# Rectangular group congruences on a semigroup

## Authors

Open AccessArticle

- First Online:
- Received:
- Accepted:

DOI: 10.1007/s00233-012-9426-y

## Abstract

We study rectangular group congruences on an arbitrary semigroup. Some of our results are an extension of the results obtained by Masat (Proc. Am. Math. Soc. 50:107–114, 1975). We show that each rectangular group congruence on a semigroup *S* is the intersection of a group congruence and a matrix congruence and vice versa, and this expression is unique, when *S* is *E*-inversive. Finally, we prove that every rectangular group congruence on an *E*-inversive semigroup is uniquely determined by its kernel and trace.

### Keywords

Rectangular group congruence Group congruence Matrix congruence## 1 Introduction and preliminaries

A groupoid *S* is called a *left* [*right*] *zero semigroup* if it satisfies the identity *xy*=*x* [*xy*=*y*]. Further, by a *rectangular band* we shall mean the direct product of a left zero and a right zero semigroup. Moreover, a semigroup *S* is said to be a *rectangular group* if it is isomorphic to the direct product *G*×*M* of a group *G* and a rectangular band *M*.

Let \(\mathcal{C}\) be a class of semigroups. We say that a congruence *ρ* on a semigroup *S* is a \(\mathcal{C}\)
*-congruence* if \(S/\rho\in\mathcal{C}\). For example, if \(\mathcal{C}\) is the class of groups, then *ρ* is called a *group* congruence on *S* if *S*/*ρ* is a group. In way of an exception, a congruence *ρ* on a semigroup *S* is said to be a *matrix* congruence if *S*/*ρ* is a rectangular band. Note that every left [right] zero semigroup is a rectangular band, so every left [right] zero congruence on *S* is a matrix congruence. Also, clearly the least matrix congruence on any semigroup *S* exists. Denote it by *ψ*. Furthermore, every group congruence and every matrix congruence is a rectangular group congruence. Hence we say that a rectangular group congruence is *proper* if it is neither a group nor a matrix congruence. We first give necessary and sufficient conditions on a semigroup *S* in order that it will have a proper rectangular group congruence. Furthermore, we show that every rectangular group congruence on *S* is the intersection of a group congruence and a matrix congruence. In addition, if *S* is *E*-inversive, then this expression is unique. Moreover, we prove that each rectangular group congruence on an *E*-inversive semigroup is uniquely determined by its kernel and trace. Before we start our study, we recall some definitions.

Let *S* be a semigroup and *a*∈*S*. The set *W*(*a*)={*x*∈*S*:*x*=*xax*} is called the set of all *weak inverses* of *a* and so the elements of *W*(*a*) will be called *weak inverse elements* of *a*. A semigroup *S* is said to be *E-inversive* if for every *a*∈*S* there exists *x*∈*S* such that *ax*∈*E*
_{
S
}, where *E*
_{
S
} (or briefly *E*) is the set of idempotents of *S* (more generally, if *A*⊆*S*, then *E*
_{
A
} denotes the set of idempotents of *A*). It is easy to see that a semigroup *S* is *E*-inversive if and only if *W*(*a*) is nonempty for all *a*∈*S*. Hence if *S* is *E*-inversive, then for every *a*∈*S* there is *x*∈*S* such that *ax*,*xa*∈*E*
_{
S
} [7, 8]. Further, by *Reg*(*S*) we shall mean the set of *regular elements* of *S* (an element *a* of *S* is called *regular* if *a*∈*aSa*) and by *V*(*a*)={*x*∈*S*:*a*=*axa*,*x*=*xax*} the set of all *inverse elements* of *a*. It is well known that an element *a* of *S* is regular if and only if *V*(*a*)≠∅, so a semigroup *S* is regular if and only if *V*(*a*)≠∅ for every *a*∈*S*. Finally, a regular semigroup *S* is said to be *orthodox* if *E*
_{
S
} forms a subsemigroup of *S*.

The following result seems to belong to folklore.

### Result 1.1

*The following conditions concerning a semigroup*

*S*

*are equivalent*:

- (i)
*S**is a rectangular band*; - (ii)
*S**is nonwhere commutative*,*i*.*e*., ∀*a*,*b*∈*S*[*ab*=*ba*⟹*a*=*b*]; - (iii)
∀

*a*,*b*∈*S*[*aba*=*a*]; - (iv)
∀

*a*,*b*,*c*∈*S*[*a*^{2}=*a*,*abc*=*ac*].

