Congruences and group congruences on a semigroup

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 ρ∨σ 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, $\mathcal{H}\cap\sigma= 1_{S}$.

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., E⊆N), closed (i.e., Nω=N, where ω:2^S→2^S is a closure operator given by Aω={s∈S:∃ a∈A [as∈A]} for all A⊆S), and self-conjugate (i.e., s ⁻¹ Ns⊆N for every s∈S). 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 ρ _N ={(a,b)∈S×S:ab ⁻¹∈N} 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)∈ρ _N if and only if ax,bx∈N for some x∈S.

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)={a∈S:a∈aSa} and by V(a) the set of inverses of a∈S, i.e., the set {x∈S:a=axa,x=xax}. Note that if a∈S is regular, say a=axa for some x∈S, then xax∈V(a). Also, S is called regular if V(a)≠∅ for every a∈S. 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 a ⁿ∈Reg(S).

Let S be a semigroup, 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 (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 E _S ≠∅ 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 E _S 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 aρ is an idempotent of S/ρ, then aρ 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 semigroup S is E-inversive.

Proof

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

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

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}. 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: ρ _n ={(k,l)∈ℕ×ℕ:n|(k−l)} (n>0). Note that E _ℕ=∅, 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

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

Lemma 2.2

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 2.3

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 .

Proof

Let (a,b)∈ρ _2,B . Then ax=yb for some x,y∈B. Also, as∈B for some s∈S, since B is dense and so sa∈B, since B is reflexive. Hence asy∈B and so (sy)a∈B. It follows that (sy)b=s(yb)=s(ax)=(sa)x∈B. Thus (sy)a, (sy)b∈B. We have just shown that ρ _2,B ⊂ρ _3,B .

Conversely, if xa,xb∈B for some x∈S, then ax,bx∈B (since B is reflexive), so a(xb)=(ax)b, where ax,xb∈B. Hence (a,b)∈ρ _2,B . Thus ρ _2,B =ρ _3,B .

Dually, ρ _1,B =ρ _4,B . Since B is reflexive, then ρ _1,B =ρ _3,B . □

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

Theorem 2.4

Let B be a dense and reflexive 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.

Proof

Let a∈S. Since B is dense, then there exists x∈S such that xa∈B. Hence ρ _B is reflexive. Obviously, ρ _B is symmetric. Also, since B is a semigroup, then ρ _B is transitive. Consequently, ρ _B is an equivalence relation on S. Moreover, ρ _B is a left congruence on S. Indeed, let (a,b)∈ρ _B ,c∈S. Then ax,bx∈B and zc∈B for some x,z∈S, so zcax,zcbx∈B. It follows that (ca)(xz),(cb)(xz)∈B, since B is reflexive. Therefore (ca,cb)∈ρ _B . By symmetry, ρ _B is a right congruence on S. Finally, S/ρ _B is a group. Indeed, let a∈S,b∈B and ax,xa∈B for some x∈S. Then bax∈B. Hence xa,x(ba)∈B, so (ba,a)∈ρ _B . Since B is dense, then S/ρ _B is a group, as required.

Since b(bb)=(bb)b for every b∈B, 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)∈ρ _B for some b∈B. Hence b ₁ s=bb ₂ for some b ₁,b ₂∈B. Thus s∈Bω, so $\operatorname{ker}\rho_{B} \subset B\omega$. Conversely, let s∈Bω. Then bs∈B for some b∈B. Since bb∈B, then (s,b)∈ρ _B , 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=Bω. 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=ρ _N , 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 Nϕ=ρ _N 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

Let B be a dense and reflexive subsemigroup of S. Then Bω◁S.

Example 2.6

Let S={a,b,c,e,f} be the semigroup with the multiplication table given below:

It is easy to see that E is a dense and reflexive subsemigroup of S but E is not closed, since ea∈E and a∉E. Also, N={a,e,f} is normal. Indeed, the group congruence ρ _E has two ρ _E -classes: N and {b,c}, since ae,ee,bb,bc∈E and (e,b)∉ρ _E . Note also that $E \subset \operatorname{ker}\rho_{N} = N, E \not= N$ and ρ _E =ρ _N . 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 ρ _H is a congruence on G. We have a corresponding result:

Let A be a closed subsemigroup of a semigroup S. Then A is normal if and only if $\rho_{A} \in \mathcal{GC}(S)$.

