Semigroup Forum

, Volume 86, Issue 2, pp 431–450

Congruences and group congruences on a semigroup

Open AccessRESEARCH ARTICLE

DOI: 10.1007/s00233-012-9425-z

Cite this article as:
Gigoń, R.S. Semigroup Forum (2013) 86: 431. doi:10.1007/s00233-012-9425-z

Abstract

We show that there is an inclusion-preserving bijection between the set of all normal subsemigroups of a semigroup S and the set of all group congruences on S. We describe also group congruences on E-inversive (E-)semigroups. In particular, we generalize the result of Meakin (J. Aust. Math. Soc. 13:259–266, 1972) concerning the description of the least group congruence on an orthodox semigroup, the result of Howie (Proc. Edinb. Math. Soc. 14:71–79, 1964) concerning the description of ρσ in an inverse semigroup S, where ρ is a congruence and σ is the least group congruence on S, some results of Jones (Semigroup Forum 30:1–16, 1984) and some results contained in the book of Petrich (Inverse Semigroups, 1984). Also, one of the main aims of this paper is to study of group congruences on E-unitary semigroups. In particular, we prove that in any E-inversive semigroup, \(\mathcal{H}\cap\sigma\subseteq\kappa\), where κ is the least E-unitary congruence. This result is equivalent to the statement that in an arbitrary E-unitary E-inversive semigroup S, \(\mathcal{H}\cap\sigma= 1_{S}\).

Keywords

Group congruenceE-inversive semigroupE-semigroupIdempotent-surjective semigroupEventually regular semigroupIdempotent pure congruenceIdempotent-separating congruenceE-unitary congruence

1 Introduction and preliminaries

Let S be an inverse semigroup with semilattice of idempotents E. Define an inverse subsemigroup N of S to be normal if it is full (i.e., EN), closed (i.e., =N, where ω:2S→2S is a closure operator given by ={sS:∃ aA [asA]} for all AS), and self-conjugate (i.e., s−1NsN for every sS). It follows from [11] (see p. 181) that there exists an inclusion-preserving bijection between the set of normal subsemigroups of S and the set of group congruences on S. In fact, the relation ρN={(a,b)∈S×S:ab−1N} is a group congruence on S and \(\operatorname{ker}\rho_{N} = N\). These results were generalized in [9] and [16]. It is easy to see that (a,b)∈ρN if and only if ax,bxN for some xS.

The main purpose of the next section is a description of group congruences on a semigroup S in the terms of some special subsemigroups of S. Our description is simpler than that of Dubreil (see 10.2 [1]) and a little more general than the description of Gomes [6] (nota bene our proof is simpler). We apply this description to determine group congruences (in particular, the least group congruence) on some special classes of semigroups; namely: E-inversive (E-)semigroups (in particular, idempotent-surjective (E-)semigroups), eventually regular semigroups.

We divide this paper into seven sections. In Sect. 2 we describe group congruences on an arbitrary semigroup S in the terms of normal subsemigroups of S (see below for the definition). In Sect. 3 we investigate group congruences on E-inversive semigroups. In particular, we show that the least group congruence on an E-inversive semigroup exists (in general, this is false: see Example 1.2). In Sect. 4 and 5 we study group congruences on E-inversive E-semigroups and E-unitary E-inversive semigroups, respectively. Further, in Sect. 6 we use the results of Sect. 2 for an easy description of all group congruences on eventually regular semigroups (in terms of full, closed and self-conjugate subsemigroups) and we give some remarks on group congruences on inverse semigroups. Finally, in Sect. 7, some remarks on the hypercore of a semigroup are given (see [8]).

Let S be a semigroup. Denote by Reg(S) the set of regular elements of S, that is, Reg(S)={aS:aaSa} and by V(a) the set of inverses of aS, i.e., the set {xS:a=axa,x=xax}. Note that if aS is regular, say a=axa for some xS, then xaxV(a). Also, S is called regular if V(a)≠∅ for every aS. Further, S is said to be eventually regular if every element a of S has a regular power. In such a case, by r(a) we shall mean the regular index of a, i.e., the least positive integer n for which anReg(S).

Let S be a semigroup, 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 (see [19, 20]). Clearly, eventually regular semigroups are E-inversive. We remark that the class of eventually regular semigroups is very wide and contains the class of regular, group-bound (in particular, periodic, finite) semigroups. In [7] Hall observed that the set Reg(S) of a semigroup S with ES≠∅ forms a regular subsemigroup of S if and only if the product of any two idempotents of S is regular. In that case, S is said to be an R-semigroup. Also, we say that S is an E-semigroup if ES is a subsemigroup of S. Evidently, every E-semigroup is an R-semigroup. Finally, [eventually] regular E-semigroups are called [eventually] orthodox.

A generalization of the concept of eventually regular will also prove convenient. Define a semigroup S to be idempotent-surjective if whenever ρ is a congruence on S and is an idempotent of S/ρ, then contains some idempotent of S. It is well known that eventually regular semigroups are idempotent-surjective [2]. Further, we have the following known result [10] (we include a simple proof for completeness).

Result 1.1

Every idempotent-surjective semigroupSisE-inversive.

Proof

Let aS. From the definition of a Rees congruence on S follows that the ideal SaS has at least one idempotent, that is, xay=eES, where x,yS. Hence exaye=e. Thus yex=(yex)a(yex), so yexW(a), as required. □

A subset A of S is said to be (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 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 [6].

By the kernel\(\operatorname{ker}\rho\) of a congruence ρ on a semigroup S we shall mean the set {xS:(x,x2)∈ρ}. Finally, denote by \(\mathcal{C}(S)\) the complete lattice of all congruences on a semigroup S.

Example 1.2

Consider the semigroup of positive integers (ℕ,+) (with respect to addition). It is well known that every group congruence on ℕ is of the following form: ρn={(k,l)∈ℕ×ℕ:n|(kl)} (n>0). Note that E=∅, so a semigroup without idempotents possesses group congruences but ℕ has not least group congruence. Also, \(\operatorname{ker}\rho_{n} = n\rho_{n} = \{n, 2n, 3n, \ldots\}\).

2 Group congruences—general case

Let S be a semigroup, \(\rho \in \mathcal{C}(S)\). We say that ρ is a group congruence if S/ρ is a group. Denote by \(\mathcal{GC}(S)\) the set of group congruences on S. Clearly, if \(\rho \in \mathcal{GC}(S)\), then \(\operatorname{ker}\rho\) is the identity of the group S/ρ. Finally, by \(\mathcal{N}(S)\) we shall mean the set of all normal subsemigroups of S.

The following two lemmas are almost evident and we omit their easy proofs.

Lemma 2.1

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

Lemma 2.2

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:
https://static-content.springer.com/image/art%3A10.1007%2Fs00233-012-9425-z/MediaObjects/233_2012_9425_Equa_HTML.gif

Lemma 2.3

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

Proof

Let (a,b)∈ρ2,B. Then ax=yb for some x,yB. Also, asB for some sS, since B is dense and so saB, since B is reflexive. Hence asyB and so (sy)aB. It follows that (sy)b=s(yb)=s(ax)=(sa)xB. Thus (sy)a, (sy)bB. We have just shown that ρ2,Bρ3,B.

Conversely, if xa,xbB for some xS, then ax,bxB (since B is reflexive), so a(xb)=(ax)b, where ax,xbB. Hence (a,b)∈ρ2,B. Thus ρ2,B=ρ3,B.

