On representations of variants of semigroups Ganna Kudryavtseva and Victor Maltcev 6 0 0 2 Abstract n a We construct a family of representations of an arbitrary variant J 4 Sa of a semigroup S, induced by a given representation of S, and 2 investigate properties of such representations and their kernels. ] R 2000 Mathematics Subject Classification 20M10, 20M30. G . h 1 Introduction t a m Let S be a semigroup and a ∈ S. Set s∗ t = sat, s,t ∈ S. The semigroup [ (S,∗) is called a variant of S and will be denoted by S . One of the mo- a 1 tivations for the study of variants is their importance to understanding the v 9 structure of an original semigroup. In particular, the notion of a regularity- 7 preserving element, which generalizes the notion of an invertible element, 5 1 is defined using variants, see [2], [3], [5]. Other situations where variants 0 naturally appear and work are discussed in [5]. 6 0 Let · : M × S → M or just (S,M) be a representation of a semigroup / S by transformations of a set M. It defines a homomorphism from S to the h t full transformation semigroup T (M); we denote the image of (m,s) under a m the function · by m·s. For a ∈ S and a decomposition a = βα, α,β ∈ S1, : we introduce a map ∗ : M ×S → M or just (S ,M;α,β), defined by v a a i X m∗s = m·(αsβ) for all s ∈ S, m ∈ M. r a Since m∗(s∗t) = m∗(sat) = m·(αsβαtβ) = (m·(αsβ))·(αtβ) = (m∗s)∗t for all s,t ∈ S, m ∈ M, it follows that (S ,M;α,β) is a representation of S on M. a a The kernel ρ of a representation of S naturally leads to the consideration of the family of congruences {ρ : b,c ∈ S1,cb = a} on S , which are b,c a 1 the kernels of the corresponding representations of S . In the case when ι a is the identity relation on S, ι and ι coincide with the congruences l a,1 1,a and r on S respectively, which were first introduced by Symons (see [8]) a for the generalized transformation semigroups and many properties of which resemble the corresponding properties of the Green’s relations L and R. The present paper is devoted to the study of the connections between a given representation (S,M) of S on M and the representations (S ,M;α,β); a and also to the study of the properties of congruences {ρ : b,c ∈ S1,cb = b,c a}. In Section 2 we prove that for a regular semigroup S, which does not contain the bicyclic semigroup, a faithful representation of S coincides with a certain (S ,M;α,β) if and only if α and β are left and right cancellable in a S respectively (Theorems 1, 2 and Proposition 1). We also show that for the bicyclic semigroup these conditions are not equivalent. The next result is Theorem 4, claiming that for a regular semigroup S each representation ∗ : M×S → M of any variant of S is induced by some · : M×S → M such a that either m∗s = m·(as) for all m ∈ M or m∗s = m·(sa) for all m ∈ M, if and only if it is either a left or right group, giving a new characterization of the class of left (right) groups. In Section 3 we investigate the properties of the family of congruences {ρ : b,c ∈ S1,cb = a}. In particular, in Theorem b,c 6 we prove that in the case when c, b are regular elements of S the map sending ρ to ρ is a homomorphism from the lattice of congruences of S to b,c the lattice of congruences of S . Without the requirement of regularity of cb b,c this map always preserves the meet, while may not preserve the join, the corresponding example is provided. Finally, the aim of Section 4 is, given a ∈ S1 and a congruence ρ on S, we define a congruence ρ which naturally a generalizes the congruence d on S , introduced in [8] and studied afterwards a in [3] and [6]. 