Indeed, let $\rho_{A} \in \mathcal{GC}(S)$. From A=Aω and the second paragraph of the proof of Theorem 2.4 we obtain that $A = \operatorname{ker}\rho_{A}$. Thus A◁S (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,N◁S but M∩N=∅.

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

Theorem 2.7

Let B be a seminormal subsemigroup of a semigroup $S, \rho \in \mathcal{C}(S)$. Then:

• (i) ρ∨ρ _B =ρ _B ρρ _B ;

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

• (iii) (x,y)∈ρ∨ρ _B if and only if (ax,yb)∈ρ for some a,b∈B.

Proof

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

(ii). This is evident.

(iii). Let (x,y)∈ρ∨ρ _B . Then (x,r)∈ρ _B ,(r,s)∈ρ and (s,y)∈ρ _B for some r,s∈S. 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,db∈B. Conversely, let (ax,yb)∈ρ for some a,b∈B. Since (x,ax),(yb,y)∈ρ _B , then (x,y)∈ρ _B ρρ _B =ρ∨ρ _B (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

Let B be a seminormal subsemigroup of a semigroup $S, \rho \in \mathcal{C}(S)$. Then $\operatorname{ker}(\rho \lor \rho_{B}) = (B\rho)\omega$. In particular, (Bρ)ω◁S.

Proof

Let $x \in \operatorname{ker}(\rho \lor \rho_{B})$. Then there exists b∈B such that (x,b)∈ρ∨ρ _B , since $B \subset \operatorname{ker}(\rho \lor \rho_{B})$. Hence (ax,bc)∈ρ for some a,c∈B (by Theorem 1.6(iii)). Thus ax∈Bρ. It follows that x∈(Bρ)ω. Conversely, if x∈(Bρ)ω, then ax∈Bρ for some a∈Bρ, so (ax,b),(a,c)∈ρ for some b,c∈B. It follows that (cx,b)∈ρ. Hence ((cc)x,cb)∈ρ. Thus (x,c)∈ρ∨ρ _B . 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 semigroup S is dually right modular element of $\mathcal{C}(S)$.

Corollary 2.10

Let B be a seminormal subsemigroup of a semigroup $S, \rho \in \mathcal{C}(S)$. Then ρ∨ρ _B =S×S if and only if (Bρ)ω=S.

Let B be a seminormal subsemigroup of a semigroup $S, \rho_{1}, \rho_{2} \in \mathcal{C}(S)$. Suppose that (x,y)∈(ρ ₁∨ρ _B )∩(ρ ₂∨ρ _B ). Then (ax)ρ ₂(yb), where a,b∈B. Moreover, ax(ρ ₁∨ρ _B )x,x(ρ ₁∨ρ _B )y,y(ρ ₁∨ρ _B )yb, so ax(ρ ₁∨ρ _B )yb. Thus (cax,ybd)∈ρ ₁, where c,d∈B. It follows that (caxd,cybd)∈ρ ₁. Moreover, (caxd,cybd)∈ρ ₂. Hence (xd,cy)∈(ρ ₁∩ρ ₂)∨ρ _B . Thus (x,y)∈(ρ ₁∩ρ ₂)∨ρ _B , since (ρ ₁∩ρ ₂)∨ρ _B 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 (ρ ₁∨ρ _B )∩(ρ ₂∨ρ _B )⊂(ρ ₁∩ρ ₂)∨ρ _B . The converse inclusion is evident. Thus we may conclude that (ρ ₁∨ρ _B )∩(ρ ₂∨ρ _B )=(ρ ₁∩ρ ₂)∨ρ _B .

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

Theorem 2.11

Let B be a seminormal subsemigroup of a semigroup S. 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 [ρ _B ,S×S] of all group congruences on S containing ρ _B .

Proof

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

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

Corollary 2.12

Let B be a seminormal subsemigroup of a semigroup S. Then ρ _B distributes over meet.

