Nil extensions of simple regular ordered semigroup

TL;DR

This paper investigates nil extensions of specific ordered semigroups, providing characterizations and decompositions for these structures, enhancing understanding of their algebraic properties and relationships.

Contribution

It introduces new characterizations of nil extensions in simple regular, left simple, and right regular ordered semigroups, and describes their semilattice decompositions.

Findings

Characterization of nil extensions in various ordered semigroups

Complete semilattice decomposition of nil extension semigroups

Enhanced understanding of algebraic structure of ordered semigroups

Abstract

In this paper, nil extensions of some special type of ordered semigroups, such as, simple regular ordered semigroups, left simple and right regular ordered semigroup. Moreover, we have characterized complete semilattice decomposition of all ordered semigroups which are nil extension of ordered semigroup.

Equations

$$(H{]}_S:=\{t\in S:t\le h,forsomeh\in H\}.$$

$$a|b\iff thereexistsx,y\in S^{1}suchthatb\le xay.$$

$$\ast :S/K\times S/K⟶S/K|(x,y)⟶(x\ast y)where$$

$$x*y:=\left\{\begin{array}[]{l l}xy&\mbox{if}~{}~{}~{}~{}\;xy\in S/K\\ 0&\mbox{if}~{}~{}~{}~{}\;xy\in K\end{array}\right.$$

Full text

Key Words and phrases: Nil extension; ordered regular semigroups,completely regular, left regular ordered semigroups; simple; left simple ordered semigroups.

2010 Mathematics subject Classification: 20M10; 06F05.

1 Introduction:

Nil extensions of a semigroup(without order), are precisely the ideal extensions by a nil semigroup. In 1984, S. Bogdanovic and S. Milic [2] characterized the semigroups(without order) which are nil extensions of completely simple semigroups, where as, a similar work was done by J. L Galbiati and M.L Veronesi[14] in 1980. Authors like S. Bogdanovic, M. Ciric, Beograd have investigated this type extensions for regular semigroup, group, periodic semigroup as well as completely regular semigroup(see [4], [5]).

The concepts of nil extensions have been extended to ordered semigroups by Y. Cao in the paper [7] with characterizing nil extensions all type of simple ordered semigroups. Cao was the first to do so. The notion of ideal extensions in ordered semigroups is actually introduced by N. Kehayopulu and M. Tsingelis in [8]. In [11] they have worked give on semigroups which are nil extensions of Archimedean ordered semigroups.

This paper inspired by the results of [4] [5] [6]. The aim of this paper is to describe all those ordered semigroups which are the nil extension of simple regular, left simple right regular and completely regular ordered semigroups.

Our paper organized as follows. The basic definitions and properties of ordered semigroups are presented in Section 2. Section 3 is devoted to characterizing the nil extensions of simple regular, left simple right regular and completely regular ordered semigroups. In Section 4 we have described complete semilattice decomposition of all ordered semigroups which are nil extension of ordered semigroups.

2 Preliminaries:

In this paper $\mathbb{N}$ will provide the set of all natural numbers. An ordered semigroup is a partiality ordered set $S$, and at the same time a semigroup $(S,.)$ such that $(\forall a,\;b,\;x\in S)\;a\leq b\Rightarrow\;xa\leq xb\;and\;ax\leq bx$. It is denoted by $(S,.,\leq)$. For an ordered semigroup $S$ and $H\subseteq S$, denote

[TABLE]

Also it is denoted by $(H]$ if there is no scope of confusion.

The relation $|$ on an ordered semigroup $S$ is defined by

[TABLE]

Let $I$ be a nonempty subset of an ordered semigroup $S$. $I$ is a left (right) ideal of $S$, if $SI\subseteq I\;(IS\subseteq I)$ and $(I]=I$. $I$ is an ideal of $S$ if $I$ is both a left and a right ideal of $S$. An (left, right) ideal $I\;ofS$ is proper if $I\neq S$. $S$ is left (right) simple if it does not contain proper left (right) ideals. $S$ is simple if it does not contain any proper ideals. The intersection of all ideals of an ordered semigroup $S$ , if nonempty, is called the kernel of $S$ and is denoted by $K(S)$.

An ordered semigroup $S$ is called a group like ordered semigroup or GLO-semigroup if for all $a,b\in S\;there\;are\;x,y\in S\;such\;that\;a\leq xb\;and\;a\leq by$. An ordered semigroup $S$ is called a left group like ordered semigroup or LGLO- semigroup if for all $a,b\in S\;there\;are\;x\in S$ such that $a\leq xb$ Similarly we define right group like ordered semigroup or RGLO-semigroup. A band is a semigroup in which every element is an idempotent. A commutative band is called a semilattice. Kehayopulu [9] defined Greens relations on an ordered semigroup $S$ as follows: $a\mathcal{L}b\;if\;L(a)=L(b),\;a\mathcal{R}b\;if\;R(a)=R(b),\;a\mathcal{J}b\;if\;I(a)=I(b),\;and\;\mathcal{H}=\;\mathcal{L}\cap\;\mathcal{R}$. These four relations $\mathcal{L},\;\mathcal{R},\;\mathcal{J}\;and\;\mathcal{H}$ are equivalence relations. In an ordered semigroup $S$, an equivalence relation $\rho$ is called left (right) congruence if for $a,b,c\in S\;a\;\rho\;b\;implies\;ca\;\rho\;cb\;(ac\;\rho\;bc)$. $\rho$ is congruence if it is both left and right congruence. A congruence $\rho$ on $S$ is called semilattice if for all $a,b\in S\;a\;\rho\;a^{2}\;and\;ab\;\rho\;ba$. A semilattice congruence $\rho$ on $S$ is called complete if $a\leq b$ implies $(a,ab)\in\rho$. The ordered semigroup $S$ is called complete semilattice of subsemigroup of type $\tau$ if there exists a complete semilattice congruence $\rho$ such that $(x)_{\rho}$ is a type $\tau$ subsemigroup of $S$. Equivalently: There exists a semilattice $Y$ and a family of subsemigroups $\{S\}_{\alpha\in Y}$ of type $\tau$ of $S$ such that:

$S_{\alpha}\cap S_{\beta}=\;\phi$ for any $\alpha,\;\beta\in\;Y\;with\;\alpha\neq\beta,$ 2. 2.

$S=\bigcup_{\alpha\;\in\;Y}\;S_{\alpha},$ 3. 3.

