We should like to construct all the semigroups that have a left univocal factorization with factors to pair (A,B) of prescribed semigroups such that their intersections consists of only one element, that is right identity of A and left identity of B

M.M. MICCOLI

MICCOLI, Maria Maddalena

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Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 34 (2014) no. 2, 157–162. doi:10.1285/i15900932v34n2p157On left univocal factorizations of semigroupsMaria Maddalena MiccoliDipartimento di Matematica e Fisica “E. De Giorgi”Università del Salento73100 Lecce, Italymaddalena.miccoli@unisalento.itReceived: 7.4.2014; accepted: 31.5.2014.Abstract. We should like to construct all the semigroups that have a left univocal factor-ization with factors to pair (A,B) of prescribed semigroups such their intersection consists ofonly one element, that is right identity of A and left identity of B.Keywords: Factorizable semigroup, Left univocal factorization.MSC 2000 classification: 20M101 IntroductionFollowing Tolo [10], a semigroup S is said to be factorizable if it can bewritten as the set product AB of proper subsemigroups A and B. In this case,we call the pair (A,B) a factorization of S, with factors A and B. As in Catino[1], a factorization (A,B) of a semigroup S is called left univocal if ab = a′b′implies a = a′, for all a, a′ ∈ A and b, b′ ∈ B. There are several papers on thissubject, for istance see [1, 2, 4, 5, 6, 7, 8, 9].The main aim of the paper is to construct a semigroup S from a givensemigroups A and B in such a way that (A,B) is a left univocal factorization ofS where A ∩B consists of only one element, that is right identity of A and leftidentity of B. The construction obtained generalizes those of Köhler [Theorem2.1, [4] ], of Krishnan [Theorem 3, [5]] on monoids and of Catino [Theorem 5,[1]] on univocal factorizations.The notations and terminology of Howie [3] will be used, with a few minorexceptions.2 Results and proofsA well–known result about left univocal factorizations of semigroups is thefollowing lemma (Lemma 1 of [1]).http://siba-ese.unisalento.it/ c© 2014 Università del Salento158 M. MiccoliLemma 1. Let (A,B) a left univocal factorization of a semigroup S. ThenA ∩B is a left zero semigroup formed by right identities of A.Then we can obtain an easy but useful result.Proposition 1. Let (A,B) a left univocal factorization of a semigroup S.If E is a non-empty subset of A∩B, then B̄ := EB is a subsemigroup of S and(A, B̄) is a left univocal factorization of S with A ∩ B̄ = E.Proof. Let E be is a non-empty subset of A ∩ B. Put B̄ := EB, it easy toshow that (A, B̄) is a left univocal factorization of S and E ⊆ A∩ B̄. Moreover,let z be an element of A ∩ B̄ and let e ∈ A and b ∈ B be such that z = eb.Then zb = ebb and thus z = e, since (A,B) is a left univocal factorization of S.Therefore A ∩ B̄ = E. QEDCorollary 1. Let (A,B) a left univocal factorization of a semigroup S ande an element of A ∩ B. Then (A, eB) is a left univocal factorization of S andA ∩ eB = {e}. In this case e is a right identity of A and a left identity of eB.Note that we may consider left univocal factorizations (A,B) with A∩B = {e}and e non left identity of B (see [1], Example of p. 168). But, of