International Journal of Algebra, Vol. 4, 2010, no. 17, 799 - 808 Isomorphism Theorems for Semigroups of Order-preserving Partial Transformations Y. Kemprasit Department of Mathematics, Faculty of Science Chulalongkorn University, Bangkok 10330, Thailand [email protected] W. Mora Department of Mathematics, Faculty of Science Prince of Songkla University, Songkhla 90110, Thailand k [email protected] T. Rungratgasame Department of Mathematics, Faculty of Science Srinakharinwirot University, Bangkok 10110, Thailand Abstract The full order-preserving transformation semigroup, the order- preserving partial transformation semigroup and the order-preserving one-to-one partial transformation semigroup on a poset X are denoted by OT(X), OP(X) and OI(X), respectively. It is well-known that for ∼ any posets X and Y , OT(X) = OT(Y ) if and only if X and Y are either order-isomorphic or order-anti-isomorphic. The purpose of this paper is to show the analogous results for OP(X) and OI(X). It is shown that ∼ ∼ for any posets X and Y , OP(X) = OP(Y )[OI(X) = OI(Y )] if and only if X and Y are either order-isomorphic or order-anti-isomorphic. From our given proofs, some isomorphism theorems on other order-preserving transformation semigroups are also provided. Mathematics Subject Classification: 20M20 Keywords: Isomophism theorems, order-preserving transformation semi- groups 800 Y. Kemprasit, W. Mora and T. Rungratgasame 1 Introduction and Preliminaries The image of a map α at x in the domain of α will be written by xα. The set of all idempotents of a semigroup S is denoted by E(S). For a set X, let T (X),P(X) and I(X) be the full transformation semi- group, the partial transformation semigroup and the one-to-one partial trans- formation semigroup (the symmetric inverse semigroup) on X, respectively. It is well-known that T (X),P(X) and I(X) are regular semigroups, and more- over, I(X) is an inverse semigroup. The domain and the range of α ∈ P (X) are denoted by dom α and ran α, respectively. We know that for α ∈ P (X), α ∈ E(P (X)) if and only if ran α ⊆ dom α and xα = x for all x ∈ ran α. For ∅ = A ⊆ X and x ∈ X, let Ax denote the constant map in P (X) whose domain and range are A and {x}, respectively. Hence Ax ∈ E(P (X)) if and only if x ∈ A. We note here that for 0 = α ∈ I(X), α ∈ E(I(X)) if and only if α is the identity map on dom α. For better understanding, we often denote a map α ∈ P (X) by bracket notation. For posets (partially ordered sets) X and Y , the map ϕ : X → Y is said to be order-preserving [order-reversing] if for all x, y ∈ X, x ≤ x im- plies xϕ ≤ xϕ [xϕ ≥ xϕ]. We call ϕ an order-isomorphism [order-anti- isomorphism] from X into Y if ϕ is a bijection from X onto Y and both ϕ and ϕ−1 are order-preserving [order-reversing]. If such a ϕ exists, then X and Y are said to be order-isomorphic [order-anti-isomorphic]. For a poset X, let OT(X)={α ∈ T (X) | α is order-preserving} , OP(X)={α ∈ P (X) | α is order-preserving} , OI(X)={α ∈ I(X) | α is order-preserving} . Then OT(X),OP(X) and I(X) are respectively subsemigroups of T (X),P(X) and I(X). Order-preserving transformation semigroups have been widely stud- ied. If X is a finite chain, then OT(X) is a regular semigroup ([2], p. 201). Kemprasit and Changphas [3] extended this result by proving that if X is a chain which is order-isomophic to Z, the set of integers under their natural order, then OT(X) is regular. Also, they have shown that for an interval X in R, the set of real numbers with the natural order, OT(X) is regular if and only if X is closed and bounded. This result was generalized by Rungrattrakoon and Kemprasit [5] as follows : For a nontrivial interval X in a subfield F of R, OT(X) is regular if and only if F = R and X is closed and bounded. Hence OT(X) is not regular for any nontrivial interval X in Q where Q is the set of rational numbers under the usual order. We note that the so called III- homomorphisms and