*A*of a semigroup

*S*is called

*left*[

*right*]

*dense*if the condition

*ab*∈

*A*implies that

*a*∈

*A*[

*b*∈

*A*] for all

*a*,

*b*∈

*S*. Further,

*A*is said to be

*quasi dense*if the following two conditions hold:

- (i)
∀

*a*∈*S*[*a*∈*A*⇔*a*^{2}∈*A*]; - (ii)
∀

*a*,*b*∈*S*[*ab*∈*A*⇔*aSb*⊆*A*].

*A*is a

*quasi ideal*of

*S*if

*AS*∩

*SA*⊆

*A*. For the connections between left [right] zero, matrix congruences on a semigroup

*S*and left dense right [right dense left] ideals, quasi dense subsemigroups of

*S*(respectively), the reader is referred to [9]. We note only some results of [9]. Firstly, denote by

*X*the set of all left dense right ideals of a semigroup

*S*and all right dense left ideals of a semigroup

*S*with the empty set included and

*S*excluded. Let 2

^{ X }be a family of all subsets of

*X*and \(\mathcal{MC}(S)\) be the set of all matrix congruences on

*S*. Define the map \(\phi: 2^{X} \to\mathcal{MC}(S)\) by \(\mathcal{X} \phi= \rho _{\mathcal{X}}\) (\(\mathcal{X} \in2^{X}\)), where

### Result 1.2

(Theorem 5 [9])

*The map*
*ϕ*
*is antitone* (*i*.*e*., \(\mathcal{X} \subseteq \mathcal{Y} \Longrightarrow\rho_{\mathcal{Y}} \subseteq\rho_{\mathcal {X}}\)) *and maps* 2^{
X
}
*onto*
\(\mathcal{MC}(S)\).

### Result 1.3

(Corollary to Theorem 5 [9])

*The relation*
*ρ*
_{
X
}
*is the least matrix congruence on a semigroup*
*S*. *Moreover*, *we may replace* (*in the present result*) *the set*
*X*
*by the set*
*Y*
*of all quasi dense subsemigroups of*
*S*.

### Result 1.4

(A part of Proposition 4, Theorem 9 [9])

*The following conditions concerning a congruence*

*ρ*

*on a semigroup*

*S*

*are equivalent*:

- (i)
*ρ**is a matrix congruence on**S*; - (ii)
*every**ρ*-*class of**S**is a quasi dense subsemigroup of**S*; - (iii)
*every**ρ*-*class of**S**is a quasi ideal of**S*.

*Conversely*, *a subsemigroup*
*A*
*of*
*S*
*is quasi dense*, *when*
*A*
*is a matrix of some*
*ψ*-*classes of*
*S*. *Thus*
*A*
*is quasi dense if and only if*
*A*
*is a*
*ρ*-*class of some matrix congruence*
*ρ*
*on*
*S*.

### Result 1.5

(Theorem 14 [9])

*Let*
*S*
*be a matrix of semigroups*
*S*
_{
ıλ
}, *where*
*ı*∈*I*, *λ*∈*Λ*, *such that every*
*S*
_{
ıλ
}
*has an identity element*
*e*
_{
ıλ
}
*and the set*
*M* (*say*) *of elements*
*e*
_{
ıλ
} (*ı*∈*I*, *λ*∈*Λ*) *forms a subsemigroup of*
*S*. *Then*
*M*
*is a rectangular band*. *Further*, *S*
_{
ıλ
}≅*S*
_{
ȷμ
}
*for all*
*ı*,*ȷ*∈*I*,*λ*,*μ*∈*Λ*
*and if we suppose that* 1∈*I*,*Λ*, *then*
*S*
*is isomorphic to the direct product*
*M*×*S*
_{11}
*of a rectangular band*
*M*
*and a semigroup*
*S*
_{11}. *Moreover*, *the semigroups*
*S*
_{
ıλ
}
*are precisely the*
*ψ*-*classes of*
*S*
*and*
*S*
_{
ıλ
}=*e*
_{
ıλ
}
*S*
_{
ıλ
}
*e*
_{
ıλ
}=*e*
_{
ıλ
}
*Se*
_{
ıλ
}
*for all*
*ı*∈*I*,*λ*∈*Λ*.

