Some results on $\mathcal {L}$-commutative semigroups

Abstract

We prove first that every $\mathcal {H}$-commutative semigroup is stable. Using this result [and some results from the standard text (Nagy, Special classes of semigroups, Kluwer, Dordrecht, 2001)], we give two equivalent conditions for a semigroup to be an archimedean $\mathcal {H}$-commutative semigroup containing an idempotent element. It turns out that this result can be partially extended to $\mathcal {L}$-commutative semigroups and quasi-commutative semigroups.

1 Introduction and preliminaries

For undefined terminology of semigroup theory, the reader is referred to the books [3, 8, 9].

One of the most important decompositions in the theory of semigroups is a decomposition of a semigroup (induced by a semilattice congruence) into archimedean semigroups. In this field, one of the biggest contributions is due to Putcha, see [9, Chapter 2] (recall that a semigroup is a Putcha semigroup if and only if it is a semilattice of archimedean semigroups). Therefore, it is worth investigating certain classes of Putcha semigroups (especially, with idempotents), see, e.g., [9, Chapters 4–12]. On the other hand, many naturally arising semigroups are semilattice unions of archimedean semigroups. A good example is the semigroup of all $n\times n$ upper triangular matrices over a field, and more generally, the Zariski closure of any solvable linear algebraic group. Moreover, semilattices of archimedean semigroups arise naturally in the theory of finite semigroups and languages, where the pseudovariety is denoted by $\mathbf{DS}$.

Denote the set of idempotents of a semigroup S by $E_S$, and the set of its regular elements by Reg(S).

Let S be a semigroup and let $A\subseteq S$. Put

$$\begin{aligned} \sqrt{A}=\{s\in S:(\exists \,n\in \mathbb {Z}_+)(s^n\in A)\}. \end{aligned}$$

Also, for each $a\in S$, let us put as usual: $J(a)=S^1aS^1.$

Recall that a semigroup S is archimedean if for every ideal I of S, we have

$$\begin{aligned} S=\sqrt{I}. \end{aligned}$$

Remark 1

Observe that if in addition $E_S\ne \emptyset$, then $E_S$ is contained in each ideal of S. Hence Reg(S) is a subset of a kernel of S (by a kernel of a semigroup A we mean (if it exists) the least ideal of A; this ideal is then denoted by $K_A$).

We say that a congruence $\rho$ on a semigroup S is a semilattice congruence if the semigroup $S/\rho$ is a semilattice. It is clear that the least semilattice congruence on an arbitrary semigroup always exists. Denote it by $\eta$. It is also clear (and well-known) that in any semigroup $\mathcal {J}\subseteq \eta$.

Let $\mathcal {C}$ be some fixed class of semigroups (call its elements $\mathcal {C}$-semigroups). A semigroup S is said to be a semilattice of $\mathcal {C}$-semigroups if there exists a semilattice congruence $\rho$ on S such that every $\rho$-class of S is a $\mathcal {C}$-semigroup (for example, S is a semilattice of groups if there is a semilattice congruence $\rho$ on S such that every $\rho$-class is a group). One can modify this definition for any variety of bands.

The following important result is due to Putcha (see [9, Theorem 2.1]).

Result 1.1

A semigroup S is a semilattice of archimedean semigroups if and only if it is a Putcha semigroup, i.e., $J(a)\subseteq \sqrt{J(a^2)}$ for all $a\in S$. In that case, the corresponding semilattice congruence is equal to $\eta$. Moreover,

$$ \begin{aligned} \eta =\{(a,b)\in S\times S:(\exists \,m,n\in \mathbb {Z}_+)(a^m\in SbS~{ \& }~b^n\in SaS)\}. \end{aligned}$$

Recall that a semigroup is unipotent if it has exactly one idempotent, and that a nil-semigroup is just a (unipotent) semigroup with zero in which every element has a power equal to zero.

It is clear that for an arbitrary ideal I of a semigroup S, the relation

$$\begin{aligned} \rho _I=(I\times I)\cup 1_{S}, \end{aligned}$$

where $1_{S}$ is the identity relation on S, is a congruence on S (called the Rees congruence induced by I). We denote the (Rees) quotient semigroup induced by this congruence by S/I. In that case, we also say that S is an ideal extension of the semigroup I by the semigroup S/I. Moreover, every extension of a semigroup S by its ideal I such that I is a retract of S is called a retract ideal extension.

