On inverse ordered semigroups

TL;DR

This paper explores the structure of inverse ordered semigroups, generalizing inverse semigroups to include order, and provides characterizations based on their idempotents and regularity.

Contribution

It introduces the concept of inverse ordered semigroups and characterizes them through their regularity, inverse properties, and idempotent elements.

Findings

An ordered semigroup is a complete semilattice of t-simple ordered semigroups if and only if it is completely regular and inverse.

Inverse ordered semigroups are characterized by their ordered idempotents.

The paper establishes equivalences and structural properties of inverse ordered semigroups.

Abstract

The purpose of this paper is to study the generalization of inverse semigroups (without order). An ordered semigroup S is called an inverse ordered semigroup if for every a 2 S, any two inverses of a are H-related. We prove that an ordered semigroup is complete semilattice of t-simple ordered semigroups if and only if it is completely regular and inverse. Furthermore characterizations of inverse ordered semigroups have been characterized by their ordered idempotents.

Equations

$${}^{\prime }{\le }^{\prime }:=\{(a,a),(e,e),(f,f)\}.$$

Full text

Key Words and phrases: ordered regular, ordered inverse element, ordered idempotent, completely regular, inverse.

2000 Mathematics subject Classification: 16Y60;20M10.

1 Introduction

An ordered semigroup is a partiality ordered set $(S,\leq)$, and at the same time a semigroup $(S,\cdot)$ such that for all $a,b,x\in S,\;a\leq b$ implies $xa\leq xb\;\textrm{and}\;ax\leq bx$. It is denoted by $(S,\cdot,\leq)$. For every subset $H$ of $S$, the downword closure subset of $H$ is denoted by $(H]$ and defined by $(H]=\{t\in S:t\leq h,\;\textrm{for some}\;h\in H\}$. Throughout this paper unless otherwise stated $S$ stands for an ordered semigroup. T. Saito [8] studied inverse semigroups by introducing partial order in it. There was much interest in the past decade in studying the inverses of an element in an ordered semigroup. Bhuniya and Hansda [2] prove that for every $a$ in a regular ordered semigroup $S$ and for two inverses $x,y$ of $a$, $x\mathcal{H}y$ if and only if for all $e,f\in E_{\leq}(S)$, $ef\in(fSe]$. Thus it is interesting to further study ordered semigroup in which any two inverses of an element are $\mathcal{H}$-related. Class of these ordered semigroups are natural generalization of class of inverse semigroups (without order). We call these ordered semigroups as inverse ordered semigroups. This paper is inspired by Clifford and Preston [1].

2 Preliminaries

An equivalence relation $\rho$ is called left (right) congruence if for $a,b,c\in S\;a\rho b\;\textrm{ implies}\;ca\rho cb\;(ac\rho bc)$. By a congruence we mean both left and right congruence. A congruence $\rho$ is called semilattice congruence on $S$ if for all $a,b\in S,\;a\rho a^{2}\;\textrm{and}\;ab\rho\;ba$. By a complete semilattice congruence on $S$ we mean a semilattice congruence $\sigma$ on $S$ such that for $a,b\in S,\;a\leq b$ implies that $a\sigma ab$. An ordered semigroup $S$ is called complete semilattice of subsemigroups of type $\tau$ if there exists a complete semilattice congruence $\rho$ such that $(x)_{\rho}$ is a type $\tau$ subsemigroup of $S$. Let $I$ be a nonempty subset of $S$. Then $I$ is called a left(right) ideal of $S$, if $SI\subseteq I(IS\subseteq I)$ and $(I]\subseteq I$. An ideal $I$ is both a left and a right ideal of $S$. We call $S$ a (left, right) simple ordered semigroup if it does not contain any proper (left, right) ideal. Following Kehayopulu [5] principal left ideal of $S$ generated by $a$ is defined by the set $L(a)=\{xa:x\in S^{1}\}$, and principal ideal generated by $a$ is defined by the set $I(a)=\{xay:x,y\in S^{1}\}$.