Dually, ρ1,B=ρ4,B. Since B is reflexive, then ρ1,B=ρ3,B. □

If B is a dense, reflexive subsemigroup of S, then we denote the above four relations by ρB. We have the following theorem.

Theorem 2.4

LetBbe a dense and reflexive 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.

Proof

Let aS. Since B is dense, then there exists xS such that xaB. Hence ρB is reflexive. Obviously, ρB is symmetric. Also, since B is a semigroup, then ρB is transitive. Consequently, ρB is an equivalence relation on S. Moreover, ρB is a left congruence on S. Indeed, let (a,b)∈ρB,cS. Then ax,bxB and zcB for some x,zS, so zcax,zcbxB. It follows that (ca)(xz),(cb)(xz)∈B, since B is reflexive. Therefore (ca,cb)∈ρB. By symmetry, ρB is a right congruence on S. Finally, S/ρB is a group. Indeed, let aS,bB and ax,xaB for some xS. Then baxB. Hence xa,x(ba)∈B, so (ba,a)∈ρB. Since B is dense, then S/ρB is a group, as required.

Since b(bb)=(bb)b for every bB, then \(B \subset \operatorname{ker}\rho_{B}\). Also, \(B\omega = \operatorname{ker}\rho_{B}\). Indeed, let \(s \in \operatorname{ker}\rho_{B}\). Then (s,b)∈ρB for some bB. Hence b1s=bb2 for some b1,b2B. Thus s, so \(\operatorname{ker}\rho_{B} \subset B\omega\). Conversely, let s. Then bsB for some bB. Since bbB, then (s,b)∈ρB, so \(s \in \operatorname{ker}\rho_{B}\). Thus \(B\omega \subset \operatorname{ker}\rho_{B}\), as exactly required. Finally, if B is normal, then B=. Hence \(B = \operatorname{ker}\rho_{B}\).

Conversely, let ρ be a group congruence on S. By Lemma 2.1, \(\operatorname{ker}\rho \lhd S\). Put \(\operatorname{ker}\rho = N\). Then by Lemma 2.2, ρ=ρN, since \(N = \operatorname{ker}\rho_{N} = \operatorname{ker}\rho\). It is now easy to see that the map \(\phi : \mathcal{N}(S) \to \mathcal{GC}(S)\), where =ρN for every \(N \in \mathcal{N}(S)\), is an inclusion-preserving bijection between the set of all normal subsemigroups of S and the set of all group congruences on S (with the inverse \(\phi^{- 1} : \mathcal{GC}(S) \to \mathcal{N}(S)\), where \(\rho \phi^{- 1} = \operatorname{ker}\rho\) for all \(\rho \in \mathcal{GC}(S)\)). Note that ϕ−1 is an inclusion-preserving mapping, too. □

Since the first part of Theorem 2.4 is true for an arbitrary dense and reflexive subsemigroup of S, then we get the following corollary.

Corollary 2.5

LetBbe a dense and reflexive subsemigroup ofS. ThenS.

Example 2.6

Let S={a,b,c,e,f} be the semigroup with the multiplication table given below:
https://static-content.springer.com/image/art%3A10.1007%2Fs00233-012-9425-z/MediaObjects/233_2012_9425_Figa_HTML.gif
It is easy to see that E is a dense and reflexive subsemigroup of S but E is not closed, since eaE and aE. Also, N={a,e,f} is normal. Indeed, the group congruence ρE has two ρE-classes: N and {b,c}, since ae,ee,bb,bcE and (e,b)∉ρE. Note also that \(E \subset \operatorname{ker}\rho_{N} = N, E \not= N\) and ρE=ρN. It follows that there is no a one-to-one correspondence between the set of all seminormal subsemigroups of S and the set of all group congruences on S.

Remark 1

Obviously, every subgroup of a group is full and unitary but not every subgroup of a group is reflexive (for example: each two element subgroup of the group of all permutations of the six-element set X is not reflexive). It is well known that a subgroup H of a group G is normal if and only if the relation ρH is a congruence on G. We have a corresponding result:

LetAbe a closed subsemigroup of a semigroupS. ThenAis normal if and only if\(\rho_{A} \in \mathcal{GC}(S)\).

Indeed, let \(\rho_{A} \in \mathcal{GC}(S)\). From A= and the second paragraph of the proof of Theorem 2.4 we obtain that \(A = \operatorname{ker}\rho_{A}\). Thus AS (Lemma 2.1). The converse of the result follows from Theorem 2.4.

The set of all group congruences on a semigroup S (in general) does not form a lattice. Indeed, let (ℝ,+) be the semigroup of real positive numbers with respect to addition. Put M=ℕ and N={x,2x,3x,…}, where x∈ℝ∖ℚ. Then M,NS but MN=∅.

We generalize now the results of Howie [12], LaTorre [16] and Hanumantha Rao and Lakshmi [9].

Theorem 2.7

LetBbe a seminormal subsemigroup of a semigroup\(S, \rho \in \mathcal{C}(S)\). Then:
  1. (i)

    ρρB=ρBρρB;

     
  2. (ii)

    \(\rho \lor \rho_{B} \in \mathcal{GC}(S)\);

     
  3. (iii)

    (x,y)∈ρρBif and only if (ax,yb)∈ρfor somea,bB.

     

Proof

(i). Since ρ,ρBρρB, ρBρρBρρB. Also, ρBρρB is a reflexive, symmetric and compatible relation on S. We show that ρBρρB is transitive. Then ρρB=ρBρρB. Indeed, let (r,s),(s,t)∈ρBρρB. Then (a) (w,s),(s,x)∈ρB; (b) (y,w),(x,z)∈ρ; (c) (r,y),(z,t)∈ρB for some w,x,y,zS. From (a) we obtain (w,x)∈ρB, so aw=xb for some a,bB. From (b) follows that (aw,ay),(xb,zb)∈ρ. Hence (ay,zb)∈ρ, since aw=xb. Finally, by (c), (r,ay),(zb,t)∈ρB, since \(B \subset \operatorname{ker}\rho_{B}\), so (r,ay)∈ρB,(ay,zb)∈ρ,(zb,t)∈ρB. Thus (r,t)∈ρBρρB, as required.

(ii). This is evident.

(iii). Let (x,y)∈ρρB. Then (x,r)∈ρB,(r,s)∈ρ and (s,y)∈ρB for some r,sS. Hence ax=rb,cs=yd for some elements a,b,c,d of B. Therefore (ca)x=c(ax)=c(rb)=(crb)ρ(csb)=(cs)b=(yd)b=y(db), where ca,dbB. Conversely, let (ax,yb)∈ρ for some a,bB. Since (x,ax),(yb,y)∈ρB, then (x,y)∈ρBρρB=ρρB (by (i)). □

Let A be a nonempty subset of a semigroup \(S, \rho \in \mathcal{C}(S)\). Put
$$A\rho = \bigl\{s \in S: \exists\, a \in A~[(s, a) \in \rho]\bigr\}. $$

Corollary 2.8

LetBbe a seminormal subsemigroup of a semigroup\(S, \rho \in \mathcal{C}(S)\). Then\(\operatorname{ker}(\rho \lor \rho_{B}) = (B\rho)\omega\). In particular, ()ωS.