Let S be a semigroup, N◁S. Put

$$\mathcal{P}(S; N) = \bigl\{A \subseteq S: A^2 \subseteq A, N \subseteq A, A\omega = A\bigr\}. $$

Also, denote S/ρ _N 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

Let S be a semigroup, N◁S. Then there exists an inclusion-preserving bijection ϕ between the set $\mathcal{P}(S; N)$ and the set $\mathcal{P}(S/N; \{N\})$. Moreover, $M \in \mathcal{P}(S; N)$ and M◁S if and only if Mϕ◁S/N.

Proposition 2.14

Let ϕ be an epimorphism of a semigroup S onto a group (G,⋅,1). Then:

• (i) Ker(ϕ)=ϕϕ ⁻¹ is a group congruence on S;

• (ii) N={1}ϕ ⁻¹◁S;

• (iii) Ker(ϕ)=ρ _N .

Conversely, if N◁S, then N is the kernel of the canonical homomorphism of S onto S/N.

Example 2.15

We now describe all normal subsemigroups of the bicyclic semigroup S=ℕ₀×ℕ₀, where (k,l)(m,n)=(k−l+max{l,m},n−m+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 E _S ={(0,0),(1,1),(2,2),…}◁S and (k,l)ρ _E (m,n) if and only if k+n=l+m, so S/ρ _E ≅(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

Let S be a semigroup; M,N◁S and M⊆N. Then:

• (i) M◁N;

• (ii) N/M◁S/M;

• (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 a∈A. Then ax∈E _S =E _A for some x∈S, so x∈Aω=A. Consequently, there is x∈A such that ax∈E _A .

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

Let S be an E-inversive semigroup, N◁S. Suppose also that a subsemigroup M of S is full and closed. Then:

• (i) M∩N◁M;

• (ii) N◁(MN)ω;

• (iii) M/(M∩N)≅(MN)ω/N.

Proof

(i). It is clear that E _S ⊂M∩N, so M∩N is a full subsemigroup of M. Let a,b∈M be such that ab∈M∩N. Then ba∈M and ba∈N (since N is reflexive in S). Hence ba∈M∩N. Hence M∩N is reflexive in M. Further, if x∈(M∩N)ω, then yx∈M∩N for some y∈M∩N, so x∈M∩N (because N and M are closed). Since M∩N is full and closed, then it is E-inversive, so it is dense in M. Thus M∩N◁M.

(ii). We show that (MN)ω is a subsemigroup of S. Let a,b∈(MN)ω. Then m ₁ n ₁ a=m ₂ n ₂ for some m ₁,m ₂∈M,n ₁,n ₂∈N. Since S is E-inversive, then $W(m_{1}) \not= \emptyset$. Hence mm ₁,m ₁ m∈E _S ⊂M for some m∈S. Thus m∈M (since M is closed), (mm ₁)n ₁ a=(mm ₂)n ₂. Therefore (n ₁ a,mm ₂)∈ρ _N , since mm ₁∈E _S ⊂N, so (a,m ₃)∈ρ _N (m ₃∈M). Similarly, (b,m ₄)∈ρ _N for some m ₄∈M. It follows that (ab,m ₅)∈ρ _N , where m ₅∈M. Hence n ₃ ab=m ₅ n ₄ for some n ₃,n ₄∈N. Thus (m ₅ n ₃)ab=(m ₅ m ₅)n ₄. Consequently, ab∈(MN)ω. Furthermore, N⊂(MN)ω. Indeed, let n∈N. Then n ₁ n=en ₂ for some e∈E _S ,n ₁,n ₂∈N. Hence we have (en ₁)n=en ₂∈MN, so n∈(MN)ω. Consequently, N◁(MN)ω (since N◁S).

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 E _S 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 σ _S . Then σ=ρ _U and $\operatorname{ker}\sigma = U$ (Theorem 2.4).

Firstly, we have the following proposition.

Proposition 3.1

Let S be an E-inversive semigroup. Then $\mathcal{GC}(S) = [\sigma, S \times S]$. Thus $\mathcal{GC}(S)$ is a complete sublattice of $\mathcal{C}(S)$.