Notice that if *S* is a rectangular group (that is, *S*≅*M*×*G*, where *M* is a rectangular band and *G* is a group), then we shall write rather *S*=*M*×*G* than *S*≅*M*×*G*. The following theorem is known but for example: Masat considered in [5, 6] a regular semigroup *S* such that *E*
_{
S
} forms a rectangular band, and he did not know that *S* is a rectangular group, and so we include a simple proof for the completeness. Green’s relations on a semigroup *S* are denoted by \(\mathcal{L}^{S}\), \(\mathcal{R}^{S}\), \(\mathcal{H}^{S}\), \(\mathcal{D}^{S}\) and \(\mathcal{J}^{S}\). For undefined terms the reader is referred to the books [3, 4].

### Theorem 1.6

*The following conditions concerning a semigroup*

*S*

*are equivalent*:

- (i)
*S**is a rectangular group*; - (ii)
*S**is completely simple and orthodox*; - (iii)
*S**is completely regular and satisfies the identity*:*x*^{−1}*yy*^{−1}*x*=*x*^{−1}*x*; - (iv)
*S**is regular and**E*_{ S }*forms a rectangular band*.

*Consequently*, *if*
*S*
*is a rectangular group*, *then*
\(S \cong E_{S} \times \mathcal{H}_{e} = E_{S} \times eSe\)
*for some* (*all*) *e*∈*E*
_{
S
}.

### Proof

\(\mathrm{(ii)} \Longrightarrow\mathrm{(i)}\). If *S* is completely simple, then *S* is a matrix of groups \(\mathcal{H}_{e}\) (*e*∈*E*
_{
S
}), since Lemma III.2.4 [4] implies that \(\mathcal{H}\) is a matrix congruence on *S*, so \(\mathcal{H} = \psi\). Clearly, every \(\mathcal {H}_{e}\) has an identity element *e*. Also, \((efe, e) \in\mathcal{H}\) for all *e*,*f*∈*E*
_{
S
}, that is, *e*=*efe* for all *e*,*f*∈*E*
_{
S
} (since *efe*∈*E*
_{
S
}). Hence *E*
_{
S
} is a rectangular band. Thus \(S \cong E_{S} \times \mathcal{H}_{e}\) for some (all) *e*∈*E*
_{
S
} (by Result 1.5).

*S*=

*M*×

*G*, where

*M*is a rectangular band and

*G*is a group (with an identity 1). Define the mapping

^{−1}:

*S*→

*S*by (

*a*,

*g*)

^{−1}=(

*a*,

*g*

^{−1}) for all (

*a*,

*g*)∈

*S*. One can easily verify that (

*S*,⋅,

^{−1}) is completely regular, that is, (

*x*

^{−1})

^{−1}=

*x*,

*xx*

^{−1}=

*x*

^{−1}

*x*and

*xx*

^{−1}

*x*=

*x*for every

*x*∈

*S*. Further, suppose that

*x*=(

*a*,

*g*),

*y*=(

*b*,

*h*)∈

*S*. Then

*e*∈

*E*

_{ S }. Since

*ee*

^{−1}=

*e*

^{−1}

*e*,(

*e*

^{−1})

^{−1}=

*e*, then

*e*

^{−1}

*ff*

^{−1}

*e*=

*e*

^{−1}

*e*, i.e.,

*efe*=

*e*for all

*e*,

*f*∈

*E*

_{ S }. Thus

*E*

_{ S }is a rectangular band. Consequently, the condition (iv) holds.

\(\mathrm{(iv)} \Longrightarrow\mathrm{(ii)}\). Clearly, each idempotent of *S* is primitive and \(\mathcal{D}^{E_{S}} = E_{S} \times E_{S}\). Since *S* is regular, then every element of *S* is \(\mathcal{D}\)-related with some of its idempotent. It follows that \(\mathcal{D}^{S} = \mathcal{J}^{S} = S \times S\). Thus *S* is completely simple and orthodox. □

By the *trace*
\(\operatorname{tr}\rho\) of a relation *ρ* on a semigroup *S* we shall mean the restriction of *ρ* to the set *E*
_{
S
}.

The following result will be useful in the proof of Theorems 2.2(iii), 2.5, 2.6.

### Result 1.7

(Corollary 2.7 [1])

*If*
*ρ*
*is a matrix congruence on an*
*E*-*inversive semigroup*
*S*, *then every*
*ρ*-*class of*
*S*
*is*
*E*-*inversive*. *Also*, *every matrix congruence on an*
*E*-*inversive semigroup is uniquely determined by its trace*.