Proof

Let \(x \in \operatorname{ker}(\rho \lor \rho_{B})\). Then there exists bB such that (x,b)∈ρρB, since \(B \subset \operatorname{ker}(\rho \lor \rho_{B})\). Hence (ax,bc)∈ρ for some a,cB (by Theorem 1.6(iii)). Thus ax. It follows that x∈()ω. Conversely, if x∈()ω, then ax for some a, so (ax,b),(a,c)∈ρ for some b,cB. It follows that (cx,b)∈ρ. Hence ((cc)x,cb)∈ρ. Thus (x,c)∈ρρB. Consequently, \(x \in\operatorname{ker}(\rho \lor \rho_{B})\). □

Also, by Theorem 1.6(i) and Proposition 2.3(ii) in [15] we obtain the following (see Corollary 3.2 [15]) corollary.

Corollary 2.9

Every group congruence on a semigroupSis dually right modular element of\(\mathcal{C}(S)\).

Corollary 2.10

LetBbe a seminormal subsemigroup of a semigroup\(S, \rho \in \mathcal{C}(S)\). ThenρρB=S×Sif and only if ()ω=S.

Let B be a seminormal subsemigroup of a semigroup \(S, \rho_{1}, \rho_{2} \in \mathcal{C}(S)\). Suppose that (x,y)∈(ρ1ρB)∩(ρ2ρB). Then (ax)ρ2(yb), where a,bB. Moreover, ax(ρ1ρB)x,x(ρ1ρB)y,y(ρ1ρB)yb, so ax(ρ1ρB)yb. Thus (cax,ybd)∈ρ1, where c,dB. It follows that (caxd,cybd)∈ρ1. Moreover, (caxd,cybd)∈ρ2. Hence (xd,cy)∈(ρ1ρ2)∨ρB. Thus (x,y)∈(ρ1ρ2)∨ρB, since (ρ1ρ2)∨ρB is a group congruence on S and \(c, d \in B \subset \operatorname{ker}((\rho_{1} \cap \rho_{2}) \lor \rho_{B})\). We have just shown that (ρ1ρB)∩(ρ2ρB)⊂(ρ1ρ2)∨ρB. The converse inclusion is evident. Thus we may conclude that (ρ1ρB)∩(ρ2ρB)=(ρ1ρ2)∨ρB.

We have the following theorem (see Theorem III.5.6 [21] and Theorem 4 [23]).

Theorem 2.11

LetBbe a seminormal subsemigroup of a semigroupS. Then the mapping\(\phi : \mathcal{C}(S) \to \mathcal{GC}(S)\), where
$$\rho \phi = \rho \lor \rho_B $$
for every\(\rho \in \mathcal{C}(S)\), is a (lattice) homomorphism of\(\mathcal{C}(S)\)onto the (modular) lattice [ρB,S×S] of all group congruences onScontainingρB.

Proof

We have just proved that (ρ1ρ2)ϕ=ρ1ϕρ2ϕ for all \(\rho_{1}, \rho_{2} \in \mathcal{C}(S)\). Clearly, (ρ1ρ2)ϕ=ρ1ϕρ2ϕ for all \(\rho_{1}, \rho_{2} \in \mathcal{C}(S)\) and evidently ϕ is onto [ρB,S×S]. □

We have the following corollary (see Theorem 4.5 [15]).

Corollary 2.12

LetBbe a seminormal subsemigroup of a semigroupS. ThenρBdistributes over meet.

Let S be a semigroup, NS. Put
$$\mathcal{P}(S; N) = \bigl\{A \subseteq S: A^2 \subseteq A, N \subseteq A, A\omega = A\bigr\}. $$
Also, denote S/ρN by S/N. In particular, \(\mathcal{P}(S/N; \{N\})\) is the set of all subgroups of the group S/N. Remark that if \(A \in \mathcal{P}(S; N)\), then A is full and dense.

The proofs of the following two propositions are standard and so we omit the proofs.

Proposition 2.13

LetSbe a semigroup, NS. Then there exists an inclusion-preserving bijectionϕbetween the set\(\mathcal{P}(S; N)\)and the set\(\mathcal{P}(S/N; \{N\})\). Moreover, \(M \in \mathcal{P}(S; N)\)andMSif and only ifS/N.

Proposition 2.14

Letϕbe an epimorphism of a semigroupSonto a group (G,⋅,1). Then:
  1. (i)

    Ker(ϕ)=ϕϕ−1is a group congruence onS;

     
  2. (ii)

    N={1}ϕ−1S;

     
  3. (iii)

    Ker(ϕ)=ρN.

     

Conversely, ifNS, thenNis the kernel of the canonical homomorphism ofSontoS/N.

Example 2.15

We now describe all normal subsemigroups of the bicyclic semigroup S=ℕ0×ℕ0, where (k,l)(m,n)=(kl+max{l,m},nm+max{l,m}). It is known that every (non-identical) homomorphic image of the bicyclic semigroup is a cyclic group. Also, it is almost evident that ES={(0,0),(1,1),(2,2),…}◁S and (k,l)ρE(m,n) if and only if k+n=l+m, so S/ρE≅(ℤ,+). It follows that (iℤ)ϕ−1={(m,n)∈S:(m)i=(n)i} for every i∈ℕ. The conclusion is that every cyclic group is a homomorphic image of the bicyclic semigroup.

We have also the following well known proposition (from group theory).

Proposition 2.16

LetSbe a semigroup; M,NSandMN. Then:
  1. (i)

    MN;

     
  2. (ii)

    N/MS/M;

     
  3. (iii)

    (S/M)/(N/M)≅S/N.

     

Every full and closed subsemigroup A of an E-inversive semigroup S is itself E-inversive. Indeed, let aA. Then axES=EA for some xS, so x=A. Consequently, there is xA such that axEA.

Finally, by way of contrast, we prove in the present section the following proposition which is valid for the class of all E-inversive semigroups.

Proposition 2.17

LetSbe anE-inversive semigroup, NS. Suppose also that a subsemigroupMofSis full and closed. Then:
  1. (i)

    MNM;

     
  2. (ii)

    N◁(MN)ω;

     
  3. (iii)

    M/(MN)≅(MN)ω/N.

     

Proof

(i). It is clear that ESMN, so MN is a full subsemigroup of M. Let a,bM be such that abMN. Then baM and baN (since N is reflexive in S). Hence baMN. Hence MN is reflexive in M. Further, if x∈(MN)ω, then yxMN for some yMN, so xMN (because N and M are closed). Since MN is full and closed, then it is E-inversive, so it is dense in M. Thus MNM.

(ii). We show that (MN)ω is a subsemigroup of S. Let a,b∈(MN)ω. Then m1n1a=m2n2 for some m1,m2M,n1,n2N. Since S is E-inversive, then \(W(m_{1}) \not= \emptyset\). Hence mm1,m1mESM for some mS. Thus mM (since M is closed), (mm1)n1a=(mm2)n2. Therefore (n1a,mm2)∈ρN, since mm1ESN, so (a,m3)∈ρN (m3M). Similarly, (b,m4)∈ρN for some m4M. It follows that (ab,m5)∈ρN, where m5M. Hence n3ab=m5n4 for some n3,n4N. Thus (m5n3)ab=(m5m5)n4. Consequently, ab∈(MN)ω. Furthermore, N⊂(MN)ω. Indeed, let nN. Then n1n=en2 for some eES,n1,n2N. Hence we have (en1)n=en2MN, so n∈(MN)ω. Consequently, N◁(MN)ω (since NS).