Also, ρ _M ∨ρ _N =ρ _M ρ _N =ρ _N ρ _M for all M,N◁S. 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(ρ _M ρ _N )b. Then (a,c)∈ρ _M ,(c,b)∈ρ _N , where c∈S. Take any x∈W(c). Then xc,cx∈E _S ,(cxa)ρ _N (bxa),(bxa)ρ _M (bxc), so (a,bxa)∈ρ _N ,(bxa,b)∈ρ _M . Hence (a,b)∈ρ _N ρ _M . Therefore ρ _M ρ _N ⊂ρ _N ρ _M . We may equally well show the opposite inclusion. Consequently, ρ _M ∨ρ _N =ρ _M ρ _N =ρ _N ρ _M . 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)∈ρ _M ρ _N for some e∈E _S . Hence (x,n)∈ρ _M , (n,e)∈ρ _N , where n∈S (in fact, $n \in \operatorname{ker}\rho_{N} = N$). Thus x∈Nρ _M . Conversely, if x∈Nρ _M , then (x,n)∈ρ _M for some n∈N. Hence (x,n)∈ρ _M ,(n,e)∈ρ _N , where e∈E _S . Thus (x,e)∈ρ _M ρ _N , 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 x∈Mρ _N . Then n ₁ x=mn ₂ for some n ₁,n ₂∈N,m∈M. Hence (mn ₁)x∈MN. Thus x∈(MN)ω. We have proved that $\operatorname{ker}(\rho_{M} \rho_{N}) \subset (MN)\omega$. Conversely, let x∈(MN)ω. Then m ₁ n ₁ x=m ₂ n ₂ for some m ₁,m ₂∈M,n ₁,n ₂∈N. Since S is E-inversive, then mm ₁=e∈E _S ⊂M for some m∈S. It follows that m∈M (since M is closed), so en ₁ x=mm ₂ n ₂. Hence (x,mm ₂)∈ρ _N . 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 =ρ _M ρ _N =ρ _N ρ _M =ρ _(NM)ω for all M,N◁S. Moreover, notice that $\operatorname{ker}(\rho_{M} \cap \rho_{N}) = \operatorname{ker}\rho_{M} \cap \operatorname{ker}\rho_{N} = M \cap N$ (M,N◁S), so ρ _M ∩ρ _N =ρ _M∩N for M,N◁S. Consequently, the lattice $(\mathcal{N}(S), \subseteq, \cap, \lor)$, where M∨N=(MN)ω for all M,N◁S, 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

Let S be an E-inversive semigroup. Then there exists a (lattice) isomorphism ϕ between the lattice $(\mathcal{N}(S), \subseteq, \cap, \lor)$, where M∨N=(MN)ω for all M,N◁S, and the lattice $(\mathcal{GC}(S), \subseteq, \cap, \circ)$. In fact, ϕ is defined by Nϕ=ρ _N for every $N \in \mathcal{N}(S)$. Moreover, ϕ is a complete lattice isomorphism.

Finally, we have the following proposition.

Proposition 3.4

Let S be an E-inversive semigroup, N◁S. Then (a,b)∈ρ _N if and only if ab ^∗∈N for some (all) b ^∗∈W(b).

Proof

(⟹). Let na=bm, where n,m∈N, and b ^∗∈W(b). Then nab ^∗=bmb ^∗. Since b ^∗ bm∈N and N is reflexive, then nab ^∗∈N. Hence ab ^∗∈Nω=N.

(⟸). Let ab ^∗=n∈N for some b ^∗∈W(b). Then a(b ^∗ b)=nb, so (a,b)∈ρ _N (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

Let S be a semigroup. The following conditions are equivalent:

• (i) S is an E-semigroup;

• (ii) ∀a,b∈S [W(b)W(a)⊆W(ab)].

Moreover, the condition (i) implies the following condition:

• (iii) ∀e∈E _S  [W(e)⊆E _S ].

If in addition S is an R-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

Let S be an E-semigroup. Then:

• (i) ∀e∈E _S  [W(e),V(e)⊆E _S ];

• (ii) ∀a∈S,a ^∗∈W(a),e∈E _S  [aea ^∗,a ^∗ ea∈E _S ];

• (iii) ∀a∈S,a ^∗∈W(a),e,f∈E _S  [ea ^∗,a ^∗ e,ea ^∗ f∈W(a)].

Proof

(i). This follows from Proposition 4.1.

