Semigroup Forum

, Volume 87, Issue 1, pp 120–128

Rectangular group congruences on a semigroup

Authors

    • Institute of Mathematics and Computer ScienceWroclaw University of Technology

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 axE S , where E S (or briefly E) is the set of idempotents of S (more generally, if AS, 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 aS. Hence if S is E-inversive, then for every aS there is xS such that ax,xaE 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 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 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:
  1. (i)

    S is a rectangular band;

     
  2. (ii)

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

     
  3. (iii)

    a,bS [aba=a];

     
  4. (iv)

    a,b,cS [a 2=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 [aAa 2A];

     
  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 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
$$\rho_{\mathcal{X}} = \bigl\{(a, b) \in S \times S : \forall A \in\mathcal {X}\ [a, b \in A\ \mbox{or}\ a, b \notin A]\bigr\}. $$

Result 1.2

(Theorem 5 [9])

The 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:
  1. (i)

    ρ is a matrix congruence on S;

     
  2. (ii)

    every ρ-class of S is a quasi dense subsemigroup of S;

     
  3. (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 ıI, λΛ, such that every S ıλ has an identity element e ıλ and the set M (say) of elements e ıλ (ıI, λΛ) forms a subsemigroup of S. Then M is a rectangular band. Further, S ıλ S ȷμ for all ı,ȷI,λ,μΛ 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 ıI,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 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:
  1. (i)

    S is a rectangular group;

     
  2. (ii)

    S is completely simple and orthodox;

     
  3. (iii)

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

     
  4. (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) eE S .

Proof

\(\mathrm{(ii)} \Longrightarrow\mathrm{(i)}\). If S is completely simple, then S is a matrix of groups \(\mathcal{H}_{e}\) (eE 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,fE S , that is, e=efe for all e,fE S (since efeE S ). Hence E S is a rectangular band. Thus \(S \cong E_{S} \times \mathcal{H}_{e}\) for some (all) eE S (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 −1 x and xx −1 x=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 eE S . Since ee −1=e −1 e,(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 −1 ff −1 e=e −1 e, i.e., efe=e for all e,fE 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,bS [abAbaA] and ∀sS ∃ x,yS [sx,ysA]. Also, define the closure operator ω 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,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 1ρ 2 if 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:
https://static-content.springer.com/image/art%3A10.1007%2Fs00233-012-9426-y/MediaObjects/233_2012_9426_Equd_HTML.gif

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 NS.)

Theorem 2.1

Let S be a semigroup. The following conditions are equivalent:
  1. (i)

    there exists a proper rectangular group congruence on S;

     
  2. (ii)

    S is a disjoint union of two or more quasi dense subsemigroups of S and contains a proper normal subsemigroup of S;

     
  3. (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 ME M×G ={(m,1):mM}. Further, for all mM, define Q m 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 ME 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 :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 E M×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 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 1ρ 2 is a proper rectangular group congruence on a semigroup S. Indeed, let aS. Since S/ρ 2 is regular, then (axa,a)∈ρ 2 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/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 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,bS. Then (aba,a)∈ρ 2 for all a,bS. 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 ρρ 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 (mM) be the preimage of {mG by the canonical epimorphism ρ from S onto S/ρ, and put N={sS:E M×G }. Moreover, denote by υ the matrix congruence on S, induced by the partition {Q m :mM} of S (see the proof of “ \(\mathrm{(i)} \Longrightarrow\mathrm{(ii)}\) ” in Theorem 2.1). Then:
  1. (i)

    ρ=υρ N ;

     
  2. (ii)

    S/ρS/υ×S/ρ N .

     
If in addition S is E-inversive, then:
  1. (iii)

    mM [NQ m Q m ];

     
  2. (iv)

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

     

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 =(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)∈ρ N (see Remark 1). Also, =(m,g)=∈{mG. Hence a,bQ m , so (a,b)∈υ. Thus (a,b)∈υρ N . Consequently, ρυρ N . Conversely, let (a,b)∈υρ N . Then ,∈{mG (=(m,g 1), =(m,g 2)), xa,xbN for some mM and xQ n , where nM (say =(n,g)), and so (xa)ρ=(nm,gg 1). On the other hand, (xa)ρM×{1 G }. Hence g 1=g −1. We may equally well show that g 2=g −1. Thus (a,b)∈ρ. Consequently, ρ=υρ N , as exactly required.

(ii). Indeed, ρ=υρ N by (i). Define the mapping ϕ:S/ρS/υ×S/ρ N by ()ϕ=(, N ) (aS). Clearly, ϕ is a monomorphism. We show that ϕ is surjective. Let (, N )∈S/υ×S/ρ N . Then aQ m , where mM. Take any element nNQ m . 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 N m =NQ 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 mM. Define the map \(\phi: Q_{m}/\rho_{N_{m}}\to S/\rho _{N}\) by \((a\rho_{N_{m}})\phi= a\rho_{N}\) (aQ m ). Clearly, ϕ is a well-defined homomorphism. Furthermore, if aS and nN m N, then nanQ 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, bxN for some xS. Hence for every nN m N, anx, bnxN. Thus n(anx)nN∩(Q m NQ m )⊆NQ m =N m and similarly: n(bnx)nN m . Since na, nb, nxnQ 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 ψσ is the least rectangular group congruence on S. In particular, in any E-inversive semigroup, ψσ 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|(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

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 υρ N , where υ is a matrix congruence on S, NS, 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:
$$(\upsilon, \rho_{N})\phi = \upsilon\cap\rho_{N} $$
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 1υ 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 1υ 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

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

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