$S$ is said to be regular (resp. Completely regular, right regular) ordered semigroup if for every $a\in S,\;a\in(aSa](a\in(a^{2}Sa^{2}],\;a\in(a^{2}S])$. Due to Kehayopulu [5] Green’s relations on a regular ordered semigroup given as follows:

$a\mathcal{L}b\;\textrm{if}\;L(a)=L(b)$, $a\mathcal{R}b\;\textrm{if}\;R(a)=R(b)$, $a\mathcal{J}b\;\textrm{if}\;I(a)=I(b)$, $\mathcal{H}=\mathcal{L}\cap\mathcal{R}$.

This four relation $\mathcal{L},\mathcal{R},\mathcal{J},\textrm{and}\;\mathcal{H}$ are equivalence relation.

A regular ordered semigroup $S$ is said to be group-like (resp. left group-like) [2] ordered semigroup if for every $a,b\in S,\;a\in(Sb]\;\textrm{and}\;b\in(aS](\textrm{resp}.\;a\in(Sb])$. Right group-like ordered semigroup can be defined dually. An element $b\in S$ is said to be an inverse of $a\in S$ if $a\leq aba$ and $b\leq bab$. The set of all inverses of an element $a$ is denoted by $V_{\leq}(a)$. By an ordered idempotent of $S$, we mean an element $e\in S$ such that $e\leq e^{2}$. The set of all ordered idempotents of $S$ are denoted by $E_{\leq}(S)$. Any two elements $a,b\in S$ are said to be $\mathcal{H}$-commutative if $ab\leq bxa$ for some $x\in S$.

For the sake of convenience of general reader we state some results of [2].

Lemma 2.1**.**

[2]** Let $S$ be a completely regular ordered semigroup. Then following statements hold in $S$:

For every $a$ there is $x\in S$ such that $a\leq axa^{2}$ and $a\leq a^{2}xa$. 2. 2.

$\mathcal{J}$* is the least complete semilattice congruence on $S$.* 3. 3.

$S$* is a complete semilattice of completely simple ordered semigroups.*

3 Inverse ordered semigroup

Let $S$ be an ordered semigroup and $\rho$ be an equivalence relation on $S$. We call an ideal $I$ of $S$ is generated by an $\rho$-unique element $b\in S$ if for any generator of $x\in I$, $b\rho x$.

Definition 3.1**.**

A regular ordered semigroup $S$ is called inverse if for every $a\in S$, any two inverses of $a$ are $\mathcal{H}$-related.

Example 3.2**.**

The ordered semigroup $S=\{a,e,f\}$ defined by multiplication and order below is an inverse ordered semigroup.

[TABLE]

[TABLE]

We present a role of ordered idempotents in an inverse ordered semigroup in the next theorem.

Theorem 3.3**.**

An ordered semigroup $S$ is inverse if and only if every principal left ideal and every principal right ideal of $S$ are generated by an $\mathcal{H}$-unique ordered idempotent.

Proof.

Suppose that $S$ is inverse. Let $I$ be a principal left ideal of $S$. Then there exists $e\in E_{\leq}(S)$ such that $I=(Se]$. If possible let $I=(Sf]$ for some $f\in E_{\leq}(S)$. Then $e\mathcal{L}f$ and thus $e\leq xf$ and $f\leq ye$ for some $x,y\in S$. Now $e\leq ee\leq eee\leq exfe$. Therefore $exf\leq exfexf$ so that $exf\in E_{\leq}(S)$. Also $exf\leq exfexf\leq exf(fe)exf$ and $fe\leq feee\leq fexfe\leq fe(exf)fe$. Therefore $fe\in V_{\leq}(exf)$. Also $exf\in V_{\leq}(exf)$. Since $S$ is inverse, we have $fe\mathcal{H}exf$. Then $e\leq ee\leq exf.fe\leq fezexf$ for some $z\in S$, and so $e\leq fz_{1}$, where $z_{1}=ezexf$. Similarly $f\leq ez_{2}$ for some $z_{2}\in S$. So $e\mathcal{R}f$. Hence $e\mathcal{H}f$. Likewise every principal right ideal of $S$ generated by $\mathcal{H}$-unique ordered idempotent.