$S_{\alpha}S_{\beta}\;\subseteq\;S_{\alpha\;\beta}$ for any $\alpha,\;\beta\in\;Y,$ 4. 4.

$S_{\beta}\cap(S_{\alpha}]\neq\phi$ implies $\beta\;\preceq\;\alpha,$ where $\preceq$ is the order of the semilattice $Y$ defined by

$\preceq:=\{(\alpha,\;\beta)\;\mid\;\alpha=\alpha\;\beta\;(\beta\;\alpha)\}$ [10].

Before going to the main results of this paper we shall illustrate some important definition and results related to nil extensions of ordered semigroups.

Lemma 2.1**.**

[7]** Let $I$ be an ideal of an ordered semigroup $S$. Then $I=K(S)$ if and only if $I$ is simple.

Definition 2.2**.**

[7]** Let $S$ be an ordered semigroup with zero $0(i.e.,\;there\;exists\;0\in S\;such\;that\;0x=x0=0\;for\;any\;x\in S)$ . An element $a\in S$ is called a nilpotent if $a^{n}\leq 0$ for some $n\in\mathbb{N}$. The set of all nilpotents of $S$ is denoted by $Nil(S)$. $S$ is called nil ordered semigroup (nilpotent) if $S=Nil(S)$.

Lemma 2.3**.**

[8]** Let $(S,.,\leq)$ be an ordered semigroup and $K$ an ideal of $S$. Let $S/K:=(S\backslash K)\;\cup\{0\}$, where [math] is an arbitrary element of $K\;(S\backslash K$ is the complement of $K\;to\;S$). Define an operation $"*"$ and an order $"\preceq"\;on\;S/K$ as follows:

[TABLE]

[TABLE]

$\preceq:=(\leq\cap[(S/K)\times(S/K)])\cup\{(0,x)|x\in S/K\}$. Then $(S/K,*,\preceq)$ is an ordered semigroup and [math] is its zero.

If there is no confusion, we denote the multiplication and the order in different ordered semigroups by same symbol.

We know that $(S/K,*)$ is a Rees semigroup where the congruence is the Rees Congruence on $S$. Now we shall proceed to define the ideal extension of ordered semigroup in Kehayopulu’s way.

Definition 2.4**.**

[8]** Let $(S,\cdot,\leq_{S})$ be an ordered semigroup, $(Q,\cdot,\leq_{Q})$ an ordered semigroup with $0,\;S\cap Q^{*}=\phi,\;where\;Q^{*}=Q\setminus\{0\}$. An ordered semigroup $(V,\cdot,\leq_{V})$ is called an ideal extension of $S$ by $Q$ if there exists an ideal $S^{\prime}$ of $V$ such that $(S^{\prime},\cdot,\leq_{S^{\prime}})\approx(S,\cdot,\leq_{S})$ and $(V/S^{\prime},*,\preceq)\approx(Q,\cdot,\leq_{Q})$, where $\leq_{S^{\prime}}=\leq_{V}\cap(S^{\prime}\times S^{\prime})$ and $"*",\;"\preceq"$ the multiplication and the order on $V/S^{\prime}$ defined in LABEL:1

So, now we are ready to consider the definition of nil extension of ordered semigroup given by Cao.

Definition 2.5**.**

[7]** Let $I$ be an ideal of an ordered semigroup $S$. Then $(S/I,.,\preceq)$ is called the Rees factor ordered semigroup of $S\;modulo\;I$, and $S$ is called an ideal extension of $I$ by ordered semigroup $S/I$. An ideal extension $S\;of\;I$ is called a nil-extension of $I$ if $S/I$ is a nil ordered semigroup.

Lemma 2.6**.**

[7]** Let $S$ be an ordered semigroup and $I$ an ideal of $S$. Then the following are equivalent:

• (i)

$S$* is a nil-extension of $I$;*

• (ii)

$(\forall a\in S)(\exists m\in\mathbb{N})\;a^{m}\in I$.

In [1], we have introduced the notion of Clifford and left Clifford ordered semigroups and characterized their structural r

Lemma 2.7**.**

Let $S$ be an ordered semigroup. Then

for $a\in Reg_{\leq}(S)$, there is $x\in Reg_{\leq}(S)$ such that $a\leq axa$. 2. 2.

for $a\in\mathbf{R}Reg_{\leq}(S)$ and any $k\in\mathbb{N},\;a^{k}\in(a^{k+n}S]$ for all $n\in\mathbb{N}$.

3 nil extension of completely regular ordered semigroups:

Theorem 3.1**.**

An ordered semigroup $S$ is nil extensions of left simple and right regular ordered semigroup if and only if for all $\;a,b\in S,\;there\;exists\;n\in\mathbb{N}\;such\;that\;a^{n}\in(a^{2n}Sb]$ and for all $a\in S,\;and\;b\in\mathbf{R}Reg_{\leq}(S)$ such that $a\leq ba$ implies $a\leq a^{2}x\;for\;some\;x\in S$.

Proof.

Suppose $S$ is a nil extension of a left simple and right regular ordered semigroup $K$. Let $a,b\in S$. Then there exists $\;m\in\mathbb{N}$ such that $a^{m},a^{m}b\in K$. Since $K$ is a right regular ordered semigroup, there is $x\in K$ such that $a^{m}\leq a^{2m}x$. The left simplicity of $K$ yields $x\leq t^{\prime}a^{m}b$ for some $t^{\prime}\in K$. So $a^{m}\in(a^{2m}Sb]$. Again let $b\in\mathbf{R}Reg_{\leq}(S)$ and $a\in S$ such that $a\leq ba$. Since $b\in\mathbf{R}Reg_{\leq}(S),\;there\;exists\;z\in K$, such that $b\leq b^{n+1}(z)^{n}$, which holds for all $n\in\mathbb{N}$. So there is some $n_{1}\in\mathbb{N}\;such\;that\;z^{n_{1}}\in K$. Since $K$ is an ideal of $S$, $b^{n_{1}+1}(z)^{n_{1}}\in K$. Thus $b\in K$, so $ba\in K$ and hence $a\in K$. Then the right regularity of $K$ yields $a\in(a\in a^{2}S]$. Thus the condition is necessary.