Further, some preliminaries about group congruences on a semigroup *S* are needed. A subset *A* of *S* is called (respectively) *full*; *reflexive* and *dense* if *E*
_{
S
}⊆*A*; ∀*a*,*b*∈*S* [*ab*∈*A*⟹*ba*∈*A*] and ∀*s*∈*S* ∃ *x*,*y*∈*S* [*sx*,*ys*∈*A*]. Also, define the *closure operator*
*ω* on *S* by *Aω*={*s*∈*S*:∃ *a*∈*A* [*as*∈*A*]} (*A*⊆*S*). We shall say that *A*⊆*S* is *closed* (in *S*) if *Aω*=*A*. Finally, a subsemigroup *N* of a semigroup *S* is said to be *normal* if it is full, dense, reflexive and closed (if *N* is normal, then we shall write *N*◁*S*). Moreover, if a subsemigroup of *S* is full, dense and reflexive, then it is called *seminormal*.

By the *kernel*
\(\operatorname{ker}\rho\) of a congruence *ρ* on a semigroup *S* we shall mean the set {*x*∈*S*:(*x*,*x*
^{2})∈*ρ*}.

The following two results follow directly from the definition of the group.

### Lemma 1.8

*Let*
*ρ*
*be a group congruence on a semigroup*
*S*. *Then*
\(\operatorname{ker}\rho\lhd S\).

### Lemma 1.9

*Let*
*ρ*
_{1},*ρ*
_{2}
*be group congruences on a semigroup*
*S*. *Then*
*ρ*
_{1}⊂*ρ*
_{2}
*if and only if*
\(\operatorname{ker}\rho_{1} \subset \operatorname{ker}\rho_{2}\).

*B*be a nonempty subset of a semigroup

*S*. Consider four relations on

*S*:

### Lemma 1.10

[2]

*Let a subsemigroup*
*B*
*of a semigroup*
*S*
*be dense and reflexive*. *Then*
*ρ*
_{1,B
}=*ρ*
_{2,B
}=*ρ*
_{3,B
}=*ρ*
_{4,B
}.

If *B* is a seminormal subsemigroup of *S*, then we denote the above four relations by *ρ*
_{
B
}.

The following theorem says that there exists an inclusion-preserving bijection between the set of all normal subsemigroups of a semigroup *S* and the set of all group congruences on *S*.

### Theorem 1.11

[2]

*Let*
*B*
*be a seminormal subsemigroup of a semigroup*
*S*.* Then the relation*
*ρ*
_{
B
}
*is a group congruence on*
*S*. *Moreover*, \(B\subseteq B\omega= \operatorname{ker}\rho_{B}\). *If*
*B*
*is normal*, *then*
\(B = \operatorname{ker}\rho_{B}\).

*Conversely*, *if*
*ρ*
*is a group congruence on*
*S*, *then there exists a normal subsemigroup*
*N*
*of*
*S*
*such that*
*ρ*=*ρ*
_{
N
} (*in fact*, \(N = \operatorname{ker}\rho\)). *Thus there exists an inclusion*-*preserving bijection between the set of all normal subsemigroups of*
*S*
*and the set of all group congruences on*
*S*.

Finally, the following remark will be useful.

### Remark 1

Denote by *σ* the least group congruence on a semigroup (if it exists). One can easily seen that if *S* is an *E*-inversive semigroup (and so *E*
_{
S
} is dense), then there exists the least normal subsemigroup of *S*. In the light of the above theorem, every *E*-inversive semigroup possesses a least group congruence.

## 2 Rectangular group congruences

The following theorem gives necessary and sufficient conditions on a semigroup *S* in order that it has a proper rectangular group congruence. (Notice that a normal subsemigroup *N* of *S* is called *proper* if *N*≠*S*.)

### Theorem 2.1

*Let*

*S*

*be a semigroup*.

*The following conditions are equivalent*:

- (i)
*there exists a proper rectangular group congruence on**S*; - (ii)
*S**is a disjoint union of two or more quasi dense subsemigroups of**S**and contains a proper normal subsemigroup of**S*; - (iii)
*there exists a non*-*universal group and a non*-*universal matrix congruence on**S*.