Recall from [9] that a semigroup S is $\mathcal {H}$-commutative if

$$\begin{aligned} (\forall a,b\in S)(\exists \,x\in S^1)(ab=bxa). \end{aligned}$$

Also, S is:

• (a) $\mathcal {R}$-commutative if

  $$\begin{aligned} (\forall a,b\in S)(\exists \,x\in S^1)(ab=bax); \end{aligned}$$

• (b) $\mathcal {L}$-commutative if

  $$\begin{aligned} (\forall a,b\in S)(\exists \,x\in S^1)(ab=xba). \end{aligned}$$

Recall further from [2] that a semigroup S is stable if for all $a,b\in S$,

• (a) $(Sa\subseteq Sab)\implies (Sa=Sab),$ and

• (b) $(aS\subseteq baS)\implies (aS=baS).$

We say that S is left stable if it meets the condition (a), and S is right stable if it satisfies the condition (b).

Recently in [1], the authors studied $\mathcal {H}$-commutative semigroups. In particular, they proved the following result (see [1, Theorem 4.10]).

Result 1.2

Every $\mathcal {H}$-commutative archimedean semigroup containing an idempotent element is saturated.

We prove first that every $\mathcal {H}$-commutative semigroup is stable. Using this result (and some results from [9] and [2]), we give two equivalent conditions for an arbitrary semigroup to be an archimedean $\mathcal {H}$-commutative semigroup containing an idempotent element. Also, a weaker such characterization for $\mathcal {L}$-commutative semigroups is contained in Section 3. The last section brings some concluding remarks on $\mathcal {H}(\mathcal {L})$-commutative regular semigroups and archimedean quasi-commutative semigroups.

Recall also that a semigroup S is intra-regular if $J(a)=J(a^2)$ for all $a\in S$. It is easily seen (and this is also well-known) that in this case, $\mathcal {J}$ is a semilattice congruence on S, whence $\eta =\mathcal {J}$ in any intra-regular semigroup. This implies that each principal factor $J_a^0$ is not null. Thus every $\mathcal {J}$-class $J_a$ of S is simple. In fact, the following well-known result of Croisot is valid (see [3, Theorem 4.4]).

Result 1.3

A semigroup S is intra-regular if and only if it is the semilattice $S/\mathcal {J}$ of the simple semigroups $J_a$ $(a\in S)$.

Recall from [8] that a semigroup is completely regular (i.e., a (disjoint) union of groups) if and only if it is a semilattice of completely simple semigroups.

Corollary 1.4

Each intra-regular stable semigroup is completely regular.

Proof

Let S be an intra-regular stable semigroup and let $a\in S$. Then $a=xaay$ for some $x,y\in S^1$. If $x=1$ and $y\ne 1$, then

$$\begin{aligned} a=aay=a(aay)y=aa^2y^2\in Sa^2S. \end{aligned}$$

The same conclusion we obtain in other cases. The statement of the corollary now follows from the above remark, Result 1.3 and [2, Corollary 1.1 (7)]. $\square$

Further, the following result will be very useful (see [9, Theorems 5.1–2]).

Result 1.5

A semigroup is both $\mathcal {R}$-commutative and $\mathcal {L}$-commutative if and only if it is $\mathcal {H}$-commutative.

Also, a semigroup S is $\mathcal {H}(\mathcal {L})$-commutative if and only if $\mathcal {H}\,(\mathcal {L})$ is a commutative congruence on S.

Finally, the following result has been proved independently in [1, Proposition 2.1] and in [6, Proposition 2.1].

Result 1.6

The set of idempotents of an arbitrary $\mathcal {H}$-commutative semigroup S is contained in the center of S. In particular, $E_S=\emptyset$ or $E_S$ forms a semilattice.

2 Some results on $\mathcal {H}$-commutative semigroups

To begin, we have the following useful proposition.

Proposition 2.1

The following three statements are valid:

• (a) Every $\mathcal {L}$-commutative semigroup is left stable;

• (b) Every $\mathcal {R}$-commutative semigroup is right stable;

• (c) Every $\mathcal {H}$-commutative semigroup is stable.

Proof

• (a) Let S be an $\mathcal {L}$-commutative semigroup and let $Sa\subseteq Sab$, where $a,b\in S$. Take also an arbitrary $x\in Sab$. Then $x=(sa)b$ for some $s\in S$. As S is $\mathcal {L}$-commutative, there is $y\in S^1$ such that $(sa)b=yb(sa)=(ybs)a$. Hence $x\in Sa$. Thus $Sa=Sab$, so S is left stable.

• (b) This follows from (a) by duality.

• (c) This follows from (a), (b) and the first sentence of Result 1.5.

$\square$

Remark 2

Let us note that the concept of stability considered in this article is stronger than that of [4, Section 6.6] (c.f. also [4, Exercise 1] for 6.6). Hence the above proposition holds under either interpretation.

