RATIO MATHEMATICA 24 (2013), 31–40 ISSN: 1592-7415 Ordered Polygroups Mahmood Bakhshi, Radjab Ali Borzooei Department of Mathematics, Bojnord University, Bojnord, Iran Department of Mathematics, Shahid Beheshti University, Tehran, Iran bakhshi@ub.ac.ir; borzooei@sbu.ac.ir Abstract In this paper, those polygroups which are partially ordered are in- troduced and some properties and related results are given. Key words: Hypergroup, Polygroup, Ordered Polygroup. MSC2010: 20N20, 06F15, 06F05. 1 Introduction and Preliminaries The notion of a hyperstructure and hypergroup, as a generalization of group, was introduced by F. Marty [5] in 1934 at the 8th congress of Scan- dinavian Mathematicians. In this definition for nonempty set H, a function · : H×H −→ P∗(H), where P∗(H) is the set of all nonempty subsets of H, is called a hyperoperation on H, and the system (H, ·) is called a hypergroupoid. If the hypergroupoid H satisfies a ·H = H ·a = H, for all a ∈ H, it is called a hypergroup. In a hypergroupoid H, for A,B ⊆ H and x ∈ H, A · B and A ·x are defined as A ·B = ⋃ a∈A,b∈B a · b,A ·x = A · {x}. An element e of hypergorupoid H is called an identity if for all a ∈ H, a ∈ a◦e∩e◦a. An element a′ ∈ H is called an inverse for a ∈ H if there is an identity e ∈ H such that e ∈ a◦a′ ∩a′ ◦a. By a subhypergroupoid of hypergroupoid H we mean a subset K of H that is closed with respect to the hyperoperation on H, and contains the unique identity of H and the inverses of its elements, provided there exist. M. Bakhshi, R. A. Borzooei Hyperstructures have many applications to several sectors of both pure and applied sciences. A short review of the theory of hyperstructures appear in [2]. In [3] a wealth of applications can be found, too. There are appli- cations to the following subjects: geometry, hypergraphs, binary relations, lattices, fuzzy set and rough sets, automata, cryptography, combinatorics, codes, artificial intelligence and probabilities. Polygroups are certain sub- classes of hypergroups which studied in 1981 by Ioulidis in [4] and are used to study colour algebra. A polygroup is a system < G, ·,−1 ,e > where e ∈ G, ‘−1’ is a unary oper- ation on G and ‘·’ is a binary hyperoperation on H satisfying the following: (1) (x ·y) ·z = x · (y ·z), (2) e ·x = x ·e = {x}, (3) x ∈ y ·z ⇔ y ∈ x ·z−1 ⇔ z ∈ y−1 ·x. In any polygroup the following hold: e ∈ x ·x−1 ∩x−1 ·x, e−1 = e, (x−1)−1 = x, (x ·y)−1 = y−1 ·x−1 where A−1 = {x−1 : x ∈ A}. Some other concepts in polygroups is as follows. A nonempty subset K of polygroup G is said to be a subpolygroup if and only if e ∈ K and < K, ·,−1 ,e > is itself a polygroup. Subpolygroup K of polygroup G is said to be normal if and only if a−1Ka ⊆ K, for all a ∈ G. From now on, in this paper, G =< G, ·,−1 ,e > will denote a polygroup. 2 Ordered hyperstructures: Definition and properties This section is devoted to introduce the concept of a compatible order on a polygroup. It is first introduced the concept of an ordered hypergroupoid and some basic notions. Then, the concept of ordered polygroups is introduced and some related results are given. For more details on compatible orders, specially ordered algebraic structures we refer to [1]. Definition 2.1. Let (H, ·) be a hypergroupoid. By a compatible order on H we mean an order “≤” with respect to which all translations x 7→ x ·y and x 7→ y ·x are isotone, that is x ≤ y implies b ·x ·a ≤ b ·y ·a, for all a,b ∈ H (2.1) 32 Ordered Polygroups where for A,B ⊆ H, A ≤ B means that for all a ∈ A there exists b ∈ B and for all b ∈ B there exists a ∈ A such that a ≤ b. Definition 2.2. By an ordered hypergroupoid we mean a hypergroupoid on which is defined a compatible order. When “·” is commutative or associative, H is said to be an ordered com- mutative hypergroupoid or an ordered semihypergroup, respectively. Example 2.3. (1) Consider R1 = [1,∞), the set of all real numbers greater than 1, as a poset with the natural ordering, and define x ·y to be the set of all upper bounds of {x,y}. Thus (R1, ·,≤) is an ordered commu- tative semihypergroup with 1 as the unique identity. (2) Consider Z, the additive group of all integers which is a chain with the natural ordering. For m,n ∈ Z, let m·n be the subgroup of Z generated by {m,n}. Then (Z, ·,≤) is an ordered commutative semihypergroup in which 0 is an identity. (3) Let (G, ·,e,≤) be an ordered group, and let x◦y = 〈{x,y}〉, the sub- group of G generated by {x,y}. Then, (G,◦,≤) is an ordered commu- tative hypergroup with an identity e. (4) Let (L;∨,∧, 0) be a lattice with the least element 0. For a,b ∈ L, let a◦b = F(a∧b), where F(x) is the principal filter generated by x ∈ L. Then, (L;◦) is an ordered hypergroup. Also, 0 is an identity, and if x ∈ L be such that x∧y = 0, for some y ∈ L, then y is an inverse of x. Definition 2.4. Let H be an ordered hypergroupoid. (1) For every x,y ∈ H with x ≤ y, the set [x,y] = {z ∈ H : x ≤ z ≤ y} is said to be an interval in H. (2) A subset A of H is said to be convex if for all a,b ∈ A, where a ≤ b, we have [a,b] ⊆ A. Definition 2.5. Let (E;≤) be an ordered set. A subset D of E is said to be a down-set if y ≤ x and x ∈ D imply y ∈ D. Down-set D is said to be principal if there exists x ∈ D such that D = {y ∈ E : y ≤ x} denoted by x↓. Definition 2.6. Let (G;◦G,≤G) and (H;◦H,≤H ) be ordered hypergroupoids and f : G −→ H be an isotone map, that is f(x) ≤H f(y) whenever x ≤G y. Then, 33 M. Bakhshi, R. A. Borzooei (1) f is said to be an order homomorphism if f is a homomorphism of hypergroupoids (G; ,◦G) and (H;◦H ), (2) f is an order isomorphism if f is an isomorphism of hypergroupiods, and f−1 is isotone, (3) the kernel of f is defined by kerf = {(x,y) ∈ G×G : f(x) = f(y)}. 3 Ordered polygroups In this section, we assume that G =< G, ·,−1 ,e > is a polygroup unless otherwise mentioned. Hereafter, in this paper, we use xy for x ·y, and a for {a}. Definition 3.1. By an ordered polygroup we mean a polygroup which is also a poset under the binary relation ≤ and in which (2.1) holds. Definition 3.2. Let H be an ordered hypergroupoid with a unique identity e. An element x ∈ H is called positive if e ≤ x. The set of all positive elements of H is called the positive cone of H and is denoted by H+. x ∈ H is called negative if x ≤ e. The set of all negative elements of H is called the negative cone of H and is denoted by H−. By an elementary consequence of translations we have Proposition 3.3. In any ordered polygroup G, for each x,y ∈ G, we have x ≤ y ⇔ x−1y ∩G+ 6= ∅ ⇔ yx−1 ∩G+ 6= ∅ ⇔ xy−1 ∩G− 6= ∅ ⇔ y−1x∩G− 6= ∅⇔ y− ≤ x−. Theorem 3.4. A subset P of a polygroup G is the positive cone with respect to some compatible order if and only if (1) P ∩P−1 = {e}, (2) P 2 = P , (3) for all x ∈ G, xPx−1 = P . Moreover, if this order is total, P ∪P−1 = G. Proof. (⇒) Let ≤ be a compatible order on G