2 Connections between representations Let S beasemigroup. Anelement u ∈ S1 will becalled left [right]cancellable if us = ut [su = tu] implies s = t for all s,t ∈ S. Fix a ∈ S and a decomposition a = βα, α,β ∈ S1. Lemma 1. The implication αsβ = αtβ ⇒ s = t for all s,t ∈ S holds if and only if α is left cancellable and β is right cancellable simultaneously. Proof. The proof is straightforward. A representation · : M×S → M is said to be faithful if the corresponding homomorphism from S to T(M) is injective. That is, · : M × S → M is 2 faithful if and only if (∀m ∈ M) [m·s = m·t] ⇒ [s = t] for all s,t ∈ S. Lemma 2. Suppose that (S ,M;α,β) is faithful. Then so is (S,M). a Proof. Indeed, by the definition we have (∀ m ∈ M) [m·(αsβ) = m·(αtβ)] ⇒ [s = t] for all s,t ∈ S. Take s,t ∈ S. If now m·s = m·t for all m ∈ M, then (m·α)·(sβ) = (m·α)·(tβ) for all m ∈ M, and so s = t. Theorem 1. Let (S,M) be a representation of a semigroup S on a set M and a ∈ S. Take α,β ∈ S1 such that βα = a. Then the following conditions are equivalent: 1. (S ,M;α,β) is faithful; a 2. (S,M) is faithful and α,β are left and right cancellable respectively. Proof. That1)implies2)isduetoLemmas1and2. Theoppositeimplication is a consequence of Lemma 1. Recall that an element u of a semigroup S is said to be a mididentity in S if sut = st for all s,t ∈ S. For the case when S is regular, we obtain the following theorem. Theorem 2. Let S be a regular semigroup and a = βα ∈ S, α,β ∈ S1. Suppose that there exist α∗ and β∗ of S1, inverses of α and β in S1 respec- ∗ ∗ tively, such that β βαα is a mididentity in S. Let also ∗ : M × S → M a be a faithful representation of S on M. Then the following conditions are a equivalent: 1. there is arepresentation (S,M) such that(S ,M) coincideswith(S ,M;α,β); a a 2. α and β are left and right cancellable respectively. Proof. Due to Theorem 1 we have that 1) implies 2). Now let α and β be left and right cancellable respectively. Let us prove that a function · : M ×S → M, given by ∗ ∗ m·s = m∗(α sβ ) for all s ∈ S, m ∈ M, 3 definesarepresentation(S,M)ofS suchthat(S ,M)coincideswith(S ,M;α,β). a a ∗ ∗ Since β βαα is a mididentity in S, we have ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (m·s)·t = (m∗(α sβ ))∗(α tβ ) = m∗(α sβ ·βα·α tβ ) = ∗ ∗ m∗(α stβ ) = m·(st) for all s,t ∈ S, m ∈ M. Thus (S,M) is indeed a representation. ∗ ∗ Note that the equality αα αs = αs implies α αs = s for all s ∈ S. ∗ Analogously, sββ = s for all s ∈ S. Then we have ∗ ∗ m·(αsβ) = m∗(α αsββ ) = m∗s for all s ∈ S, m ∈ M. The latter means that (S ,M) coincides with (S ,M;α,β). That is, the a a function · defines the required representation (S,M). The following example shows that the condition of Theorem 2 that there are α∗ and β∗ in S1, such that β∗βαα∗ is a mididentity in S, is essential. Example 1. Let B = ha,b | ba = 1i be the bicyclic semigroup. Consider the Cayley representation of B: m∗s = ms for all m,s ∈ B. Then there is no faithful