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Cyclic Groups Obtained as Quotient Hypergroups

Cyclic Groups Obtained as Quotient Hypergroups We introduce a strongly regular equivalence relation on the hyperA H group H, such that in a particular case the quotient is a cyclic group. Then by A using the notion of -parts, we investigate the transitivity condition of A . Finally, a A characterization of the derived hypergroup Dc (H) has been considered. Mathematics Subject Classification 2010: 20N20. Key words: hypergroup, strongly regular relation, complete parts. 1. Introduction Hyperstructure theory was born in 1934 when Marty [16] introduced hypergroups as a generalization of groups at the 8th congress of Scandinavian Mathematicions. After that, he proved the utility of hypergroups in solving some problems of groups, algebraic functions and rational fractions. Complete enough surveys of the algebraic hyperstructure theory are the books by Corsini [2], Corsini and Leoreanu [3], Vougiouklis [19]. Some relations in groupoids and hyperstructures which have been considered since the beginning are A, An , , n , and n , see Corsini [1, 4], Davvaz and Karimian [5, 6, 12], Freni [8, 9], Koskas [10, 11], Leoreanu [13, 14], Romeo [17], Vougiouklis [18], Zhan [20] and others. Using the relation , complete parts are introduced by Koskas [11] and studied by Corsini, Migliorato and Sureau [2, 3, 15]. This research has been supported financially by Mahani Mathematical Research Cen- ter. We present now some basic definitions and results about hypergroups (see [3]). Let H be a non-empty set and P (H) be the set of all non-empty subsets of H. Let be a hyperoperation or a join operation on H, that is, is a function from H × H into P (H) . If (a, b) H × H, then its image under in P (H) is denoted by a b or ab. The join operation is extended to subsets of H in a natural way, that is A B = {ab | a A, b B}. The notation aA is used for {a}A and Aa for A{a}. Generally, the singleton {a} is identified with its member a. The structure (H, ) is called a semihypergroup if a(bc) = (ab)c for all a, b, c H and is called a hypergroup if it is semi-hypergroup and aH = Ha = H for all a H. Let H be a hypergroup and K be a non-empty subset that is closed under the hyperoperation in H. If K is itself a hypergroup under the hyperoperation in H, then K is said to be a subhypergroup of H. Suppose (H, ) and (H , ) are two semi-hypergroups. A function f : H H is called a homomorphism if f (a b) f (a) f (b) for all a and b in H. We say f is a good homomorphism if for all a and b in H, f (a b) = f (a) f (b). If (H, ) is a hypergroup and R H × H is an equivalence relation, we set = A R B a R b, a A, b B, for all pairs (A, B) of non-empty subsets of H. The relation R is said to be left (right) strongly regular if x R y a x R a y (resp. x R y x a R y a), for all (x, y, a) H 3 . Moreover, R is called strongly regular if it is both left and right strongly regular. If (H, ) is a semi-hypergroup (hypergroup) and R is strongly regular, then the quotient H is a semigroup (group respectively) under the operation: R R(x) R(y) = R(z), z x y. = = For all n > 1 define the relatio on a semi-hypergroup H, as follows: a n b (x1 , ..., xn ) H n : {a, b} i=1 xi and = , where 1 = {(x, x) | x H} is the diagonal relation i=1 H. If is the transitive closure of then is an equivalence relation (see [2]). If H is a hypergroup and we consider the canonical projection : H H/ then the heart of H is the inverse image through of the identity of the group H/ and it is denoted by H . Hence H = -1 (eH/ ). We call the kernel of a homomorphism f , the set K(f ) = f -1 (H ), where H is the heart of H . The following result was shown by Freni [7]; another proof can be found in [2]. Theorem 1.1. If H is hypergroup then = . A non-empty subset A of H is called a complete part of H if for all n and for all (x1 , x2 , ..., xn ) H n we have the following implication: xi i=1 A= i=1 xi A. A subhypergroup K of a hypergroup (H, ·) is called · closed if from K aK = or K Ka = , where a H, it follows that a K; · invertible if from a bK it follows that b aK and from a Kb it follows that b Ka, where a, b H; · normal if for all a H, aK = Ka. Remark 1.2. The kernel of every homomorphism is a complete part and a normal subhypergroup. Proposition 1.3. Let K be a subhypergroup of a hypergroup H. (i) If K is a complete part, then it is invertible; (ii) If K is a invertible, then it is closed. In [8], Freni introduced the relation = n1 n , where 1 is the diagonal relation and for every integer n > 1, n is the relation defined as follows: x n y (z1 , ..., zn ) H n , Sn , : x i=1 zi , y i=1 z(i) , where Sn is the symmetric group of order n. In [5], Davvaz and Karimian introduced the concept of -parts and studied some properties of C (A), where C (A) is the intersection of all parts of a semi-hypergroup H. The transitive closure of is the smalest strongly regular equivalence such that the quotient H/ is commutative semigroup. Moreover, if H is hypergroup then is transitive (see [8]). Throughout this paper, H is a hypergroup and C = aA Ha , where = A H, Ha = an , n N and for all a, b A if a = b, then Ha Hb = . 