### Proof

\(\mathrm{(i)} \Longrightarrow\mathrm{(ii)}\). Let *ρ* be a proper rectangular group congruence on *S*, say *S*/*ρ* is equal *M*×*G*, where *M* is a rectangular band, *G* is a group (with identity 1). Note that *M*≅*E*
_{
M×G
}={(*m*,1):*m*∈*M*}. Further, for all *m*∈*M*, define *Q*
_{
m
} to be the preimage of {*m*}×*G* by the canonical epimorphism *ρ*
^{♮} from *S* onto *S*/*ρ*. It follows easily from Result 1.1(iv) that {*m*}×*G* is a quasi dense subsemigroup of *M*×*G*. Thus the preimage of {*m*}×*G* by *ρ*
^{♮} is also such a subsemigroup of *S* (by *M*≅*E*
_{
M×G
}). Since *ρ* is not a group congruence, then |*M*|>1, and so *S* has a proper matrix congruence (Result 1.3). Hence *S* is a disjoint union of two or more quasi dense subsemigroups of *S*, see Result 1.4 (notice that *S*=⋃{*Q*
_{
m
}:*m*∈*M*}, where the union is disjoint, and this decomposition of *S* induced, by the first part of Result 1.4, a matrix congruence on *S*). Let \(N = \operatorname{ker}\rho\). Then *a*∈*N* if and only if *aρ*∈*E*
_{
M×G
}, that is, if and only if *aρ*∈*M*×{1}. Clearly, *N* is a full subsemigroup of *S*. Also, *M*×{1} is reflexive in *M*×*G*, so *N* is reflexive in *S*. Furthermore, *N* is dense, since *S*/*ρ* is *E*-inversive. Finally, *N* is closed in *S*, since *M*×{1} is closed in *M*×*G*. Consequently, *N* is a normal subsemigroup of *S* and since *S*/*ρ* is not a rectangular band, then *N* is proper.

\(\mathrm{(ii)} \Longrightarrow\mathrm{(iii)}\). This follows from Result 1.4 and Theorem 1.11.

\(\mathrm{(iii)} \Longrightarrow\mathrm{(i)}\). Let *ρ*
_{1} be a non-universal matrix congruence on *S* and *ρ*
_{2} be a non-universal group congruence on *S*. We show that *ρ*=*ρ*
_{1}∩*ρ*
_{2} is a proper rectangular group congruence on a semigroup *S*. Indeed, let *a*∈*S*. Since *S*/*ρ*
_{2} is regular, then (*axa*,*a*)∈*ρ*
_{2} for some *x*∈*S*, so (*axa*,*a*)∈*ρ*. Therefore *S*/*ρ* is regular. Clearly, *xρ*
_{2}, where \(x \in \operatorname{ker}\rho_{2}\) is an identity of the group *S*/*ρ*
_{2} and so (*xyx*,*x*)∈*ρ*
_{2} for all \(x, y \in \operatorname{ker}\rho_{2}\). Hence (*xyx*,*x*)∈*ρ* for all \(x, y \in \operatorname{ker}\rho _{2}\). On the other hand, if *xρ*∈*E*
_{
S/ρ
}, then \(x \in \operatorname{ker}\rho_{2}\). It follows that *E*
_{
S/ρ
} forms a rectangular band, therefore, *S*/*ρ* is a rectangular group (Theorem 1.6(iv)). Finally, suppose by way of contradiction that *ρ* is a matrix congruence on *S*, that is, (*aba*,*a*)∈*ρ* for all *a*,*b*∈*S*. Then (*aba*,*a*)∈*ρ*
_{2} for all *a*,*b*∈*S*. Hence *S*/*ρ*
_{2} must be a trivial group. Thus *ρ*
_{2}=*S*×*S*, a contradiction from the assumption that *ρ*
_{2} is a non-universal congruence on *S*. Similarly, if *ρ* is a group congruence, then *ρ*
_{1} is a group congruence (since *ρ*⊆*ρ*
_{1}), so *S*/*ρ*
_{1} must be trivial. Hence *ρ*
_{1} is the universal relation, but this is impossible. Consequently, *ρ* is a proper rectangular group congruence on *S*. □

We have just seen that the intersection of a group congruence on a semigroup *S* and a matrix congruence on *S* is a rectangular group congruence on *S*. Conversely, the part (i) of the following theorem (together with Theorem 2.1) implies that any rectangular group congruence on *S* can be expressed in this way.

### Theorem 2.2

*Let*

*ρ*

*be a rectangular group congruence on a semigroup*

*S*(

*and so*

*S*/

*ρ*=

*M*×

*G*,

*where*

*M*

*is a rectangular band*,

*G is a group*).