representation ◦ : B ×B → B such that m◦(asb) = m∗s = ms for all m,s ∈ B. Since B is an inverse semigroup with the identity element, 1, its Cay- ley representation is faithful. That ba = 1 implies B = B. Note that ba b−1baa−1 = a·ba·b = ab is not a mididentity in B, as 1·ab·1 = ab 6= 1. Assume that there is a faithful representation ◦ : B ×B → B such that m◦(asb) = m∗s = ms for all m,s ∈ B. Then m◦(ab) = m∗1 = m for all m ∈ B. So we have m◦a = (m◦a)◦(ab) = m◦(a2b) = m∗a = ma and m◦b = (m◦(ab))◦b = m◦(ab2) = m∗b = mb for all m ∈ B. But this leads to a contradiction: 1 = 1◦(ab) = (1◦a)◦b = a◦b = ab. Let us recall (see Corollary 1.32 of [1]) that every homomorphic image of B is either a cyclic group or isomorphic to B. Using this fact, we are going to prove that in the case when S is a regular semigroup, which does not contain a subsemigroup, isomorphic to B, one obtains that the conditions 1) and 2) of Theorem 2 are equivalent. 4 Proposition 1. Let S be a regular semigroup, which does not contain a subsemigroup isomorphic to B. Let a = βα ∈ S, α,β ∈ S1. Let also ∗ : M ×S → M be a faithful representation of S on M. Then the conditions a a 1) and 2) of Theorem 2 are equivalent. Proof. We have only to prove that 2) implies 1). ∗ Let now α and β be left and right cancellable respectively. Take α and β∗, inverses of α and β in S1 respectively. Then the equality αα∗αs = αs ∗ ∗ ∗ implies α αs = s for all s ∈ S. Also sββ β = sβ implies sββ = s for all s ∈ S. It follows that α∗α and ββ∗ are left and right identity elements of S1 respectively. Consider three possible cases. ∗ ∗ Case 1. Let α,β ∈ S. Then α α and ββ are both in S. It follows that ∗ ∗ ∗ ∗ S has the identity 1 = α α = ββ . Assume that αα 6= 1. Then hα,α i is a homomorphic image of B = ha,b | ba = 1i under a map, which sends a to α ∗ ∗ ∗ and b to α . Since hα,α i is not isomorphic to a cyclic group, then hα,α i ∗ is a subsemigroup, isomorphic to B. We get a contradiction. Thus αα = 1. ∗ ∗ ∗ Analogously, one shows that β β = 1. Then β βαα is a mididentity in S. Now due to Theorem 2, we have that 1) holds. Case 2. Let β ∈/ S and α = a ∈ S. Arguments are similar to those of ∗ ∗ Case 1. Then αα = α α is again a mididentity in S and to prove that 1) holds, it remains to use Theorem 2. ∗ ∗ Case 3. Let α ∈/ S and β = a ∈ S. Analogously to Case 2, β β = ββ is a mididentity in S. Thus, according to Theorem 2, 1) holds. Recall (see [1]) that a right [left] group is a semigroup which is right [left] simple and left [right] cancellative. We will need the following known fact. Theorem 3 (Theorem 1.27 of [1]). The following conditions for a semi- group S are equivalent: 1. S is a left [right] group; 2. S is right [left] simple and contains an idempotent; 3. S is a direct product G×U of a group G and a left [right] zero semigroup U. Among all the decompositions a = βα, α,β ∈ S1, two are rather special, namely a = a · 1 and a = 1 · a. The main result of the following theorem is the characterization of semigroups S such that for all a ∈ S and for each representation ∗ : M×S → M there exists a representation · : M×S → M a such that either m∗s = m·(as) for all m ∈ M or m∗s = m· (sa) for all m ∈ M. 5 Theorem 4. Let S be a regular semigroup with the set of idempotents E. Then the following conditions are