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, \(\mathcal{H}\cap\sigma= 1_{S}\).
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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:2S→2S 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
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
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
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
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
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,
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,
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
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)
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,
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
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:
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,
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
Moreover, if in addition E S forms a semilattice, then
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
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
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
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. □
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Communicated by Marcel Jackson.
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Gigoń, R.S. Congruences and group congruences on a semigroup. Semigroup Forum 86, 431–450 (2013). https://doi.org/10.1007/s00233-012-9425-z
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DOI: https://doi.org/10.1007/s00233-012-9425-z