Conversely, let us assume that the given condition holds in $S$. Choose $a\in S$ arbitrarily. Then by the given condition $\;there\;exists\;m\in\mathbb{N}\;and\;y\in S$ such that $a^{m}\leq a^{2m}yb$. Clearly $\mathbf{R}Reg_{\leq}(S)\neq\phi$. Say $T=\mathbf{R}Reg_{\leq}(S)$. It is evident that for each $b\in S,\;there\;exists\;k\in\mathbb{N}$ such that $b^{k}\in T$. Consider $s\in S$ and $a\in T$. Then there exists $y\in S\;such\;that\;a\leq a^{2}y$. That is $a\leq a^{3}y^{2}\leq........................\leq a^{n+1}(y)^{n},\;\;for\;all\;n\in\mathbb{N}$. Then $sa\leq saa^{n}y^{n},\;for\;all\;n\in\mathbb{N}$. From the given condition there are $m_{1}\in\mathbb{N}\;and\;t_{1}\in S$ such that $y^{m_{1}}\leq y^{2m_{1}}t_{1}sa$, whence $sa\leq sa(a^{m}y^{2m_{1}}t_{1})sa$. Set $t=a^{m}y^{2m_{1}}t_{1}$, Then $sa\leq satsa$. Now $sat\leq(sat)^{2}sat$, which shows that $sat\in T$. Using the given second condition to $sa\in S\;and\;sat\in T;\;sa\leq satsa$ yields that $sa\in((sa)^{2}S]$, that is, $sa\in T$.

In a similar approach it can be easily obtain that $as\leq a^{m_{2}+1}z^{2m_{2}}t_{2}aas$ for some $m_{2}\in\mathbb{N}$. Now since $a^{m_{2}+1}z^{2m_{2}}t_{2}\in S$, by preceding conclusion $a^{m_{2}+1}z^{2m_{2}}t_{2}a\in S$. Hence it follows given condition that, $as\leq(as)^{2}a^{m_{2}+1}z^{2m_{2}}t_{2}aas$, that is $as\in((as)^{2}S]$. Therefore $sa,\;as\in T$. And so $T$ is a right regular subsemigroup of $S$.

Consider $a\in S$ and $b\in T$ such that $a\leq b$. Then for some $u\in\mathbb{N},\;a\leq b^{n+1}u^{n},\;for\;all\;n\in\mathbb{N}$. Then there exists $m_{3}\in\mathbb{N}$ such that $a\leq(b^{m_{3}+1}u^{m_{3}}a)a$. Since $b^{m_{3}+1}u^{m_{3}}a\in T$, the given condition yields that $a\leq a^{2}t_{4}\;for\;some\;t_{4}\in S$.

Finally, choose $c,d\in T$. Then $c\leq c^{n+1}v^{n}\;for\;all\;n\in\mathbb{N}$ and for some $v\in S$. Now for $v,d\in S,\;there\;exists\;t_{5}\in S$ such that $v^{m_{4}}\leq v^{2m_{4}}t_{5}d$ for some $m_{4}\in\mathbb{N}$. Since $c\in T,\;c^{m_{4}}t_{5}\in T$ and so $c\in(Td]$. Hence $T$ is left simple. Thus $T$ is left simple and right regular ordered semigroup. Hence $S$ is nil extension of a left simple and right regular ordered semigroup $T$. ∎

In [7] authors have characterized ordered semigroup which are nil extensions of simple(order) semigroups. In the next result we have described ordered semigroups which are nil extensions of both simple and ordered regular semigroups.

Theorem 3.2**.**

An ordered semigroup $S$ is nil extensions of simple and regular ordered semigroup if and only if the following conditions are holds

• (i)

for all $a,b\in S$, there exists $n\in\mathbb{N}$ such that $a^{n}\in(a^{n}SbSa^{n}]$.

• (ii)

for $a\in S$, and $b\in Reg_{\leq}(S)$ such that $a\leq ba$ implies that $a\in(aSbSa]$..

• (iii)

for $a\in S$, and $b\in Reg_{\leq}(S)$ such that $a\leq ab$ implies that $a\in(aSbSa]$.

• (iv)

for $a\in S$, and $b\in Reg_{\leq}(S)$ such that $a\leq b$ implies that $a\in Reg_{\leq}(S)$.

Proof.

Suppose that $S$ is a nil extension of a simple and regular ordered semigroup $K$.

$(i)$ Choose $a,b\in S$. Then there exists $m\in\mathbb{N}$ such that $a^{m},a^{m}b\in K$. Since $K$ is regular ordered semigroup, there is some $x\in K$ such that $a^{m}\leq a^{m}xa^{m}\leq a^{m}(xa^{m})xa^{m}$, whence, by the simplicity of $K$ there is

$y\in K$ such that $a^{m}\leq a^{m}(ya^{m})bxa^{m}\in(a^{m}SbSa^{m}]$.

$(ii)$ Let $b\in Reg_{\leq}(S)$ and $a\in S$ such that $a\leq ba$. Since $b\in Reg_{\leq}(S),\;there\;exists\\ \;z\in S$ such that $b\leq b(zb)^{n}\;for\;all\;n\in\mathbb{N}$. Then for some $n_{1}\in\mathbb{N}$ such that $(zb)^{n_{1}}\in K$. This gives $b(zb)^{n_{1}}\in K$, which gives that $b\in K$ and so $ba\in K$ and finally $a\in K$. Since $K$ is a simple and regular ordered semigroup, for $a,b\in K,\;a\in(aKbKa]\subseteq(aSbSa]$.

$(iii)$ and $(iv)$ follows similarly.

Thus the given conditions are obtained.

Conversely, let us assume that given condition holds in $S$. Choose $a\in S$ arbitrarily. By the given condition, there is $m\in\mathbb{N}$ and some $x\in S$ such that $a^{m}\leq a^{m}xa^{m}$. So $Reg_{\leq}(S)\neq\phi$. Say $T=Reg_{\leq}(S)$. It is evident that, for every $a\in S,\;there\;exists\;m\in\mathbb{N}$ such that $a^{m}\in T$. Consider $s\in S$ and $a\in T$. Then for some $h\in S,\;a\leq aha$, whence $sa\leq sa(ha)^{n},\;and\;as\leq(ah)^{n}as\;for\;all\;n\in\mathbb{N}$. Then there is $m_{1}\in\mathbb{N}$ such that $(ha)^{m_{1}}\in T$. So using the condition(ii) to $sa\leq sa(ha)^{m_{1}}$ yields that $sa\in((sa)S(ha)^{m_{1}}Ssa]$, that is $sa\in T$. On the otherhand $as\in T$ follows similarly by using condition(ii). Hence by condition(iv) $T$ is an ideal of $S$. Also by the given condition(iv) it can be easily seen that $T$ is simple in $S$. Hence $S$ is nil extension of a simple and regular ordered semigroup $T$. ∎

Theorem 3.3**.**

An ordered semigroup $S$ is nil extensions of completely regular ordered semigroup if and only if S is completely $\pi$-regular and for all $a\in S,\;b\in Reg_{\leq}(S)$ such that $a\leq ba$ and $a\leq ab$, both separately implies $a\leq(a^{2}SbSa^{2}]\;for\;some\;x\in S$.