equivalent: 1. for all a ∈ S and for each representation ∗ : M ×S → M there exists a a representation · : M ×S → M such that either m∗s = m·(as) for all m ∈ M or m∗s = m·(sa) for all m ∈ M; 2. for all e ∈ E there are a faithful representation ∗ : M ×S → M and e a representation · : M ×S → M such that either m∗s = m·(es) for all m ∈ M or m∗s = m·(se) for all m ∈ M; 3. every element of E is either a left or a right identity in S; 4. S is either a left or a right group. Proof. Obviously, 1) implies 2). Assume that 2) holds. Take an idempotent e ∈ E. The first case is that there are a set M, a faithful representation ∗ : M × S → M and a e representation · : M × S → M such that m ∗ s = m · (es) for all m ∈ M. Then m∗s = m∗(es) for all m ∈ M and s ∈ S and so s = es for all s ∈ S. In the second case we obtain s = se for all s ∈ S. Thus, 2) implies 3). Let us prove now that 3) implies 4). Note that E is nonempty as S is regular. If S contains both left and right identities then it contains the identity 1, whence each idempotent coincides with 1. Hence, S is a regular semigroup with the unique idempotent 1. It follows now from Lemma II.2.10 from [7] that S is a group and so also a left group. If every idempotent e ∈ E is a left [right] unit in S, then sS1 = sS = ss′S = S [S1s = Ss = Ss′s = S] ′ for all s ∈ S and for each s, inverse of s, and so S is right [left] simple. But due to Theorem 3 we obtain now that S is a right [left] group. Finally, assume that S is, for instance, a left group. Then by Theorem 3 there are a group G and a left zero semigroup U such that S = G × U. Take (g,u) ∈ S and a representation ∗ : M ×S → M. Put m·(h,v) = (g,u) m ∗ (hg−1,v) for all (h,v) ∈ S. Then m · ((h,v)(g,u)) = m ∗ (h,v) for all (h,v) ∈ S. We are left to prove that · is a representation of S. Indeed, we have m·(h h ,v v ) = m∗(h h g−1,v v ) = m∗((h g−1,v )∗(h g−1,v )) = 1 2 1 2 1 2 1 2 1 1 2 2 (m·(h ,v ))·(h ,v ) for all (h ,v ),(h ,v ) ∈ S, m ∈ M. 1 1 2 2 1 1 2 2 This completes the proof. Let S be a semigroup and · : M ×S → M a representation of S. Denote by M · α the set {m · α : m ∈ M} and by m · S the set {m · s : s ∈ S}. 6 In the case when S 6= S1 set M ·1 = M. A representation (S,M) is said to be cyclic (see [1]) if there is a generating element m ∈ M for (S,M), i.e., m·S = M. Proposition 2. Let S be an arbitrary semigroup and a ∈ S, a = βα, α,β ∈ S1. Let (S,M) be a representation of S on M. The following conditions are equivalent: 1. (S ,M;α,β) is cyclic; a 2. (S,M) is cyclic, M ·β = M and M ·α contains a generating element for (S,M). Proof. Suppose that 1) holds. It follows immediately from the definition of cyclic representations that (S,M) is cyclic. If m ∈ M is a generating element for (S ,M;α,β) then m · (αSβ) = M which implies M · β = M. a Since (m·α)·S ⊇ m·(αSβ) = M, we have that M ·α contains a generating element for (S,M). Suppose now that 2) holds. Let m ∈ M · α be a generating element for (S,M) such that m = m α for some m ∈ M. Let us prove that m 0 0 0 is a generating element for (S ,M;α,β). Indeed, take m ∈ M. Then a 1 there exists m ∈ M such that m = m · β. Also there is s ∈ S such that 1 m = (m α)·s. Then we obtain m ∗s = m ·(αsβ) = ((m α)·s)·β = m . 