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

(iii). Let a∈S,a ^∗∈W(a),e,f∈E _S . Since e∈W(e) and f∈W(f), then ea ^∗∈W(e)W(a)⊆W(ae). Hence ea ^∗=ea ^∗ aeea ^∗=(ea ^∗)a(ea ^∗). Therefore ea ^∗∈W(a). Similarly, a ^∗ e∈W(a). Finally, ea ^∗ f∈W(e)W(a)W(f)⊆W(fae) and so ea ^∗ f=ea ^∗ ffaeea ^∗ f=(ea ^∗ f)a(ea ^∗ f). Hence ea ^∗ f∈W(a). □

Proposition 4.3

Let S be an E-inversive E-semigroup. Then

$$\rho_{1, E} = \rho_{2, E} = \rho_{3, E} = \rho_{4, E}. $$

Proof

Let (a,b)∈ρ _2,E and a ^∗∈W(a). Then ae=fb for some e,f∈E. Moreover, a ^∗ f∈W(a) (Corollary 4.2(iii)), so (a ^∗ f)a,a(a ^∗ f)∈E. Further, a ^∗ fb=a ^∗ ae∈E. We have just shown that xa,ax,xb∈E for some x∈S. Thus ρ _2,E ⊂ρ _4,E .

On the other hand, if xa,xb∈E for some x∈S, say xa=e,xb=f, then (efx)a(efx)=ef(xa)efx=efx, so efx∈W(a). Also, fxbfx=f(xb)fx=fx, i.e., fx∈W(b). Hence efx∈W(b) (Corollary 4.2(iii)). Thus $W(a) \cap W(b) \not= \emptyset$. It follows that ay,by,ya,yb∈E for some y∈S. Dually, if ax,bx∈E for some x∈S, then ay,by,ya,yb∈E for some y∈S. Thus ρ _4,E =ρ _1,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 x∈W(a)∩W(b), then a(xb)=(ax)b and xb,ax∈E. Thus ρ _2,E =ρ _4,E =ρ _1,E . We may equally well show that ρ _3,E =ρ _4,E =ρ _1,E . Consequently, ρ _1,E =ρ _2,E =ρ _3,E =ρ _4,E . □

Lemma 4.4

Let S be an E-inversive E-semigroup. Then:

• (i) ∀a∈S ∃ e,f∈E _S  [ea,af∈Reg(S)];

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

Proof

Let a∈S,x∈W(a). Then (ax)a,a(xa)∈Reg(S), where ax,xa∈E _S , so (i) holds. Also, r=axa∈Reg(S) and xrx=x. Thus x∈W(a)∩W(r). □

Denote the above four relations from Proposition 4.3 by ρ _E . 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 any E-inversive E-semigroup, σ=ρ _E . Moreover, $\operatorname{ker}\sigma = E_{S}\omega$. Thus E _S ω◁S.

Proof

It is clear that ρ _E is an equivalence relation on S. Let (a,b)∈ρ _E ,c∈S. Then x∈W(a)∩W(b). Take any y∈W(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)∈ρ _E . Thus ρ _E is a left congruence on S. We may equally well show that ρ _E is a right congruence on S. Also, if e,f∈E _S , then ee,ef∈E _S . Consequently, (e,f)∈ρ _E for all e,f∈E _S . Lemma 4.4(ii) says that every ρ _E -class of S contains a regular element. This implies that S/ρ _E 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 E _S ω◁S (Theorem 2.4). Finally, ρ _E ⊆ρ _N for ever N◁S. Indeed, E _S ⊆N. Hence E _S ω⊆Nω=N. Thus $\rho_{E} = \rho_{E_{S}\omega} \subseteq \rho_{N}$ (Theorem 2.4). Consequently, σ=ρ _E . □

Corollary 4.6

The least group congruence σ on an E-inversive E-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 “∃ e∈E _S  [eae=ebe]” from the above corollary is equivalent to the apparently weaker condition “∃ s∈S [sas=sbs]”.

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

Theorem 4.7

In any idempotent-surjective E-semigroup, σ=ρ _E .

Let S be a semigroup. A congruence ρ on S is called idempotent pure if eρ⊆E _S for every e∈E _S . 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: {E _S ,S∖E _S }. 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 τ⊆σ. Indeed, let (a,b)∈τ and b ^∗∈W(b). Then bb ^∗∈E _S ,(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

If ρ is a congruence on an idempotent-surjective E-semigroup S, then $\operatorname{ker}(\rho \lor \sigma) = (\operatorname{ker}\rho)\omega$. In particular, τ⊆σ.