*Also*,

*let*

*Q*

_{ m }(

*m*∈

*M*)

*be the preimage of*{

*m*}×

*G*

*by the canonical epimorphism*

*ρ*

^{♮}

*from*

*S*

*onto*

*S*/

*ρ*,

*and put*

*N*={

*s*∈

*S*:

*sρ*∈

*E*

_{ M×G }}.

*Moreover*,

*denote by*

*υ*

*the matrix congruence on S*,

*induced by the partition*{

*Q*

_{ m }:

*m*∈

*M*}

*of*

*S*(

*see the proof of “*\(\mathrm{(i)} \Longrightarrow\mathrm{(ii)}\)

*” in Theorem*2.1).

*Then*:

- (i)
*ρ*=*υ*∩*ρ*_{ N }; - (ii)
*S*/*ρ*≅*S*/*υ*×*S*/*ρ*_{ N }.

*If in addition*

*S*

*is*

*E*-

*inversive*,

*then*:

- (iii)
∀

*m*∈*M*[*N*∩*Q*_{ m }◁*Q*_{ m }]; - (iv)
∀

*m*∈*M*[\(S/\rho_{N} \cong Q_{m}/\rho_{(N \, \cap\: Q_{m})}\)].

### Proof

Firstly, notice that *N* is the preimage of *M*×{1_{
G
}} by *ρ*
^{♮}. Secondly, every *Q*
_{
m
} is a quasi dense subsemigroup of *S* (see Result 1.4). Also, if *S* is *E*-inversive, then each *Q*
_{
m
} is an *E*-inversive subsemigroup of *S* (Result 1.7).

(i). Let (*a*,*b*)∈*ρ* and *aρ*=(*m*,*g*), where (*m*,*g*)∈*M*×*G*. Take *x*=(*m*,*g*
^{−1}), where *g*
^{−1} is a group inverse of *g* in *G*. Then clearly *xa*,*xb*∈*N* and so (*a*,*b*)∈*ρ*
_{
N
} (see Remark 1). Also, *aρ*=(*m*,*g*)=*bρ*∈{*m*}×*G*. Hence *a*,*b*∈*Q*
_{
m
}, so (*a*,*b*)∈*υ*. Thus (*a*,*b*)∈*υ*∩*ρ*
_{
N
}. Consequently, *ρ*⊂*υ*∩*ρ*
_{
N
}. Conversely, let (*a*,*b*)∈*υ*∩*ρ*
_{
N
}. Then *aρ*,*bρ*∈{*m*}×*G* (*aρ*=(*m*,*g*
_{1}), *bρ*=(*m*,*g*
_{2})), *xa*,*xb*∈*N* for some *m*∈*M* and *x*∈*Q*
_{
n
}, where *n*∈*M* (say *xρ*=(*n*,*g*)), and so (*xa*)*ρ*=(*nm*,*gg*
_{1}). On the other hand, (*xa*)*ρ*∈*M*×{1_{
G
}}. Hence *g*
_{1}=*g*
^{−1}. We may equally well show that *g*
_{2}=*g*
^{−1}. Thus (*a*,*b*)∈*ρ*. Consequently, *ρ*=*υ*∩*ρ*
_{
N
}, as exactly required.

*ρ*=

*υ*∩

*ρ*

_{ N }by (i). Define the mapping

*ϕ*:

*S*/

*ρ*→

*S*/

*υ*×

*S*/

*ρ*

_{ N }by (

*aρ*)

*ϕ*=(

*aυ*,

*aρ*

_{ N }) (

*a*∈

*S*). Clearly,

*ϕ*is a monomorphism. We show that

*ϕ*is surjective. Let (

*aυ*,

*bρ*

_{ N })∈

*S*/

*υ*×

*S*/

*ρ*

_{ N }. Then

*a*∈

*Q*

_{ m }, where

*m*∈

*M*. Take any element

*n*∈

*N*∩

*Q*

_{ m }. Then

(iii). Let *m*∈*M*. Put *N*
_{
m
}=*N*∩*Q*
_{
m
}. Evidently, *N*
_{
m
} is a full, reflexive and closed subsemigroup of *Q*
_{
m
} (even if *S* is an arbitrary semigroup). By Result 1.7, *N*
_{
m
} is dense in *Q*
_{
m
}. Thus *N*
_{
m
}◁*Q*
_{
m
}.

