Pure Mathematical Sciences, Vol. 1, 2012, no. 3, On CS-Algebras. Kyung Ho Kim

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1 Pure Mathematical Sciences, Vol. 1, 2012, no. 3, On CS-Algebras Kyung Ho Kim Department of Mathematics Korea National University of Transportation Chungju , Korea Abstract In this paper, we introduce the notion of CS-algebra, and some fundamental properties to CS-algebras are discussed. Mathematics Subject Classification: 06F35, 03G25, 08A30 Keywords: CI-algebra, self-distributive, right (resp. (resp. left) deductive system left) stable, right 1 Introduction Y. Imai and K. Iséki introduced two classes of abstract algebras: BCK-algebras and BCI-algebras([3, 4]). It is known that the class of BCK-algebras is a proper subclass of the class of BCI-algebras. In [1, 2], Q. P. Hu and X. Li introduced a wide class of abstracts: BCH-algebras. They have shown that the class of BCI-algebras is a proper subclass of the class of BCH-algebras. In [5], B. L. Meng introduced the notion of an CI-algebra as a generation of a BE-algebra. In this paper, In this paper, we introduce the notion of CS-algebra, and some fundamental properties to CS-algebras are discussed. 2 Preliminaries In what follows, let X denote an CS-algebra unless otherwise specified. By an CI-algebra we mean an algebra (X;, 1) of type (2, 0) with a single binary operation that satisfies the following identities: for any x, y, z X, (CI1) x x = 1 for all x X, (CI2) 1 x = x for all x X,

2 116 Kyung Ho Kim (CI3) x (y z) =y (x z) for all x, y, z X. We introduce a relation onx by x y imply x y =1. An CI-algebra (X,, 1) is said to be self-distributive if x (y z) =(x y) (x z) for all x, y, z X. A non-empty subset S of an CI-algebra X is said to be a subalgebra of X if x y S whenever x, y S. In an CI-algebra, the following identities are true: (p1) y ((y x) x) =1. (p2) (x 1) (y 1) = (x y) 1. Let a X be an element of a CI-algebra X. a is said to be an atom if for any x X, a x = 1 implies a = x. Denote the set of all atoms in X by A(X) which is called the singular part of X. Obviously, 1 A(X), and so A(X) φ. Define a set G(X) byg(x) ={x X x 1=x}. The set G(X) is called a G-part of a CI-algebra X. 3 CS-Algebras Definition 3.1. By an CS-algebra (X,,, 1) with two binary operations and that satisfies the following axioms: (CS1) S(X) =(X, ) is a semigroup, (CS2) C(X) =(X, ) isaci- algebra, (CS3) x (y z) =x y x z and (x y) z = x z y z for any x, y, z X. Example 3.2. Let X = {1,a,b,c} in which and are defined by a 1 1 b c b 1 a 1 c a b c 1 a b c Then it is easy to check that (X,, ) isancs-algebra. Example 3.3. Let X = {1,a,b,c} in which and are defined by

3 On CS-Algebras 117 a 1 1 b c b 1 a 1 c a b b c 1 1 b c Then it is easy to check that (X,,, 1) is an CS-algebra. Proposition 3.4. Let X be an CS-algebra. Then the following identities hold. (1) x 1=1and 1 x =1for all x X, (2) x y implies a x a y and x a y a for all x, y, a X, (3) x (y z) =x z y z for all x, y, z X. Proof. (1) x 1 =x (1 1) = x 1 x 1 = 1 and 1 x =(1 1) x =1 x 1 x =1. (2) Let x y. Then we have x y =1, and so a x a y = a (x y) =a 1=1. Hence a x a y. Similarly, we have xa y a. (3) x (y z) = x ((z y) y) = x (z y) x y =(x z x y) x y = x y x z. Definition 3.5. A nonempty subset A of a CS-algebra X is called to be left (resp. right) stable if x a A (resp. ax A) whenever x X and a A. Proposition 3.6. Let X be an CS-algebra. Then, a X is an atom in X if and only if it satisfies the equation x X, a =(a x) x for any x X. Proof. Let a be an atom in X and x X. We have a =(a x) x from a ((a x) x) =1. Conversely, suppose that a X satisfies a =(a x) x for any x X. If a x =1, we get a =(a x) x =1 x = x, which implies that a is an atom in X. This completes the proof. Proposition 3.7. Let X be an CS-algebra. Then, A(X) and G(X) are stable subsets of S(X). Proof. For any x S(X) and a A(X), we have and hence (x a) 1=x a x 1=x (a 1), ((x a) 1) 1=(x (a 1)) x 1=x ((a 1) 1) = x a, which implies that xa A(X) from Proposition Similarly, a x A(x) and so A(X) is stable. Now, let x S(X) and b G(X). Then we have b x 1 =b x 1 x =(b 1) x = b x and x b 1 =x b x 1 =x (b 1) = x b, which implies b x, x b G(X).

