Semigroup Forum

, Volume 87, Issue 1, pp 120–128

# Rectangular group congruences on a semigroup

Open Access
RESEARCH ARTICLE

DOI: 10.1007/s00233-012-9426-y

Gigoń, R.S. Semigroup Forum (2013) 87: 120. 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 aS. The set W(a)={xS: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 aS there exists xS such that axES, where ES (or briefly E) is the set of idempotents of S (more generally, if AS, then EA 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 aS. Hence if S is E-inversive, then for every aS there is xS such that ax,xaES [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 aaSa) and by V(a)={xS: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 aS. Finally, a regular semigroup S is said to be orthodox if ES forms a subsemigroup of S.

The following result seems to belong to folklore.

### Result 1.1

The following conditions concerning a semigroupSare equivalent:
1. (i)

Sis a rectangular band;

2. (ii)

Sis nonwhere commutative, i.e., ∀a,bS [ab=baa=b];

3. (iii)

a,bS [aba=a];

4. (iv)

a,b,cS [a2=a,abc=ac].

Recall from [9] that a nonempty subset A of a semigroup S is called left [right] dense if the condition abA implies that aA [bA] for all a,bS. Further, A is said to be quasi dense if the following two conditions hold:
1. (i)

aS [aAa2A];

2. (ii)

a,bS [abAaSbA].

Finally, we say that A is a quasi ideal of S if ASSAA. 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 2X 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
$$\rho_{\mathcal{X}} = \bigl\{(a, b) \in S \times S : \forall A \in\mathcal {X}\ [a, b \in A\ \mbox{or}\ a, b \notin A]\bigr\}.$$

### Result 1.2

(Theorem 5 [9])

The mapϕis antitone (i.e., $$\mathcal{X} \subseteq \mathcal{Y} \Longrightarrow\rho_{\mathcal{Y}} \subseteq\rho_{\mathcal {X}}$$) and maps 2Xonto$$\mathcal{MC}(S)$$.

### Result 1.3

(Corollary to Theorem 5 [9])

The relationρXis the least matrix congruence on a semigroupS. Moreover, we may replace (in the present result) the setXby the setYof all quasi dense subsemigroups ofS.

### Result 1.4

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

The following conditions concerning a congruenceρon a semigroupSare equivalent:
1. (i)

ρis a matrix congruence onS;

2. (ii)

everyρ-class ofSis a quasi dense subsemigroup ofS;

3. (iii)

everyρ-class ofSis a quasi ideal ofS.

Conversely, a subsemigroupAofSis quasi dense, whenAis a matrix of someψ-classes ofS. ThusAis quasi dense if and only ifAis aρ-class of some matrix congruenceρonS.

### Result 1.5

(Theorem 14 [9])

LetSbe a matrix of semigroupsSıλ, whereıI, λΛ, such that everySıλhas an identity elementeıλand the setM (say) of elementseıλ (ıI, λΛ) forms a subsemigroup ofS. ThenMis a rectangular band. Further, SıλSȷμfor allı,ȷI,λ,μΛand if we suppose that 1∈I,Λ, thenSis isomorphic to the direct productM×S11of a rectangular bandMand a semigroupS11. Moreover, the semigroupsSıλare precisely theψ-classes ofSandSıλ=eıλSıλeıλ=eıλSeıλfor allıI,λΛ.

Notice that if S is a rectangular group (that is, SM×G, where M is a rectangular band and G is a group), then we shall write rather S=M×G than SM×G. The following theorem is known but for example: Masat considered in [5, 6] a regular semigroup S such that ES 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 semigroupSare equivalent:
1. (i)

Sis a rectangular group;

2. (ii)

Sis completely simple and orthodox;

3. (iii)

Sis completely regular and satisfies the identity: x−1yy−1x=x−1x;

4. (iv)

Sis regular andESforms a rectangular band.

Consequently, ifSis a rectangular group, then$$S \cong E_{S} \times \mathcal{H}_{e} = E_{S} \times eSe$$for some (all) eES.

### Proof

$$\mathrm{(ii)} \Longrightarrow\mathrm{(i)}$$. If S is completely simple, then S is a matrix of groups $$\mathcal{H}_{e}$$ (eES), 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,fES, that is, e=efe for all e,fES (since efeES). Hence ES is a rectangular band. Thus $$S \cong E_{S} \times \mathcal{H}_{e}$$ for some (all) eES (by Result 1.5).

$$\mathrm{(i)} \Longrightarrow\mathrm{(iii)}$$. Let S=M×G, where M is a rectangular band and G is a group (with an identity 1). Define the mapping −1:SS 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−1x and xx−1x=x for every xS. Further, suppose that x=(a,g),y=(b,h)∈S. Then
$$x^{- 1}yy^{- 1}x = \bigl(ab^2a, 1\bigr) = \bigl(a^2, 1\bigr) = \bigl(a, g^{- 1}\bigr) (a, g) = x^{- 1}x.$$
$$\mathrm{(iii)} \Longrightarrow\mathrm{(iv)}$$. Let eES. Since ee−1=e−1e,(e−1)−1=e, then
$$e = ee^{-1}e = e^{- 1}e = \bigl(e^{- 1}e \bigr)^2 = \bigl(e^{- 1}e\bigr) \bigl(ee^{- 1}\bigr) = e^{- 1}ee^{- 1} = e^{- 1}.$$
Hence e−1ff−1e=e−1e, i.e., efe=e for all e,fES. Thus ES 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 ES.