0 0 0 0 1 Thus, m ∗S = M. 0 3 Congruences ρ b,c Let ρ be a congruence on a semigroup S and a ∈ S, a = cb, b,c ∈ S1. Define a relation ρ on S as follows: b,c a (s,t) ∈ ρ if and only if (bsc,btc) ∈ ρ for all s,t ∈ S. b,c It is straightforward that ρ is a congruence on S . If now we have a b,c a representation · : M ×S → M of S on M, then the congruence ν, related to it, is given by (s,t) ∈ ν if and only if m·s = m·t for all m ∈ M. Then one can easily see that the congruence on S , related to the represen- a tation (S ,M;b,c), coincides with ν . a b,c Denote by Cong(S) the set of all congruences on a semigroup S. Set ρ = ρ for all ρ ∈ Cong(S). In the case when S 6= S1, set also S = S. 1,1 1 7 Proposition 3. Let S be a semigroup, b,c ∈ S1, ρ ∈ Cong(S). Then ∼ S /ρ = bSc/ρ∩(bSc×bSc). cb b,c Proof. Define a map ϕ : S → bSc/ρ∩(bSc×bSc) as follows: cb ϕ(x) = (bxc)ρ for all x ∈ S. One can easily show that ϕ is an onto homomorphism and ϕ ◦ ϕ−1 = ρ . b,c These two facts complete the proof. Let S 6= S1 and ρ ∈ Cong(S). Then we identify (1)ρ with the identity element of (S/ρ)1. Theorem 5. Let S be a semigroup, b,c,b ,c ∈ S1, ρ,σ ∈ Cong(S). Then 1 1 1. if bρ and b ρ are L-related, cρ and c ρ are R-related in (S/ρ)1, then 1 1 ρ = ρ ; b,c b1,c1 2. ρ ⊆ ρ ; b,c 3. ρ = ρ if and only if bρ and cρ are left and right cancellable in (S/ρ)1 b,c respectively; 4. if ρ ⊆ σ then ρ ⊆ σ . If bσ and cσ are left and right cancellable in b,c b,c (S/σ)1 then ρ ⊆ σ if and only if ρ ⊆ σ . b,c b,c Proof. Statements 1) and 2) follow immediately from the definition of ρ . b,c Statement 3) follows from Lemma 1 and the fact that ρ = ρ is equivalent b,c to the implication (bρ)x(cρ) = (bρ)y(cρ) ⇒ [x = y] for all x,y ∈ S/ρ. Finally, let us prove 4). If ρ ⊆ σ then (x,y) ∈ ρ implies (bxc,byc) ∈ ρ ⊆ σ b,c which, in turn, implies (x,y) ∈ σ . Thus, if ρ ⊆ σ, then ρ ⊆ σ . b,c b,c b,c Suppose now that bσ and cσ are left and right cancellable in (S/σ)1 and ρ ⊆ σ . Take (x,y) ∈ ρ. Then (bxc,byc) ∈ ρ which implies (x,y) ∈ ρ ⊆ b,c b,c b,c σ or just (bσ)(xσ)(cσ) = (bσ)(yσ)(cσ), whence xσ = yσ. Thus, we obtain b,c that ρ ⊆ σ. This completes the proof. The converse statement of 1) of Theorem 5 is not true in general as the following easy example shows. Example 2. Consider a semilattice E = {a,b}, where a ≤ b. Let ρ be the identity relation on E. Then ρ = ρ but (a,b) ∈/ R. a,a a,b 8 Set ρr = ρ and ρl = ρ for all congruences ρ on a semigroup S and a 1,a a a,1 a ∈ S1. The following proposition shows that the converse statement of 1) of Theorem 5 is true in the case when b = b = 1 [c = c = 1] and S/ρ is 1 1 inverse. Proposition 4. Let S be a semigroup, b,c ∈ S1, ρ ∈ Cong(S). Suppose that S/ρ is inverse. Then ρr = ρr [ρl = ρl] if and only if bρ and cρ are R-related b c b c [L-related] in (S/ρ)1. Proof. The sufficiency follows from 1) of Theorem 5. Let now assume that ρr = ρr. Set bρ = u and cρ = v. Then we have xu = yu if and only if b c xv = yv for all x,y ∈ S/ρ. In particular, xu = xuu−1u gives us xv = xuu−1v for all x ∈ S/ρ. Hence, v = vv−1v = vv−1uu−1v = uu−1v, whence vv−1 = uu−1vv−1. Analogously, uu−1 = vv−1uu−1. Thus, vv−1 = uu−1 which is well-known to be equivalent to uRv (see [4]). Take b,c ∈ S1. Set a map ϕ : Cong(S) → Cong(S ) as follows: b,c cb ϕ (ρ) = ρ for all ρ ∈ Cong(S). b,c b,c It is well-known that if one has ρ,σ ∈ Cong(S) then ρ∨σ coincides with the