(iv). Let *m*∈*M*. Define the map \(\phi: Q_{m}/\rho_{N_{m}}\to S/\rho _{N}\) by \((a\rho_{N_{m}})\phi= a\rho_{N}\) (*a*∈*Q*
_{
m
}). Clearly, *ϕ* is a well-defined homomorphism. Furthermore, if *a*∈*S* and *n*∈*N*
_{
m
}⊆*N*, then *nan*∈*Q*
_{
m
} and \(((nan)\rho_{N_{m}})\phi= (nan)\rho_{N} = a\rho_{N}\). Thus *ϕ* is surjective. Finally, we show that *ϕ* is injective. Let \(a, b \in Q_{m}, (a\rho_{N_{m}})\phi= (b\rho _{N_{m}})\phi\). Then (*a*,*b*)∈*ρ*
_{
N
}, so *ax*, *bx*∈*N* for some *x*∈*S*. Hence for every *n*∈*N*
_{
m
}⊆*N*, *anx*, *bnx*∈*N*. Thus *n*(*anx*)*n*∈*N*∩(*Q*
_{
m
}
*NQ*
_{
m
})⊆*N*∩*Q*
_{
m
}=*N*
_{
m
} and similarly: *n*(*bnx*)*n*∈*N*
_{
m
}. Since *na*, *nb*, *nxn*∈*Q*
_{
m
}, then \((na, nb) \in\rho_{N_{m}}\), and so \((a, b) \in\rho_{N_{m}}\), because \(n \in N_{m} = \operatorname{ker}\rho_{N_{m}}\). □

### Corollary 2.3

*If the least group congruence exists on a semigroup*
*S*, *then the relation*
*ψ*∩*σ*
*is the least rectangular group congruence on*
*S*. *In particular*, *in any*
*E*-*inversive semigroup*, *ψ*∩*σ*
*is the least rectangular group congruence*.

### Remark 2

If *S* is not *E*-inversive, then the least rectangular group congruence on a semigroup *S* may not exist. Indeed, consider the additive semigroup of non-negative integers ℕ. It is well known that every group congruence on ℕ is of the following form: *ρ*
_{
n
}={(*k*,*l*)∈ℕ×ℕ:*n*|(*k*−*l*)} (*n*>0). Further, since ℕ has identity, then the least matrix congruence on ℕ is the universal relation, so any rectangular group congruence on ℕ is a group congruence (Theorem 2.2(i)). Consequently, ℕ has no least rectangular group congruence.

Let \(\mathcal{C}\) be a class of semigroups which is closed under homomorphic images. Note that if the least \(\mathcal {C}\)-congruence \(\rho_{\mathcal{C}}\) on a semigroup *S* exists, then the interval [\(\rho_{\mathcal{C}}\), *S*×*S*] consists of all \(\mathcal{C}\)-congruences on *S* and it is a complete sublattice of the complete lattice \(\mathcal{C}(S)\) of congruences on *S*. Evidently, the class of all groups [rectangular bands] is closed under homomorphic images. Using Theorem 1.6(iv) one can prove without difficulty that the class of all rectangular groups has this property. Denote by *θ* the least rectangular group congruence on an *E*-inversive semigroup. In particular, the intervals [*ψ*, *S*×*S*], [*σ*, *S*×*S*], [*θ*, *S*×*S*] consist of all matrix, group, rectangular group congruences on an *E*-inversive semigroup *S*, respectively, and they are complete sublattices of \(\mathcal{C}(S)\). Denote them by \(\mathcal{MC}(S)\), \(\mathcal{GC}(S)\), \(\mathcal{RGC}(S)\), respectively. Clearly, the direct product \(\mathcal{MC}(S) \times\mathcal{GC}(S)\) is a complete sublattice of \(\mathcal{C}(S) \times\mathcal{C}(S)\) (see [3, p. 37]).

For terminology and elementary facts about lattices the reader is referred to the book [10, Sect. I.2]. The following simple result will be useful (see Lemma I.2.8 and Exercise I.2.15(iii) in [10]).

### Result 2.4

*If*
*ϕ*
*is an order isomorphism of a lattice*
*L*
*onto a lattice*
*M*, *then*
*ϕ*
*is a lattice isomorphism*. *Moreover*, *every lattice isomorphism of complete lattices is a complete lattice isomorphism*.

We show that each rectangular group congruence on an *E*-inversive semigroup can be expressed as the unique intersection of a group and a matrix congruence.