2. The relation A In this section we introduce the relation A on a hypergroup H, which we use in order to obtain a cyclic group as a quotient structure of H. For all n 1 define A as follows A : = A A A , where : = A : = n A def def yi ) | Sn , i=1 yi = i=1 x(i) , def i=1 i=1 yi ) | a A, {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = , a Sn , yj {y1 , ..., yn } Ha yj = xa (j) and A : = n def i=1 i=1 def yi ) | {x1 , ..., xn } C = {y1 , ..., yn } C = . Notice that A : = {({x}, {y}) | {x, y} C = or x = y} and 1 A : = 2 def (xz, yz), (xz, zy), (zx, zy), (zx, yz) | ({x}, {y}) A , z H . 1 For every n 1, define the relatio on H by xn y (A, B) A , x A A n A, y B. Notice that for n = 1 we obtain x1 y ({x}, {y}) A or A 1 x = y C. Obviously, for every n 1, the relations n are symmetric, A and the relation A = n 1 n is reflexive and symmetric. Let be the A A transitive closure of A . Theorem 2.1. The relation is a strongly regular relation. A Proof. Notice that is an equivalence relation. In order to prove that A it is strongly regular, we have to show first that: (2.1) xA y xz A yz, zx A zy. = = for every z H. Suppose that xA y, thus there exists n 1 such that xn y. A Therefore there exists (B, C) A such that x B and y C. We obtai the following three cases: Case 1. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi and there exist Sn such that i=1 yi = i=1 x(i) . Set z = yn+1 = xn+1 and let be the permutation of Sn+1 such that: (i) = (i), n + 1, if i {1, 2, ..., n}; if i = n + 1. So Cz = n+1 x (i) and hence (Bz, Cz) A . +1 Case 2. Suppose that (B, C) A so we have B = n C = i=1 yi and there exists a A such that {x1 , ..., xn } Ha = {y1 , ..., yn } i=1 Ha = and a Sn such that for all 1 j n if yj Ha , then yj = xa (j) . If z C, then it is easy to see that (Bz, Cz) A . If z C, then there / n+1 exists b A such that z Hb . Set z = yn+1 = xn+1 and let a be the following permutation of Sn+1 : a (i) = a (i), n + 1, if i {1, 2, ..., n}; if i = n + 1. So for 1 j n + 1 if yj Ha , then yj = xa (j) . Therefore (Bz, Cz) A n+1 and hence (Bz, Cz) A . n+1 Case 3. Suppose that (B, C) A so we have B = n C = i=1 A / i=1 yi , {x1 , ..., xn }C = {y1 , ..., yn }C = . If z C, then (Bz, Cz) n+1 A . and if z C, then (Bz, Cz) n+1 Therefore for all v xz and for all y yz, we have v xz Bz and u yz Cz, so vm u for some m N, because (Bz, Cz) A for some A m = m N and hence vA u. Thus xz A yz. In the same way we obtain = that xA y zx A zy. Moreover, if x y, then there exists m N and A (u0 = x, u1 , ..., um = y) H m such that x = u0 A u1 A ...A um-1 A um = y, whence, by (2.1), we obtain that xz = u0 z A u1 z A u2 z A ... A um-1 A um z = yz. = = = = = Finally, for all v xz = u0 z and for all u um z = yz, taking z1 u1 z, z2 u2 z, ..., zm-1 um-1 z, we have vA z1 A z2 A ...A zm-1 A u, and so v u. Therefore x y xz yz. A A A Similarly we can prove that x y zx zy, hence is strongly A A A regular. H Theorem 2.2. The quotient is an abelian group with the generators A { (x0 ), (a) | x0 (H - C), a A}. A A Proof. By Theorem 2.1, is a strongly regular equivalence, so the A H quotient is a group under the following operation (x) (y) = A A A (z), z xy. A H Since , we conclude that is an abelian group. For all (x, y) A (H - C)2 since {x, y} C = , we have ({x}, {y}) A and hence x y A 1 so (x) = (y). Now suppose that (h) is given. If h (H - C), then A A A (h) = (x0 ) and if h C, then there exists a A such that h Ha . A A Therefore (h) = [ (a)]k for some k N. A A ow suppose that A = {a}, so C = Ha . Put a : = n , a = A and a : = . So we have the following corollary: A def def n 1 n a Corollary 2.3. The quotient H a is a cyclic group. Proof. By the proof of Theorem 2.2, we conclude that the equivalence classes determined by of all elements of (H - C) coincide and the A equivalence class of every element of C is a power of a (a). If H - C = , and x0 H - C then there exists u H - C such that x0 au since H is a hypergroup. We obtain a (x0 ) = a (a) a (u) = a (a) a (x0 ) whence a (a) = e H the identity element of the group H . >. Moreover, since a x0 v for some v H - C, we =< Hence (x ) = ( (x ))-1 which means that the order of (x ) is obtain that a 0 a 0 a 0 two, so in this case H is a cyclic group of order two. a Now if H - C = , which means that H is a cyclic hypergroup, then H =< a (a) > . H a a (x0 ) Therefore H a is a cyclic group. Example 2.4. Let S3 be the permutation group of order 3, i.e., S3 = {(1), (12), (13), (23), (123), (132)} and a = (123) be a cycle of order 3 in S3 . Then S3 Z2 . a = Proof. It is easy to see that A = {(x, y) | x, y are odd permutations or 1 x = y} and A = {(z, t) | z, t are odd or z, t are even permutations} and 2 A = A = a = , so S3 Z2 . n = 2 Theorem 2.5. The relation a is the least strongly regular equivalence such that the quotient H is a cyclic group and the equivalence classes of a all elements of (H - C) are equal. Proof. Let R be a strongly regular equivalence such that H is a cyclic R group and the equivalence classes of R of all elements of (H - C) are equal. Suppose that : H H is the canonical projection. is a good homoR morphism. We show that a R. n y. So there exists (B, C) A such that x B and Let n 1 and x a n y C . We have three cases: Case 1. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , and there exist Sn such that i=1 yi = i=1 x(i) . Since (x) = n (xi ) and (y) = n (yi ) = n (x(i) ), it follows that i=1 i=1 i=1 (x) = (y), because H is commutative. Therefore xRy. R Case 2. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = and there exists Sn such that for all 1 j n if yj Ha , then yj = x(j) . Renumber of the elements of the sets {x1 , ..., xn } and {y1 , ..., yn } such that {y1 , ..., ym } C, where 1 m n and xk , yk C for all m + 1 k n. So we have (x) = / (xk )) and (y) = (m (x(j) ))(n (m (xj ))(n j=1 j=1 k=m+1 (yk )). k=m+1 H For all m + 1 t, l n, (xt ) = (yl ) and since R is a commutative group, we have (x) = (y) and hence xRy. Case 3. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , {x1 , ..., xn } C = {y1 , ..., yn } C = . So for all 1 i, j n, (xi ) = (yj ), thus (x) = (y) and hence xRy. In all cases we have xRy and hence x a y xRy whence x a y xRy R. by transitivity of R. Therefore a 3. Transitivity condition of A In this section we introduce the concept of -part of a hypergroup H A and we determine necessary and sufficient conditions such that the relation A to be transitive. For this purpose, we will follow arguments similar to those exploited by Freni in [8] in the study of fundamental relation in hypergroups. Definition 3.1. Let M be a non-empty subset of H and n xi M = i=1 . We say that M is a -part of H if the following conditions hold: A (P1 ) For all Sn , n x(i) M ; i=1 (P2 ) If a A, {x1 , ..., xn } Ha = implies that for all n yi such i=1 that {y1 , ..., yn } Ha = {x1 , ..., xn } Ha there exists a Sn for which yj {y1 , ..., yn } Ha , yj = xa (j) , the yi M ; i=1 (P3 ) If {x1 , ..., xn }C = then for all n yi such that {y1 , ..., yn }C = i=1 , we have n yi M . i=1 Proposition 3.2. Let M be a non-empty subset of a hypergroup H. The following conditions are equivalent: (i) M is a -part; A (ii) x M, xA y y M ; (iii) x M, x y y M . A Proof. (i) (ii) Suppose that the pair (x, y) H 2 is such that x M and xA y. Then there exists (B, C) A such that x B and y C. So n we have three cases. Case 1. Let (B, C) A so we have B = n C = n yi and i=1 there exists Sn such that n yi = n x(i) . Since x n xi M , i=1 i=1 i=1 by (P1 ) we have n x(i) M and hence y M . i=1 Case 2. Let (B, C) A . So B = n C = n yi , there exists i=1 a A , such that {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = and there exists a Sn such that for all 1 j n if yj Ha , then yj = xa (j) . Since x n xi M , by (P2 ) we have n yi M and hence y M . i=1 i=1 Case 3. Suppose that (B, C) A so we have B = n C = i=1 yi , {x1 , ..., xn } C = {y1 , ..., yn } C = . Since x n xi M , by i=1 i=1 (P3 ) we have n yi M and hence y M . i=1 (ii) (iii) Suppose that (x, y) H 2 is such that x M and x y. A So there exist m N and (w0 = x, w1 , ..., wm-1 , wm = y) H m such that x = w0 A w1 A w2 ...A wm-1 A wm = y. Since x M , applying (ii) m times, we obtain y M . (iii) (i) Let x n xi M . We shall check the conditions (P1 ), i=1 (P2 ) and (P3 ). (P1 ) Let Sn be given, thus ( n n x(i) ) A . So for all i=1 y n x(i) we have x y and hence y M . Therefore n x(i) M . i=1 A i=1 (P2 ) Let {x1 , ..., xn } Ha = for some a A. Now suppose that for all n i=1 yi if {y1 , ..., yn } Ha = {x1 , ..., xn } Ha there exists a Sn such that yj {y1 , ..., yn } Ha , yj = xa (j) . Thus ( n n yi ) A and hence i=1 for all y n yi we have x y. Therefore y M and so n yi M . i=1 A i=1 (P3 ) Let {x1 , ..., xn }C = and n yi be such that {y1 , ..., yn }C = . i=1 Thus ( n n yi ) A and hence for all y n yi we have x y. i=1 i=1 i=1 A n Therefore y M and so n yi M . otation 3.3. Let x be an arbitrary element of a hypergroup H. For n 1, set Pn (x) Pn (x) Pn (x), where x(i) | x xi ; {y1 , ..., yn } Ha = {x1 , ..., xn } Ha , i=1 yi | x i=1 a Sn , yj {y1 , ..., yn } Ha yj = xa (j) ; i=1 yi | x i=1 {y1 , ..., yn } C = {x1 , ..., xn } C = . Denote P (x) = n 1 Pn (x). Notice that if x C, then P1 (x) = H - C and / P2 (x) = zH dz, zd | d (H - C), b (H - C), x bz zb , while if x an C for some n N, then Pm (x) = an for all m 1. We have the following: Proposition 3.4. For every x H, P (x) = {y H | xA y}. Proof. Suppose that x H and y P (x) are given. Then there exist B and C such that x B, y C and: (i) y Pn (x) (B, C) A ; n (ii) y Pn (x) (B, C) A ; n (iii) y Pn (x) (B, C) A . n Therefore xA y and so P (x) {y H | xA y}. The proof of the reverse of the inclusion is similar to the above. Lemma 3.5. Suppose that H is a hypergroup and M is a -part of A H. If x M , then P (x) M . Proof. It follows by Definition 3.1. Theorem 3.6. Let H be a hypergroup. The following conditions are equivalent: (i) A is transitive; (ii) for every x H, (x) = P (x); A (iii) for every x H, P (x) is a -part of H. A Proof. (i) (ii) By Proposition 3.4, for all pair (x, y) H 2 we have: y (x) x y xA y y P (x). A A (ii) (iii) By Proposition 3.2, if M is a non-empty subset of H, then M is a -part of H if and only if it is a union of equivalence classes modulo A . Particularly, every equivalence class modulo is a -part of H. A A A (iii) (i) Suppose that xA y and yA z so x P (y) and y P (z). By Lemma 3.5, we have P (y) P (z) and hence x P (z). By Proposition 3.4 it follows that xA z and the proof is complete. 