The proof of the condition (iii) is standard. □

3 Group congruences on an E-inversive semigroup

Note that if a semigroup S is E-inversive, then every full subsemigroup of S is dense (since ES is dense), so a subsemigroup A of S is normal if and only if A is full, reflexive and closed. It follows that S has a least normal subsemigroup U. Thus the least group congruence on an arbitrary E-inversive semigroup exists. Denote it by σ or σS. Then σ=ρU and \(\operatorname{ker}\sigma = U\) (Theorem 2.4).

Firstly, we have the following proposition.

Proposition 3.1

LetSbe anE-inversive semigroup. Then\(\mathcal{GC}(S) = [\sigma, S \times S]\). Thus\(\mathcal{GC}(S)\)is a complete sublattice of\(\mathcal{C}(S)\).

Also, ρMρN=ρMρN=ρNρMfor allM,NS. Hence the lattice
$$\bigl(\mathcal{GC}(S), \subseteq, \cap, \circ\bigr) $$
is modular.

Proof

The first part of the above proposition is clear. We show its second part. Let a(ρMρN)b. Then (a,c)∈ρM,(c,b)∈ρN, where cS. Take any xW(c). Then xc,cxES,(cxa)ρN(bxa),(bxa)ρM(bxc), so (a,bxa)∈ρN,(bxa,b)∈ρM. Hence (a,b)∈ρNρM. Therefore ρMρNρNρM. We may equally well show the opposite inclusion. Consequently, ρMρN=ρMρN=ρNρM. In the light of Proposition I.8.5 [11], the lattice \((\mathcal{GC}(S), \subseteq, \cap, \circ)\) is modular. □

Let M,N be normal subsemigroups of a semigroup S. From Proposition 3.1 and Corollary 2.8 we obtain that \(\operatorname{ker}(\rho_{M} \rho_{N}) = \operatorname{ker}(\rho_{N} \rho_{M}) = (M\rho_{N})\omega = (N\rho_{M})\omega\). In fact, if S is E-inversive, then \(\operatorname{ker}(\rho_{M} \rho_{N}) = \operatorname{ker}(\rho_{N} \rho_{M}) = M\rho_{N} = N\rho_{M}\). Indeed, let \(x \in \operatorname{ker}(\rho_{M} \rho_{N})\). Then (x,e)∈ρMρN for some eES. Hence (x,n)∈ρM, (n,e)∈ρN, where nS (in fact, \(n \in \operatorname{ker}\rho_{N} = N\)). Thus xM. Conversely, if xM, then (x,n)∈ρM for some nN. Hence (x,n)∈ρM,(n,e)∈ρN, where eES. Thus (x,e)∈ρMρN, that is, \(x \in \operatorname{ker}(\rho_{M} \rho_{N})\), so \(\operatorname{ker}(\rho_{M} \rho_{N}) = N\rho_{M}\). Similarly, \(\operatorname{ker}(\rho_{N} \rho_{M}) = M\rho_{N}\). This implies the required equalities. Also, \(\operatorname{ker}(\rho_{M} \rho_{N}) = (MN)\omega\). Indeed, let xN. Then n1x=mn2 for some n1,n2N,mM. Hence (mn1)xMN. Thus x∈(MN)ω. We have proved that \(\operatorname{ker}(\rho_{M} \rho_{N}) \subset (MN)\omega\). Conversely, let x∈(MN)ω. Then m1n1x=m2n2 for some m1,m2M,n1,n2N. Since S is E-inversive, then mm1=eESM for some mS. It follows that mM (since M is closed), so en1x=mm2n2. Hence (x,mm2)∈ρN. Thus \(x \in M\rho_{N} = \operatorname{ker}(\rho_{M} \rho_{N})\), so \((MN)\omega \subset \operatorname{ker}(\rho_{M} \rho_{N})\), as exactly required.

In fact, we have just shown that in an arbitrary E-inversive semigroup S, ρ(MN)ω=ρMρN=ρNρM=ρ(NM)ω for all M,NS. Moreover, notice that \(\operatorname{ker}(\rho_{M} \cap \rho_{N}) = \operatorname{ker}\rho_{M} \cap \operatorname{ker}\rho_{N} = M \cap N\) (M,NS), so ρMρN=ρMN for M,NS. Consequently, the lattice \((\mathcal{N}(S), \subseteq, \cap, \lor)\), where MN=(MN)ω for all M,NS, is isomorphic to the lattice \((\mathcal{GC}(S), \subseteq, \cap, \circ)\) (by the inclusion-preserving bijection ϕ, see the proof of Theorem 2.4). Note also that the lattice \((\mathcal{N}(S), \subseteq, \cap, \lor)\) is complete (since it has the greatest element S and the intersection of any nonempty family of normal subsemigroups of S is a normal subsemigroup of S).

For terminology and elementary facts about lattices the reader is referred to the book [21] (Sect. I.2). The following result will be useful (see Exercise I.2.15(iii) in [21]).

Lemma 3.2

Every lattice isomorphism of complete lattices is a complete lattice isomorphism.

From the above consideration we obtain the following theorem.

Theorem 3.3

LetSbe anE-inversive semigroup. Then there exists a (lattice) isomorphismϕbetween the lattice\((\mathcal{N}(S), \subseteq, \cap, \lor)\), whereMN=(MN)ωfor allM,NS, and the lattice\((\mathcal{GC}(S), \subseteq, \cap, \circ)\). In fact, ϕis defined by=ρNfor every\(N \in \mathcal{N}(S)\). Moreover, ϕis a complete lattice isomorphism.

Finally, we have the following proposition.

Proposition 3.4

LetSbe anE-inversive semigroup, NS. Then (a,b)∈ρNif and only ifabNfor some (all) bW(b).

Proof

(⟹). Let na=bm, where n,mN, and bW(b). Then nab=bmb. Since bbmN and N is reflexive, then nabN. Hence ab=N.

(⟸). Let ab=nN for some bW(b). Then a(bb)=nb, so (a,b)∈ρN (by Lemma 2.3). □

4 Group congruences on an E-semigroup

First, we “generalize” some results from orthodox semigroups to E-semigroups (see Theorem VI.1.1 [11]).

Proposition 4.1

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

    Sis anE-semigroup;

     
  2. (ii)

    a,bS [W(b)W(a)⊆W(ab)].

     
Moreover, the condition (i) implies the following condition:
  1. (iii)

    eES [W(e)⊆ES].

     

If in additionSis anR-semigroup, then the conditions (i)(iii) are equivalent.

Proof

The proof is closely similar to the proof of Theorem VI.1.1 [11]. □

Corollary 4.2

LetSbe anE-semigroup. Then:
  1. (i)

    eES [W(e),V(e)⊆ES];

     
  2. (ii)

    aS,aW(a),eES [aea,aeaES];

     
  3. (iii)

    aS,aW(a),e,fES [ea,ae,eafW(a)].

     

Proof

(i). This follows from Proposition 4.1.

(ii). This follows from the proof of Proposition VI.1.4 [11].