Conversely assume that given conditions hold in $S$. Let $a\in S$ and $a^{\prime},a^{\prime\prime}\in V_{\leq}(a)$. Clearly $(Sa]=(Sa^{\prime}a]=(Sa^{\prime\prime}a]$. Since $a^{\prime}a,a^{\prime\prime}a\in E_{\leq}(S)$ we have that $a^{\prime}a\mathcal{H}a^{\prime\prime}a$, by given condition. Then there are $s,t\in S$ such that $a^{\prime}\leq a^{\prime\prime}asa^{\prime}$ and $a^{\prime\prime}\leq a^{\prime}ata^{\prime\prime}$. Thus $a^{\prime}\mathcal{R}a^{\prime\prime}$. Likewise $a^{\prime}\mathcal{L}a^{\prime\prime}$, that is $a^{\prime}\mathcal{H}a^{\prime\prime}$. Hence $S$ is an inverse ordered semigroup. ∎

In the following we show that an ordered semigroup $S$ is inverse if and only if any two ordered idempotents of $S$ are $\mathcal{H}$-commutative.

Theorem 3.4**.**

The following conditions are equivalent on an ordered semigroup $S$.

$S$* is an inverse semigroup;* 2. 2.

$S$* is regular and its idempotents are $\mathcal{H}$-commutative;* 3. 3.

For every $e,f\in E_{\leq}(S)$, $e\mathcal{L}f(e\mathcal{R}f)$ implies $e\mathcal{H}f$.

Proof.

$(1)\Rightarrow(2)$: Obviously $S$ is regular. Let us assume that $a\in S$ and $a^{\prime},a^{\prime\prime}\in V_{\leq}(a)$. Consider $e,f\in E_{\leq}(S)$. Since $S$ is regular, so there is $x\in S$ such that $x\in V_{\leq}(ef)$. Now $x\leq xefx$ implies that $fxe\leq fxe(ef)fxe$ and $ef\leq efxef$ implies $ef\leq ef(fxe)ef$. Thus $ef\in V_{\leq}(fxe)$. Also $fxe\leq fxefxe$ that is $fxe\in E_{\leq}(S)$. So $fxe\in V_{\leq}(fxe)$. Since $S$ is inverse, so $fxe\mathcal{H}ef$. Then there are $s_{1},s_{2}\in S$ such that $ef\leq fxes_{1}$ and $ef\leq s_{2}fxe$. Now $ef\leq efxef$ implies that $ef\leq f(xes_{1}xs_{2}fx)e$. Therefore $ef\leq fye$, where $y=xes_{1}xs_{2}fx$. Similarly there is $z\in S$ such that $fe\leq ezf$. Hence any two idempotents are $\mathcal{H}$-commutative.

$(2)\Rightarrow(3)$: Let $e,f\in E_{\leq}(S)$ be such that $e\mathcal{L}f$. Then $e\leq xf$ and $f\leq ye$ for some $x,y\in S$. Now $e\leq xf$ implies $e\leq exf$, and so $e\leq ee\leq exfe$ which implies that $exf\leq exfexf$. So $exf\in E_{\leq}(S)$. Similarly $fye\in E_{\leq}(S)$. Now $e\leq exf\leq exff\leq exffye$. Since $exf,fye\in E_{\leq}(S)$, by condition (2) we have $exffye\leq(fye)z(exf)$ for some $z\in S$ . Hence $e\leq ft$, where $t=yezexf$. Similarly $f\leq ew$ for some $w\in S$, so that $e\mathcal{R}f$. Hence $e\mathcal{H}f$. If $e\mathcal{R}f$ then $e\mathcal{H}f$ can be done dually.