transitiveclosure(ρ∪σ)t. DenotebySL(S)thelowersemilattice(Cong(S);⊆ ,∩) of congruences on S and by L(S) the lattice (Cong(S);⊆,∩,∨) of con- gruences on S. Theorem 6. Let S be a semigroup and b,c ∈ S1. Then 1. ϕ is a homomorphism from SL(S) to SL(S ); b,c cb 2. ρ ∨σ ⊆ (ρ∨σ) for all ρ,σ ∈ Cong(S); b,c b,c b,c 3. if b and c are regular in S1 then ϕ is a homomorphism from L(S) to b,c L(S ); cb 4. if bSc = S then ϕ is injective homomorphism from L(S) to L(S ). b,c cb Proof. Takeρ,σ ∈ Cong(S). Then(x,y) ∈ ρ ∩σ ifandonlyif(bxc,byc) ∈ b,c b,c ρ∩σ, which is, in turn, equivalent to (x,y) ∈ (ρ∩σ) for all x,y ∈ S. Thus, b,c ρ ∩σ = (ρ∩σ) , which completes the proof of 1). b,c b,c b,c To prove 2), we note that due to ρ ⊆ (ρ∨σ) and σ ⊆ (ρ∨σ) , we b,c b,c b,c b,c have that ρ ∨σ ⊆ (ρ∨σ) . b,c b,c b,c Let now b and c be regular in S1 and b′, c′ be certain inverses of b and c in S1 respectively. To prove 3), in view of what has already been done, we 9 are left to show that (ρ∨σ) ⊆ ρ ∨σ . Suppose that (x,y) ∈ (ρ∨σ) . b,c b,c b,c b,c Then (bxc,byc) ∈ ρ∨σ and so there are p ,...,p ∈ S such that 1 m (bxc,p ) ∈ ρ∪σ,...,(p ,p ) ∈ ρ∪σ,...,(p ,byc) ∈ ρ∪σ. (1) 1 i i+1 m Clearly, ρ∪σ is left and right compatible, and therefore ′ ′ (bxc,b·bp c ·c) ∈ ρ∪σ,..., 1 ′ ′ ′ ′ ′ ′ (b·bp c ·c,b·bp c ·c) ∈ ρ∪σ,...,(b·bp c ·c,byc) ∈ ρ∪σ, i i+1 m which yields ′ ′ ′ ′ ′ ′ (x,bp c) ∈ ρ ∪σ ,...,(bp c,bp c) ∈ ρ ∪σ ,..., 1 b,c b,c i i+1 b,c b,c ′ ′ (bp c,y) ∈ ρ ∪σ , m b,c b,c whence (x,y) ∈ ρ ∨σ . Thus, 3) is proved. b,c b,c Finally, assume that bSc = S. To prove that ϕ is a homomorphism, it b,c is enough to show that (ρ∨σ) ⊆ ρ ∨σ . Take (x,y) ∈ (ρ∨σ) . Then b,c b,c b,c b,c there are p ,...,p ∈ S such that (1) holds. Then due to bSc = S and the 1 m fact that (bsc,btc) ∈ ρ ∪ σ if and only if (s,t) ∈ ρ ∪ σ for all s,t ∈ S, b,c b,c we obtain that (x,y) ∈ ρ ∨σ . It remains to prove that ϕ is injective. b,c b,c b,c Suppose that ρ = σ . Take (s,t) ∈ ρ. There are s and t of S such that b,c b,c 1 1 s = bs c and t = bt c. Then (s ,t ) ∈ ρ which implies (s ,t ) ∈ σ , which, 1 1 1 1 b,c 1 1 b,c in turn, is equivalent to (s,t) ∈ σ. Thus, ρ ⊆ σ. Analogously, σ ⊆ ρ. So ρ = σ and ϕ is an injective homomorphism. b,c We note that the converse inclusion of one in 2) of Theorem 6, namely (ρ∨σ) ⊆ ρ ∨σ forρ,σ ∈ Cong(S), isnot true ingeneralasthe following b,c b,c b,c example illustrates. Example 3. Consider the free semigroup {a,b}+ over the alphabet {a,b}. Let I be the ideal consisting of all words from {a,b}+ of length not less than 3. Set S = {a,b}+/I to be the Rees quotient semigroup. Set ρ = (ba,ab)∪(ab,ba)∪ι and σ = (ab,bb)∪(bb,ab)∪ι, where ι denotes the identity relation on S. It follows immediately from the construction of S that ρ,σ ∈ Cong(S). Now we observe that (a,b) ∈ (ρ∨σ) . Indeed, we b,1 have (ba,ab) ∈ ρ and (ab,bb) ∈ σ, whence (ba,bb) ∈ ρ∨σ which is equivalent to (a,b) ∈ (ρ∨σ) . But (a,b) ∈/ ρ ∨σ . Indeed, otherwise there would b,1 b,1 b,1 exist t ,...,t ∈ S such that 1 n (ba,bt ) ∈ ρ∪σ,...,(bt ,bt ) ∈ ρ∪σ,...,(bt ,bb) ∈ ρ∪σ. 1 i i+1 n It follows from the construction of ρ and σ that (ba,bt ) ∈ ρ ∪ σ implies 1 t = a. Now inductive arguments show that t = a for all possible i. In 1 i particular, t = a. Then (ba,bb) ∈ ρ∪σ, and we get a contradiction. Thus, n (ρ∨σ) * ρ ∨σ . b,1 b,1 b,1 10