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])

Ifρis a matrix congruence on anE-inversive semigroupS, then everyρ-class ofSisE-inversive. Also, every matrix congruence on anE-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 ESA; ∀a,bS [abAbaA] and ∀sS ∃ x,yS [sx,ysA]. Also, define the closure operatorω on S by ={sS:∃ aA [asA]} (AS). We shall say that AS is closed (in S) if =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 NS). 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 {xS:(x,x2)∈ρ}.

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

### Lemma 1.8

Letρbe a group congruence on a semigroupS. Then$$\operatorname{ker}\rho\lhd S$$.

### Lemma 1.9

Letρ1,ρ2be group congruences on a semigroupS. Thenρ1ρ2if and only if$$\operatorname{ker}\rho_{1} \subset \operatorname{ker}\rho_{2}$$.

Let B be a nonempty subset of a semigroup S. Consider four relations on S:

### Lemma 1.10

[2]

Let a subsemigroupBof a semigroupSbe dense and reflexive. Thenρ1,B=ρ2,B=ρ3,B=ρ4,B.

If B is a seminormal subsemigroup of S, then we denote the above four relations by ρB.

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]

LetBbe a seminormal subsemigroup of a semigroupS. Then the relationρBis a group congruence onS. Moreover, $$B\subseteq B\omega= \operatorname{ker}\rho_{B}$$. IfBis normal, then$$B = \operatorname{ker}\rho_{B}$$.

Conversely, ifρis a group congruence onS, then there exists a normal subsemigroupNofSsuch thatρ=ρN (in fact, $$N = \operatorname{ker}\rho$$). Thus there exists an inclusion-preserving bijection between the set of all normal subsemigroups ofSand the set of all group congruences onS.

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 ES 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 NS.)

### Theorem 2.1

LetSbe a semigroup. The following conditions are equivalent:
1. (i)

there exists a proper rectangular group congruence onS;

2. (ii)

Sis a disjoint union of two or more quasi dense subsemigroups ofSand contains a proper normal subsemigroup ofS;

3. (iii)

there exists a non-universal group and a non-universal matrix congruence onS.

### 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 MEM×G={(m,1):mM}. Further, for all mM, define Qm to be the preimage of {mG by the canonical epimorphism ρ from S onto S/ρ. It follows easily from Result 1.1(iv) that {mG is a quasi dense subsemigroup of M×G. Thus the preimage of {mG by ρ is also such a subsemigroup of S (by MEM×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=⋃{Qm:mM}, 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 aN if and only if EM×G, that is, if and only if 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 ρ1 be a non-universal matrix congruence on S and ρ2 be a non-universal group congruence on S. We show that ρ=ρ1ρ2 is a proper rectangular group congruence on a semigroup S. Indeed, let aS. Since S/ρ2 is regular, then (axa,a)∈ρ2 for some xS, so (axa,a)∈ρ. Therefore S/ρ is regular. Clearly, 2, where $$x \in \operatorname{ker}\rho_{2}$$ is an identity of the group S/ρ2 and so (xyx,x)∈ρ2 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 ES/ρ, then $$x \in \operatorname{ker}\rho_{2}$$. It follows that ES/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,bS. Then (aba,a)∈ρ2 for all a,bS. Hence S/ρ2 must be a trivial group. Thus ρ2=S×S, a contradiction from the assumption that ρ2 is a non-universal congruence on S. Similarly, if ρ is a group congruence, then ρ1 is a group congruence (since ρρ1), so S/ρ1 must be trivial. Hence ρ1 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

Letρbe a rectangular group congruence on a semigroupS (and soS/ρ=M×G, whereMis a rectangular band, G is a group). Also, letQm (mM) be the preimage of {mGby the canonical epimorphismρfromSontoS/ρ, and putN={sS:EM×G}. Moreover, denote byυthe matrix congruence on S, induced by the partition {Qm:mM} ofS (see the proof of “$$\mathrm{(i)} \Longrightarrow\mathrm{(ii)}$$” in Theorem 2.1). Then:
1. (i)

ρ=υρN;

2. (ii)

S/ρS/υ×S/ρN.