4. A characterization of a new derived hypergroup of a hypergroup Let H be a hypergroup and A be an invertible subhypergroup of H. For all x H, there exists y H such that x Ay. Therefore the relation A E on H defined by: x A E y x Ay is a regular equivalence and a conH gruence. Moreover the quotient A E = {Ax | x H} is a hypergroup with respect to the hyperoperation Ax Ay = {Az | z xAy}. and the canonH ical projection : H A E is an almost strong homomorphism (see [2]). This means that if (x) = (a) (t), then there exist a , t H such that (a) = (a ), (t) = (t ) and x a t . In this section we suppose that A = {a} and a (a) = e H . Proposition 4.1. For all x (H - C), a (x) = e Proof. Suppose that a (x) = e H a H a =< a (a) >, so a (x) = a (t) such that t an for some n N. Therefore x t and hence there exist the elements x = x1 , ..., xn-2 , xn-1 , xn = t in H such that x = x1 a ... xn-2 a xn-1 a xn = t. Thus xn-1 a t and by Proposition 3.4 it follows that xn-1 P (t). Since t an , P (t) = an and hence xn-1 an . Similarly from xn-2 a xn-1 we obtain xn-2 an . After (n - 1) steps, we conclude that x = x1 an C and it is a contradiction. H a . By Theorem 2.5, For all pair (a, b) H 2 and for all pair (A, B) of non-empty subsets of H, we set: a/b = {x H | a xb}, a\b = {x H | b ax}, A/B = aA,bB a/b, A\B = aA,bB a\b. Denote D1 = (x,y)H 2 xy/yx, D2 = (x,y)H 2 xy\yx, D3 = H - C and Dc = D1 D2 D3 . Define the derived hypergroup Dc (H) as the intersection of all subhypergroups, that are complete parts and contain Dc . Set D = D1 D2 . Since H D Dc (H), it follows that D (H) E is a commutative group (see [8]). Proposition 4.2. If H is a hypergroup, then a Dc (H) E. Proof. Let x, y (H - C), so x, y Dc (H). Since H is a hypergroup, there exists d H such that x dy. Moreover Dc (H) is a complete part, so it is closed by Proposition 1.3 and hence d Dc (H). Therefore x Dc (H) E y and so x and y have the same equivalence class under the relation Dc (H) E. Now suppose that x an C. If n = 1, then x Dc (H) E a. Let n > 1, H we have (x) = (a)n , where : H E is the canonical projection. Thus (x) = (a) (t), where t an-1 . Since is almost strong, there exist a , t H such that (a) = (a ), (t) = (t ) and x a t , so a t Dc (H)aDc (H)t. Since Dc (H) is the kernel of , Dc (H) is normal and hence a t Dc (H)at. From at an , there exists s an such that x Dc (H)s. Therefore Dc (H) E(x) =Dc (H) E(s), for some s an . H Hence D (H) E is a cyclic group. By Theorem 2.5 it follows that c a Dc (H) E. Theorem 4.3. Let H be a hypergroup. If c is the canonical projection H - c H a , Dc (H) then Dc (H) = -1 (e c H a ). H a Proof. Let z Dc , so z D or z D3 . If z D, then a (z) = e (see [8]). If z D3 , then z (H - C). By Proposition 4.1 we have a (z) = H a complete part subhypergroup of H, whence Dc (H) -1 (e c Conversely, for all x -1 (e c H a H a . Therefore Dc -1 (e c H a ). Since a is strongly regular, -1 (e c H a ) is a ). ) we have c () = e Dc (H). So we obtain a (x) = a () and hence x a . From Proposition 4.2 it follows that x Dc (H) E and hence x Dc (H). Since Dc (H) is complete part, by Proposition 1.3 we have Dc (H)x = Dc (H) whence x Dc (H). Thus -1 (e H ) Dc (H) and the proof is complete. c H a = c (x), where Proposition 4.4. If M is a non-empty subset of a hypergroup H, then (i) -1 (c (M )) = Dc (H)M = M Dc (H); c (ii) M is an a -part if and only if -1 (c (M )) = M . c Proof. (i) For every x Dc (H)M , there exists a pair (d, m) Dc (H)× M such that x dm and so c (x) = c (d) ( m) = e H c (m) by Theorem 4.3. So c (x) = c (m) c (M ) and hence x -1 (c (M )). c Conversely, for all x -1 (c (M )), an element m M exists such that c c (x) = c (m). Moreover, there exists d H such that x dm, so c (x) = c (m) = c (d) c (m), whence c (d) = e H , and d -1 (e H ) = Dc (H). c Therefore x Dc (H)M . (ii) Let M be an a -part of H and set x -1 (c (M )). Then there c exists m M such that c (x) = c (m). So m a x and by Proposition 3.2 from m M it follows that x M and hence -1 (c (M )) M . It is c obvious that M -1 (c (M )) and so the proof is complete. c Theorem 4.5. If H is a hypergroup, then a is transitive. Proof. By Theorem 3.6, it is enough to show that for all x H, P (x) is an a -part of H. We check that -1 (c (P (x))) = P (x) and then we apply c Proposition 4.4. Set z -1 (c (P (x))), so there exists k P (x) such that c (z) = c (k) c and hence a (z) = a (k). By Proposition 3.4 from k P (x) it follows that (k) = (x) and so (z) = (x). Therefore there exist x a k. Thus a a a a z = x1 , x2 , ..., xn-1 , xn = x such that: (4.1) We have two cases: z = x1 a x2 ... a xn-1 a xn = x. a a a Case 1. Set x (H - C), so (H - C) P (x). From (4.1) we have z a x2 whence by Proposition 3.4, x2 P (z). If we suppose that z an C, then P (z) = an thus x2 an . Using again (4.1) we would obtain x an C and it is a contradiction. Therefore z (H - C) and hence z P (x). Case 2. Set x an C, so P (x) = an . By Proposition 3.4 and (4.1), we have z an and hence P (z) = an . Therefore z an = P (x). Hence -1 (c (P (x))) P (x) and since the converse inclusion holds obvic ously, it follows that P (x) = -1 (c (P (x))) and thus the proof is complete. c http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