Proof.

Suppose that $S$ is a nil extension of an completely regular ordered semigroup $K$ and $a,b\in S$. Then there is $\;m\in\mathbb{N}$ such that $a^{m}\in K$. By the complete regularity of $K$,it is possible to get some $m\in\mathbb{N}$ and $x\in K$ such that $a^{m}\leq a^{2m}xa^{2m}\leq a^{m+1}xa^{m+1}$ Thus $S$ is completely $\pi$-regular.

Again let $b\in Reg(S)$ and $a\in S$ such that $a\leq ba$ and $a\leq ab$. Since $b\in Reg(S),\;there\;is\;z\in S$ such that $b\leq b(zb)^{n}\;for\;all\;n\in\mathbb{N}$. Then there is $n_{1}\in\mathbb{N}$ such that $(zb)^{n_{1}}\in K$. That is $b(zb)^{n_{1}}\in K$, which gives that $b\in K$ and so $ba\in K$ and finally $a\in K$. Since $K$ is an ordered completely regular ordered semigroup, by theorem ()[1], it follows that $K$ is complete semilattice of ordered completely simple semigroups and $\mathcal{J}$ is a complete semilattice congruence on $K$. So for $a,ba\in K\;such\;that\;a\leq ba\;yields\;(a)_{\mathcal{J}}=(ba)_{\mathcal{J}}$. Thus $a\in(a^{2}KbKa^{2}]\subseteq(a^{2}SbSa^{2}]$. Thus the given conditions are obtained.

Conversely, let us assume that given condition holds in $S$. Let $a\in S$ be arbitrary. Since $S$ is completely $\pi$-regular, there exists $m\in\mathbb{N}$ such that $a^{m}\leq a^{m+1}xa^{m+1}$, for some $x\in S$. So $Gr_{\leq}(S)\neq\phi$. Say $T=Gr(S)$. Also it is clear that, for each $a\in S,\;there\;exists\;m\in\mathbb{N}$ such that $a^{m}\in T$. Now choose $s\in S$ and $a\in T$. Then by lemma(), there exists $h\in S\;such\;that\;a\leq a^{2}ha^{2}\leq a(ah)a^{2}\leq a(a^{2}h)^{2}a^{2}\leq a^{2}(a^{2}h)^{3}a^{2}\leq a^{n-1}(a^{2}h)^{n}a^{2},\;\;for\;all\;n\in\mathbb{N}$. Now $\exists m_{1}\in\mathbb{N}$ such that $a^{n-2}(a^{2}x)^{n}a^{2}\in T$. Again $sa\leq sa^{n-1}(a^{2}h)^{n}a^{2}=saa^{n-2}(a^{2}h)^{n}a^{2},\;for\;all\;n\in\mathbb{N}$. Then using the given condition and the preceding observation $sa\leq saa^{n-2}(a^{2}h)^{n}a^{2}\leq satsa\;where\;t=s^{\prime}a^{n-2}(a^{2}h)^{m_{1}}a^{2}t^{\prime}\in S\;and\;for\;some\;s^{\prime},t^{\prime}\in S$. Now consider $a\in S$ and $b\in T$ such that $a\leq b$. Then by the given condition, $a\leq at_{1}bt_{2}a\;for\;some\;t_{1},t_{2}\in S$. Also $T$ is ordered completely regular semigroup such that $a^{m}\in T\;for\;all\;a\in S$ and for some $m\in\mathbb{N}$. Hence $S$ is nil extension of a completely regular ordered semigroup $T$. ∎

4 complete semilattice of nil extensions simple ordered regular semigroups:

Theorem 4.1**.**

Let $S$ be an ordered semigroup. Then the following conditions are equivalent on $S$:

• (i)

$S$* is complete semilattice of nil extensions simple ordered regular semigroups;*

• (ii)

for all $a,b\in S$ there is $n\in\mathbb{N}$ such that $(ab)^{n}\in((ab)^{n}Sa^{2}S(ab)^{n}]$;

• (iii)

$S$* is *$\pi-$regular, and is complete semilattice of Archimedean semigroups;

• (iv)

$S$* is *$\pi-$regular, and for all $a\in S$, $e\in E_{\leq}(S)$, $a\mid e$ implies $a^{2}\mid e$;

• (v)

$S$* is *$\pi-$regular, and for all $a,b\in S$, and $e\in E_{\leq}(S)$, $a\mid e\;and\;b\mid e$ implies $ab\mid e$;

• (vi)

$S$* is *$\pi-$*regular, and every *$\mathcal{J}-$class of $S$ containing an ordered idempotent is a subsemigroup of $S$;

• (vii)

$S$* is intra *$\pi-$*regular and every *$\mathcal{J}-$class of $S$ containing an intra-regular element is a regular subsemigroup of $S$;

• (viii)

$S$* is complete semilattice of nil extensions simple semigroups and $Intra_{\leq}(S)=Reg_{\leq}(S)$.*

Proof.

$(i)\Rightarrow(ii)$: Suppose that $S$ is complete semilattice $Y$ of semigroups $S_{\alpha}(\alpha\in Y)$, where each $S_{\alpha}$ is nil extensions of a simple ordered regular semigroup $K_{\alpha}$. Let $\rho$ be the corresponding complete semilattice congruence on $S$. Let $a,b\in S$. Then there are $\alpha,\beta\in Y$ such that $a\in S_{\alpha}$ and $b\in S_{\beta}$, so that there is $\gamma\in Y$ such that $ab,a^{2}b\in S_{\gamma}$. Let $S_{\gamma}$ be the nil extensions of $K_{\gamma}$. Then there are $m,n\in\mathbb{N}$ such that $(ab)^{m},(a^{2}b)^{n}\in K_{\gamma}$. Since $K_{\gamma}$ is ordered regular, $(ab)^{m}\leq(ab)^{m}z(ab)^{m}$ for some $z\in K_{\gamma}$. Then for $z,(a^{2}b)^{n}\in K_{\gamma}$, the simplicity of $K_{\gamma}$ yields that $z\in(K_{\gamma}(a^{2}b)^{n}K_{\gamma}]$. Hence $(ab)^{m}\in((ab)^{m}Sa^{2}S(ab)^{m}]$.