(iii). Let aS,aW(a),e,fES. Since eW(e) and fW(f), then eaW(e)W(a)⊆W(ae). Hence ea=eaaeea=(ea)a(ea). Therefore eaW(a). Similarly, aeW(a). Finally, eafW(e)W(a)W(f)⊆W(fae) and so eaf=eaffaeeaf=(eaf)a(eaf). Hence eafW(a). □

Proposition 4.3

LetSbe anE-inversiveE-semigroup. Then
$$\rho_{1, E} = \rho_{2, E} = \rho_{3, E} = \rho_{4, E}. $$

Proof

Let (a,b)∈ρ2,E and aW(a). Then ae=fb for some e,fE. Moreover, afW(a) (Corollary 4.2(iii)), so (af)a,a(af)∈E. Further, afb=aaeE. We have just shown that xa,ax,xbE for some xS. Thus ρ2,Eρ4,E.

On the other hand, if xa,xbE for some xS, say xa=e,xb=f, then (efx)a(efx)=ef(xa)efx=efx, so efxW(a). Also, fxbfx=f(xb)fx=fx, i.e., fxW(b). Hence efxW(b) (Corollary 4.2(iii)). Thus \(W(a) \cap W(b) \not= \emptyset\). It follows that ay,by,ya,ybE for some yS. Dually, if ax,bxE for some xS, then ay,by,ya,ybE for some yS. Thus ρ4,E=ρ1,E. In fact, we get \(\rho_{4, E} = \rho_{1, E} = \{(a, b) \in S \times S: W(a) \cap W(b) \not= \emptyset \}\). Finally, if xW(a)∩W(b), then a(xb)=(ax)b and xb,axE. Thus ρ2,E=ρ4,E=ρ1,E. We may equally well show that ρ3,E=ρ4,E=ρ1,E. Consequently, ρ1,E=ρ2,E=ρ3,E=ρ4,E. □

Lemma 4.4

LetSbe anE-inversiveE-semigroup. Then:
  1. (i)

    aS ∃ e,fES [ea,afReg(S)];

     
  2. (ii)

    \(\forall a \in S~\exists\, r \in \mathit{Reg}(S)~[W(a) \cap W(r) \not= \emptyset]\).

     

Proof

Let aS,xW(a). Then (ax)a,a(xa)∈Reg(S), where ax,xaES, so (i) holds. Also, r=axaReg(S) and xrx=x. Thus xW(a)∩W(r). □

Denote the above four relations from Proposition 4.3 by ρE. Recall that from the proof of Proposition 4.3 follows that \(\rho_{E} = \{(a, b) \in S \times S: W(a) \cap W(b) \not= \emptyset \}\).

Theorem 4.5

In anyE-inversiveE-semigroup, σ=ρE. Moreover, \(\operatorname{ker}\sigma = E_{S}\omega\). ThusESωS.

Proof

It is clear that ρE is an equivalence relation on S. Let (a,b)∈ρE,cS. Then xW(a)∩W(b). Take any yW(c). In the light of Proposition 4.1,
$$xy \in W(a)W(c) \cap W(b)W(c) \subseteq W(ca) \cap W(cb). $$
Hence (ca,cb)∈ρE. Thus ρE is a left congruence on S. We may equally well show that ρE is a right congruence on S. Also, if e,fES, then ee,efES. Consequently, (e,f)∈ρE for all e,fES. Lemma 4.4(ii) says that every ρE-class of S contains a regular element. This implies that S/ρE is a group.
Furthermore,
$$ x \in \operatorname{ker}\rho_E\ \Leftrightarrow\ \exists\, e \in E_S\ [(x, e) \in \rho_E]\ \Leftrightarrow\ \exists\, e, f, g \in E_S\ [fx = eg]\ \Leftrightarrow\ x \in E_S\omega, $$
so \(\operatorname{ker}\sigma = E_{S}\omega\). Thus ESωS (Theorem 2.4). Finally, ρEρN for ever NS. Indeed, ESN. Hence ESω=N. Thus \(\rho_{E} = \rho_{E_{S}\omega} \subseteq \rho_{N}\) (Theorem 2.4). Consequently, σ=ρE. □

Corollary 4.6

The least group congruenceσon anE-inversiveE-semigroup is given by
$$\sigma = \bigl\{(a, b) \in S \times S: \exists\, e \in E_S~[eae = ebe]\bigr\}. $$

Remark 2

Note that the condition “∃ eES [eae=ebe]” from the above corollary is equivalent to the apparently weaker condition “∃ sS [sas=sbs]”.

From Result 1.1 and Theorem 4.5 we obtain the following theorem.

Theorem 4.7

In any idempotent-surjectiveE-semigroup, σ=ρE.

Let S be a semigroup. A congruence ρ on S is called idempotent pure if ES for every eES. Note that if S is idempotent-surjective, then ρ is idempotent pure if and only if \(\operatorname{ker}\rho = E_{S}\). Let \(\mathcal{E}\) be an equivalence relation on S induced by the partition: {ES,SES}. Then \(\mathcal{E}^{\flat}\) (defined in [13], see p. 27) is the greatest idempotent pure congruence on S. Put \(\tau = \mathcal{E}^{\flat}\). Then (see [13], p. 28)
$$\tau = \bigl\{(a, b)\in S \times S: \forall x, y\in S^1~[xay\in E_S \iff xby\in E_S]\bigr\}. $$
Finally, if S is E-inversive, then τσ. Indeed, let (a,b)∈τ and bW(b). Then bbES,(ab,bb)∈τ. Hence \(ab^{*} \in E_{S} \subseteq \operatorname{ker}\sigma\). In the light of Proposition 3.4, (a,b)∈σ, as exactly required. In the following corollary we give an alternative proof of this fact.

Corollary 4.8

Ifρis a congruence on an idempotent-surjectiveE-semigroupS, then\(\operatorname{ker}(\rho \lor \sigma) = (\operatorname{ker}\rho)\omega\). In particular, τσ.

Proof

By Corollary 2.8, \(\operatorname{ker}(\rho \lor \sigma) = (E_{S}\rho)\omega = (\operatorname{ker}\rho)\omega\). In particular,
$$\operatorname{ker}(\tau \lor \sigma) = E_S\omega \subseteq \operatorname{ker}\sigma. $$
Hence τσ=σ. Thus τσ. □
Let ρ be a congruence on a semigroup S. By the trace\(\operatorname{tr}\rho\) of ρ we shall mean the restriction of ρ to ES. Also, we say that ρ is idempotent-separating if \(\operatorname{tr}\rho = 1_{E_{S}}\). Edwards in [3] shows that if S is an eventually regular semigroup, then the relation \(\theta = \{(\rho_{1}, \rho_{2}) \in \mathcal{C}(S) \times \mathcal{C}(S): \operatorname{tr}\rho_{1} = \operatorname{tr}\rho_{2}\}\) is a complete congruence on \(\mathcal{C}(S)\) and proves that every θ-class ρθ is a complete sublattice of \(\mathcal{C}(S)\) with the maximum element
$$\mu (\rho) = \bigl\{(a, b) \in S \times S: (a\rho, b\rho) \in \mu_{S/\rho}\bigr\} $$
and the minimum element 1(ρ). Edwards generalizes some of these results for the class of all idempotent-surjective semigroups [4]. In fact, if S is an arbitrary idempotent-surjective semigroup, then every θ-class ρθ is the interval [1(ρ),μ(ρ)], where μ is the maximum idempotent-separating congruence on S (see [4] for more details).