and P = G+, the associ- ated positive cone. (1) If x ∈ P∩P−1, on the one hand e ≤ x, and on the other hand x = y−1, for some y ∈ P . Since, e ≤ y, then x = y−1 ≤ e proves that x = e. 34 Ordered Polygroups (2) Since e ∈ P , P = Pe ⊆ PP = P 2. Now, let x,y ∈ P . Then e ≤ x and e ≤ y and so e ≤ xy which implies that xy ⊆ P . Hence, P 2 ⊆ P . (3) Let y ∈ P , and x ∈ G. Then, e ≤ y implies that e ∈ xex−1 ≤ xyx−1 proves that xyx−1 ⊆ P . Since, this follows for all x ∈ G, replacing x by x−1, we have x−1Px ⊆ P and so P ⊆ xPx−1, complete the proof. (⇐) Let P be a subset of G that satisfies properties (1)-(3), and define the relation ≤ on G by x ≤ y ⇔ yx−1 ∩P 6= ∅. Since, e ∈ P , by (3), xx−1 = xex−1 ⊆ xPx−1 = P implies that x ≤ x and so ≤ is reflexive. Suppose that x ≤ y and y ≤ x, for x,y ∈ G. Then yx−1∩P 6= ∅ and xy−1 ∩P 6= ∅ whence xy−1 ∩P−1 ∩P 6= ∅, implies that e ∈ xy−1, i.e., x = y proving ≤ is antisymmetric. Now, assume that x ≤ y and y ≤ z, for x,y,z ∈ G. Then yx−1 ∩ P 6= ∅ and zy−1 ∩ P 6= ∅. Let u ∈ yx−1 ∩ P and v ∈ zy−1 ∩P . Then uv ⊆ P 2 = P . On the other hand, ∈ zy−1 and v ∈ yx−1 imply y−1 ∈ z−1u and y ∈ vx whence e ∈ y−1y ⊆ z−1(uv)x. Then, there is t ∈ uv and s ∈ tx such that e ∈ z−1s. This implies that z = s ∈ tx. Hence, t ∈ zx−1, i.e., uv ∩ zx−1 6= ∅ whence zx−1 ∩ P 6= ∅ proving ≤ is transitive. Thus, ≤ is an order. For compatibility, we first prove that Px = xP , for all x ∈ G. Let z ∈ G. Then z ∈ Px ⇒ z ∈ yx for some y ∈ P ⇒ x−1z ⊆ x−1yx = x−1y(x−1)−1 ⊆ P ⇒ z ∈ xP, i.e., Px ⊆ xP . By a similar way, we can prove that xP ⊆ Px. Hence, xP = Px, for all x ∈ G. Now, assume that x ≤ y and a,b ∈ G. Since, ≤ is reflexive, by (3) ayb(axb)−1 = aybb−1x−1a−1 ⊆ ayPx−1a−1 ⊆ aPyx−1a−1 ⊆ aP 2a−1 = aPa−1 = P which shows that axb ≤ ayb. By the definition of ≤ we get x ∈ P if and only if e ≤ x and so P = G+. If G is totally ordered, then x ≤ e or e ≤ x, for all x ∈ G. So, e ∈ xx−1 ≤ ex−1 = x−1 and so x ∈ P or x ∈ P−1, observe that x = (x−1)−1. Thus, G = P ∪P−1. 2 Proposition 3.5. If G is an ordered polygroup with |G| > 1, then G can not have a top element or a bottom element. Proof. Let G = {e,a}. If e < a or a < e, then a = a−1 < e or e < a−1 = a, respectively, which is a contradiction. Now, assume that 35 M. Bakhshi, R. A. Borzooei |G| > 2, t be the top element of G and e 6= a ∈ G. Then a ≤ t and so ta ≤ t whence t ∈ te ⊆ taa−1 ≤ ta−1. Hence, t ∈ ta−1. Likewise, we conclude that t ∈ a−1t. By the uniqueness, we get a = e which is a contradiction. The proof of the other case is concluded as well. 2 Definition 3.6. An element x of G is said to be of order n, n ∈ N, if e ∈ xn where xn = (· · ·(( n times︷ ︸︸ ︷ x◦x) ◦x) ◦ · · ·) ◦x). If such a natural number does not exist, we say that x is of infinite order. Theorem 3.7. Suppose that G is an ordered polygroup in which G+ 6= {e}. Then every element of G+ \{e} is of infinite order. Proof. Suppose that x ∈ G+ \{e}. We first observe that if x = x−1, x can not belong to G+. Then, e < x implies that e < x = ex < x2. Moreover, this implies that e 6∈ x2. Similarly, we conclude that e < x3 and e 6∈ x3. Continuing this process we get e < xn and e 6∈ xn, for all n ∈ N, proving x is not of finite order. 2 Corollary 3.8. Any ordered polygroup in which every nontrivial element is of finite order is an antichain. Proof. Let G be an ordered polygroup satisfying the hypothesis. By Theorem 3.7, we know that G+ = {e}. Now, if a,b ∈ G be such that a ≤ b, then e ∈ a−1a ≤ a−1b and so e ≤ u, for some u ∈ a−1b. This implies that u ∈ G+ and so u = e. Thus, e ∈ a−1b whence a = b. This means that G is an antichain. 