III-congruences on posets have been recently introduced by Shum and others in [6]. In fact, their concept of III-homomorphism gen- eralizes the concept of a strong homomorphism for semilattices. In addition, Isomorphism theorems for semigroups 801 Fernandes [1] showed that if X is a finite chain, then |OI(X)| =2|OT(X)|. It is noteworthy that Kemprasit and Changphas [3] also proved that for any chain X, OP(X) and OI(X) are regular semigroups. It then follows that for any chain X, OI(X) is an inverse semigroup since I(X) is inverse. An in- teresting well-known isomorphism theorem on order-preseving transformation semigroups is the following. ∼ Theorem 1.1. ([4], page 222 and 223) For any posets X and Y ,ifOT(X) = OT(Y ), then X and Y are either order-isomorphic or order-anti-isomorphic. The converse direction of Theorem 1.1 is easily seen from the following obvious fact which will be used after. Proposition 1.2. Let X and Y be posets. If ϕ : X → Y is an order- isomorphism or order-anti-isomorphism, then the map α → ϕ−1αϕ is an iso- morphism from OT(X),OP(X) and OI(X) onto OT(Y ),OP(Y ) and OI(Y ), respectively. Our work in this paper is motivated by Theorem 1.1. We achieve our ∼ purpose by showing that for any posets X and Y , OP(X) = OP(Y ) if and only if X and Y are either order-isomorphic or order-anti-isomorphic, and the same result holds for OI(X) and OI(Y ). Our provided proofs give remarkable isomorphism theorems for some other order-preserving transformation semi- groups. 2 Isomorphism Theorems For a poset X and x ∈ X, let C(X, x) denote the subset of OP(X) defined by C(X, x)={Ax |∅= A ⊆ X}. Note that C(X, s) ∩ C(X, t)=∅ for all distinct s, t ∈ X. To obtain the first main result, the following lemma is needed. Lemma 2.1. Let X and Y be posets and ϕ : OP(X) → OP(Y ) an isomor- phism. Then the following statements hold. (i) For every x ∈ X, there is an element x ∈ Y such that C(X, x)ϕ = C(Y,x). (ii) The map x → x is a bijection from X onto Y . 802 Y. Kemprasit, W. Mora and T. Rungratgasame Proof. (i) Let x ∈ X. Since Xx ∈ E(OP(X)), we have Xxϕ ∈ E(OP(Y ))\{0}. Let x ∈ ran (Xxϕ). Then x ∈ dom (Xxϕ) and x(Xxϕ)=x. Hence Yx(Xxϕ)= −1 −1 −1 Yx, and so ((Yx)ϕ ) Xx =(Yx)ϕ . This implies that ran ((Yx)ϕ )={x}. −1 Hence (Yx)ϕ = Zx for some ∅ = Z ⊆ X with x ∈ Z.Thus (Xxϕ)Yx =(Xxϕ)(Zxϕ)=(XxZx)ϕ = Xxϕ. It follows that ran (Xxϕ)={x}. To show that C(X, x)ϕ = C(Y,x), let ∅ = A ⊆ X. Since (Axϕ)(Xxϕ)=(AxXx)ϕ = Axϕ and ran (Xxϕ)={x}, we have that ran (Axϕ)={x}. Hence Axϕ = Bx for some ∅ = B ⊆ Y . This proves that C(X, x)ϕ ⊆ C(Y,x). Since ϕ−1 is an isomorphism from OP(Y )ontoOP(X), from the previous proof, we have C(Y,x)ϕ−1 ⊆ C(X, z) for some z ∈ X. Therefore we have C(X, x)ϕ ⊆ C(Y,x) ⊆ C(X, z)ϕ which implies that C(X, x) ⊆ C(X, z) since ϕ : OP(X) → OP(Y ) is injective. Hence x = z. Consequently, C(X, x)ϕ = C(Y,x). (ii) From (i), the map x → x is clearly an injective map from X into Y . Let y ∈ Y . Since ϕ−1 : OP(Y ) → OP(X) is an isomorphism, from (i), there is an element z ∈ X such that C(Y,y)ϕ−1 = C(X, z). Hence C(Y,z)=C(X, z)ϕ = C(Y,y) which implies that z = y. Hence (ii) is proved. ∼ Theorem 2.2. For posets X and Y , OP(X) = OP(Y ) if and only if X and Y are either order-isomorphic or order-anti-isomorphic. Proof. Let ϕ : OP(X) → OP(Y ) be an isomorphism. For each x ∈ X, let x ∈ Y be as in Lemma 2.1. Define θ : X → Y by xθ = x for all x ∈ X.By Lemma 2.1(ii), θ is a bijection from X onto Y . Let a, b ∈ X be such that a<b. To show that a andb are comparable, suppose instead thata and b are not a b a b OP Y ϕ−1 ∈ OP X comparable. Then b a is an element of ( ), b a ( ) and a b a b Y Y Y Y . a b a = b and b b a = a Isomorphism theorems for semigroups 803 Since C(X, a)ϕ = C(Y,a) and C(X, b)ϕ = C(Y,b), we have that Aaϕ = Ya and Bbϕ = Yb for some nonempty subsets A and B of X. Then a b a b a b A ϕ−1 A ϕ ϕ−1 Y ϕ−1 Y ϕ−1 B , a b a = ( a ) b a = a b a = b = b a b a b a b B ϕ−1 B ϕ ϕ−1 Y ϕ−1 Y ϕ−1 A .