Cyclic Groups Obtained as Quotient Hypergroups

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Abstract

We introduce a strongly regular equivalence relation on the hyperA H group H, such that in a particular case the quotient is a cyclic group. Then by A using the notion of -parts, we investigate the transitivity condition of A . Finally, a A characterization of the derived hypergroup Dc (H) has been considered. Mathematics Subject Classification 2010: 20N20. Key words: hypergroup, strongly regular relation, complete parts. 1. Introduction Hyperstructure theory was born in 1934 when Marty [16] introduced hypergroups as a generalization of groups at the 8th congress of Scandinavian Mathematicions. After that, he proved the utility of hypergroups in solving some problems of groups, algebraic functions and rational fractions. Complete enough surveys of the algebraic hyperstructure theory are the books by Corsini [2], Corsini and Leoreanu [3], Vougiouklis [19]. Some relations in groupoids and hyperstructures which have been considered since the beginning are A, An , , n , and n , see Corsini [1, 4], Davvaz and Karimian [5, 6, 12], Freni [8, 9], Koskas [10, 11], Leoreanu [13, 14], Romeo [17], Vougiouklis [18], Zhan [20] and others. Using the relation , complete parts are introduced by Koskas [11] and studied by Corsini, Migliorato and Sureau [2, 3, 15]. This research has been supported financially by Mahani Mathematical Research Cen- ter. We present now some basic definitions and results about hypergroups (see [3]). Let H be a non-empty set and P (H) be the set of all non-empty subsets of H. Let be a hyperoperation or a join operation on H, that is, is a function from H × H into P (H) . If (a, b) H × H, then its image under in P (H) is denoted by a b or ab. The join operation is extended to subsets of H in a natural way, that is A B = {ab | a A, b B}. The notation aA is used for {a}A and Aa for A{a}. Generally, the singleton {a} is identified with its member a. The structure (H, ) is called a semihypergroup if a(bc) = (ab)c for all a, b, c H and is called a hypergroup if it is semi-hypergroup and aH = Ha = H for all a H. Let H be a hypergroup and K be a non-empty subset that is closed under the hyperoperation in H. If K is itself a hypergroup under the hyperoperation in H, then K is said to be a subhypergroup of H. Suppose (H, ) and (H , ) are two semi-hypergroups. A function f : H H is called a homomorphism if f (a b) f (a) f (b) for all a and b in H. We say f is a good homomorphism if for all a and b in H, f (a b) = f (a) f (b). If (H, ) is a hypergroup and R H × H is an equivalence relation, we set = A R B a R b, a A, b B, for all pairs (A, B) of non-empty subsets of H. The relation R is said to be left (right) strongly regular if x R y a x R a y (resp. x R y x a R y a), for all (x, y, a) H 3 . Moreover, R is called strongly regular if it is both left and right strongly regular. If (H, ) is a semi-hypergroup (hypergroup) and R is strongly regular, then the quotient H is a semigroup (group respectively) under the operation: R R(x) R(y) = R(z), z x y. = = For all n > 1 define the relatio on a semi-hypergroup H, as follows: a n b (x1 , ..., xn ) H n : {a, b} i=1 xi and = , where 1 = {(x, x) | x H} is the diagonal relation i=1 H. If is the transitive closure of then is an equivalence relation (see [2]). If H is a hypergroup and we consider the canonical projection : H H/ then the heart of H is the inverse image through of the identity of the group H/ and it is denoted by H . Hence H = -1 (eH/ ). We call the kernel of a homomorphism f , the set K(f ) = f -1 (H ), where H is the heart of H . The following result was shown by Freni [7]; another proof can be found in [2]. Theorem 1.1. If H is hypergroup then = . A non-empty subset A of H is called a complete part of H if for all n and for all (x1 , x2 , ..., xn ) H n we have the following implication: xi i=1 A= i=1 xi A. A subhypergroup K of a hypergroup (H, ·) is called · closed if from K aK = or K Ka = , where a H, it follows that a K; · invertible if from a bK it follows that b aK and from a Kb it follows that b Ka, where a, b H; · normal if for all a H, aK = Ka. Remark 1.2. The kernel of every homomorphism is a complete part and a normal subhypergroup. Proposition 1.3. Let K be a subhypergroup of a hypergroup H. (i) If K is a complete part, then it is invertible; (ii) If K is a invertible, then it is closed. In [8], Freni introduced the relation = n1 n , where 1 is the diagonal relation and for every integer n > 1, n is the relation defined as follows: x n y (z1 , ..., zn ) H n , Sn , : x i=1 zi , y i=1 z(i) , where Sn is the symmetric group of order n. In [5], Davvaz and Karimian introduced the concept of -parts and studied some properties of C (A), where C (A) is the intersection of all parts of a semi-hypergroup H. The transitive closure of is the smalest strongly regular equivalence such that the quotient H/ is commutative semigroup. Moreover, if H is hypergroup then is transitive (see [8]). Throughout this paper, H is a hypergroup and C = aA Ha , where = A H, Ha = an , n N and for all a, b A if a = b, then Ha Hb = . 