$(3)\Rightarrow(1)$: Let $a\in S$ and $a^{\prime},a^{\prime\prime}\in V_{\leq}(a)$. Now $aa^{\prime}\leq aa^{\prime\prime}aa^{\prime}$ and $aa^{\prime\prime}\leq aa^{\prime}aa^{\prime\prime}$. So $aa^{\prime}\mathcal{R}aa^{\prime\prime}$ which implies that $aa^{\prime}\mathcal{H}aa^{\prime\prime}$, by the condition (3). Also $a^{\prime}a\mathcal{H}a^{\prime\prime}a$. Then $a^{\prime}\leq a^{\prime}aa^{\prime}$ gives that $a^{\prime}\leq a^{\prime\prime}axa$ for some $x\in S$. Therefore $a^{\prime}\leq a^{\prime\prime}t$ where $t=axa$. In similar way it is possible to obtained $u,v,w\in S$ such that $a^{\prime}\leq ua^{\prime\prime}$, $a^{\prime\prime}\leq a^{\prime}v$ and $a^{\prime\prime}\leq wa^{\prime}$. So $a^{\prime}\mathcal{H}a^{\prime\prime}$. Hence $S$ is an inverse ordered semigroup.

∎

Lemma 3.5**.**

Let $S$ be an inverse ordered semigroup. Then following statements hold in $S$.

$a\mathcal{L}b$* if and only if $a^{\prime}a\mathcal{H}b^{\prime}b$ for some $a,b\in S$ and $a^{\prime}\in V_{\leq}(a)$ $b^{\prime}\in V_{\leq}(b)$;* 2. 2.

$a\mathcal{R}b$* if and only if $aa^{\prime}\mathcal{H}bb^{\prime}$ for some $a,b\in S$ and $a^{\prime}\in V_{\leq}(a)$ $b^{\prime}\in V_{\leq}(b)$;* 3. 3.

for any $a\in S$ and $e\in E_{\leq}(S)$ there are $x,y\in S$ such that $aexa^{\prime},a^{\prime}eya\in E_{\leq}(S)$; where $a^{\prime}\in V_{\leq}(a)$. 4. 4.

for any $a,b\in S$ there are $x,y\in S$ such that $ab\leq abb^{\prime}xa^{\prime}ab$ and $b^{\prime}a^{\prime}\leq b^{\prime}a^{\prime}aybb^{\prime}a^{\prime}$, where $a^{\prime}\in V_{\leq}(a)$ and $b^{\prime}\in V_{\leq}(b)$.

Proof.

(1): Let $a,b\in S$ be such that $a\mathcal{L}b$. Let $a^{\prime}\in V_{\leq}(a)$, $b^{\prime}\in V_{\leq}(b)$. Since $a\leq aa^{\prime}a$ and $a^{\prime}a\leq a^{\prime}aa^{\prime}a$, we have $a\mathcal{L}a^{\prime}a$ which implies that $b\mathcal{L}a^{\prime}a$. Also $b\mathcal{L}b^{\prime}b$. Hence $a^{\prime}a\mathcal{L}b^{\prime}b$. Since $a^{\prime}a,b^{\prime}b\in E_{\leq}(S)$ and $S$ is inverse we have $a^{\prime}a\mathcal{H}b^{\prime}b$, by Theorem 3.4(3).

Conversely suppose that given condition holds in $S$. Let $a,b\in S$ with $a^{\prime}\in V_{\leq}(a)$ and $b^{\prime}\in V_{\leq}(b)$. Then by given condition $aa^{\prime}\mathcal{H}bb^{\prime}$. Also we have $a\mathcal{L}a^{\prime}a$ and $b\mathcal{L}b^{\prime}b$ so that $a\mathcal{L}b$.

(2): This is similar to (1).

(3): Let $a\in S$ and $e\in E_{\leq}(S)$. Also $a^{\prime}a\in E_{\leq}(S)$. Since $S$ is an inverse, there is an $x\in S$ such that $a^{\prime}ae\leq exa^{\prime}a$ by Theorem 3.4(2). Now $aexa^{\prime}\leq aa^{\prime}aeexa^{\prime}\leq aexa^{\prime}aexa^{\prime}$. So $aexa^{\prime}\in E_{\leq}(S)$. Likewise $a^{\prime}eya\in E_{\leq}(S)$; for some $y\in S$.