2 Corollary 3.9. Every finite ordered polygroup is an antichain. Example 3.10. Let G = {e,a}. Then G is a polygroup where the hyperop- eration is given by the following table: ◦ e a e e a a a {e,a} in which a−1 = a i.e., a is an idempotent. Now, if a is a positive element, so G+ = {e,a} and hence (G+)−1 ∩ G+ 6= {e}. This contradicts Theorem 3.4. This example shows that the converse of Theorem 3.7 does not hold in general. Definition 3.11. If G is an ordered polygroup, by a convex subgroup of G we shall mean a subgroup which is also a convex subset, under the order of G. 36 Ordered Polygroups Definition 3.12. A nonempty subset H of G is said to be S-reflexive if xy ∩H 6= ∅ implies that xy ⊆ H, for all x,y ∈ G. Theorem 3.13. If H is a subpolygroup of an ordered polygroup G then H+ = H ∩ G+. Moreover, if H+ is S-reflexive, the following statements are equivalent: (1) H is convex; (2) H+ is a down-set of G+. Proof. Since, eH = eG, it is clear that H + = H ∩G+. (1) ⇒ (2) Suppose that eH ≤ y ≤ x where eH,x ∈ H+ ⊆ H. Then (1) gives y ∈ H ∩G+ = H+ and so H+ is a down-set of G+. (2) ⇒ (1) Suppose now that x ≤ y ≤ z where x,z ∈ H. Then x−1x ≤ x−1y ≤ x−1z. Thus, x−1z ⊆ H+ and so there is a ∈ H+ such that a ∈ x−1z. Hence, there is b ∈ x−1y such that b ≤ a ∈ H+, and since H+ is a down-set of G+, b ∈ H+, i.e., x−1y∩H+ 6= ∅. Since, H+ is S-reflexive, so x−1y ⊆ H+ ⊆ H whence y ∈ xH = H, proving H is convex. 2 If G is an ordered polygroup and H is a normal subpolygroup of G, then a natural candidate for a positive cone of G/H is \H (G +), where \H : G −→ G/H is the canonical projection. Precisely when this occurs is the substance of the following result. Theorem 3.14. Let G be an ordered polygroup and let H be a normal sub- polygroup of G. Then \H (G +) = {pH : p ∈ G+} is the positive cone of a compatible order on the quotient polygroup G/H if and only if H is convex. Proof. Suppose that Q = {pH : p ∈ G+} is the positive cone of a compatible order on G/H. To show that H is convex, suppose that c ≤ b ≤ a with c,a ∈ H. Then (bH)−1 = (bH)−1 · aH = b−1aH. On the other hand, b ≤ a implies that b−1a∩G+ 6= ∅. Hence (bH)−1∩Q 6= ∅ and so bH∩Q−1 6= ∅. Similarly, we have bH = bH · c−1H = bc−1H and since bc−1 ∩ G+ 6= ∅, bH ∩Q 6= ∅. Thus, bH ∩ (Q∩Q−1) 6= ∅ whence bH = H, i.e., b ∈ H. Conversely, suppose that H is convex and let Q = {pH : p ∈ G+}. It is clear that Q2 = Q. Suppose now that xH ∈ Q ∩ Q−1. Then xH = pH = q−1H where p,q ∈ G+. These equalities also give pq ∩ H 6= ∅. Now, since p ≤ pq, then eH ≤ p ≤ u, where u ∈ pq∩H whence the convexity of H gives p ∈ H. It follows that xH = pH = H and hence Q∩Q−1 = {H}. Finally, since G+ is a normal subsemihypergroup of G it is clear that Q = \H (G +) is a normal subsemihypergroup of G/H. It now follows by Theorem 3.4 that Q is the positive cone of a compatible order on G/H. 2 37 M. Bakhshi, R. A. Borzooei If H is a convex normal subpolygroup of an ordered polygroup G then the order ≤H on G/H that corresponds to the positive cone {pH : p ∈ G+} can be described as in the proof of Theorem 3.4. We have xH ≤H yH ⇒ yx−1H ⊆ Q ⇒ (∀a ∈ yx−1)(∃p ∈ G+)aH = pH ⇒ (∀a ∈ yx−1)(∃p ∈ G+)(∃h ∈ H)a ∈ ph ≥ h ⇒ (∀a ∈ yx−1)(∃h ∈ H) a ≥ h ⇒ yx−1 ≥ h. From the last inequality and that y ∈ ye ⊆ yx−1x it