Recall that a semigroup is simple if it has no proper ideals. It is clear that if a semigroup has a simple ideal, then it is its kernel. Note finally that a kernel of any semigroup is a simple semigroup.

Observe that any group is $\mathcal {H}$-commutative.

Corollary 2.2

The kernel of an arbitrary $\mathcal {H}$-commutative semigroup S is a group.

Proof

Indeed, we know that S is stable. In the light of [2, Corollary 2.5], $K_S$ is completely simple. On the other hand, by Result 1.6, $K_S$ contains exactly one idempotent. Thus $K_S$ is a group. $\square$

Recall from [8] that a semigroup S with zero (denoted by 0) is a 0-group if $S\setminus \{0\}$ is a group.

Notice further that in an arbitrary $\mathcal {H}$-commutative semigroup S, we have

$$\begin{aligned} aS=Sa \end{aligned}$$

for all $a\in S$ ( [1, Proposition 2.2], c.f. also [9, Lemma 5.2] and the first sentence of Result 1.5).

Proposition 2.3

Every 0-simple semigroup S, in which for any $a\in S$:

$$\begin{aligned} aS=Sa, \end{aligned}$$

is a 0-group, provided that S has at least three elements.

Proof

Let $a\in S\setminus \{0\}$. Then $aS=Sa$ is a non-zero ideal of S (otherwise, $\{0,a\}$ is a proper ideal of S). Hence $aS=Sa=S$ for all $a\in S\setminus \{0\}$. Thus S is a 0-group by [8, Proposition 1.1.1]. $\square$

Theorem 2.4

The following conditions on a semigroup S are equivalent:

• (a) S is an archimedean $\mathcal {H}$-commutative semigroup with $E_S\ne \emptyset ;$

• (b) S is a retract ideal extension of a group by a commutative nil-semigroup;

• (c) S is a subdirect product of a group and a commutative nil-semigroup.

Proof

$(a)\Longrightarrow (b)$. Indeed, by Remark 1 and Corollary 2.2, S has a kernel which is a group. It is clear that the semigroup $S/K_S$ is an $\mathcal {H}$-commutative nil-semigroup. As in any nil-semigroup, $\mathcal {H}$ is the identity relation, $S/K_S$ is a commutative nil-semigroup (see the second sentence of Result 1.5). Finally, take any $a\in S$ and denote the identity of the group $K_S$ by e. Then $ea,ae\in K_S$. Hence

$$\begin{aligned} ae=e(ae)=(ea)e=ea. \end{aligned}$$

Thus the map $a\rightarrow ea$ $(a\in S)$ is a retraction of S onto $K_S$, as required.

$(b)\Longrightarrow (c)$. Let $\varphi$ be a retraction of S onto its group kernel G. Denote the congruence induced by this retraction by $\rho$. Then

$$\begin{aligned} S/\rho \cong G,~~\rho \cap \rho _G=1_S. \end{aligned}$$

Hence S is a subdirect product of the group G and the commutative nil-semigroup S/G.

$(c)\Longrightarrow (a)$. Let S be a subdirect product of a group A with identity 1 and a commutative nil-semigroup X with zero 0 (we may assume that $S\subseteq A\times X$). Then there is $a\in A$ such that $(a,0)\in S$. Also, $(a^{-1}, x)\in S$ for some $x\in X$. Hence $(1,0)\in S$. Consequently,

$$\begin{aligned} A\cong A\times \{0\}\subseteq S \end{aligned}$$

is a kernel of S. It is clear that S is archimedean. As $(a,x)\,\mathcal {H}^S\,(b,y)$ if and only if $a\,\mathcal {H}^A\,b$ and $x\,\mathcal {H}^X\,y$, $\mathcal {H}$ is a commutative congruence on S (since in any group Green’s relation $\mathcal {H}$ is the universal relation and X is commutative). Thus (a) follows from the above and the second sentence of Result 1.5. $\square$

3 Some results on $\mathcal {L}$-commutative semigroups

In this section, we give some equivalent condition for an $\mathcal {L}$-commutative semigroup S to be an archimedean semigroup with $E_S\ne \emptyset$.

First of all, the following result will be useful.

Proposition 3.1

In an arbitrary $\mathcal {L}$-commutative semigroup S,

$$\begin{aligned} \mathcal {L}=\mathcal {D}=\mathcal {J},~~~~\mathcal {H}=\mathcal {R}. \end{aligned}$$

Proof

Indeed, let $a,b\in S$ and $a\ne b$. Suppose also that $a=xby$ for some $x,y\in S^1$. If $y=1$ or $by=yb$, then $a\in Sb$; otherwise $by=zyb$ for some $z\in S$. Hence $a=xzyb\in Sb$. We may conclude that $\mathcal {J}=\mathcal {L}$. The rest of the result follows from the structure of the lattice of Green’s relations. $\square$

We are now able to show the following result.