course, also inthis case we may consider a left univocal factorization (A, B̄) in which the onlyelement of the intersection of two factors is a left identity of B̄.We should now like to construct all semigroups that have a left univocalfactorization with factors isomorphic to a pair (A,B) of prescribed semigroupswhere A∩B consists of only one element, that is a right identity of A and a leftidentity of B.Let (A,B) a left univocal factorization of a semigroup S and let e be a rightidentity element of A and a left identity of B such that A ∩B = {e}.Firstly, for each a ∈ A define a relation θa on B byb θa b′ :⇐⇒ ab = ab′for all elements b and b′ of B. Then θa is an equivalence relation on B, even aright congruence. Moreover, if |l denotes left divisibility on A, i.e. a |l a′ if thereexists c ∈ A such that ca = a′, then we have∀a, a′ ∈ A a |l a′ =⇒ θa ⊆ θa′ (1)θe = idB (2)For all a ∈ A and b ∈ B, let b ⊲ a be the element of A such that ba = (b ⊲ a)b′,for some b′ ∈ B. Then, by the Axiom of Choise, there exists a mapping ⊳ :A×B −→ B such that ba = (b ⊲ a)((a, b)⊳). If we denote (a, b)⊳ by b ⊳ a, thenba = (b ⊲ a)(b ⊳ a).On left univocal factorizations of semigroups 159Without loss of generality we may assume thate ⊳ a = e and b ⊳ e = be (3)for all a ∈ A and b ∈ B. Moreover, we havee ⊲ a = ea and b ⊲ e = e (4)Now, let a1, a2 ∈ A and let b1, b2 ∈ B. Then((b1b2) ⊲ a1)((b1b2) ⊳ a1) = b1b2a1 = b1(b2 ⊲ a1)(b2 ⊳ a1)= (b1 ⊲ (b2 ⊲ a1))(b1 ⊳ (b2 ⊲ a1))(b2 ⊳ a1)and(b1 ⊲ (a1a2))(b1 ⊳ (a1a2)) = b1a1a2 = (b1 ⊲ a1)(b1 ⊳ a1)a2= (b1 ⊲ a1)((b1 ⊳ a1) ⊲ a2)((b1 ⊳ a1) ⊳ a2))It follows that(b1b2) ⊲ a1 = b1 ⊲ (b2 ⊲ a1) (5)b1 ⊲ (a1a2) = (b1 ⊲ a1)((b1 ⊳ a1) ⊲ a2) (6)(b1 ⊳ (a1a2)) θb1⊲(a1a2) ((b1 ⊳ a1) ⊳ a2) (7)((b1b2) ⊳ a1) θ(b1b2)⊲a1 ((b1 ⊳ (b2 ⊲ a1))(b2 ⊳ a1) (8)for all a1, a2 ∈ A and b1, b2 ∈ B.Finally, let a1, a2 ∈ A and b1, b′1, b2, b′2 ∈ B such that a1b1 = a1b′1 and a2b2 =a2b′2. Thena1(b1 ⊲ a2)(b1 ⊳ a2)b2 = a1b1a2b2 = a1b′1a2b′2= a1(b′1 ⊲ a2)(b′1 ⊳ a2)b′2Thus, we have that if b1 θa1 b′1 and b2 θa2 b′2, thena1(b1 ⊲ a2) = a1(b′1 ⊲ a2) (9)((b1 ⊳ a2)b2) θa1(b1⊲a2) ((b′1 ⊳ a2)b′2) (10)Now, let A be a semigroup with right identity e1 and let B be a semigroupwith left identity e2 such that A ∩ B = {e}, and e := e1 = e2. Denote by E(B)the lattice of equivalence relation on B, let θ : A −→ E(B), ⊲ : A × B −→ Band ⊳ : B×A −→ B. If a ∈ A and b ∈ B, then denote (a, b)⊲ by a ⊲ b, (b, a)⊳ byb ⊳ a and aθ by θa. The quintet (A,B, θ, ⊲, ⊳) is called admissible if conditions(1)–(10) hold.So far we have seen that any left univocal fattorization (A,B) of a semigroupS as above gives an admissible quintet. In the following we will show that thesedata suffice to reconstruct S from A and B.160 M. MiccoliTheorem 1. Given a admissible quintet (A,B, θ, ⊲, ⊳) there is a semigroupS with a left univocal factorization (Ā, B̄) such that Ā and B̄ are isomorphicto A and B, respectively. Moreover, Ā ∩ B̄ has only element ē that is a rightidentity of Ā and a left identity of B̄.Proof. Define on A×B the following equivalence relation(a1, b1) ∼ (a2, b2) :⇐⇒ a1 = a2 and b1θa1b2for all a1, a2 ∈ A, b1, b2 ∈ B and denote by 〈a, b〉 the equivalence class of (a, b),for all a ∈ A and b ∈ B.We define on S := (A×B)/ ∼ the following multiplication〈a1, b1〉〈a2, b2〉 := 〈a1(b1 ⊲ a2), (b1 ⊳ a2)b2〉for all a1, a2 ∈ A, b1, b2 ∈ B. By conditions (9) and (10) this in fact is well-defined.To prove associativity, let a, a1, a2 ∈ A, b, b1, b2 ∈ B. Then(〈a, b〉〈a1, b1〉)〈a2, b2〉 = 〈a(b ⊲ a1)(b ⊳ a1)b1〈a2, b2〉= 〈a(b ⊲ a1)(((b ⊳ a1)b1) ⊲ a2), (((b ⊳ a1)b1) ⊲ a2)b2〉and〈a, b〉(〈a1, b1〉〈a2, b2〉) = 〈a2, b2〉〈a(b1 ⊲ a2)(b1 ⊳ a2)b2= 〈a(b ⊲ (a1(b1 ⊲ a2))), (b ⊳ (a1(b1 ⊲ a2)))(b1 ⊳ a2)b2〉By (5) and (6) we havea(b ⊲ a1)(((b ⊳ a1)b1) ⊲ a2) = a(b ⊲ (a1(b1 ⊲ a2))).We remark that by (10), applied to a2 := e, and by (3),(4) we have that θa1 isa right congruence, for all a1 ∈ A.By this and by (8),(1) we have(((b ⊳ a1)b1) ⊲ a2)b2 θa(b⊲a1)(((b⊳a1)b1)⊲a2) (((b ⊳ a1) ⊳ (b1 ⊲ a2))(b1 ⊳ a2))b2.Similary, by (7) and (1) we have(b ⊳ (a1(b1 ⊲ a2)))(b1 ⊳ a2)b2 θa(b⊲(a1(b1⊲a2))) (((b ⊳ a1) ⊳ (b1 ⊲ a2))(b1 ⊳ a2))b2.This finally shows(b ⊳ (a1(b1 ⊲ a2)))(b1 ⊳ a2)b2 θa(b⊲(a1(b1⊲a2))) (b ⊳ (a1(b1 ⊲ a2)))(b1 ⊳ a2)b2.On left univocal factorizations of semigroups 161Thus the above multiplication is associative.It is easily seen by (2),(3) and (4) that the mappingsα : A −→ S, a 7−→ 〈a, e〉 and β : B −→ S, b 7−→ 〈e, b〉are monomorphisms. Put Ā = Aα, B̄ = Bβ and ē = 〈e, e〉, using (3) and (4) iteasy to shows that (Ā, B̄) is a left univocal factorization of S such that Ā andB̄ are isomorphic to A and B, respectively, Ā∩ B̄ = {ē} and ē is a right identityof Ā and a left identity of B̄. QEDFinally, following Köhler [4], a left univocal factorization (A,B) of a semi-group S is a quasi-decomposition of S if Ba ⊆ aB, for all a ∈ A. If A∩B = {e}with e be a right identity element of A and a left identity of B such thatA∩B = {e}, then e is an identity of A. We note the e is not necessarily an iden-tity of B. In this regard, consider S := {a, b, c, d, e} with the following Cayleymoltiplication tablea b c d ea b b a a ab b b b b bc a b c d ed a b c d ee a b e e eThen S is a semigroup that admits the left univocal factorization (A,B) withA := {a, b, c} and B := {c, d, e} in which c is a left identity of B but not a rightidentity.References[1] F. Catino: Factorizable semigroups, Semigroup Forum 36 (1987), no. 2, 167–174.[2] S.Y. Chen, S.C. Hsieh: Factorizable inverse semigroups, Semigroup Forum 8 (1974),283–297.[3] J.M. Howie: An Introduction to Semigroup Theory, Academic Press, London, 1976.[4] P. Köhler: Quasi-decompositions of semigroups, Houston J. Math., 5 (1979), no. 4, 525– 542.[5] V.S. Krishnan: Quasi resolved semigroups and quintets, Semigroup Forum 17 (1979),no. 3, 267–277.[6] T.T. Peng, X.J. Guo, K.P. Shum: Factorizable rpp monoids, International Mathemat-ical Forum 2 (2007), 143–153.[7] T.T. Peng, X.J. Guo: Factorizable quasi-adequate semigroups, Chin. Quart. J. of Math.26(1) (2011), 93–94.162 M. Miccoli[8] S. Qui, M. Hartman, X.J. Guo, K.P. Shum: Almost factorizable IC quasi-adequatsemigroups, Comm. Algebra 39 (2011), no.5, 1796–1808.[9] J. Schmidt: Quasi-decompositions, exact sequences and triple sums of semigroups, I.General theory, II Applications, Colloquia Math. Soc. János Bolyai, Contributions toUniversal Algebra, Szeged, 17 (1975), 365–397 (I), 399–428 (II).[10] K. Tolo: Factorizable semigroups, Pacific J. Math. 31 (1969), 523–535.

On left univocal factorizations of semigroups

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