(4): Let $a,b\in S$ with $a^{\prime}\in V_{\leq}(a)$, $b^{\prime}\in V_{\leq}(b)$. So $a^{\prime}a,b^{\prime}b\in E_{\leq}(S)$. Now $ab\leq aa^{\prime}abb^{\prime}b\leq$ and $a^{\prime}abb^{\prime}\leq b^{\prime}bxa^{\prime}a$, by Theorem 3.4(2). Thus $ab\leq abb^{\prime}xa^{\prime}ab$. Likewise $b^{\prime}a^{\prime}\leq b^{\prime}a^{\prime}aybb^{\prime}a^{\prime}$; for some $y\in S$. ∎

In the following theorem an inverse ordered semigroup has been characterized by the inverse of an element of the set $(eSf]$.

Theorem 3.6**.**

Let $S$ be an ordered semigroup and $e,f\in E_{\leq}(S)$. Then $S$ is inverse if and only if for every $x\in(eSf]$ implies $x^{\prime}\in(fSe]$, where $x^{\prime}\in V_{\leq}(x)$.

Proof.

First suppose that $S$ is an inverse ordered semigroup and $x\in(eSf]$. Then $x\leq es_{1}f$ for some $s_{1}\in S$. Let $x^{\prime}\in V_{\leq}(x)$. Now $x^{\prime}\leq x^{\prime}xx^{\prime}\leq x^{\prime}es_{1}fx^{\prime}$, and so $es_{1}fx^{\prime}\leq es_{1}fx^{\prime}es_{1}fx^{\prime}$. Hence $es_{1}fx^{\prime}\in E_{\leq}(S)$. Similarly $x^{\prime}es_{1}f\in E_{\leq}(S)$. Now there is $s_{2}\in S$ such that $x^{\prime}es_{1}fx^{\prime}\leq x^{\prime}es_{1}ffx^{\prime}\leq fs_{2}x^{\prime}es_{1}fx^{\prime}$, by Theorem 3.4(2) . Also $fs_{2}x^{\prime}es_{1}fx^{\prime}\leq fs_{2}x^{\prime}ees_{1}fx^{\prime}\leq fs_{2}x^{\prime}es_{1}fx^{\prime}s_{3}e$, for some $s_{3}\in S$. Then $x^{\prime}\leq x^{\prime}xx^{\prime}$ implies that $x^{\prime}\leq fs_{2}x^{\prime}es_{1}fx^{\prime}\leq fs_{2}x^{\prime}es_{1}fx^{\prime}s_{3}e$. Hence $x^{\prime}\in(fSe]$.

Conversely assume that the given conditions hold in $S$. First consider a left ideal $L$ of $S$ such that $L=(Se]=(Sf]$ for $e,f\in E_{\leq}(S)$. Then $e\mathcal{L}f$, so that $e\leq ee\leq ezf$ for some $z\in S$. Therefore $e\in(eSf]$. Since $e\in V_{\leq}(e)$ we have $e\in(fSe]$, by given condition. Likewise $f\in(eSf]$. This implies that $e\mathcal{R}f$ and so $e\mathcal{H}f$. Similarly it can be shown that every principal right ideal of $S$ generated by $\mathcal{H}$-unique ordered idempotent. Thus by Theorem 3.3, $S$ is an inverse ordered semigroup. ∎

Corollary 3.7**.**

The following conditions are equivalent on a regular ordered semigroup $S$.

$S$* is an inverse ordered semigroup;* 2. 2.

for any $a\in S$ and for any $a^{\prime}\in V_{\leq}(a)$, $aa^{\prime},a^{\prime}a$ are $\mathcal{H}$-commutative; 3. 3.

for any $e\in E_{\leq}(S)$, any two inverses of $e$ are $\mathcal{H}$-related; 4. 4.

for any $e\in E_{\leq}(S)$ and all its inverses are $\mathcal{H}$-commutative; 5. 5.

for any $e\in E_{\leq}(S)$ and $e^{\prime}\in V_{\leq}(e)$, $ee^{\prime}$ and $e^{\prime}e$ are $\mathcal{H}$-commutative.

Proof.

$(1)\Rightarrow(2)$, $(2)\Rightarrow(3)$, $(3)\Rightarrow(4)$, and $(4)\Rightarrow(5)$: These are obvious.