Theorem 3.2

The following conditions on a semigroup S are equivalent:

• (a) S is an archimedean $\mathcal {L}$-commutative semigroup with $E_S\ne \emptyset ;$

• (b) S is an ideal extension of a left group by a commutative nil-semigroup.

Proof

$(a)\Longrightarrow (b)$. Indeed, by Remark 1 and Proposition 3.1, S has a kernel which is a left group. It is clear that the semigroup $S/K_S$ is $\mathcal {L}$-commutative nil-semigroup. As in any nil-semigroup, $\mathcal {L}$ is the identity relation, $S/K_S$ is a commutative nil-semigroup (see the second sentence of Result 1.5).

$(b)\Longrightarrow (a)$. Let S be an ideal extension of a left group A by a commutative nil-semigroup B. As A is a simple ideal of S, $A=K_S$. This implies that S is archimedean. We show that S is $\mathcal {L}$-commutative. For this, let $a,b\in S$. Suppose first that $a,b\notin K_S$. If $ab\notin K_S$, then $ba\notin K_S$ and $ab=ba$ (as B is commutative); otherwise $ab,ba\in K_S$. As $K_S$ is left simple $ab=xba$ for some $x\in K_S$. Hence S is $\mathcal {L}$-commutative. On the other hand, if a or b belongs to $K_S$, then $ab,ba\in K_S$ and so (as above) S is $\mathcal {L}$-commutative. $\square$

Remark 3

Observe that in any archimedean semigroup A with a kernel,

$$\begin{aligned} \mathcal {J}=(K_A\times K_A)\cup 1_A. \end{aligned}$$

Thus the following statements are also valid:

S is an archimedean semigroup with $E_S\ne \emptyset$ and $\mathcal {L}=\mathcal {J}$ if and only if S is an ideal extension of a left group by a nil-semigroup.

S is an archimedean semigroup with $E_S\ne \emptyset$ and $\mathcal {H}=\mathcal {J}$ if and only if S is a retract ideal extension of a group by a nil-semigroup.

Some well-known facts on epigroups will be helpful. First of all, a semigroup S is an epigroup if for every $a\in S$ there is a positive integer n and $e\in E_S$ such that $a^n\in H_e$. In that case, denote by $\textsf {g}(a)$ the least positive integer n for which $a^n\in H_e$ and recall that $a^m\in H_e$ for all natural numbers $m\ge \textsf {g}(a)$. Finally, for any $e\in E_S$, the set

$$\begin{aligned} K_e=\{a\in S:a^{\textsf {g}(a)}\in H_e\} \end{aligned}$$

is called the unipotency class of S generated by e. By Green’s Theorem, $E_S\cap K_e=\{e\}$, and by above, $\{K_e:e\in E_S\}$ is a partition of S. Denote the equivalence relation induced by this partition by $\mathcal {K}$ and the only idempotent in $K_a$ by $e_a$ ($e_a=e_b$ if and only if $a\,\mathcal {K}\,b$).

The following example shows that there is an archimedean $\mathcal {L}$-commutative semigroup S with $E_S\ne \emptyset$ which is not a retract extension of a left group by a commutative nil-semigroup.

Example 3.3

Indeed, consider the following well-known semigroup:

$$\begin{aligned} L_{3,1}=\langle a,f:a^2f=a^2,fa=f^2=f\rangle =\{e=a^2,f,g=af,a\}. \end{aligned}$$

This semigroup has the following multiplication table:

$\cdot$	e	f	g	a
e	e	e	e	e
f	f	f	f	f
g	g	g	g	g
a	e	g	e	e

Clearly, $K_{L_{3,1}}=E_{L_{3,1}}$, so $L_{3,1}$ is archimedean. Observe also that $\mathcal {L}$ is a commutative congruence on $L_{3,1}$ since $L_{3,1}/\mathcal {L}=\{\{e,f,g\},\{a\}\}$. Hence $L_{3,1}$ is $\mathcal {L}$-commutative. On the other hand, $K_{L_{3,1}}$ is not a retract ideal of $L_{3,1}$ since $\mathcal {K}$ is not a congruence on $L_{3,1}$.

Further, we have the following theorem.

Theorem 3.4

The following conditions on a semigroup S are equivalent:

• (a) S is a retract ideal extension of a left group by a commutative nil-semigroup;

• (b) S is a subdirect product of a left group and a commutative nil-semigroup;

• (c) S is an $\mathcal {L}$-commutative left zero band of unipotent epigroups and $\,Reg (S)$ is an ideal of S.

