Rectangular group congruences on a semigroup

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.

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].

Recall from [9] that a nonempty subset 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].

Finally, we say that 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

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

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:

• (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).

$\mathrm{(i)} \Longrightarrow\mathrm{(iii)}$. Let S=M×G, where M is a rectangular band and G is a group (with an identity 1). Define the mapping ⁻¹: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

$$x^{- 1}yy^{- 1}x = \bigl(ab^2a, 1\bigr) = \bigl(a^2, 1\bigr) = \bigl(a, g^{- 1}\bigr) (a, g) = x^{- 1}x. $$

$\mathrm{(iii)} \Longrightarrow\mathrm{(iv)}$. Let e∈E _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,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 $\operatorname{ker}\rho_{1} \subset \operatorname{ker}\rho_{2}$.

Let 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.

(ii). Indeed, 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

$$\bigl((nbn)\rho\bigr)\phi= \bigl((nbn)\upsilon, (nbn)\rho_{N}\bigr) = (n\upsilon, b\rho_{N}) = (a\upsilon, b\rho_{N}). $$

(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:

$$(\upsilon, \rho_{N})\phi = \upsilon\cap\rho_{N} $$

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 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

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 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. □