$(ii)\Rightarrow(iii)$ and $(iii)\Rightarrow(iv)$: These follow from Theorem 2.8 [10]

$(iv)\Rightarrow(v)$: This follows from Lemma 2.6.

$(v)\Rightarrow(vi)$: Consider a $\mathcal{J}-$class $J$ of $S$ that contains an ordered idempotent $e\in E_{\leq}(S)$. Choose $a,b\in J$ arbitrarily. Then there are $x,y,z,w\in S^{1}$ such that $e\leq xay$ and $e\leq zbw$. Now it is obvious that $ab\in(SeS]$. Moreover by the given condition, $ab\mid e$. Thus $ab\mathcal{J}e$. Hence $J$ is subsemigroup of $S$.

$(i)\Rightarrow(vii)$: Suppose that $S$ is complete semilattice $Y$ of semigroups $S_{\alpha}(\alpha\in Y)$, where each $S_{\alpha}$ is nil extensions of simple ordered regular semigroup $K_{\alpha}$. Moreover let $\rho$ be the corresponding complete semilattice congruence on $S$. Consider $a\in S$ then there are $\alpha\in Y$ and $m\in\mathbb{N}$ such that $a^{m}\in K_{\alpha}$, where $S_{\alpha}$ is nil extensions of a simple ordered regular semigroup $K_{\alpha}$. So by the ordered regularity of $K_{\alpha}$ there is $x\in K_{\alpha}$ such that $a^{m}\leq a^{m}xa^{m}$. Since $x,a^{2m}\in K_{\alpha}$; the simplicity of $K_{\alpha}$ yields that $a^{m}\leq a^{m}ya^{2m}za^{m}$ for some $y,z\in K_{\alpha}$. This shows that $S$ is intra $\pi-$regular.

Next consider a $\mathcal{J}-$class $J\;of\;S$ that contains a intra-regular element $a$. Then $a\leq xa^{2}y$ for some $x,y\in S^{1}$ and there is some $\alpha\in Y$ such that $a\in S_{\alpha}$. Now $a\leq xa^{2}y\leq xaxa^{2}y^{2}$ implies that $(a)_{\rho}=(xay)_{\rho}=(xa)_{\rho}$, so $xa\in S_{\alpha}$. Likewise, $ay\in S_{\alpha}$. It is evident that $a\leq(xa)^{n}ay^{n}\;for\;all\;n\in\mathbb{N}$. For $xa\in S_{\alpha}$ there is $m^{\prime}\in\mathbb{N}$ such that $(xa)^{m^{\prime}}\in K_{\alpha}$, so $(xa)^{m^{\prime}}ay^{m^{\prime}}\in K_{\alpha}$, that is, $a\in K_{\alpha}$.

Now it is obvious that $a\mathcal{J}a^{2}$. Choose $b,c\in J$. Since $a\mathcal{J}b\;and\;a\mathcal{J}c$ there are $z_{1},w_{1},x_{1},y_{1}\in S^{1}$ such that $a\leq x_{1}by_{1}$ and $a\leq z_{1}cw_{1}$. This implies that $a^{2}\leq z_{1}(cw_{1}x_{1}b)y_{1}$, which together with the completeness of $\rho$ gives that $(a)_{\rho}=(a^{2})_{\rho}=(az_{1}(cw_{1}x_{1}b)y_{1})_{\rho}=(az_{1}(cw_{1}x_{1}b)^{m}y_{1})_{\rho}$ for all $m\in\mathbb{N}$. Then for all $m\in\mathbb{N},\;az_{1}(cw_{1}x_{1}b)^{m}y_{1}\in S_{\alpha}$. Since $S_{\alpha}$ is nil extension of $K_{\alpha}$ for some $n^{\prime}\in\mathbb{N}$ we have that $(az_{1}(cw_{1}x_{1}b)^{m}y_{1})^{n^{\prime}}\in K_{\alpha}$. Then for $a,(az_{1}(cw_{1}x_{1}b)^{m}y_{1})^{n^{\prime}}\in K_{\alpha}$ the simplicity of $K_{\alpha}$ yields that $a\leq t_{1}(az_{1}(cw_{1}x_{1}b)^{m}y_{1})^{n^{\prime}}t_{2}$ for some $t_{1},t_{2}\in K_{\alpha}$. This shows that $a\leq s_{1}(cw_{1}x_{1}b)^{m}s_{2}$ for some $s_{1},s_{2}\in S$ and for all $m\in\mathbb{N}$. Since $S$ is intra $\pi-$regular, there are $u_{1},u_{2}\in S$ and $m\in\mathbb{N}$ such that $a\leq s_{1}u_{1}(cw_{1}x_{1}b)^{2m}u_{2}s_{2}$, that is, $a\leq s_{1}u_{1}(cw_{1}x_{1}b)^{2}(cw_{1}x_{1}b)^{2m-2}u_{2}s_{2}\leq s_{1}u_{1}cw_{1}x_{1}(bc)w_{1}x_{1}b(cw_{1}x_{1}b)^{2m-2}u_{2}s_{2}\leq z_{1}bcz_{2}$, where $z_{1}=s_{1}u_{1}cw_{1}x_{1}$ and $z_{2}=w_{1}x_{1}b(cw_{1}x_{1}b)^{2m-2}u_{2}s_{2}$. Also, it is evident that $bc\in(S^{1}aS^{1}]$. Thus $a\mathcal{J}bc$, which shows that $J$ is subsemigroup of $S$. So what we have seen is that every element of $J$ is intra-regular. Also we have seen earlier that, $a$ being an intra-regular element $a\in K_{\alpha}$. This clearly shows that $J=K_{\alpha}$. Hence every $\mathcal{J}-$ class of $S$ containing an intra-regular element is an ordered regular subsemigroup of $S$.