$(5)\Rightarrow(1)$: Let $e,f\in E_{\leq}(S)$ and $x\in V_{\leq}(ef)$. So $ef\leq efxef\leq effxeef$ and $x\leq xefx$ implies that $fxe\leq fxeeffxe$. So $ef\in V_{\leq}(fxe)$. Also $fxe\in E_{\leq}(S)$. Now $ef\leq efxef\leq effxeef\leq effxefxeef\leq fxez_{1}efz_{2}fxe$, for some $z_{1},z_{2}\in S$, by the given condition. So $ef\leq fz_{3}e$ where $z_{3}=xemefnfx$. Similarly $fe\leq ez_{4}f$, for some $z_{4}\in S$. So $e,f$ are $\mathcal{H}$-commutative. Hence by Theorem 3.4 $S$ is inverse ordered semigroup. ∎

We study inverse ordered semigroup together with complete regularity in the following theorem.

Theorem 3.8**.**

The following conditions are equivalent on a regular ordered semigroup $S$.

$S$* is inverse and completely regular;* 2. 2.

$S$* is a complete semilattice of group like ordered semigroups;* 3. 3.

$ab\mathcal{H}ba$* whenever $ab,ba\in E_{\leq}(S)$;* 4. 4.

any ordered idempotent of $S$ is $\mathcal{H}$-commutative to any element of $S$; 5. 5.

for any $e,f\in E_{\leq}(S)$ $e\mathcal{J}f$ implies $e\mathcal{H}f$; 6. 6.

$\mathcal{H}=\mathcal{L}=\mathcal{R}=\mathcal{J}$.

Proof.

$(1)\Rightarrow(2)$: Let $S$ be a completely regular and inverse ordered semigroup. Then by Theorem 2.1, $\mathcal{J}$ is the complete semilattice congruence on $S$ and every $\mathcal{H}$-class is a group-like ordered semigroup. We now prove $\mathcal{H}=\mathcal{J}$. Let $a,b\in S$ be such that $a\mathcal{J}b$. So there are $x,y,u,v\in S$ such that $a\leq xby$ and $b\leq uav$. Since $S$ is completely regular, so there are $h,g,f\in S$ such that $x\leq x^{2}hx$, $b\leq b^{2}gb$, $b\leq bgb^{2}$, $y\leq yfy^{2}$. Now $a\leq x^{2}hxb^{2}gbyfy^{2}\leq x^{2}hxb^{2}gbgb^{2}yfy^{2}$. Let $p\in V_{\leq}(x^{2}hxb^{2}g)$. So $x^{2}hxb^{2}g\leq x^{2}hxb^{2}gpx^{2}hxb^{2}g\leq x^{2}hxb^{2}g(b^{2}gpx^{2}h)x^{2}hxb^{2}g$ and $b^{2}gpx^{2}h\leq b^{2}gpx^{2}hxb^{2}gpx^{2}h\leq b^{2}gpx^{2}h(x^{2}hxb^{2}g)b^{2}gpx^{2}h$. This shows that $b^{2}gpx^{2}h\in V_{\leq}(x^{2}hxb^{2}g)$. Also $x^{2}hxb^{2}g\leq x^{2}hxb^{2}gpx^{2}hxb^{2}g\leq x^{2}hxb^{2}g(x^{2}hxb^{2}gp^{2})x^{2}hxb^{2}g$ and $x^{2}hxb^{2}gp^{2}\leq x^{2}hxb^{2}gpx^{2}hxb^{2}gp^{2}\leq x^{2}hxb^{2}gp^{2}(x^{2}hxb^{2}g)x^{2}hxb^{2}gp^{2}$, which implies that $x^{2}hxb^{2}gp^{2}\in V_{\leq}(x^{2}hxb^{2}g)$. Similarly $p^{2}x^{2}hxb^{2}g\in V_{\leq}(x^{2}hxb^{2}g)$. Since $b^{2}gpx^{2}h,x^{2}hxb^{2}gp^{2}\in V_{\leq}(x^{2}hxb^{2}g)$ and $S$ is inverse, so there is $t\in S$ such that $x^{2}hxb^{2}gp^{2}\leq b^{2}gpx^{2}ht$. Thus $x^{2}hxb^{2}g\leq x^{2}hxb^{2}gpx^{2}hxb^{2}g\leq x^{2}hxb^{2}gp^{2}(x^{2}hxb^{2}g)^{2}$ implies that $x^{2}hxb^{2}g\leq b^{2}gpx^{2}hxt(x^{2}hxb^{2}g)^{2}=bs$ where $s=bgpx^{2}ht(x^{2}hxb^{2}g)^{2}$. Similarly there is $s_{1}\in S$ such that $b^{2}gyfy^{2}\in s_{1}b$. Hence $a\leq x^{2}hxb^{2}gbyfy^{2}\leq bsbyfy^{2}=bs_{2}$, where $s_{2}=sbyfy^{2}$. Similarly $a\leq s_{3}b$ for some $s_{3}\in S$. Likewise $b\leq s_{4}a$ and $b\leq as_{5}$, for some $s_{4},s_{5}\in S$. So $a\mathcal{H}b$. Thus $\mathcal{J}\subseteq\mathcal{H}$. Also $\mathcal{H}\subseteq\mathcal{J}$, and Hence $\mathcal{J}=\mathcal{H}$. Therefore $S$ is complete semilattice of group-like ordered semigroups.