Proof

$(a)\Longrightarrow (b)$. This is (now) clear.

$(b)\Longrightarrow (c)$. Let S be a subdirect product of a left group A and a commutative nil-semigroup X with zero 0 (we may assume that $S\subseteq A\times X$). Take further an arbitrary idempotent e of A. Then there exists $x\in X$ such that $(e,x)\in S$. Also, $x^n=0$ for some positive integer n. Hence

$$\begin{aligned} (e,0)=(e^n,x^n)=(e,x)^n\in S. \end{aligned}$$

Thus $E_S\cong E_A$. As, $ae_a=a$ for any $a\in A$ and Reg$(X)=\{0\}$, we conclude that

$$\begin{aligned} A\cong A\times \{0\}=\text {Reg}(S)\subseteq S \end{aligned}$$

is a kernel of S (since A is simple). It is now obvious that S is archimedean. Hence S is an epigroup. Finally, the proof that $\mathcal {L}$ is a commutative congruence on S is closely similar to the corresponding proof in Theorem 2.4. Consequently, S is $\mathcal {L}$-commutative (Result 1.5).

Next, define the relation $\rho$ on S by

$$\begin{aligned} (a,x)\,\rho \,(b,y)\iff a\,\mathcal {H}^A\,b. \end{aligned}$$

Since $\mathcal {H}^A$ is a left zero band congruence on A, $\rho$ is a left zero band congruence on S. Note also that $(a,x)\,\rho \, (e_a,0)$. From the above, the definition of $\rho$ and Green’s Theorem follows that every $\rho$-class is unipotent epigroup (as $(a,x)\,\rho \, (a^n,x^n)$ for every positive integer n).

$(c)\Longrightarrow (a)$. First of all, every idempotent of S is primitive. Hence, by [7, Theorem 2.5],

$$\begin{aligned} \text {Reg}(S)=K_S \end{aligned}$$

is a completely simple semigroup. Also, $E_S$ is a left zero semigroup. Indeed, ($\rho$ is below the corresponding left zero congruence on S from the above condition (c)) $e\,\rho \,ef$ for all $e,f\in E_S$ and $ef\in \text {Reg}(S)$. Hence

$$\begin{aligned} (ef)^2=(ef\cdot e)f=eff=ef=e. \end{aligned}$$

It is now clear that S is a nil-extension of the left group $K_S$ by the commutative nil-semigroup $S/K_S$. Finally, $K_S$ is a retract ideal of S by [6, Theorem 3.3]. $\square$

Denote the least band congruence on an arbitrary semigroup by $\beta$ and observe that if an epigroup is a band of unipotent semigroups, then the corresponding band congruence is equal to $\mathcal {K}$, i.e., $\mathcal {K}=\beta$ (as clearly $\mathcal {K}\subseteq \beta$ in any epigroup, whence $\beta =\mathcal {K}^\sharp$, here $\mathcal {K}^\sharp$ denotes the least congruence containing the relation $\mathcal {K}$).

We have shown above that an archimedean $\mathcal {L}$-commutative semigroup S with $E_S\ne \emptyset$ is an epigroup. For brevity, any epigroup in which $\mathcal {K}$ is a congruence will be called a $\mathcal {K}$-epigroup.

Corollary 3.5

The following conditions on a semigroup S are equivalent:

• (a) S is an archimedean $\mathcal {L}$-commutative $\mathcal {K}$-epigroup; 

• (b) S is a retract ideal extension of a left group by a commutative nil-semigroup;

• (c) S is a subdirect product of a left group and a commutative nil-semigroup;

• (d) S is an $\mathcal {L}$-commutative left zero band of unipotent epigroups and $\,Reg (S)$ is an ideal of S;

• (e) S is an $\mathcal {L}$-commutative left zero band of unipotent epigroups.

Proof

$(a)\Longrightarrow (b)$. We know that S is an ideal extension of the left group $K_S$ by a commutative nil-semigroup. Further, define the map from S onto $K_S$ by $a\rightarrow e_aa$. As $E_S$ is a left zero band and $e_aa=ae_a$, we have $(a,b\in S)$:

$$\begin{aligned} (e_aa)(e_bb)=(ae_a)(e_bb)=a(e_ae_b)b=ae_ab=e_aab=(e_ae_b)ab, \end{aligned}$$

so S is a retract ideal extension (since $\mathcal {K}$ is a congruence on S) of the left group $K_S$ by the commutative nil-semigroup $S/K_S$.

$(b)\Longrightarrow (c)\Longrightarrow (d)\Longrightarrow (b)$. This follows from the above proofs.

