is not an L2-transposition regular AG-groupoid; see Example 20;

E represents a strong R2/L3-transposition regular AG-groupoid that is not a strong L2-transposition regular AG-groupoid; see Example 15;

F represents an R2-transposition regular AG-groupoid that is not a strong R2-transposition regular AG-groupoid; see Example 13;

G represents an L3-transposition regular AG-groupoid that is not a strong L3-transposition regular AG-groupoid; see Example 17.

Then, A + B is an L1/R1-transposition regular AG-groupoid, A + C is an L2-transposition regular AG-groupoid, A + C + D is an R3-transposition regular AG-groupoid, A + E is a strong R2/L3-transposition regular AG-groupoid, A + E + F is an R2-transposition regular AG-groupoid, and A + E + G is an L3-transposition regular AG-groupoid.

6. Transposition Regular AG-Groupoid Semigroup

In [20], they introduced the definition of an AG-groupoid semigroup, that is, algebra that is both semigroup and AG-groupoid. In this paper, we reproved that the theorem of a regular AG-groupoid semigroup is a commutative semigroup. We also introduce transposition regularity into the AG-groupoid semigroup, and we discuss the relations among various transpositions of regular AG-groupoid semigroups, commutative Clifford semigroups, and Abelian groups as well as the decomposition theorem of AG-groupoid semigroups.

Definition 15

([21]). Let R be a algebra. Subalgebra A is an anti-ideal of R, if for any $x\in R\backslash A$, there is, $xA\cap A\cap Ax=⌀$.

Definition 16

([22]). Let $(G,\ast )$ be a generalized group. Then, for any $x,y\in G$, if $x\ast y=y\ast x$, then G is a group.

Definition 17

([23]). S is a Clifford semigroup, if it is a completely regular semigroup, and for any $x,y\in S$,

$$\left(x{x}^{-1}\right)\left(y{y}^{-1}\right)=\left(y{y}^{-1}\right)\left(x{x}^{-1}\right).$$

For any element in S, the element c in S is a central element, if for any $s\in S$, there is $cs=sc$. The set of all central elements in S is the center in S.

Theorem 14.

Let S be an AG-groupoid semigroup. If S is regular, then S is a commutative semigroup.

Proof. 

Let S be a regular AG-groupoid semigroup. Then, for any elements $a,b\in S$ such that $aba=a$. Then, $\forall c\in S$, according to medial law and left invertive law as well as associative law, there is,

$$ac=\left(aba\right)c=\left(ab\right)\left(ac\right)=\left(aa\right)\left(bc\right)=\left(a\right(ab\left)\right)c=c\left(ab\right)a=c\left(aba\right)=ca.$$

Because S is regular, then for any regular elements m in S, for any element n in S, there is $mn=nm$. That is, S is a commutative semigroup. □

The following shows the concepts of various transposition regular AG-groupoid semigroups as well as the connections among them and the commutative Clifford semigroup as well as Abelian group.

In [19], they introduced various transposition regular semigroups and prove that L1/R1/LR-transposition regular semigroups and a completely regular semigroup are equivalent to each other.

Definition 18.

Let G be a semigroup. G is an L1-transposition regular AG-groupoid semigroup if it is an L1-transposition regular AG-groupoid. Similarly, there is a R1/LR, L2/R2/L3/R3-transposition regular AG-groupoid semigroup.

According to Definition 18, L1/R1/LR-transposition regular AG-groupoid semigroups are equivalent to each other.

Theorem 15.

Let G be an AG-groupoid semigroup. It is an L1-transposition regular AG-groupoid semigroup if and only if it is a commutative Clifford semigroup.

Proof. 

(⇒) Let G be an L1-transposition regular AG-groupoid semigroup. Clearly, it is regular, according to Theorem 14, and G is a commutative semigroup. Then, for any $x,y\in G$, there is, $\left(x{x}^{-1}\right)\left(y{y}^{-1}\right)=\left(y{y}^{-1}\right)\left(x{x}^{-1}\right)$. Then, G is a commutative Clifford semigroup.

(⇐) Let G be a commutative Clifford semigroup. According to the definition of a Clifford semigroup, G is a completely regular semigroup. Then, G is an L1-transposition regular semigroup. $\forall a,b,c\in G$, there is,

$$\left(ab\right)c=c\left(ab\right)=c\left(ba\right)=\left(cb\right)a,$$

Then, G satisfies the left invertive law. Then, G is an L1-transposition regular AG-groupoid semigroup. □

Theorem 16.