2. The relation A In this section we introduce the relation A on a hypergroup H, which we use in order to obtain a cyclic group as a quotient structure of H. For all n 1 define A as follows A : = A A A , where : = A : = n A def def yi ) | Sn , i=1 yi = i=1 x(i) , def i=1 i=1 yi ) | a A, {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = , a Sn , yj {y1 , ..., yn } Ha yj = xa (j) and A : = n def i=1 i=1 def yi ) | {x1 , ..., xn } C = {y1 , ..., yn } C = . Notice that A : = {({x}, {y}) | {x, y} C = or x = y} and 1 A : = 2 def (xz, yz), (xz, zy), (zx, zy), (zx, yz) | ({x}, {y}) A , z H . 1 For every n 1, define the relatio on H by xn y (A, B) A , x A A n A, y B. Notice that for n = 1 we obtain x1 y ({x}, {y}) A or A 1 x = y C. Obviously, for every n 1, the relations n are symmetric, A and the relation A = n 1 n is reflexive and symmetric. Let be the A A transitive closure of A . Theorem 2.1. The relation is a strongly regular relation. A Proof. Notice that is an equivalence relation. In order to prove that A it is strongly regular, we have to show first that: (2.1) xA y xz A yz, zx A zy. = = for every z H. Suppose that xA y, thus there exists n 1 such that xn y. A Therefore there exists (B, C) A such that x B and y C. We obtai the following three cases: Case 1. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi and there exist Sn such that i=1 yi = i=1 x(i) . Set z = yn+1 = xn+1 and let be the permutation of Sn+1 such that: (i) = (i), n + 1, if i {1, 2, ..., n}; if i = n + 1. So Cz = n+1 x (i) and hence (Bz, Cz) A . +1 Case 2. Suppose that (B, C) A so we have B = n C = i=1 yi and there exists a A such that {x1 , ..., xn } Ha = {y1 , ..., yn } i=1 Ha = and a Sn such that for all 1 j n if yj Ha , then yj = xa (j) . If z C, then it is easy to see that (Bz, Cz) A . If z C, then there / n+1 exists b A such that z Hb . Set z = yn+1 = xn+1 and let a be the following permutation of Sn+1 : a (i) = a (i), n + 1, if i {1, 2, ..., n}; if i = n + 1. So for 1 j n + 1 if yj Ha , then yj = xa (j) . Therefore (Bz, Cz) A n+1 and hence (Bz, Cz) A . n+1 Case 3. Suppose that (B, C) A so we have B = n C = i=1 A / i=1 yi , {x1 , ..., xn }C = {y1 , ..., yn }C = . If z C, then (Bz, Cz) n+1 A . and if z C, then (Bz, Cz) n+1 Therefore for all v xz and for all y yz, we have v xz Bz and u yz Cz, so vm u for some m N, because (Bz, Cz) A for some A m = m N and hence vA u. Thus xz A yz. In the same way we obtain = that xA y zx A zy. Moreover, if x y, then there exists m N and A (u0 = x, u1 , ..., um = y) H m such that x = u0 A u1 A ...A um-1 A um = y, whence, by (2.1), we obtain that xz = u0 z A u1 z A u2 z A ... A um-1 A um z = yz. = = = = = Finally, for all v xz = u0 z and for all u um z = yz, taking z1 u1 z, z2 u2 z, ..., zm-1 um-1 z, we have vA z1 A z2 A ...A zm-1 A u, and so v u. Therefore x y xz yz. A A A Similarly we can prove that x y zx zy, hence is strongly A A A regular. H Theorem 2.2. The quotient is an abelian group with the generators A { (x0 ), (a) | x0 (H - C), a A}. A A Proof. By Theorem 2.1, is a strongly regular equivalence, so the A H quotient is a group under the following operation (x) (y) = A A A (z), z xy. A H Since , we conclude that is an abelian group. For all (x, y) A (H - C)2 since {x, y} C = , we have ({x}, {y}) A and hence x y A 1 so (x) = (y). Now suppose that (h) is given. If h (H - C), then A A A (h) = (x0 ) and if h C, then there exists a A such that h Ha . A A Therefore (h) = [ (a)]k for some k N. A A ow suppose that A = {a}, so C = Ha . Put a : = n , a = A and a : = . So we have the following corollary: A def def n 1 n a Corollary 2.3. The quotient H a is a cyclic group. Proof. By the proof of Theorem 2.2, we conclude that the equivalence classes determined by of all elements of (H - C) coincide and the A equivalence class of every element of C is a power of a (a). If H - C = , and x0 H - C then there exists u H - C such that x0 au since H is a hypergroup. We obtain a (x0 ) = a (a) a (u) = a (a) a (x0 ) whence a (a) = e H the identity element of the group H . >. Moreover, since a x0 v for some v H - C, we =< Hence (x ) = ( (x ))-1 which means that the order of (x ) is obtain that a 0 a 0 a 0 two, so in this case H is a cyclic group of order two. a Now if H - C = , which means that H is a cyclic hypergroup, then H =< a (a) > . H a a (x0 ) Therefore H a is a cyclic group. Example 2.4. Let S3 be the permutation group of order 3, i.e., S3 = {(1), (12), (13), (23), (123), (132)} and a = (123) be a cycle of order 3 in S3 . Then S3 Z2 . a = Proof. It is easy to see that A = {(x, y) | x, y are odd permutations or 1 x = y} and A = {(z, t) | z, t are odd or z, t are even permutations} and 2 A = A = a = , so S3 Z2 . n = 2 Theorem 2.5. The relation a is the least strongly regular equivalence such that the quotient H is a cyclic group and the equivalence classes of a all elements of (H - C) are equal. Proof. Let R be a strongly regular equivalence such that H is a cyclic R group and the equivalence classes of R of all elements of (H - C) are equal. Suppose that : H H is the canonical projection. is a good homoR morphism. We show that a R. n y. So there exists (B, C) A such that x B and Let n 1 and x a n y C . We have three cases: Case 1. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , and there exist Sn such that i=1 yi = i=1 x(i) . Since (x) = n (xi ) and (y) = n (yi ) = n (x(i) ), it follows that i=1 i=1 i=1 (x) = (y), because H is commutative. Therefore xRy. R Case 2. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = and there exists Sn such that for all 1 j n if yj Ha , then yj = x(j) . Renumber of the elements of the sets {x1 , ..., xn } and {y1 , ..., yn } such that {y1 , ..., ym } C, where 1 m n and xk , yk C for all m + 1 k n. So we have (x) = / (xk )) and (y) = (m (x(j) ))(n (m (xj ))(n j=1 j=1 k=m+1 (yk )). k=m+1 H For all m + 1 t, l n, (xt ) = (yl ) and since R is a commutative group, we have (x) = (y) and hence xRy. Case 3. Suppose that (B, C) A so we have B = n C = i=1 i=1 yi , {x1 , ..., xn } C = {y1 , ..., yn } C = . So for all 1 i, j n, (xi ) = (yj ), thus (x) = (y) and hence xRy. In all cases we have xRy and hence x a y xRy whence x a y xRy R. by transitivity of R. Therefore a 3. Transitivity condition of A In this section we introduce the concept of -part of a hypergroup H A and we determine necessary and sufficient conditions such that the relation A to be transitive. For this purpose, we will follow arguments similar to those exploited by Freni in [8] in the study of fundamental relation in hypergroups. Definition 3.1. Let M be a non-empty subset of H and n xi M = i=1 . We say that M is a -part of H if the following conditions hold: A (P1 ) For all Sn , n x(i) M ; i=1 (P2 ) If a A, {x1 , ..., xn } Ha = implies that for all n yi such i=1 that {y1 , ..., yn } Ha = {x1 , ..., xn } Ha there exists a Sn for which yj {y1 , ..., yn } Ha , yj = xa (j) , the yi M ; i=1 (P3 ) If {x1 , ..., xn }C = then for all n yi such that {y1 , ..., yn }C = i=1 , we have n yi M . i=1 Proposition 3.2. Let M be a non-empty subset of a hypergroup H. The following conditions are equivalent: (i) M is a -part; A (ii) x M, xA y y M ; (iii) x M, x y y M . A Proof. (i) (ii) Suppose that the pair (x, y) H 2 is such that x M and xA y. Then there exists (B, C) A such that x B and y C. So n we have three cases. Case 1. Let (B, C) A so we have B = n C = n yi and i=1 there exists Sn such that n yi = n x(i) . Since x n xi M , i=1 i=1 i=1 by (P1 ) we have n x(i) M and hence y M . i=1 Case 2. Let (B, C) A . So B = n C = n yi , there exists i=1 a A , such that {x1 , ..., xn } Ha = {y1 , ..., yn } Ha = and there exists a Sn such that for all 1 j n if yj Ha , then yj = xa (j) . Since x n xi M , by (P2 ) we have n yi M and hence y M . i=1 i=1 Case 3. Suppose that (B, C) A so we have B = n C = i=1 yi , {x1 , ..., xn } C = {y1 , ..., yn } C = . Since x n xi M , by i=1 i=1 (P3 ) we have n yi M and hence y M . i=1 (ii) (iii) Suppose that (x, y) H 2 is such that x M and x y. A So there exist m N and (w0 = x, w1 , ..., wm-1 , wm = y) H m such that x = w0 A w1 A w2 ...A wm-1 A wm = y. Since x M , applying (ii) m times, we obtain y M . (iii) (i) Let x n xi M . We shall check the conditions (P1 ), i=1 (P2 ) and (P3 ). (P1 ) Let Sn be given, thus ( n n x(i) ) A . So for all i=1 y n x(i) we have x y and hence y M . Therefore n x(i) M . i=1 A i=1 (P2 ) Let {x1 , ..., xn } Ha = for some a A. Now suppose that for all n i=1 yi if {y1 , ..., yn } Ha = {x1 , ..., xn } Ha there exists a Sn such that yj {y1 , ..., yn } Ha , yj = xa (j) . Thus ( n n yi ) A and hence i=1 for all y n yi we have x y. Therefore y M and so n yi M . i=1 A i=1 (P3 ) Let {x1 , ..., xn }C = and n yi be such that {y1 , ..., yn }C = . i=1 Thus ( n n yi ) A and hence for all y n yi we have x y. i=1 i=1 i=1 A n Therefore y M and so n yi M . otation 3.3. Let x be an arbitrary element of a hypergroup H. For n 1, set Pn (x) Pn (x) Pn (x), where x(i) | x xi ; {y1 , ..., yn } Ha = {x1 , ..., xn } Ha , i=1 yi | x i=1 a Sn , yj {y1 , ..., yn } Ha yj = xa (j) ; i=1 yi | x i=1 {y1 , ..., yn } C = {x1 , ..., xn } C = . Denote P (x) = n 1 Pn (x). Notice that if x C, then P1 (x) = H - C and / P2 (x) = zH dz, zd | d (H - C), b (H - C), x bz zb , while if x an C for some n N, then Pm (x) = an for all m 1. We have the following: Proposition 3.4. For every x H, P (x) = {y H | xA y}. Proof. Suppose that x H and y P (x) are given. Then there exist B and C such that x B, y C and: (i) y Pn (x) (B, C) A ; n (ii) y Pn (x) (B, C) A ; n (iii) y Pn (x) (B, C) A . n Therefore xA y and so P (x) {y H | xA y}. The proof of the reverse of the inclusion is similar to the above. Lemma 3.5. Suppose that H is a hypergroup and M is a -part of A H. If x M , then P (x) M . Proof. It follows by Definition 3.1. Theorem 3.6. Let H be a hypergroup. The following conditions are equivalent: (i) A is transitive; (ii) for every x H, (x) = P (x); A (iii) for every x H, P (x) is a -part of H. A Proof. (i) (ii) By Proposition 3.4, for all pair (x, y) H 2 we have: y (x) x y xA y y P (x). A A (ii) (iii) By Proposition 3.2, if M is a non-empty subset of H, then M is a -part of H if and only if it is a union of equivalence classes modulo A . Particularly, every equivalence class modulo is a -part of H. A A A (iii) (i) Suppose that xA y and yA z so x P (y) and y P (z). By Lemma 3.5, we have P (y) P (z) and hence x P (z). By Proposition 3.4 it follows that xA z and the proof is complete. 