$(2)\Rightarrow(3)$: Suppose that $S$ is a complete semilattice $Y$ of group like ordered semigroups $\{S_{\alpha}\}_{\alpha\in Y}$. Let $a,b\in S$ such that $ab,ba\in E_{\leq}(S)$. Let $\rho$ be the corresponding semilattice congruence on $S$. Then there is $\alpha\in Y$ such that $ab,ba\in S_{\alpha}$ . Since $S_{\alpha}$ is group like ordered semigroups so $ab\mathcal{H}ba$.

$(3)\Rightarrow(4)$: Let $a\in S$ and $e\in E_{\leq}(S)$. Since $S$ is regular there is an $x\in S$ such that $a\leq axa$. Clearly $ax,xa\in E_{\leq}(S)$. Thus by condition (3) $ax\mathcal{H}xa$. So $xa\leq axu$ and $ax\leq vxa$, for some $u,v\in S$. Then we have $a\leq axa\leq axaxa\leq axaxaxa\leq aaxuxvxaa=a^{2}ta^{2}$, where $t=xuxvx$. Now $a\leq a^{2}ta^{2}\leq a(a^{2}ta^{2}ta^{2}ta^{2})a\leq a^{2}(a^{2}ta^{2}ta^{2}ta^{2}ta^{2})a$, that is $a\leq a^{2}ya$, where $y=a^{2}ta^{2}ta^{2}ta^{2}ta^{2}$. Similarly $a\leq aya^{2}$. Clearly $a^{2}y,ya^{2}\in E_{\leq}(S)$.

Let $e,f\in E_{\leq}(S)$ and $x\in V_{\leq}(ef)$. Then we have $x\leq xefx$. So $fxe\leq fxefxe\leq fxeeffxe$ and $ef\leq efxef\leq effxeef$. So $ef\in V_{\leq}(fxe)$. Also $ef\leq effxeef$ implies that $effxe\leq effxeeffxe$, and $fxeef\leq fxeeffxeef$. So $effxe,fxeef\in E_{\leq}(S)$ and thus $effxe\mathcal{H}fxeef$, by the condition(3). Then there are $u,v\in S$ such that $effxe\leq fxeefu$ and $fxeef\leq veffxe$. Now $ef\leq effxefxeef\leq fxeefuveffxe=fce$; where $c=xe^{2}fuvef^{2}x$. Likewise $fe\leq edf$, for some $d\in S$.