follows that y ≥ u, for some u ∈ hx. Conversely, assume that there exists h ∈ H and u ∈ hx such that y ≥ u, and let a ∈ yx−1. From yx−1 ≥ yx−1 it follows that a ≥ t, for some t ∈ ux−1 and hence at−1 ≥ tt−1. This implies that v ≥ e, for some v ∈ at−1 and so vH ∈ at−1H ∩Q. (3.1) Now, t ∈ ux−1 implies that t−1 ∈ xu−1 ⊆ xx−1h−1 ⊆ xx−1H and so at−1 ⊆ axx−1H = axHx−1 = aH. Thus, at−1H ⊆ aH. Combining (3.1), we get {aH}∩Q 6= ∅, i.e., aH ∈ Q and so aH = pH, for some p ∈ G+. This implies yx−1H ⊆ Q and hence xH ≤H yH, completes the proof. Thus we see that ≤H can be described by xH ≤H yH ⇔ (∃h ∈ H)(∃u ∈ hx) y ≥ u. In referring to the ordered quotient polygroup G/H we shall implicitly infer that the order is ≤H as described above. Here we give a characterization of polygroup homomorphisms that are isotone. Theorem 3.15. Let G and H be ordered polygroups. If f : G −→ H is a polygroup homomorphism, f is isotone if and only if f(G+) ⊆ H+. Proof. Assume that f is isotone. If x ∈ G+, i.e., x ≥ e then f(x) ≥ f(eG) = eH means that f(x) ∈ H+. Conversely, assume that x ≤ y in G. Then yx−1 ⊆ G+ and so f(y)f(x)−1 = f(yx−1) ⊆ f(G+) ⊆ H+. This implies that f(y) ≥ f(x) proving f is isotone. 2 Corollary 3.16. If G is an ordered polygroup and H is a convex normal subpolygroup of G, then the natural homomorphism \H : G −→ G/H is isotone. 38 Ordered Polygroups Proof. By Theorem 3.15, it is enough to prove that \(G+) ⊆ (G/H)+. For this, let yH ∈ \(G+). Then yH = gH, for some g ∈ G+ whence y ∈ gh ≥ h for some h ∈ H. This implies that eH ≤H yH and so yH ∈ (G/H)+. 2 Definition 3.17. Let G and H are ordered polygroups. A mapping f : G −→ H is said to be exact if f(G+) = H+. Definition 3.18. Two ordered polygroups G and H are said to be isomorphic if there is a polygroup isomorphism f : G −→ H that is also an order isomorphism. If two ordered polygroups G and H are isomorphic we write G '̇ H. Theorem 3.19. For ordered polygroups G and H, the following are equiva- lent: (1) G'̇H, (2) there is an exact polygroup isomorphism f : G −→ H. Proof. (1) ⇒ (2) If G and H are isomorphic, there is a polygroup isomorphism f : G −→ H which is also an order isomorphism. By Theorem 3.15, f(G+) ⊆ H+. Let g = f−1. Obviously, g satisfies the conditions of Theorem 3.15. Hence, g(H+) ⊆ G+ whence H+ = f(g(H+)) ⊆ f(G+). Thus H+ = f(G+) and so (2) holds. (2) ⇒ (1) It is obvious. 2 Theorem 3.20. Let G and H be ordered polygroups and f : G −→ H be an exact polygroup homomorphism. Then Imf '̇ G/kerf. Proof. We first observe that kerf is a convex normal subpolygroup of G and so G/kerf is an ordered polygroup. By first isomorphism theorem of polygroups there is an isomorphism φ : G/kerf ' Imf which φ(xK) = f(x) where K = kerf. It remains that we prove φ is exact. Let xK ∈ (G/K)+. Then eGK ≤K xK whence k ≤ x, for some k ∈ K, and so eH = f(k) ≤ f(x) whence φ(xK) = f(x) ∈ (Imf)+. Conversely, if f(x) ∈ (Imf)+ ⊆ H+, since f is exact, there exists g ∈ G+ such that f(x) = f(g). Consequently, xK = gK and so x ∈ gk ≥ k, for some k ∈ K. Thus, xK ∈ (G/K)+ ⇔ φ(xK) = f(x) ∈ (Imf)+ proving φ is exact. It now follows by Theorem 3.19 that G/kerf '̇ Imf. 2 39 M. Bakhshi, R. A. Borzooei References [1] T. S. Blyth, Lattices and Ordered Algebraic Structures, Springer-Verlag, London, 2005. [2] P. Corsini, Prolegomena of hypergroup theory, 2nd edition, Aviani editor, 1993. [3] P. Corsini and V. 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