1. (iii)

mM [NQmQm];

2. (iv)

mM [$$S/\rho_{N} \cong Q_{m}/\rho_{(N \, \cap\: Q_{m})}$$].

### Proof

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

(i). Let (a,b)∈ρ and =(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,xbN and so (a,b)∈ρN (see Remark 1). Also, =(m,g)=∈{mG. Hence a,bQm, so (a,b)∈υ. Thus (a,b)∈υρN. Consequently, ρυρN. Conversely, let (a,b)∈υρN. Then ,∈{mG (=(m,g1), =(m,g2)), xa,xbN for some mM and xQn, where nM (say =(n,g)), and so (xa)ρ=(nm,gg1). On the other hand, (xa)ρM×{1G}. Hence g1=g−1. We may equally well show that g2=g−1. Thus (a,b)∈ρ. Consequently, ρ=υρN, as exactly required.

(ii). Indeed, ρ=υρN by (i). Define the mapping ϕ:S/ρS/υ×S/ρN by ()ϕ=(,N) (aS). Clearly, ϕ is a monomorphism. We show that ϕ is surjective. Let (,N)∈S/υ×S/ρN. Then aQm, where mM. Take any element nNQm. Then
$$\bigl((nbn)\rho\bigr)\phi= \bigl((nbn)\upsilon, (nbn)\rho_{N}\bigr) = (n\upsilon, b\rho_{N}) = (a\upsilon, b\rho_{N}).$$

(iii). Let mM. Put Nm=NQm. Evidently, Nm is a full, reflexive and closed subsemigroup of Qm (even if S is an arbitrary semigroup). By Result 1.7, Nm is dense in Qm. Thus NmQm.

(iv). Let mM. Define the map $$\phi: Q_{m}/\rho_{N_{m}}\to S/\rho _{N}$$ by $$(a\rho_{N_{m}})\phi= a\rho_{N}$$ (aQm). Clearly, ϕ is a well-defined homomorphism. Furthermore, if aS and nNmN, then nanQm 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)∈ρN, so ax, bxN for some xS. Hence for every nNmN, anx, bnxN. Thus n(anx)nN∩(QmNQm)⊆NQm=Nm and similarly: n(bnx)nNm. Since na, nb, nxnQm, 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 semigroupS, then the relationψσis the least rectangular group congruence onS. In particular, in anyE-inversive semigroup, ψσis 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: ρn={(k,l)∈ℕ×ℕ:n|(kl)} (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

Ifϕis an order isomorphism of a latticeLonto a latticeM, thenϕis 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 anE-inversive semigroupSis of the formυρN, whereυis a matrix congruence onS, NS, and this expression is unique.

Moreover, there exists an inclusion-preserving bijectionϕbetween the complete lattice$$\mathcal{MC}(S) \times\mathcal{GC}(S)$$and the complete lattice$$\mathcal{RGC}(S)$$. In fact, ϕis defined by:
$$(\upsilon, \rho_{N})\phi = \upsilon\cap\rho_{N}$$
for every$$(\upsilon, \rho_{N}) \in\mathcal{MC}(S) \times\mathcal{GC}(S)$$. Furthermore, ϕ −1is an inclusion-preserving bijection (by the proof of Theorem 2.2(i)), soϕis an order isomorphism of the complete lattice$$\mathcal{MC}(S) \times\mathcal{GC}(S)$$onto the complete lattice$$\mathcal{RGC}(S)$$. Consequently, ϕis 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 υi is a matrix congruence on S, NiS (i=1,2). Let (a,b)∈υ1. Since υ1υ2 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)∈υ2, i.e., υ1υ2. We may equally well show the opposite inclusion. Put υ1=υ2=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 anE-inversive semigroupSis uniquely determined by its kernel and trace.

### Proof

Let ρ1, $$\rho_{2} \in\mathcal{GC}(S), \upsilon_{1}$$, $$\upsilon_{2} \in\mathcal{MC}(S)$$ 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
$$\operatorname{ker}(\upsilon_1 \cap\rho_1) = \operatorname{ker}\upsilon_1 \cap \operatorname{ker}\rho_1 = S \cap\operatorname{ker}\rho_1 = \operatorname{ker}\rho_1 \subset \operatorname{ker}\rho_2.$$
In the light of Lemma 1.9, ρ1ρ2. Similarly, we obtain that $$\operatorname{tr}\upsilon_{1} \subset \operatorname{tr}\upsilon_{2}$$. Hence υ1υ2 (this follows from the proof of Result 1.7, see [1]). Thus υ1ρ1υ2ρ2. This implies the thesis of the theorem. □

## Authors and Affiliations

1. 1.Institute of Mathematics and Computer ScienceWroclaw University of TechnologyWroclawPoland