$(i)\Rightarrow(iii)$: Let $S$ be complete semilattice $Y$ of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$ and $\rho$ be the corresponding complete semilattice congruence on $S$. Choose $a,b\in S_{\alpha}$ such that $a\mid b$. Then there are $x,y\in S^{1}$ such that $b\leq xay$. Now there is $n,m\in\mathbb{N}$ such that $a^{n},b^{m}\in K_{\alpha}$, where $S_{\alpha}$ is nil extension of simple regular semigroup $K_{\alpha}$. Since $a^{n},b^{m}\in K_{\alpha}$, the simplicity of $K_{\alpha}$ implies that $b^{m}\leq z_{1}a^{n}z_{2}$ for some $z_{1},z_{2}\in K_{\alpha}\subseteq S_{\alpha}$. Hence $S_{\alpha}$ is an Archimedean ordered semigroup. This shows that $S$ is complete semilattice of Archimedean ordered semigroups.

Also it is obvious that $S$ is $\pi-$regular.

$(i)\Rightarrow(viii)$: Suppose that $S$ is a complete semilattice of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$ and $\rho$ be the corresponding complete semilattice congruence on $S$. Let for $\alpha\in Y,\;S_{\alpha}$ is nil extension of simple ordered regular semigroup $K_{\alpha}$. Here we only have to show that $Intra_{\leq}(S)=Reg_{\leq}(S)$. For this, consider $a\in Reg_{\leq}(S)$. Then there is $\alpha\in Y$ such that $a\in S_{\alpha}$ and there is $s\in S_{\alpha}$ such that $a\leq asa$. Also by the completeness of $\rho,\;(a)_{\rho}=(sa)_{\rho}$, so $sa\in E_{\leq}(S_{\alpha})$. Then there is some $m^{\prime}\in\mathbb{N}$ such that $(sa)^{m^{\prime}}\in K_{\alpha}$. Since $K_{\alpha}$ is an ideal of $S_{\alpha}\;and\;for\;all\;n\in\mathbb{N}\;a\leq a(sa)^{n}$, it follows that $a\in K_{\alpha}$. So finally the simplicity and the ordered regularity of $K_{\alpha}$ yields that $a\in Intra_{\leq}(S)$.

Conversely, let $b\in Intra_{\leq}(S)$. Then there are $x,y\in S^{1}$ and $\gamma\in Y$ such that $b\leq xb^{2}y$ and $b\in S_{\gamma}$. Now for all $n\in\mathbb{N},b\leq(xb)^{n}by^{n}$. Also by completeness of $\rho$ we have that $(b)_{\rho}=(xby)_{\rho}=(xbxbby^{2})_{\rho}=(xbxby)_{\rho}=(xb)_{\rho}(xby)_{\rho}=(xb)_{\rho}(b)_{\rho}=(xb)_{\rho}$. Since $xb\in S_{\gamma}$, there is $m^{{}^{\prime\prime}}\in\mathbb{N}$ such that $(xb)^{m^{{}^{\prime\prime}}}\in K_{\gamma}$, where $S_{\gamma}$ is nil extension of simple ordered regular semigroup $K_{\gamma}$. Thus $b\in K_{\gamma}$. Therefore by the ordered regularity of $K_{\gamma}$ it follows that $a\in Reg_{\leq}(S)$. Hence $Intra_{\leq}(S)=Reg_{\leq}(S)$.

$(viii)\Rightarrow(i)$ Suppose that $S$ is a complete semilattice of semigroups $S_{\alpha}(\alpha\in Y)$, where $S_{\alpha}$ is nil extension of simple semigroup $K_{\alpha}$. Let $a\in K_{\alpha}$. Then by the simplicity of $K_{\alpha},\;a\in Intra_{\leq}(S)=Reg_{\leq}(S)$. This implies that $a\leq axa\leq a(xa)^{n}xa$ for some $x\in S$ and for all $n\in\mathbb{N}$. The completeness of $\rho$ gives that $(a)_{\rho}=(xa)_{\rho}$. Thus $xa\in S_{\alpha}$ and so for some $m\in\mathbb{N},\;(xa)^{m}\in K_{\alpha}$, that is $(xa)^{m}x\in K_{\alpha}$. Hence $K_{\alpha}$ is ordered regular and so $S$ is complete semilattice of nil extensions simple ordered regular semigroups. ∎

Theorem 4.2**.**

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

• (i)

$S$* is complete semilattice of nil extensions of left group like ordered semigroups;*

• (ii)

$S$* is complete semilattice of nil extensions of left simple and right regular ordered semigroups;*

• (iii)

$S$* is complete semilattice $Y$ of nil extensions of completely regular ordered semigroups $K_{\alpha}(\alpha\in Y)$ and every *$\mathcal{L}-$class of $S$ that contains a completely regular element $a\;(a\in S_{\alpha})$ is $K_{\alpha}$;

• (iv)

$S$* is completely *$\pi-$regular and is a complete semilattice of left Archimedean ordered semigroups.

• (v)

for all $a,b\in S$ there is $n\in\mathbb{N}$ such that $(ab)^{n}\in((ab)^{n}aSb(ba)^{n}]$;

• (vi)

$S$* is *$\pi-$regular, and for all $a\in S$, $e\in E_{\leq}(S)$, $a\mid e$ implies $a\mid_{l}e$;

• (vii)

$S$* is complete semilattice $Y$ of nil extensions of completely regular ordered semigroups $K_{\alpha}(\alpha\in Y)$ and every *$\mathcal{L}-$class of $S$ that contains an ordered idempotent $e\;(e\in S_{\alpha})$ is $K_{\alpha}$;

• (viii)

$S$* is complete semilattice of nil extensions of completely regular semigroups $K_{\alpha}$ and every *$\mathcal{L}-$class of $S$ that contains an ordered regular element $a\;(a\in S_{\alpha})$ is $K_{\alpha}$;

• (ix)

$S$* is complete semilattice of nil extensions left simple semigroups and $LReg_{\leq}(S)=Reg_{\leq}(S)$;*

• (x)

$S$* is complete semilattice of nil extensions left Clifford ordered semigroups $K_{\alpha}$ and every *$\mathcal{L}-$class of $S$ containing a regular element $a\;(a\in S_{\alpha})$ is $K_{\alpha}$.

Proof.

$(ii)\Rightarrow(i)$: Suppose that $S$ is complete semilattice $Y$ of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$ and let $S$ be a nil extension of a left simple and right regular semigroup $K_{\alpha}$. Choose $a\in K_{\alpha}$. Since $K_{\alpha}$ is right regular, there is some $x\in K_{\alpha}$ such that $\;a\leq a^{2}x$. For $ax,a\in K_{\alpha}$, the simplicity of $k_{\alpha}$ yields that $a\in(aK_{\alpha}a]$. Thus $K_{\alpha}$ is ordered regular. Hence $S$ is nil extension of left group like ordered semigroup.