Now $ae\leq a^{2}yae$. Let $z\in V_{\leq}(a^{2}yae)$. So $a^{2}yae\leq a^{2}yaeza^{2}yae\leq a^{2}yae(eza^{2}y)a^{2}yae$. Clearly $a^{2}yaeeza^{2}y,eza^{2}ya^{2}yae\in E_{\leq}(S)$ and thus $a^{2}yaeeza^{2}y\mathcal{H}eza^{2}ya^{2}yae$, by condition (3). Now $ae\leq a^{2}yae\leq a^{2}yaeeza^{2}ya^{2}yae\leq eza^{2}ys_{1}a^{2}yaea^{2}yae$, for some $s_{1}\in S$. So $ae\leq es_{2}ae$, where $s_{2}=za^{2}ys_{1}a^{2}yaea^{2}y$. Again $ae\leq es_{2}aya^{2}e\leq es_{2}aes_{3}ya^{2}$, for some $s_{3}\in S$, since $ya^{2},e\in E_{\leq}(S)$. That is $ae\leq es_{4}a$, for some $s_{4}\in S$. Similarly $ea\leq as_{5}e$, for some $s_{5}\in S$. So $a,e$ are $\mathcal{H}$-commutative.

$(4)\Rightarrow(5)$: Let $e,f\in E_{\leq}(S)$ such that $e\mathcal{J}f$. Then there are $x,y,z,u\in S$ such that $e\leq xfy$ and $f\leq zeu$. Now $e\leq xfy$ implies that $e\leq fhxy$ and $e\leq xykf$ by the given condition for some $h,k\in S$. Similarly $f\leq zeu$ gives $f\leq es_{1}zu$ and $f\leq zus_{2}e$ for some $s_{1},s_{2}\in S$. Hence $e\mathcal{H}f$.

$(5)\Rightarrow(6)$: Let $a,b\in S$ such that $a\mathcal{J}b$. Then there are $s,t,u,v\in S$ such that $a\leq sbt$ and $b\leq uav$. Since $S$ is regular so $a\leq axa$ and $b\leq byb$ for some $x,y\in S$ so that $ax\leq axax$ and $by\leq byby$. Now $axax\leq axsbtx\leq axsbybtx$ that is $ax\leq axsbybtx$. Likewise $by\leq byuaxavy$. Thus $ax\mathcal{J}by$, so from given condition $ax\mathcal{H}by$. Similarly $xa\mathcal{H}yb$. So there is $c\in S$ such that $ax\leq byc$, that is $a\leq byca=bd$, for some $d=yca\in S$. Likewise $a\leq pb$, $b\leq qa$ for some $p,q\in S$. Thus $a\mathcal{H}b$. So $\mathcal{H}=\mathcal{J}$. Now $\mathcal{J}=\mathcal{H}=\mathcal{L}\cap\mathcal{R}$ gives $\mathcal{J}\subseteq\mathcal{L}$ and $\mathcal{J}\subseteq\mathcal{R}$. Therefore $\mathcal{L}=\mathcal{J}=\mathcal{R}$.

$(6)\Rightarrow(1)$: Let $a\in S$ Since $S$ is regular so there exists $a^{\prime}\in V_{\leq}(a)$. Clearly $a\mathcal{L}a^{\prime}a$ and $a\mathcal{R}aa^{\prime}$. So by the given condition $a\mathcal{R}a^{\prime}a$ and $a\mathcal{L}aa^{\prime}$. Now $a\leq aa^{\prime}a\leq aa^{\prime}aa^{\prime}a\leq aa^{\prime}aa^{\prime}aa^{\prime}a\leq aas_{1}a^{\prime}s_{2}aa$ for some $s_{1},s_{2}\in S$. So $a\leq a^{2}pa^{2}$ where $p=s_{1}a^{\prime}s_{2}$. So $S$ is completely regular.

Also let $a^{\prime},a^{\prime\prime}\in V_{\leq}(a)$. Now $a\mathcal{L}a^{\prime}a\mathcal{L}a^{\prime\prime}a$ implies that $a\mathcal{R}a^{\prime}a\mathcal{R}a^{\prime\prime}a$. Also by the given condition we can show that $a\mathcal{L}aa^{\prime}\mathcal{L}a^{\prime\prime}a$. So it is to check that $a^{\prime}\mathcal{R}a^{\prime\prime}$ and $a^{\prime}\mathcal{L}a^{\prime\prime}$. So $a^{\prime}\mathcal{H}a^{\prime\prime}$. Hence $S$ is inverse ordered semigroup. ∎