Let G be an AG-groupoid semigroup. Then, the folllowing conditions are equivalent:

(1) G is an L2/R2-transposition regular AG-groupoid semigroup;

(2) G is an L3/R3-transposition regular AG-groupoid semigroup;

(3) G is an Abelian group.

Proof. 

(1) ⇒ (2) Let G be an L2/R2-transposition regular AG-groupoid semigroup. According to the decomposition theorem of them, G is a group. Clearly, G is regular; then, G is a commutative and Abelian group. That is, the L2/R2-transposition regular AG-groupoid semigroup is an Abelian group. $\forall a\in G$, there exists a unique $e,a^{-1}\in G$, such that $ae=a=ea$, $a^{-1}a=e=a{a}^{-1}$. Then, $\left(a^{-1}a\right)a=ea=a$. Then, G is an L3-transposition regular AG-groupoid semigroup. Similarly, $a\left(a{a}^{-1}\right)=ae=a$. Then, G is an R3-transposition regular AG-groupoid semigroup.

(2) ⇒ (3) Let G be an L3-transposition regular AG-groupoid semigroup. Then, it is a generalized group. According to Theorem 14, G is commutative. So, it is a group. Then, G is an Abelian group. Similarly, the R3-transposition regular AG-groupoid semigroup is a generalized group, so the L3-transposition regular AG-groupoid semigroup is an Abelian group.

(3) ⇒ (1) Let G be an Abelian group. Then, G is an L2/R2-transposition regular AG-groupoid semigroup. $\forall a,b,c\in G$, there is,

$$\left(ab\right)c=c\left(ab\right)=c\left(ba\right)=\left(cb\right)a,$$

Then, G satisfies the left invertive law. Then, G is an L2/R2-transposition regular AG-groupoid semigroup. □

Figure 3 shows the relation among them.

In the following, we introduce the properties and decomposition theorem of an AG-groupoid semigroup.

Theorem 17.

Let G be an AG-groupoid semigroup. Let H be the subset of G, and $H=\left\{a\right|aba=a,a\in G,b\in G\}$. Then,

(1) H is the subsemigroup of G;

(2) H is the center;

(3) H is a Clifford semigroup.

Proof. 

(1) Let G be an AG-groupoid semigroup, $a,x\in H$. Then, a, x is the regular element of G. Then, there exists $m\in G$, s.t. $ama=a$. Then, $a\left(mam\right)a=\left(ama\right)ma=ama=a$, $\left(mam\right)a\left(mam\right)=m\left(ama\right)mam=mamam=mam.$ Then, $mam$ is a regular element. Let $b=mam$, there is $aba=a$, $bab=b$, and there exists $y\in H$ such that $xyx=x$, $yxy=y$. Then,

$$\begin{array}{ccc}\hfill ax& =& \left(aba\right)\left(xyx\right)=\left(ab\right)\left(ax\right)\left(yx\right)=\left(\right(\left(ax\right)b\left)a\right)\left(yx\right)\hfill \\ & =& \left(ax\right)\left(\right(ba\left)\right(yx\left)\right)=\left(ax\right)\left(\right(by\left)\right(ax\left)\right)=\left(ax\right)\left(by\right)\left(ax\right).\hfill \end{array}$$

That is, for $ax\in G$, there exists $by\in G$ s.t. $\left(ax\right)\left(by\right)\left(ax\right)=ax$. That is, $ax$ is a regular element. Then, $ax\in H$. So H is a subsemigroup of G.

(2) For any regular elements a in H, assume that there exists $b\in G$ s.t. $aba=a$. Then, for any element x that is not regular in G, there is,

$$\begin{array}{ccc}\hfill xa& =& x\left(aba\right)=\left(\right(xa\left)b\right)a=\left(\right(ba\left)x\right)a=\left(ax\right)\left(ba\right)=\left(\right(ax\left)b\right)a\hfill \\ & =& \left(\right(bx\left)a\right)a=\left(aa\right)\left(bx\right)=\left(ab\right)\left(ax\right)=\left(aba\right)x=ax.\hfill \end{array}$$

Then, a is a central element; that is, any regular element in G is a central element. Then, H is the center.