4. A characterization of a new derived hypergroup of a hypergroup Let H be a hypergroup and A be an invertible subhypergroup of H. For all x H, there exists y H such that x Ay. Therefore the relation A E on H defined by: x A E y x Ay is a regular equivalence and a conH gruence. Moreover the quotient A E = {Ax | x H} is a hypergroup with respect to the hyperoperation Ax Ay = {Az | z xAy}. and the canonH ical projection : H A E is an almost strong homomorphism (see [2]). This means that if (x) = (a) (t), then there exist a , t H such that (a) = (a ), (t) = (t ) and x a t . In this section we suppose that A = {a} and a (a) = e H . Proposition 4.1. For all x (H - C), a (x) = e Proof. Suppose that a (x) = e H a H a =< a (a) >, so a (x) = a (t) such that t an for some n N. Therefore x t and hence there exist the elements x = x1 , ..., xn-2 , xn-1 , xn = t in H such that x = x1 a ... xn-2 a xn-1 a xn = t. Thus xn-1 a t and by Proposition 3.4 it follows that xn-1 P (t). Since t an , P (t) = an and hence xn-1 an . Similarly from xn-2 a xn-1 we obtain xn-2 an . After (n - 1) steps, we conclude that x = x1 an C and it is a contradiction. H a . By Theorem 2.5, For all pair (a, b) H 2 and for all pair (A, B) of non-empty subsets of H, we set: a/b = {x H | a xb}, a\b = {x H | b ax}, A/B = aA,bB a/b, A\B = aA,bB a\b. Denote D1 = (x,y)H 2 xy/yx, D2 = (x,y)H 2 xy\yx, D3 = H - C and Dc = D1 D2 D3 . Define the derived hypergroup Dc (H) as the intersection of all subhypergroups, that are complete parts and contain Dc . Set D = D1 D2 . Since H D Dc (H), it follows that D (H) E is a commutative group (see [8]). Proposition 4.2. If H is a hypergroup, then a Dc (H) E. Proof. Let x, y (H - C), so x, y Dc (H). Since H is a hypergroup, there exists d H such that x dy. Moreover Dc (H) is a complete part, so it is closed by Proposition 1.3 and hence d Dc (H). Therefore x Dc (H) E y and so x and y have the same equivalence class under the relation Dc (H) E. Now suppose that x an C. If n = 1, then x Dc (H) E a. Let n > 1, H we have (x) = (a)n , where : H E is the canonical projection. Thus (x) = (a) (t), where t an-1 . Since is almost strong, there exist a , t H such that (a) = (a ), (t) = (t ) and x a t , so a t Dc (H)aDc (H)t. Since Dc (H) is the kernel of , Dc (H) is normal and hence a t Dc (H)at. From at an , there exists s an such that x Dc (H)s. Therefore Dc (H) E(x) =Dc (H) E(s), for some s an . H Hence D (H) E is a cyclic group. By Theorem 2.5 it follows that c a Dc (H) E. Theorem 4.3. Let H be a hypergroup. If c is the canonical projection H - c H a , Dc (H) then Dc (H) = -1 (e c H a ). H a Proof. Let z Dc , so z D or z D3 . If z D, then a (z) = e (see [8]). If z D3 , then z (H - C). By Proposition 4.1 we have a (z) = H a complete part subhypergroup of H, whence Dc (H) -1 (e c Conversely, for all x -1 (e c H a H a . Therefore Dc -1 (e c H a ). Since a is strongly regular, -1 (e c H a ) is a ). ) we have c () = e Dc (H). So we obtain a (x) = a () and hence x a . From Proposition 4.2 it follows that x Dc (H) E and hence x Dc (H). Since Dc (H) is complete part, by Proposition 1.3 we have Dc (H)x = Dc (H) whence x Dc (H). Thus -1 (e H ) Dc (H) and the proof is complete. c H a = c (x), where Proposition 4.4. If M is a non-empty subset of a hypergroup H, then (i) -1 (c (M )) = Dc (H)M = M Dc (H); c (ii) M is an a -part if and only if -1 (c (M )) = M . c Proof. (i) For every x Dc (H)M , there exists a pair (d, m) Dc (H)× M such that x dm and so c (x) = c (d) ( m) = e H c (m) by Theorem 4.3. So c (x) = c (m) c (M ) and hence x -1 (c (M )). c Conversely, for all x -1 (c (M )), an element m M exists such that c c (x) = c (m). Moreover, there exists d H such that x dm, so c (x) = c (m) = c (d) c (m), whence c (d) = e H , and d -1 (e H ) = Dc (H). c Therefore x Dc (H)M . (ii) Let M be an a -part of H and set x -1 (c (M )). Then there c exists m M such that c (x) = c (m). So m a x and by Proposition 3.2 from m M it follows that x M and hence -1 (c (M )) M . It is c obvious that M -1 (c (M )) and so the proof is complete. c Theorem 4.5. If H is a hypergroup, then a is transitive. Proof. By Theorem 3.6, it is enough to show that for all x H, P (x) is an a -part of H. We check that -1 (c (P (x))) = P (x) and then we apply c Proposition 4.4. Set z -1 (c (P (x))), so there exists k P (x) such that c (z) = c (k) c and hence a (z) = a (k). By Proposition 3.4 from k P (x) it follows that (k) = (x) and so (z) = (x). Therefore there exist x a k. Thus a a a a z = x1 , x2 , ..., xn-1 , xn = x such that: (4.1) We have two cases: z = x1 a x2 ... a xn-1 a xn = x. a a a Case 1. Set x (H - C), so (H - C) P (x). From (4.1) we have z a x2 whence by Proposition 3.4, x2 P (z). If we suppose that z an C, then P (z) = an thus x2 an . Using again (4.1) we would obtain x an C and it is a contradiction. Therefore z (H - C) and hence z P (x). Case 2. Set x an C, so P (x) = an . By Proposition 3.4 and (4.1), we have z an and hence P (z) = an . Therefore z an = P (x). Hence -1 (c (P (x))) P (x) and since the converse inclusion holds obvic ously, it follows that P (x) = -1 (c (P (x))) and thus the proof is complete. c

Journal

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

Published: Jan 1, 2015

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