$(ii)\Rightarrow(iii)$: Let $S$ be complete semilattice of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$ and let $S_{\alpha}$ be a nil extension of a left simple and right regular semigroup $K_{\alpha}$. Let $a\in K_{\alpha}$. Since $K_{\alpha}$ is right regular, $a\in(a^{2}K_{\alpha}]$. Also as $K_{\alpha}$ is left simple we have that $a\in(a^{2}K_{\alpha}a^{2}]$. Thus $K_{\alpha}$ is a completely regular ordered semigroup.

Next, consider a $\mathcal{L}-$class $L$ of $S$ that contains a complete regular element $a$, where $a\in S_{\alpha}$ for some $\alpha\in Y$. By Lemma 2.8 it follows that $a\in K_{\alpha}$ and $L\subseteq K_{\alpha}$.

Now to show that $K_{\alpha}\subseteq L$ let us choose $y\in K_{\alpha}$. By the simplicity of $K_{\alpha}$ we have that $y\leq za$ for some $z\in K_{\alpha}$. Since $za\in K_{\alpha}$ it follows that $\;y\in K_{\alpha}$.

$(iii)\Rightarrow(iv)$: Suppose that $S$ is complete semilattice $Y$ of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$. Let $a\in S$. Then there is $\alpha\in Y$ such that $a\in S_{\alpha}$. Let $S_{\alpha}$ be a nil extension of completely regular ordered semigroup $K_{\alpha}$. Then there is $m\in\mathbb{N}$ such that $a^{m}\in(a^{2m}S_{\alpha}a^{2m}]\subseteq(a^{m+1}Sa^{m+1}]$. This shows that $S$ is completely $\pi-$regular ordered semigroup.

Now choose $a,b\in S_{\alpha}$. Then there is $m^{\prime}\in\mathbb{N}$ such that $a^{m^{\prime}},a^{m^{\prime}}b\in K_{\alpha}$. Consider the $\mathcal{L}-$class $L_{a^{m^{\prime}}}$ of $a^{m^{\prime}}\in S$. Since $a^{m^{\prime}}\in Reg_{(}S)$, using the given condition we have that $L_{a^{m^{\prime}}}=K_{\alpha}$. So for $a^{m^{\prime}},a^{m^{\prime}}b\in L_{a^{m^{\prime}}},\;a^{m^{\prime}}\in(K_{\alpha}b]\subseteq(S_{\alpha}b]$. Hence $S$ is complete semilattice of left Archimedean semigroups.

$(iv)\Rightarrow(v)$: This is obvious.

$(iv)\Rightarrow(vi)$: Let $S$ satisfies the given condition. Then clearly $S$ is $\pi-$regular. Let $\rho$ be the complete semilattice congruence on $S$. Choose $a\in S$ and $e\in E_{\leq}(S)$ such that $a\mid e$. Then there are $x,y\in S^{1}$ such that $e\leq xay$. This implies that $(e)_{\rho}=(exya)_{\rho}$. Since $(e)_{\rho}$ is left Archimedean and $e,\;exya\in(e)_{\rho}$, there is $m^{{}^{\prime\prime}}\in\mathbb{N}$ such that $e^{m^{{}^{\prime\prime}}}\in((e)_{\rho}exya]$. Thus $e\in(Sa]$. That is $a\mid_{l}e$

$(iv)\Rightarrow(vi)$ Let $L$ be $\mathcal{L}-$class of $S$ containing an ordered idempotent $e\in S_{\alpha}$. Then by lemma(2.8), $e\in K_{\alpha}\;and\;K_{\alpha}\subseteq S_{\alpha}$. Again by the simplicity of $K_{\alpha},K_{\alpha}\subseteq L$. Hence $L=K_{\alpha}$.

$(iii)\Leftrightarrow(viii)$: Here we just have to show the later part. Let consider an $\mathcal{L}-$class $L$ of $S$, so that $L$ contains a regular element $a\in S_{\alpha}$. Then there is $m\in\mathbb{N}$ such that $a^{m}\in K_{\alpha}$. By the regularity of $a$, there is $x\in S$ such that $a\leq a(xa)^{n}$ for all $n\in\mathbb{N}$. So for some $l\in\mathbb{N},\;(xa)^{l}\in K_{\alpha}$. Now $a^{m}\in Gr_{\leq}(S)$, and $K_{\alpha}$ is the $\mathcal{L}-$class of $S$ containing the complete regular element $a^{m}$, which also contains the regular element $a$.

Converse is obvious.

$(viii)\Rightarrow(ix)$: Suppose that $S$ is complete semilattice $Y$ of semigroups $S_{\alpha}\;(\alpha\in Y)$ and $S_{\alpha}$ is a nil extension of complete regular semigroup $K_{\alpha}$. Let $\rho$ be the corresponding complete semilattice. Consider $a,b\in K_{\alpha}$ and $L_{a},\;L_{b}$ are the $\mathcal{L}-$classes of $a,b$ respectively. Using the given condition we have $L_{a}=K_{\alpha}=L_{b}$. This yields that $a\in L_{b}$, that is, $a\in(Sb]$. Since $K_{\alpha}$ is completely regular ordered semigroup, there is $y\in K_{\alpha}$ such that $b\leq byb$. Also $a\in L_{b}$ implies that $a\leq xb$ for some $x\in S$. So $a\leq(xby)b$. Then by the completeness of $\rho$ we have that $(a)_{\rho}=(axbyb)_{\rho}=(xbyb)_{\rho}=(xb)_{\rho}$. Therefore $xb\in S_{\alpha}$. Since $K_{\alpha}$ is an ideal of $S_{\alpha},\;xby\in K_{\alpha}$. Hence $a\in(K_{\alpha}b]$ so that $K_{\alpha}$ is left simple. Therefore $S$ is complete semilattice of a nil extension of left simple semigroups.