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 τ∨σ=σ. Thus τ⊆σ. □

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 E _S . 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

Let ρ be a congruence on an idempotent-surjective E-semigroup S, a,b∈S. Then (aρ,bρ)∈σ in S/ρ implies (a,b)∈σ if and only if ρ⊆σ.

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 (aρ,bρ)∈σ. In the light of Lemma 4.9, if ρ⊆σ, then (a,b)∈σ if and only if (aρ,bρ)∈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 ρ preserves 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-surjective E-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 ρ⊆S×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 as∈A [sa∈A] implies s∈A for every a∈A,s∈S. 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 E _S is unitary.

Proposition 5.1

Let S be a semigroup with $E_{S} \not= \emptyset$. The following conditions are equivalent:

• (i) S is E-unitary;

• (ii) E _S is left unitary;

• (iii) E _S is right unitary.

Also, if S is an E-unitary E-inversive semigroup, then S is an E-semigroup.

Proof

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

$\mathrm{(ii)} \Longrightarrow \mathrm{(iii)}$. Let s∈S,e∈E _S . If se=f∈E _S , then fsef=f and so we get (efs)(efs)=efs, that is, efs∈E _S . Hence fs∈E _S . Thus s∈E _S .

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

Finally, let S be an E-unitary E-inversive semigroup. If e,f∈E _S ,x∈W(ef), then xef∈E _S . Hence xef,x∈E _S . Thus ef∈E _S . □

Corollary 5.2

Let S be an E-inversive semigroup. Then the following conditions are equivalent:

• (i) S is E-unitary;

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

• (iii) τ=σ.

In particular, if S is an E-unitary E-inversive semigroup, then E _S ◁S.

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, E _S is closed. Thus $\operatorname{ker}\sigma = E_{S}$.

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

$\mathrm{(iii)} \Longrightarrow \mathrm{(i)}$. Let a∈S,e,f∈E _S . If ea=f, then $a \in \operatorname{ker}\sigma = \operatorname{ker}\tau = E_{S}$, that is, E _S 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 F _X ¹ be the free monoid on the set X. Then F _X ¹ is E-unitary but τ is induced by the partition {F _X ,{1}}. Thus τ is not a group congruence.

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

Proposition 5.3

Let S be an E-unitary E-inversive semigroup. Then (a,b)∈σ if and only if ab ^∗∈E _S for some (all) b ^∗∈W(b).

Corollary 5.4

Let A be an E-inversive subsemigroup of an E-unitary E-inversive semigroup S. Then σ _A =σ _S ∩(A×A).

Proof

Clearly, σ _A ⊂σ _S ∩(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

Let S be an E-unitary E-inversive semigroup. Then $\sigma \cap \mathcal{H} = 1_{S}$. Moreover, if in addition E _S forms 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 E _S forms a semilattice. Then E _S is normal (Corollary 5.2), so if $(a, b) \in \sigma \cap \mathcal{L}$, then ax=e, bx=f∈E _S for some x∈S (see Proposition 5.3) and sa=b,tb=a for some s,t∈S. Hence se=sax=bx=f∈E _S ,tf=tbx=ax=e∈E _S . Thus s,t∈E _S (since E _S 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 E _S 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,h∈E _S . 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,N◁S but M∩N={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,x ²,…,x ⁶}. Also, the cyclic subgroup K _x of S with the unit e is equal {x ⁴,x ⁵,x ⁶=e}. Since x ³ e=x ⁷ x ²=x ⁴ x ²=e, then S is not E-unitary. On the other hand, σ is induced by the partition: {{x,x ⁴},{x ²,x ⁵},{x ³,e}} and $\mathcal{H}$ by the partition: {K _x ,{x},{x ²},{x ³}}. Thus $\sigma \cap \mathcal{H} = 1_{S}$.

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

Corollary 5.6

Let S be an E-unitary E-inversive semigroup. Then

$$\sigma \cap \mathcal{H} = \tau \cap \mathcal{H} = 1_S. $$

Moreover, if in addition E _S forms 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 κ _ρ on S containing ρ [5].