(3) By (1), $ax=\left(ax\right)\left(by\right)\left(ax\right)$. Then,

$$\left(by\right)\left(ax\right)\left(by\right)=\left(\right(ba\left)\right(yx\left)\right)\left(by\right)=\left(\right(ba\left)b\right)\left(\right(yx\left)y\right))=by.$$

Furthermore, $\left(ax\right)\left(by\right)=\left(by\right)\left(ax\right)$. So, H is a completely regular semigroup. According to Theorem 14, the regular AG-groupoid semigroup is a commutative Clifford semigroup. Then, H is commutative, so for any $x,y\in H$, there is $\left(x{x}^{-1}\right)\left(y{y}^{-1}\right)=\left(y{y}^{-1}\right)\left(x{x}^{-1}\right)$. Then, H is a commutative Clifford semigroup. □

Theorem 18.

Let G be an AG-groupoid semigroup and B be a center of G. Then,

(1) B is an ideal of G;

(2) $G\backslash B$ is an anti-ideal of G.

Proof. 

(1) Clearly, B is a center. For any element a, b of B, there is $ab=ba$. $\forall y\in G$, there is, $ay=ya$, $by=yb$. Then,

$$\left(ab\right)y=\left(yb\right)a=y\left(ba\right)=y\left(ab\right).$$

Then, $ab$ is a central element. That is, $ab\in B$.

For any element a in B, any element x in G. Because a is a central element, then $ax=xa$. $\forall y\in G$, there is,

$$\left(ax\right)y=\left(ya\right)x=y\left(xa\right)=y\left(ax\right).\left(xa\right)y=\left(ya\right)x=y\left(ax\right)=y\left(xa\right).$$

Then, $ax$ and $xa$ are central elements, that is, $ax\in B$, $xa\in B$. Because a and x represent any element of B and G, respectively, $BG\subseteq B$, $GB\subseteq B$. Then, B is an ideal of G.

(2) Let A be a set of elements that are not central elements in G. Then, $A\cap B=⌀$. So, $A=G\backslash B$. For any element x in B, any element y in A, there is, $xy\in B$, $yx\in B$. Then, $xA\subseteq B$, $Ax\subseteq B$. So, $xA\cap A\cap Ax=⌀$. Then, A is anti-ideal. □

The following example shows that not every central element is a regular element.

Example 21.

Let $G=\{a,b,c,d\}$ be an AG-groupoid semigroup, the operation on G as shown in

Table 10. AG-groupoid semigroup that is not commutative.

*	a	b	c	d
a	a	a	a	a
b	a	a	a	a
c	a	a	a	b
d	a	a	a	b

, where a, b are central elements, but b is not a regular element.

7. Discussion

In Example 15, we show the relationship between Table 8 and a connected directed unweighted graph. So, we can discuss applications in the graph for AG-groupoids. In this paper, we mainly talk about the various transposition regular AG-groupoids. We can discuss the relationships among transposition regular AG-groupoids, hypersemigroups and T2CA-groupoids as well as non-classical logical algebras (see [24,25,26,27,28]).

8. Conclusions

In this paper, we introduce transposition regularity into AG-groupoids, and various transposition regular AG-groupoids are obtained. Their properties and structures are discussed by studying their decomposition theorem: (1) an L1-transposition regular AG-groupoid is equivalent to an R1-transposition regular AG-groupoid and they are a disjoint union of Abelian subgroups; (2) an LR/strong LR-transposition regular AG-groupoid satisfying a certain condition is equivalent to the L1/R1-transposition regular AG-groupoid; (3) a strong L2-transposition regular AG-groupoid and strong R3-transposition regular AG-groupoid are equivalent to each other and can be decomposed into the union of Abelian subgroups; (4) a strong L3-transposition regular AG-groupoid and strong R2-transposition regular AG-groupoid are equivalent to each other and can be decomposed into the union of AG subgroups. Finally, we introduce transposition regularity into the AG-groupoid semigroup and prove that: (1) L1/R1/LR-transposition regular AG-groupoid semigroups are equivalent to commutative Clifford semigroups; (2) L2/R2/L3/R3-transposition regular AG-groupoid semigroups are equivalent to Abelian groups; (3) an AG-groupoid semigroup is a union of ideal and anti-ideal.