$(e)\Longrightarrow (a)$. Indeed, any idempotent of S is primitive. Hence, by [7, Theorem 2.5],

$$\begin{aligned} \text {Reg}(S)=K_S \end{aligned}$$

is a completely simple semigroup. It is now obvious that S is an archimedean $\mathcal {K}$-epigroup. $\square$

Remark 4

Let S be a left zero band of the semigroups $S_i$ $(i\in I)$. Then each $S_i$ is a right ideal of S.

Moreover, we have the following proposition [see above conditions (d), (e)].

Proposition 3.6

Let S be an $\mathcal {L}$-commutative left zero band $S/\rho$ of the semigroups $a\rho$ $(a\in S)$. Then each $\rho$-class is an $\mathcal {L}$-commutative right ideal of S.

Proof

Suppose that S is $\mathcal {L}$-commutative and let $c\in S$; $a,b\in c\rho$. Then either $ab=ba$ or $ab=xba$ for some $x\in S$. As $ab=xba\,\rho \,x$, $x\in c\rho$. The statement now follows from the above remark. $\square$

Further, from the above two results we obtain the following corollary.

Corollary 3.7

Every archimedean $\mathcal {L}$-commutative $\mathcal {K}$-epigroup S is a left zero band of $\mathcal {L}$-commutative retract ideal extensions of a left group by a commutative nil-semigroup.

Remark 5

One can modify Corollary 3.5 for an archimedean $\mathcal {K}$-epigroup in which $\mathcal {L}=\mathcal {J}$ (see Remarks 3, 1).

4 Concluding remarks

We give a simpler proof of the following result (see [1, Theorem 2.6(a)]).

Result 4.1

The following conditions on a semigroup S are equivalent:

• (a) S is $\mathcal {H}$-commutative and regular;

• (b) S is $\mathcal {H}$-commutative and $S/\mathcal {H}$ is regular;

• (c) S is a semilattice of groups;

• (d) $\mathcal {H}=\eta$.

Proof

$(a)\Longrightarrow (b)$. This is clear.

$(b)\Longrightarrow (c)$. As $T=S/\mathcal {H}$ is commutative and regular, T is a semilattice of groups. Thus T is a semilattice (since it is combinatorial). The thesis follows now from Green’s Theorem.

$(c)\Longrightarrow (d)$. In the light of Green’s Theorem, it is clear (and well-known) that $\mathcal {H}\subseteq \eta$ (in an arbitrary semigroup). The converse is obvious.

$(d)\Longrightarrow (a)$. Let $a,b\in S$. Again in a view of Green’s Theorem, S is a semilattice of groups. Hence the elements ab, ba are contained in some subgroup of S (say G). Thus $ab=xba$ and $ab=bay$ for some $x,y\in G$. Consequently, (a) holds by the first sentence of Result 1.5. $\square$

Remark 6

Note that [1, Theorem 2.6 (b)] has been shown independently in [5, Corollary 3.5]. In fact, the author have proved the following more general result (see [5, Proposition 3.4]).

Result 4.2

The set of regular elements of an arbitrary $\mathcal {R}$-commutative semigroup S is either empty or a semilattice of right groups. In the latter case, $E_S$ is a right regular band.

Remark 7

[1, Theorem 2.10] is a particular case of Result 1.1 (above); c.f. [1, Proposition 2.2] and [9, Theorem 5.3] with its proof on page 71.

Furthermore, notice that in any Putcha semigroup (see again Result 1.1) with central idempotents, each $\eta$-class contains at most one idempotent (see [6, Propositions 2.2]).

Recall the well-known fact due to Munn that any simple epigroup is completely simple (i.e., it is simple and its idempotents are primitive), see [8, Theorem 3.2.11].

It is also well-known that a semigroup S is completely simple if and only if it is the rectangular band $S/\mathcal {H}$ of the groups $H_e (e\in E_S)$.

Using the above methods, we are able to prove the following theorem (c.f.  [10, Theorem 3.39]).

Theorem 4.3

The following conditions on a semigroup S are equivalent:

• (a) S is an archimedean $\mathcal {K}$-epigroup;

• (b) S is a retract ideal extension of a completely simple semigroup by a nil-semigroup;

• (c) S is a subdirect product of a completely simple semigroup and a nil-semigroup;

• (d) S is a rectangular band of unipotent epigroups and $\,Reg (S)$ is an ideal of S;

• (e) S is a rectangular band of unipotent epigroups.