Let S be an idempotent-surjective semigroup, N◁S. 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

Let S be an idempotent-surjective semigroup, $N \lhd S, \rho \in \mathcal{C}(S)$. Then the elements ρ,κ _ρ ,ρ∨ρ _N are $\hat{\rho_{N}}$-equivalent and ρ⊆κ _ρ ⊆ρ∨ρ _N . Moreover, the element ρ∨ρ _N is the largest in the $\hat{\rho_{N}}$-class $\rho \hat{\rho_{N}}$.

Proof

Since κ _ρ is the least E-unitary congruence containing ρ and clearly ρ∨ρ _N is E-unitary, then ρ⊆κ _ρ ⊆ρ∨ρ _N . Hence ρ∨ρ _N ⊆κ _ρ ∨ρ _N ⊆ρ∨ρ _N . Therefore ρ∨ρ _N =κ _ρ ∨ρ _N . 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

Let B be the least seminormal subsemigroup of an E-inversive semigroup S. If ϕ is an epimorphism of S onto an E-unitary semigroup T, then Bϕ⊆E _T .

Proof

Put A=(E _T )ϕ ⁻¹. Clearly, A is a full subsemigroup of S, so A is dense. Further, if xy∈A, then E _T ∋(xy)ϕ=xϕ⋅yϕ=yϕ⋅xϕ=(yx)ϕ (since E _T is reflexive), so yx∈A. Hence B⊆A. Thus Bϕ⊆Aϕ⊆((E _T )ϕ ⁻¹)ϕ⊆E _T . □

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

Theorem 5.9

In any E-inversive semigroup S, $\mathcal{H}\cap\sigma\subseteq\kappa$. If in addition E _S forms a semilattice, then $\mathcal{L}\cap\sigma\subseteq\kappa$ and $\mathcal{R}\cap\sigma\subseteq\kappa$.

Proof

Indeed, σ=ρ _B , 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,b∈B. In the light of Proposition 5.8, (aκ)(xκ)=(yκ)(bκ), where aκ,bκ∈E _S/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 x ^r(x)−1(x ^r(x))^∗ Ax⊆A and xAx ^r(x)−1(x ^r(x))^∗⊆A for all x∈S,(x ^r(x))^∗∈V(x ^r(x)). We say that A is self-conjugate if the former condition holds.

Lemma 6.1

Let N be a subsemigroup of an eventually regular semigroup S. Then N is normal if and only if N is full, self-conjugate and closed.

Proof

Let N be normal, x∈S,(x ^r(x))^∗∈V(x ^r(x)). Then N is full and closed. Also, x ^r(x)(x ^r(x))^∗ N⊆EN⊆N, so x ^r(x)−1(x ^r(x))^∗ Nx⊆N, since N is reflexive.

Let N be full, self-conjugate and closed, xy∈N,(x ^r(x))^∗∈V(x ^r(x)). Then x ^r(x)−1(x ^r(x))^∗(xy)x∈x ^r(x)−1(x ^r(x))^∗ Nx⊆N, i.e., (x ^r(x)−1(x ^r(x))^∗ x)(yx)∈N, where x ^r(x)−1(x ^r(x))^∗ x∈E _S ⊆N. Hence yx∈Nω=N, so N is reflexive. Thus N◁S. □

Lemma 6.2

Let S be an eventually regular semigroup, N◁S. 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)∈ρ _N and (b ^r(b))^∗∈V(b ^r(b)). Then na=bm for some n,m∈N. Hence nab ^r(b)−1(b ^r(b))^∗=bmb ^r(b)−1(b ^r(b))^∗. Also, since b ^r(b)−1(b ^r(b))^∗ b∈E _S , then mb ^r(b)−1(b ^r(b))^∗ b∈NE _S ⊆N, so nab ^r(b)−1(b ^r(b))^∗=bmb ^r(b)−1(b ^r(b))^∗∈N, since N is reflexive. Consequently, ab ^r(b)−1(b ^r(b))^∗∈Nω=N.

Conversely, let a,b∈S,(b ^r(b))^∗∈V(b ^r(b)) and ab ^r(b)−1(b ^r(b))^∗=n∈N. Then a(b ^r(b)−1(b ^r(b))^∗ b)=nb, where b ^r(b)−1(b ^r(b))^∗ b∈E _S ⊆N. Hence (a,b)∈ρ _N . □

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

Corollary 6.3

Let S be an eventually regular semigroup, N◁S. 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 on S.