Now consider $a\in LReg_{\leq}(S)$. Then for all $n\in\mathbb{N}$ $\;a\leq xa^{2}\leq..........\leq x^{n}a^{n+1}$ for some $x\in\S$. Then for some $l\in\mathbb{N},\;x^{l}a^{l+1}\in K_{\alpha}$ and since $K_{\alpha}$ is an ideal of $S_{\alpha}$, $a\in K_{\alpha}$. Thus $LReg_{\leq}(S)\subseteq Reg_{\leq}(S)$. On the other hand, choose $b\in Reg_{\leq}(S)$. Then $b\in S_{\beta}$ for some $\beta\in Y$, where $S_{\beta}$ is nil extension of left simple semigroups $K_{\beta}$. Since $b\in Reg_{\leq}(S)$, there is $z\in S$ such that $b\leq b(zb)^{n}$ for all $n\in\mathbb{N}$. Now by the completeness of $\rho,\;zb\in S_{\beta}$, which follows that $(zb)^{m_{1}}\in K_{\beta}$ for some $m_{1}\in\mathbb{N}$. So $b\in K_{\beta}$. Since $K_{\beta}$ is left simple, $b\in(K_{\beta}b^{2}]\subseteq(Sb^{2}]$. Thus $Reg_{\leq}(S)=LReg_{\leq}(S)$. This completes the proof.

$(ix)\Rightarrow(i)$: Let $S$ be complete semilattice $Y$ of left simple semigroups $K_{\alpha}\;(\alpha\in Y)$ and $\rho$ be the corresponding complete semilattice congruence on $S$. To show that each $K_{\alpha}(\alpha\in Y)$ are left group like ordered semigroups, first choose $a\in K_{\alpha}$. Then by the left simplicity of $K_{\alpha},\;a\in(Sa^{2}]$. Therefore $a\in LReg_{\leq}(S)=Reg_{\leq}(S)$. And so there is $y\in S$ such that $a\leq aya$ and obviously $a\leq a(ya)^{n}ya$ for all $n\in\mathbb{N}$. Now by the completeness of $\rho,\;(a)_{\rho}=(a^{2}ya)_{\rho}=(ya)_{\rho}$, so $ya\in S_{\alpha}$. Thus for some $m\in\mathbb{N},\;(ya)^{m}\in K_{\alpha}$ and so $(ya)^{m}y\in K_{\alpha}$. This implies that $a\leq a(ya)^{m}ya$, where $(ya)^{m}y\in K_{\alpha}$. Therefore $K_{\alpha}$ is ordered regular subsemigroup. Also $K_{\alpha}$ is left simple. Hence $K_{\alpha}$ is left group like ordered semigroup.

$(iii)\Rightarrow(ii)$: Let $S$ be a complete semilattice $Y$ of semigroups $\{S_{\alpha}\}_{\alpha\in Y}$ and let each $S_{\alpha}$ are nil extension of complete regular ordered semigroup $K_{\alpha}$.

Consider $a\in K_{\alpha}$ and $L_{a}$ be the $\mathcal{L}-$class of $a\in S$. Since $a\in Gr_{\leq}(S)$ by the given condition we have that $L_{a}=K_{\alpha}$. This implies that $K_{\alpha}$ is left simple. Also, by the complete regularity of $a,$ it is evident that $a$ is right regular. Hence $S$ is complete semilattice congruence of left simple and right regular ordered semigroups.

$(vii)\Rightarrow(iii)$: Here we are only to show that every $\mathcal{L}-$class of $S$ that contains a completely regular element $a$ (where $a\in S_{\alpha}$) is $K_{\alpha}$. For this let us consider a $\mathcal{L}-$class $L$ of $S$ that contains a completely regular element $a$ where $a\in S_{\alpha}$ for some $\alpha\in Y$. Since $a\in Reg_{\leq}(S)$ we have that $a\in K_{\alpha}$, by Lemma 2.8. Since $K_{\alpha}$ is a completely regular ordered semigroup there is $h\in K_{\alpha}$ such that $a\leq aha$. Clearly $ha\in E_{\leq}(S)$. Also $a\mathcal{L}ha$. Therefore $ha\in L$. Thus $L$ is a $\mathcal{L}-$class of $S$ that contains an ordered idempotent $ha\in S_{\alpha}$. Therefore by the given condition $L=K_{\alpha}$.

$(i)\Rightarrow(ix)$: First suppose that $S$ is complete semilattice $Y$ left group like ordered semigroups $K_{\alpha}(\alpha\in Y)$. So $S$ is complete semilattice of nil extension of left simple ordered semigroups $K_{\alpha}$.

Now choose $a\in LReg_{\leq}(S)$. By Lemma 2.8, it follows that $a\in K_{\alpha}$, that is $a\in Reg_{\leq}(S)$.

Conversely, let $a\in Reg_{\leq}(S)$. By Lemma 2.8, it is evident that $a\in K_{\alpha}$. Then by the simplicity of $K_{\alpha},\;a\in(K_{\alpha}a^{2}]\subseteq(Sa^{2}]$. Hence $LReg_{\leq}(S)=Reg_{\leq}(S)$.

$(i)\Leftrightarrow(x)$: Let $S$ be a complete semilattice $Y$ left group like ordered semigroups $K_{\alpha}(\alpha\in Y)$ and $\rho$ be the corresponding complete semilattice congruence on $S$. Choose $a,b\in K_{\alpha}$. Then for $ab,a\in K_{\alpha}$, the left simplicity of $K_{\alpha}$, yields that $ab\in(K_{\alpha}a]$. Hence $S$ is complete semilattice of left Clifford ordered semigroups.

Now, let $L$ be a $\mathcal{L}-$class of $S$ containing a regular element $a$. Then by Lemma 2.8, $a\in K_{\alpha}$. Choose $y\in K_{\alpha}$. Then by the left simplicity of $K_{\alpha}$ together with $a\in K_{\alpha}$ yields that $y\mathcal{L}a,\;so\;y\in L$, that is, $L\subseteq K_{\alpha}$. Also by Lemma 2.8, $L\subseteq K_{\alpha}$. Hence the condition follows.

Conversely, let $S$ be complete semilattice $Y$ of nil extension of left Clifford ordered semigroups $K_{\alpha}(\alpha\in Y)$.let $a,b\in K_{\alpha}$. Since $K_{\alpha}$ is a left Clifford ordered semigroup, $\;ab\leq sa$, for some $s\in K_{\alpha}$. Next consider the $\mathcal{L}-$class $L_{a}$ that contains $a$, obviously then $L_{a}=K_{\alpha}$. Thus $ab,a\in L_{a}$. So $a\mathcal{L}ab$. Then there is $z\in S$ such that $a\leq zab$. Since $a\in K_{\alpha},\;za\in K_{\alpha}$. This implies that $K_{\alpha}$ is left simple. Hence $S$ is complete semilattice of left group like ordered semigroup. ∎