It is easily seen that the class of idempotent-surjective semigroups is closed under homomorphic images [10]. Using the obvious terminology we show next that every homomorphism of idempotent-surjective E-semigroups can be factored into a homomorphism preserving the maximal group homomorphic images and an idempotent-separating homomorphism. Firstly, we have need the following lemma.

Lemma 4.9

Letρbe a congruence on an idempotent-surjectiveE-semigroupS, a,bS. Then (,)∈σinS/ρimplies (a,b)∈σif and only ifρσ.

Proof

The proof is closely similar to the proof of Lemma III.5.9 [21]. □

Let S be an idempotent-surjective E-semigroup, \(\rho \in \mathcal{C}(S)\). Clearly, (a,b)∈σ implies (,)∈σ. In the light of Lemma 4.9, if ρσ, then (a,b)∈σ if and only if (,)∈σ. Hence S/σ≅(S/ρ)/σ, that is, S and S/ρ have isomorphic maximal group homomorphic images. In that case, we may say that ρpreserves the maximal group homomorphic images. Since for any congruence ρ on S we have 1(ρ)⊆ρ, then we obtain the following factorization:
$$S \to S/1(\rho) \to S/\rho \cong \bigl(S/1(\rho)\bigr)\big/\bigl(\rho/1(\rho)\bigr). $$

The following proposition generalizes Proposition III.5.10 [21].

Proposition 4.10

Every homomorphism of idempotent-surjectiveE-semigroups can be factored into a homomorphism preserving the maximal group homomorphic images and an idempotent-separating homomorphism.

Proof

Let ρ be any congruence on an idempotent-surjective E-semigroup S. Since ρS×S, then 1(ρ)⊆1(S×S). Clearly, σ∈[1(S×S),S×S] and so 1(ρ)⊆σ. It follows that the canonical epimorphism of S onto S/1(ρ) preserves the maximal group homomorphic images. Finally, an epimorphism ϕ:S/1(ρ)→S/ρ (defined by the obvious way) is idempotent-separating, since \(\operatorname{tr}\rho = \operatorname{tr}(1(\rho))\). The thesis of the proposition is a consequence of the above factorization. □

5 Group congruences on an E-unitary semigroup

A nonempty subset A of a semigroup S is called left [right] unitary if asA [saA] implies sA for every aA,sS. Also, we say that A is unitary if it is both left and right unitary. Finally, a semigroup S with \(E_{S} \not= \emptyset\) is said to be E-unitary if ES is unitary.

Proposition 5.1

LetSbe a semigroup with\(E_{S} \not= \emptyset\). The following conditions are equivalent:
  1. (i)

    SisE-unitary;

     
  2. (ii)

    ESis left unitary;

     
  3. (iii)

    ESis right unitary.

     

Also, ifSis anE-unitaryE-inversive semigroup, thenSis anE-semigroup.

Proof

\(\mathrm{(i)} \Longrightarrow \mathrm{(ii)}\). This is trivial.

\(\mathrm{(ii)} \Longrightarrow \mathrm{(iii)}\). Let sS,eES. If se=fES, then fsef=f and so we get (efs)(efs)=efs, that is, efsES. Hence fsES. Thus sES.

\(\mathrm{(iii)} \Longrightarrow \mathrm{(i)}\). We may equally well show like above that ES is left unitary. Thus the condition (i) holds.

Finally, let S be an E-unitary E-inversive semigroup. If e,fES,xW(ef), then xefES. Hence xef,xES. Thus efES. □

Corollary 5.2

LetSbe anE-inversive semigroup. Then the following conditions are equivalent:
  1. (i)

    SisE-unitary;

     
  2. (ii)

    \(\operatorname{ker}\sigma = E_{S}\);

     
  3. (iii)

    τ=σ.

     

In particular, ifSis anE-unitaryE-inversive semigroup, thenESS.

Proof

\(\mathrm{(i)} \Longrightarrow \mathrm{(ii)}\). In the light of Proposition 5.1 and Theorem 4.5, \(\operatorname{ker}\sigma = E_{S}\omega\). Also, S is left unitary, that is, ES is closed. Thus \(\operatorname{ker}\sigma = E_{S}\).

\(\mathrm{(ii)} \Longrightarrow \mathrm{(iii)}\). We have mentioned above that τσ. On the other hand, the main assumption implies that σ is idempotent pure. Hence στ. Thus τ=σ.

\(\mathrm{(iii)} \Longrightarrow \mathrm{(i)}\). Let aS,e,fES. If ea=f, then \(a \in \operatorname{ker}\sigma = \operatorname{ker}\tau = E_{S}\), that is, ES is left unitary. In the light of Proposition 5.1, S is E-unitary. □

Remark 3

Notice that if a semigroup is not E-inversive, then Corollary 5.2 is false. Indeed, let FX1 be the free monoid on the set X. Then FX1 is E-unitary but τ is induced by the partition {FX,{1}}. Thus τ is not a group congruence.

From Proposition 3.4 and Corollary 5.2 we obtain the following proposition.

Proposition 5.3

LetSbe anE-unitaryE-inversive semigroup. Then (a,b)∈σif and only ifabESfor some (all) bW(b).

Corollary 5.4

LetAbe anE-inversive subsemigroup of anE-unitaryE-inversive semigroupS. ThenσA=σS∩(A×A).

Proof

Clearly, σAσS∩(A×A). The converse follows from Proposition 5.3. □

In [14] Howie and Lallement showed that \(\sigma \cap \mathcal{H} = 1_{S}\), when S is an E-unitary regular semigroup. We prove a corresponding result.

Theorem 5.5

LetSbe anE-unitaryE-inversive semigroup. Then\(\sigma \cap \mathcal{H} = 1_{S}\). Moreover, if in additionESforms a semilattice, then\(\sigma \cap \mathcal{L} = \sigma \cap \mathcal{R} = 1_{S}\).

Proof

Let S be an E-unitary E-inversive semigroup. Suppose also that ES forms a semilattice. Then ES is normal (Corollary 5.2), so if \((a, b) \in \sigma \cap \mathcal{L}\), then ax=e, bx=fES for some xS (see Proposition 5.3) and sa=b,tb=a for some s,tS. Hence se=sax=bx=fES,tf=tbx=ax=eES. Thus s,tES (since ES is unitary), so since idempotents commute and ta=tb,
$$a = tb = t(sa) = (ts)a = (st)a = s(ta) = s(tb) = sa = b. $$

We may equally well show that \(\sigma \cap \mathcal{R} = 1_{S}\).

If S is E-unitary, then ES is normal, too. Let \((a, b) \in \sigma \cap \mathcal{H}\). By the above proof and its dual we conclude that a=eb=bf and b=ga=ah for some e,f,g,hES. In the light of Proposition 2 in [18], a=b. □

Remark 4

The assumption that S is an E-inversive semigroup is important. Indeed, let S=(ℝ0,+) be the semigroup of nonnegative real numbers with respect to addition. Then S is an E-unitary commutative semigroup. Put M=ℕ0 and N={0,x,2x,3x,…} (where x∈ℝ∖ℚ). Then M,NS but MN={0} is not normal, so S has no least group congruence.