Proof

$(a)\Longrightarrow (b)$. By Remark 1, S has a kernel that contains Reg(S). Let $a\in K_S$. Then the pseudo-inverse $(e_aa)^{-1}$ of a (see [10, Section 2.1]) belongs to $K_S$ (since $e_aa\in H_{e_a}$). Hence $K_S$ is a simple epigroup. Thus $K_S$ is completely simple by the above remark. This implies that $S/K_S$ is a nil-semigroup. Also, $\mathcal {K}$ is the least rectangular band congruence on S. Consequently, S is a retract ideal extension of a completely simple semigroup by a nil-semigroup (see [6, Theorem 3.3]).

$(b)\Longrightarrow (c)$. This is clear.

$(c)\Longrightarrow (d)$. The proof is closely similar to the proof of implication “$(b)\Longrightarrow (c)$” of Theorem 3.4.

$(e)\Longrightarrow (a)$. The proof is essentially identical to the corresponding proof of Corollary 3.5. $\square$

Further, we have the following result (c.f.  [1, Theorem 2.5 (b)]).

Proposition 4.4

Let S be a semigroup and let $\mathcal {X}\in \{\mathcal {H},\mathcal {L},\mathcal {R},\mathcal {J}\}$ be a congruence on S. Then $S/\mathcal {X}$ is $\mathcal {X}$-trivial. Moreover, if $\rho$ is any congruence on S such that $S/\rho$ is $\mathcal {X}$-trivial, then

$$\begin{aligned} \mathcal {X}\subseteq \rho . \end{aligned}$$

Proof

Let $H_a\,\mathcal {H}^{S/\mathcal {H}}\, H_b$ ($a,b\in S$). As $\mathcal {H}$ is a congruence on S, there are elements $x,y\in S^1$ such that $a\,\mathcal {H}\,xb$ and $a\,\mathcal {H}\,by$. This implies that $a\in S^1b$ and $a\in bS^1$. By symmetry, we may conclude that $H_a=H_b$. The second part of the proposition is clear.

The proof of remaining cases is similar. $\square$

Remark 8

Let $\mathcal {X}\in \{\mathcal {H},\mathcal {L},\mathcal {R},\mathcal {J}\}$ be a commutative congruence on a semigroup S. Then $S/\mathcal {X}$ is a commutative semigroup in which all Green’s relations are trivial.

Consider an arbitrary completely simple semigroup S and let x, y be arbitrary idempotents of S such that

$$\begin{aligned} xy=x. \end{aligned}$$

Then $yx\in E_S$ and $yx=yxy\in ySy=H_y$. Thus

$$\begin{aligned} yx=y. \end{aligned}$$

In fact, using [7, Theorem 2.5], one can show (as above) the following proposition (recall that a semigroup S is E-inversive if for every $a\in S$ there exists $x\in E_S$ such that $ax\in E_S$).

Proposition 4.5

Let e be an arbitrary idempotent and let f be an arbitrary primitive idempotent of an E-inversive semigroup S. Then

$$\begin{aligned} (ef=e)\implies (fe=f). \end{aligned}$$

In that case, $e\,\mathcal {L}\,f$. In particular, $e\in K_S=J(f)=J_f$ and e is primitive.

Note that the dual of the above proposition is valid.

Moreover, we have the following theorem.

Theorem 4.6

The following conditions on a semigroup S are equivalent:

• (a) S is $\mathcal {L}$-commutative and regular;

• (b) S is $\mathcal {L}$-commutative, every $\eta$-class of S contains an idempotent and $S/\mathcal {L}$ is regular;

• (c) S is a semilattice of left groups;

• (d) $\mathcal {L}=\eta$ and every $\eta$-class of S contains an idempotent element;

• (e) S is completely regular and $efe=ef$ for all $e,f\in E_S$ (i.e., $E_S$ is a left regular band);

• (f) S is completely regular and $efe=ef$ for each $e,f\in E_A$, where A is an arbitrary $\eta$-class of S; 

• (g) S is regular and $aS\subseteq Sa$ for any $a\in S.$

Proof

$(a)\Longrightarrow (b)$. This is clear.

$(b)\Longrightarrow (d)$. As above, $S/\mathcal {L}$ is a semilattice of groups. Hence $S/\mathcal {L}$ is a semilattice (since it is combinatorial as $\mathcal {L}$-trivial). Thus $\mathcal {L}=\eta .$

$(d)\Longrightarrow (c)$. Indeed, S is the semilattice $S/\eta$ of left simple semigroups (see [3, Section 4.1]). Consider an arbitrary $\eta$-class A of S. As A is left simple and contains an idempotent, A is a regular semigroup and $E_A$ is a left zero semigroup. Consequently, A is a left group.

$(c)\Longrightarrow (a)$. This is clear. We have just proved that the conditions (a), (b), (c), (d) are equivalent.