### Theorem 2.5

*Every rectangular group congruence on an*
*E*-*inversive semigroup*
*S*
*is of the form*
*υ*∩*ρ*
_{
N
}, *where*
*υ*
*is a matrix congruence on*
*S*, *N*◁*S*, *and this expression is unique*.

*Moreover*,

*there exists an inclusion*-

*preserving bijection*

*ϕ*

*between the complete lattice*\(\mathcal{MC}(S) \times\mathcal{GC}(S)\)

*and the complete lattice*\(\mathcal{RGC}(S)\).

*In fact*,

*ϕ*

*is defined by*:

*for every*\((\upsilon, \rho_{N}) \in\mathcal{MC}(S) \times\mathcal{GC}(S)\).

*Furthermore*,

*ϕ*

^{ −1}

*is an inclusion*-

*preserving bijection*(

*by the proof of Theorem*2.2(i)),

*so*

*ϕ*

*is an order isomorphism of the complete lattice*\(\mathcal{MC}(S) \times\mathcal{GC}(S)\)

*onto the complete lattice*\(\mathcal{RGC}(S)\).

*Consequently*,

*ϕ*

*is a complete lattice isomorphism between the lattices*\(\mathcal{MC}(S) \times\mathcal{GC}(S)\)

*and*\(\mathcal{RGC}(S)\),

*respectively*.

### Proof

Let *ρ* be a rectangular group congruence on *S*. Then *ρ* is the intersection of some matrix and some group congruence on *S* (Theorem 2.2(i)). Next, suppose that \(\rho= \upsilon_{1} \cap \rho_{N_{1}} = \upsilon_{2} \cap \rho_{N_{2}}\), where *υ*
_{
i
} is a matrix congruence on *S*, *N*
_{
i
}◁*S* (*i*=1,2). Let (*a*,*b*)∈*υ*
_{1}. Since *υ*
_{1}∩*υ*
_{2} is a matrix congruence on *S*, then there are idempotents *e*, *f* of *S* such that \((a, e) \in\upsilon_{1} \cap\upsilon_{2}, (e, f) \in\rho_{N_{1}}, (f, b) \in \upsilon_{1} \cap\upsilon_{2}\) (Result 1.7), so \((e, f) \in\upsilon_{1} \cap\rho_{N_{1}} = \upsilon_{2} \cap\rho_{N_{2}} \subseteq\upsilon_{2}\). Hence (*a*,*b*)∈*υ*
_{2}, i.e., *υ*
_{1}⊂*υ*
_{2}. We may equally well show the opposite inclusion. Put *υ*
_{1}=*υ*
_{2}=*υ*, so that \(\rho= \upsilon\cap \rho_{N_{1}} = \upsilon\cap \rho_{N_{2}}\). If \((a, b) \in \rho_{N_{1}}\), then \((aab, abb) \in\upsilon\cap\rho_{N_{1}} \subset\rho _{N_{2}}\), so \((a, b) \in\rho_{N_{2}}\) (by cancellation). Hence \(\rho _{N_{1}} \subset\rho_{N_{2}}\). By symmetry, \(\rho_{N_{2}} \subset\rho _{N_{1}}\). Thus \(\rho_{N_{1}} = \rho_{N_{2}}\), as exactly required.

The second part of the theorem follows directly from the above considerations and Result 2.4. □

Finally, from Result 1.7 we obtain the following theorem.

### Theorem 2.6

*Every rectangular group congruence on an*
*E*-*inversive semigroup*
*S*
*is uniquely determined by its kernel and trace*.

### Proof

*ρ*

_{1}, \(\rho_{2} \in\mathcal{GC}(S), \upsilon_{1}\), \(\upsilon_{2} \in\mathcal{MC}(S)\) be such that \(\operatorname{ker}(\upsilon_{1} \cap\rho _{1}) \subset \operatorname{ker}(\upsilon_{2} \cap\rho_{2})\) and \(\operatorname{tr}(\upsilon_{1} \cap\rho _{1}) \subset \operatorname{tr}(\upsilon_{2} \cap\rho_{2})\). Then

*ρ*

_{1}⊂

*ρ*

_{2}. Similarly, we obtain that \(\operatorname{tr}\upsilon_{1} \subset \operatorname{tr}\upsilon_{2}\). Hence

*υ*

_{1}⊂

*υ*

_{2}(this follows from the proof of Result 1.7, see [1]). Thus

*υ*

_{1}∩

*ρ*

_{1}⊂

*υ*

_{2}∩

*ρ*

_{2}. This implies the thesis of the theorem. □

### Open Access

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