The converse of Theorem 4.15 is not valid (in general). Indeed, let S=〈x〉, where x=(2 3 4 5 6 7 5) is a mapping of \(\mathcal{T}(\{1, 2, \ldots, 7\})\). Then S=M(4,3) is the monogenic semigroup with index 4 and period 3, say S={x,x2,…,x6}. Also, the cyclic subgroup Kx of S with the unit e is equal {x4,x5,x6=e}. Since x3e=x7x2=x4x2=e, then S is not E-unitary. On the other hand, σ is induced by the partition: {{x,x4},{x2,x5},{x3,e}} and \(\mathcal{H}\) by the partition: {Kx,{x},{x2},{x3}}. Thus \(\sigma \cap \mathcal{H} = 1_{S}\).

From Theorem 5.5 and Corollary 5.2 we have the following corollary.

Corollary 5.6

LetSbe anE-unitaryE-inversive semigroup. Then
$$\sigma \cap \mathcal{H} = \tau \cap \mathcal{H} = 1_S. $$
Moreover, if in additionESforms a semilattice, then
$$\sigma \cap \mathcal{L} = \tau \cap \mathcal{L} = \sigma \cap \mathcal{R} = \tau \cap \mathcal{R} = 1_S. $$

Recall that a congruence ρ on a semigroup S is E-unitary if S/ρ is E-unitary. In [5] the author described the least E-unitary congruence κ on an idempotent-surjective semigroup. Also, for every congruence ρ on an idempotent-surjective semigroup S there exists the least E-unitary congruence κρ on S containing ρ [5].

Let S be an idempotent-surjective semigroup, NS. Define the relation \(\hat{\rho_{N}}\) on \(\mathcal{C}(S)\) by the following rule: \((\rho_{1}, \rho_{2}) \in \hat{\rho_{N}} \Leftrightarrow \rho_{1} \lor \rho_{N} = \rho_{2} \lor \rho_{N}\) (\(\rho_{1}, \rho_{2} \in \mathcal{C}(S)\)). Then \(\hat{\rho_{N}}\) is a congruence on \(\mathcal{C}(S)\), since \(\phi \phi^{- 1} = \hat{\rho_{N}}\) (see Theorem 2.11).

Also, we prove the following proposition.

Proposition 5.7

LetSbe an idempotent-surjective semigroup, \(N \lhd S, \rho \in \mathcal{C}(S)\). Then the elementsρ,κρ,ρρNare\(\hat{\rho_{N}}\)-equivalent andρκρρρN. Moreover, the elementρρNis the largest in the\(\hat{\rho_{N}}\)-class\(\rho \hat{\rho_{N}}\).

Proof

Since κρ is the least E-unitary congruence containing ρ and clearly ρρN is E-unitary, then ρκρρρN. Hence ρρNκρρNρρN. Therefore ρρN=κρρN. Thus \((\rho, \kappa_{\rho}) \in \hat{\rho_{N}}\). Evidently, \((\rho, \rho \lor \rho_{N}) \in \hat{\rho_{N}}\). This implies the first part of the proposition. The second part is clear. □

Remark 5

Recall from [22] that in the class of inverse semigroups not every \(\hat{\sigma}\)-class has a least element.

Finally, it is easy to see that the least E-unitary congruence κ on an arbitrary E-inversive semigroup exists, too. We show that \(\mathcal{H}\cap\sigma\subseteq\kappa\) in any E-inversive semigroup. Firstly, we have need the following useful proposition.

Proposition 5.8

LetBbe the least seminormal subsemigroup of anE-inversive semigroupS. Ifϕis an epimorphism ofSonto anE-unitary semigroupT, thenET.

Proof

Put A=(ET)ϕ−1. Clearly, A is a full subsemigroup of S, so A is dense. Further, if xyA, then ET∋(xy)ϕ===(yx)ϕ (since ET is reflexive), so yxA. Hence BA. Thus ⊆((ET)ϕ−1)ϕET. □

We may now prove the following equivalent theorem to Theorem 5.5.

Theorem 5.9

In anyE-inversive semigroupS, \(\mathcal{H}\cap\sigma\subseteq\kappa\). If in additionESforms a semilattice, then\(\mathcal{L}\cap\sigma\subseteq\kappa\)and\(\mathcal{R}\cap\sigma\subseteq\kappa\).

Proof

Indeed, σ=ρB, where B is the least seminormal subsemigroup of S. Let \((a, b)\in\mathcal{H}\cap\sigma\). Then clearly \((a\kappa, b\kappa)\in\mathcal{H}^{S/\kappa}\). Also, ax=yb for some a,bB. In the light of Proposition 5.8, ()()=()(), where ,ES/κ. Hence \((a\kappa, b\kappa)\in\mathcal{H}^{S/\kappa}\cap\sigma_{S/\kappa}= 1_{S/\kappa}\) (Theorem 5.5). Thus \(\mathcal{H}\cap\sigma\subseteq\kappa\), as required. □

6 Group congruences on an eventually regular semigroup

Group congruences on eventually regular semigroups were described in [9] by Hanumantha Rao and Lakshmi. In the paper [9] the following definition was introduced: a subset A of S is called self-conjugate if xr(x)−1(xr(x))AxA and xAxr(x)−1(xr(x))A for all xS,(xr(x))V(xr(x)). We say that A is self-conjugate if the former condition holds.

Lemma 6.1

LetNbe a subsemigroup of an eventually regular semigroupS. ThenNis normal if and only ifNis full, self-conjugate and closed.

Proof

Let N be normal, xS,(xr(x))V(xr(x)). Then N is full and closed. Also, xr(x)(xr(x))NENN, so xr(x)−1(xr(x))NxN, since N is reflexive.

Let N be full, self-conjugate and closed, xyN,(xr(x))V(xr(x)). Then xr(x)−1(xr(x))(xy)xxr(x)−1(xr(x))NxN, i.e., (xr(x)−1(xr(x))x)(yx)∈N, where xr(x)−1(xr(x))xESN. Hence yx=N, so N is reflexive. Thus NS. □

Lemma 6.2

LetSbe an eventually regular semigroup, NS. Then
$$\rho_N = \bigl\{(a, b) \in S \times S: \exists\, \bigl({b^{r(b)}}\bigr)^{*} \in V\bigl(b^{r(b)}\bigr)~\bigl[ab^{r(b) -1}\bigl({b^{r(b)}}\bigr)^{*} \in N\bigr]\bigr\}. $$

Proof

Let (a,b)∈ρN and (br(b))V(br(b)). Then na=bm for some n,mN. Hence nabr(b)−1(br(b))=bmbr(b)−1(br(b)). Also, since br(b)−1(br(b))bES, then mbr(b)−1(br(b))bNESN, so nabr(b)−1(br(b))=bmbr(b)−1(br(b))N, since N is reflexive. Consequently, abr(b)−1(br(b))=N.

Conversely, let a,bS,(br(b))V(br(b)) and abr(b)−1(br(b))=nN. Then a(br(b)−1(br(b))b)=nb, where br(b)−1(br(b))bESN. Hence (a,b)∈ρN. □

We have the following corollary (see Theorem 1 [9]).

Corollary 6.3

LetSbe an eventually regular semigroup, NS. Then
$$\rho_N = \bigl\{(a, b) \in S \times S: \exists\, \bigl({b^{r(b)}}\bigr)^{*} \in V\bigl(b^{r(b)}\bigr)~\bigl[ab^{r(b) -1}\bigl({b^{r(b)}}\bigr)^{*} \in N\bigr]\bigr\} $$
is a group congruence onS.