$(c)\Longrightarrow (e)$. As any left group is a union of groups, S is completely regular. Finally, S is $\mathcal {L}$-commutative, so $efe=ef$ for all $e,f\in E_S$ by the dual of Result 4.2.

$(e)\Longrightarrow (f)$. This is trivial.

$(f)\Longrightarrow (c)$. Let $e,f\in E_A$, where A is an arbitrary $\eta$-class of S (note that A is a completely simple semigroup). Then $ef,e\in E_A$ and $ef\cdot e=ef$. In the light of the second part of Remark 8, $e(ef)=e$, i.e., $ef=e$ for all $e,f\in E_A$. Hence A is a left group.

$(g)\Longrightarrow (a)$. Indeed, let $a,b\in S$. As S is regular, $ab\in abS\subseteq Sab$. Hence S is $\mathcal {L}$-commutative. The converse implication is clear. $\square$

Recall from [9, Chapter 8] that a semigroup S is:

• (a) left quasi-commutative if

  $$\begin{aligned} (\forall a,b\in S)(\exists \,n\in \mathbb {Z}_+)(ab=b^na); \end{aligned}$$

• (b) right quasi-commutative if

  $$\begin{aligned} (\forall a,b\in S)(\exists \,n\in \mathbb {Z}_+)(ab=ba^n); \end{aligned}$$

• (c) $\sigma$-reflexive if for all $a,b\in S$ and any subsemigroup A of S, the following implication is valid:

  $$\begin{aligned} (ab\in A)\implies (ba\in A). \end{aligned}$$

Remark 9

Observe that any left (right) quasi-commutative semigroup is $\mathcal {L}(\mathcal {R}$)-commutative. Recall also from [9, Theorem 8.2] that the above three conditions are equivalent in an arbitrary semigroup, and that a semigroup is quasi-commutative if meets one (and hence all) of the above conditions. Consequently, every quasi-commutative semigroup is $\mathcal {H}$-commutative. In particular, $\mathcal {H}=\mathcal {J}$ in any quasi-commutative semigroup.

Also, recall from [9, Chapter 8] that a non-commutative group is called hamiltonian if every its subgroup is normal (c.f. also [9, Theorem 8.1]). As mentioned in the proof of [9, Theorem 8.1] (last line on page 111), each hamiltonian group is periodic (i.e., torsion).

Moreover, it is obvious that any subsemigroup of a quasi-commutative semigroup is quasi-commutative. By the above remarks, an arbitrary subsemigroup of a quasi-commutative group G is a normal subgroup of G.

Further, it is clear (and well-known) that a subgroup H of a group G is normal if and only if for all $a,b\in G$, $ab\in H$ implies that $ba\in H$.

Finally, note that each non-commutative quasi-commutative semigroup contains an idempotent element (see [9, Lemma 8.3]).

From the above remark and Proposition 2.1 we have the following result.

Corollary 4.7

Each quasi-commutative semigroup is stable.

Also, we have the following theorem (c.f.  [9, Theorem 8.5]).

Theorem 4.8

The following conditions on a semigroup S are equivalent:

• (a) S is an archimedean quasi-commutative semigroup;

• (b) S is a retract ideal extension of a hamiltonian group by a commutative nil-semigroup.

Proof

$(a)\Longrightarrow (b)$. This follows from Remarks 9, 3.

$(b)\Longrightarrow (a)$. It is clear that S is archimedean. Consider further an arbitrary subsemigroup A of S and let $a,b\in S$ with $ab\in A$. If

• (1) $a,b\in K_S$, as $H=A\cap K_S$ is a normal subgroup of $K_S$, then $ba\in A$ (we use here the above remarks);

• (2) $a\in K_S$, $b\notin K_S$, then $ab\in H=A\cap K_S$ and $ba\in K_S$. Let $\varphi$ be the given retraction of S onto $K_S$. Then

  $$\begin{aligned} ab=(ab)\varphi =(a\varphi )(b\varphi )\in H, \end{aligned}$$

  so (as $K_S$ is $\sigma$-reflexive)

  $$\begin{aligned} ba=(ba)\varphi =(b\varphi )(a\varphi )\in H; \end{aligned}$$

• (3) $a,b\notin K_S$. Suppose first that $ab\notin K_S$. Then $ba\notin K_S$ and $ab=ba$ (since $S/K_S$ is commutative). On the other hand, if $ab\in K_S$, then $ba\in K_S$ (as $S/K_S$ is commutative). Hence $ba\in A$ [see the proof of (2)].$\square$

Corollary 4.9

Each non-commutative archimedean component of a quasi-commutative semigroup is a periodic retract ideal extension of a hamiltonian group by a commutative nil-semigroup.