4 118 Kyung Ho Kim Definition 3.8. A non-empty set F of an CS-algebra X is said to be left (resp. right) deductive system if it satisfies the following axioms: (DS1) F is a left (resp. right) stable subset of S(X). (DS2) For any x, y C(X), x y F and x F imply y F. In a CS-algebra X, we have x 1=1 x = 1 for all x X. If F is a deductive system of X, then 1 = 1 a F for any a F. Example 3.9. Let X = {1,a,b,c} in which and are defined by a 1 1 b c b c a 1 a 1 1 b 1 1 b c c 1 1 c b Then X is a CS-algebra. It is easy to check that F = {1,a} is a deductive system of X. Definition Let X be an CS-algebra. A non-empty subset S of X is called a subalgebra of X if x y S and x y S for all x, y S Let us define the center of a CS-algebra X, denoted by cent(x), to be the set cent (X) ={x X a x = x a for all a X}. Let x, y cent (X). Then x a = a x and y b = b y for all a, b X. Thus (x y) a = x a y a = a x a y = a (x y) for all a X. This implies that x y cent (X). showing that cent (X) is a subalgebra of a CS-algebra X. Next since x, y cent(x), we have x a = a x and y a = a y. Thus (x y) a = x (y a) =x (a y) =(x a) y =(a x) y = a (x y) for all a X. The following theorems are obvious. Theorem For any CS-algebra X, cent (X) is a subalgebra of X. Theorem Let X be a CS-algebra X and a X. Then the set C(a) = {x X a x = x a} is a subalgebra, and cent (X) = C(a). Lemma Suppose that F is a deductive system of CS-algebra X and x F. If x y, then y F. Proof. x y implies x y =1 F. Combining x F and using Definition 3.8 (DS2), we obtain y F, proving the theorem. a X

5 On CS-Algebras 119 Definition A deductive system F of an CS-algebra X is said to be closed if x F implies x 1 F. Theorem A deductive system of an CS-algebra X is closed if and only if it is a subalgebra of CS-algebra X. Proof. Suppose that a deductive system F is a closed and x, y F. Since y (x y) =x (y y) =x 1 F, we have x y F by Definition 3.8 (DS2). Hence F is a subalgebra of X. Conversely, assume that a deductive system F is a subalgebra of X. For all x F, we have x 1 F since since 1 F. so F is closed. Definition A near CS-algebra is a CS-algebra X that satisfies the following condition : for each x, y X, x y = x y y. The element e is called an unity in an CS-algebra if e x = x e = x for all x X. Example Let X = {1,a,b,c} in which and are defined by a 1 1 b b b 1 a 1 a a 1 a 1 a b 1 1 b b c 1 a b c It is easy to check that (X,, ) is a near CS-algebra. Proposition Let X be a near CS-algebra. Then (1) x x y for all x, y X, (2) x y, x, y X if and only if x y y. Proof. (1) For any x, y X, x x y = x (x y) x y =(x x) y x y = (x x x) y =(x x) y =1 y =1. (2) It is easy to show (2) from the definition of near CS-algebra and the above (1). Proposition Let X be a near CS-algebra. Then the following properties hold. (1) If x y, then x y y x for all x, y X.

6 120 Kyung Ho Kim (2) If x y =1, then x y = y. Proof. (1) Let x y. Then we have x y = x y y =1, and so (x y) (y x) =(x y) (y x) (y x)=((x y) y y) x =(x y y) x =1 x =1, which implies x y y x. (2) Let x y =1. Then x y = x y y =1 y = y. Theorem Let X be a near CS-algebra and A a subset of X. If y A and x y imply x A, then A is a stable subset of X. Proof. Suppose that x A and x y imply y A. If s X and a A, then by Proposition 3.18 (1), a s a, hence s a A. Since a s, we have a s s from Proposition 3.18 (2) and a a s = a (a s) a s = a (a s s) =a 1=1, and a a s, and hence a s A. This completes the proof. Definition An element a( 1)inanCS-algebra X is said to be a left (resp. right) unit divisor if ( b( 1) X) (a b = 1 (resp. b a =1.)) An unit divisor is an element of X which is both a left and a right unit divisor. Theorem Let (X,,, 1) be an CS-algebra. If it satisfies the left (resp. right) cancellation law for the operation, i.e., ( x( 1),y,z X) (x y = x z (resp.y x = z x) y = z), then X contains no left (resp. right) unit divisors in X. Proof. Let (X,,, 1) satisfy the left cancellation law for the operation and assume that x y = 1 where x 1. Then x y =1=x 1, which implies y =1. Similarly, it holds for the right case. Hence there is no left (resp. right) unit divisors in X. Theorem Let (X,,, 1) be a CS-algebra in which there are no left (resp. right) unit divisors. Then it satisfies the left (resp. right) cancellation law for the operation. Proof. Let x, y, z X be such that x y = x z and x 1. Then x (y z) =(x y) (x z) =1 and x (z y) =(x z) (x y) =1. Since X has no left unit divisor, it follows that that y z =1=z y so that y = z. The argument is the same for the right case.

7 On CS-Algebras 121 Definition Let X and Y be CS-algebras. A mapping f : X Y is called a CS-algebra homomorphism (briefly, homomorphism) iff(x y) = f(x) f(y) and f(x y) = f(x) f(y) for all x, y X. Proposition Let f : X Y be a CS-algebra homomorphism. Then (1) f(1) = 1, (2) f(x y) =f(x) f(y) for all x, y X. (3) f(x 1) = f(x) 1 for all x X. (4) If f(x) =1, then f(x y) =1for all x, y X. (5) If f(y) =1, then f(x y) =1for all x, y X. Proof. It is easy to prove the proposition, and so we omit the proof. References [1] Q. P. Hu and X. Li, On BCH-algebras, Math. Seminar Notes 11 (1983), [2] Q. P. Hu and X. Li, On proper BCH-algebras, Math Japonicae 30 (1985), [3] K. Iseki and S. Tanaka, An introduction to theory of BCK-algebras, Math Japonicae 23 (1978), [4] K. Iseki, On BCI-algebras, Math. Seminar Notes 8 (1980), [5] Biao Long Meng, CI-algebras, Sci. Math. Japo. Online, e-2009, Received: March, 2012