Finally, we give some remarks concerning group congruences on inverse semigroups. Firstly, consider the following result (see Exercise 7(ii) [11], p. 181).

Statement 6.4

An inverse subsemigroupNof an inverse semigroupSis normal if and only if (Nx)ω=(xN)ωfor everyxS.

This result is false. Indeed, let S be a Clifford semigroup. Put \(N = \mathcal{Z}(S)\), where \(\mathcal{Z}(S) = \{s \in S: \forall a \in S~[sa = as]\}\). Clearly, N is a full subsemigroup of S. Also, N is self-conjugate. If the result is valid, then N is normal (since Nx=xN for every xS). Hence ρN=S×S=ρS, when S=S0. It follows that every Clifford semigroup is commutative, a contradiction. Consequently, we conclude that the above result is false. Moreover, the assumptions of the result and the conditions: “N is full” and “N is self-conjugate” do not imply that (Nx)ω=(xN)ω for every xS.

It is clear that every subgroup of a group is full and closed. We prove now a correct version of the above statement.

Proposition 6.5

A full and closed inverse subsemigroupNof an inverse semigroupSis normal if and only if (Nx)ω=(xN)ωfor everyxS.

Proof

It is easy to see that if N is normal, then (Nx)ω=(xN)ω for every xS.

Conversely, let (Nx)ω=(xN)ω for every xS. It is easy to check that two relations ρ1={(a,b)∈S×S:ab−1N} and ρ2={(a,b)∈S×S:a−1bN} are equivalences on S and that 1=(Nx)ω, 2=(xN)ω for every xS. Also, ρ1 is right compatible and ρ2 is left compatible. Indeed, we show first that the equality (A())ω=(AB)ω holds for all A,BS. Recall from [11] that
$$H\omega = \bigl\{s \in S: \exists\, h \in H~[h \leq s]\bigr\}\quad(H \subseteq S), $$
where ≤ is the so-called natural partial order on (an inverse semigroup) S (i.e., ab⇔∃ eES [a=eb]). Notice that ≤ is compatible. Let x∈(A())ω. Then ayx for some aA,y (that is, by for some bB). Hence abayx. Thus x∈(AB)ω. We have just proved that (A())ω⊂(AB)ω. The opposite inclusion is clear. Let now (a,b)∈ρ2,cS. Then (aN)ω=(bN)ω and so (c(aN)ω)ω=(c(bN)ω)ω. Therefore (caN)ω=(cbN)ω. Thus ρ2 is a left congruence on S. We may equally well show that ρ1 is a right congruence on S. Since (Nx)ω=(xN)ω and 1=(Nx)ω, 2=(xN)ω for every xS, then ρ1=ρ2 is a congruence on S. Put for simplicity ρ=ρ1=ρ2. Finally, if eES, then ESN==(eN)ω. Hence ρ is a group congruence on S and \(\operatorname{ker}\rho = N\). Thus NS, as required. □

Corollary 6.6

A Clifford semigroupSis commutative if and only if\(\mathcal{Z}(S)\)is closed inS (i.e., if and only if for everysSthere exists\(z \in \mathcal{Z}(S)\)such thatzs).

Lemma 6.7

LetSbe a finite inverse semigroup with semilattice of idempotentsE. Then=Sif and only ifShas zero.

Proof

It is clear that if S has zero, then =S. Conversely, let =S. Since E is finite, then E has the least idempotent with respect to the natural partial order, say 0. Let sS=. Then e=fs and e=sg for some e,f,gE (see Proposition V.2.2 in [11]). Hence 0=0s=s0. Thus S=S0, as required. □

By an analogy to groups we may introduce the concept of a σ-simple inverse semigroup in the class of finite inverse semigroups without 0. From Lemma 6.7 follows that every finite inverse semigroup S without zero has at least one non-universal group congruence, so S has exactly one non-universal group congruence if and only if S/ is a simple group. Hence we may say that a finite inverse semigroup S without zero is σ-simple if S/ is a simple group. This definition is equivalent to the following definition: S is σ-simple if S has exactly two normal subsemigroups, namely: and S.

Example 6.8

Let (E,≤) be a chain with the least element 0. Put S=E∪{a}, where aE and aaa=a. Assume also that aa=0. Hence a=aaa=0a=a0. It is easy to see that if a binary operation on S is associative, then ea=ae=a for every eES. For example, ea=e(0a)=(e0)a=0a=a. Conversely, it is straightforward to verify that such defined binary operation is associative. Thus S is a semigroup. Since a=a−1, then S is an inverse semigroup. Finally, E=, so S/E={E,{a}}.

7 The hypercore of a semigroup

In [8] Hall and Munn studied the hypercore of a semigroup. In this section we give some remarks on the hypercore of E-inversive E-semigroups and inverse semigroups.

Let S be a semigroup with \(E_{S} \not= \emptyset\). Denote by ℘S the set of all subsemigroups A of S such that A has no cancellative congruences except the universal congruence. Note that {e}∈℘S for every eES. Define the hypercore\(\operatorname{hyp}(S)\) of S, as follows: \(\operatorname{hyp}(S) = \langle \bigcup \{A: A \in \wp_{S} \} \rangle\) [8]. Furthermore, by the core\(\operatorname{core}(S)\) of an E-inversive semigroup S we shall mean \(\operatorname{ker}\sigma\).

In [8] the authors showed the following two results.

Result 7.1

LetSbe anE-inversive semigroup. Then:
  1. (i)

    \(\operatorname{hyp}(S) \in \wp_{S}\);

     
  2. (ii)

    \(\operatorname{hyp}(S)\)is full and unitary;

     
  3. (iii)

    \(\forall \rho \in \mathcal{GC}(S)~[\operatorname{hyp}(S) \subseteq \operatorname{ker}\rho]\).

     

Result 7.2

In anyE-inversive semigroupS, \(\operatorname{hyp}(S)\)is the greatestE-inversive subsemigroup ofSwith no non-universal group congruence.

Let U be the least full unitary subsemigroup of an E-inversive semigroup S. Clearly, \(U \subseteq \operatorname{hyp}(S) \subseteq \operatorname{core}(S)\).

Finally, we have the following proposition.

Proposition 7.3

LetSbe anE-inversiveE-semigroup such that\(1_{S} \notin \mathcal{GC}(S)\). Then\(U = \operatorname{hyp}(S) = \operatorname{core}(S) = E_{S}\omega\). In particular, ESωhas no non-universal group congruence.

If in additionSis an inverse semigroup andESωis finite, thenESωis an inverse semigroup with zero. In particular, every finite inverse semigroupS (which is not a group) contains exactly one normal inverse subsemigroup with zero.

Proof

Let S be an E-inversive E-semigroup. Then \(\operatorname{core}(S) = E_{S}\omega\) (Theorem 4.5). Since ESU and U is closed, then ESωU, so \(U = \operatorname{hyp}(S) = \operatorname{core}(S) = E_{S}\omega\). In the light of Result 7.2, ESω has no non-universal group congruence.

If S is an inverse semigroup, then obviously \(U = \operatorname{hyp}(S) =\operatorname{core}(S) = E_{S}\omega\) has no non-universal group congruence. Finally, if ESω is finite, then ESω has zero (Lemma 6.7). The rest of the proposition is now immediate. □

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Authors and Affiliations

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