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 subsemigroup N of an inverse semigroup S is normal if and only if (Nx)ω=(xN)ω for every x∈S.

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 x∈S). Hence ρ _N =S×S=ρ _S , when S=S ⁰. 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 x∈S.

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 subsemigroup N of an inverse semigroup S is normal if and only if (Nx)ω=(xN)ω for every x∈S.

Proof

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

Conversely, let (Nx)ω=(xN)ω for every x∈S. It is easy to check that two relations ρ ₁={(a,b)∈S×S:ab ⁻¹∈N} and ρ ₂={(a,b)∈S×S:a ⁻¹ b∈N} are equivalences on S and that xρ ₁=(Nx)ω, xρ ₂=(xN)ω for every x∈S. Also, ρ ₁ is right compatible and ρ ₂ is left compatible. Indeed, we show first that the equality (A(Bω))ω=(AB)ω holds for all A,B⊆S. 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., a≤b⇔∃ e∈E _S  [a=eb]). Notice that ≤ is compatible. Let x∈(A(Bω))ω. Then ay≤x for some a∈A,y∈Bω (that is, b≤y for some b∈B). Hence ab≤ay≤x. Thus x∈(AB)ω. We have just proved that (A(Bω))ω⊂(AB)ω. The opposite inclusion is clear. Let now (a,b)∈ρ ₂,c∈S. Then (aN)ω=(bN)ω and so (c(aN)ω)ω=(c(bN)ω)ω. Therefore (caN)ω=(cbN)ω. Thus ρ ₂ is a left congruence on S. We may equally well show that ρ ₁ is a right congruence on S. Since (Nx)ω=(xN)ω and xρ ₁=(Nx)ω, xρ ₂=(xN)ω for every x∈S, then ρ ₁=ρ ₂ is a congruence on S. Put for simplicity ρ=ρ ₁=ρ ₂. Finally, if e∈E _S , then E _S ⊆N=Nω=(eN)ω. Hence ρ is a group congruence on S and $\operatorname{ker}\rho = N$. Thus N◁S, as required. □

Corollary 6.6

A Clifford semigroup S is commutative if and only if $\mathcal{Z}(S)$ is closed in S (i.e., if and only if for every s∈S there exists $z \in \mathcal{Z}(S)$ such that z≤s).

Lemma 6.7

Let S be a finite inverse semigroup with semilattice of idempotents E. Then Eω=S if and only if S has zero.

Proof

It is clear that if S has zero, then Eω=S. Conversely, let Eω=S. Since E is finite, then E has the least idempotent with respect to the natural partial order, say 0. Let s∈S=Eω. Then e=fs and e=sg for some e,f,g∈E (see Proposition V.2.2 in [11]). Hence 0=0s=s0. Thus S=S ⁰, 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/Eω is a simple group. Hence we may say that a finite inverse semigroup S without zero is σ-simple if S/Eω is a simple group. This definition is equivalent to the following definition: S is σ-simple if S has exactly two normal subsemigroups, namely: Eω and S.

Example 6.8

Let (E,≤) be a chain with the least element 0. Put S=E∪{a}, where a∉E 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 e∈E _S . 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=Eω, 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 e∈E _S . 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

Let S be an E-inversive semigroup. Then:

• (i) $\operatorname{hyp}(S) \in \wp_{S}$;

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

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

Result 7.2

In any E-inversive semigroup S, $\operatorname{hyp}(S)$ is the greatest E-inversive subsemigroup of S with 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

Let S be an E-inversive E-semigroup such that $1_{S} \notin \mathcal{GC}(S)$. Then $U = \operatorname{hyp}(S) = \operatorname{core}(S) = E_{S}\omega$. In particular, E _S ω has no non-universal group congruence.

If in addition S is an inverse semigroup and E _S ω is finite, then E _S ω is an inverse semigroup with zero. In particular, every finite inverse semigroup S (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 E _S ⊆U and U is closed, then E _S ω⊆U, so $U = \operatorname{hyp}(S) = \operatorname{core}(S) = E_{S}\omega$. In the light of Result 7.2, E _S ω 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 E _S ω is finite, then E _S ω has zero (Lemma 6.7). The rest of the proposition is now immediate. □