ON MEDIAL B-ALGEBRAS

Abstract

In this paper we introduce the notion of medial B-algebras, and we obtain a fundamental theorem of B-homomorphism for B-algebras.

1. Introduction

Y. Imai and K. Iséki introduced two classes of abstract algebras: BCK-algebras and BCI-algebras [4,5]. It is known that the class of BCK-algebras is a proper subclass of the class of BCI-algebras. In [2,3] Q. P. Hu and X. Li introduced a wide class of abstract algebras: BCH-algebras. They have shown that the class of BCI-algebras is a proper subclass of the class of BCH-algebras. J. Neggers and H. S. Kim [8] introduced the notion of d-algebras, i.e., (I) x ∗ x = 0;(V) 0 ∗ x = 0; (VI) x ∗ y = 0 and y ∗ x = 0 implay x = y, which is another useful generalization of BCK-algebras, and then they investigated several relations between d-algebras and oriented digraphs. Recently, Y. B. Jun, E. H. Roh and H. S. Kim [6] introduced a new notion, called an BH-algebra, i.e., (I),(II) x ∗ 0 = 0 and (IV), which is a generalization of BCH/BCI/BCK-algebras. They also defined the notions of ideals and boundedness in BH-algebras, and showed that there is a maximal ideal in bounded BH-algebras. J. Neggers and H. S. Kim[9] introduced and investigated a class of algebras, i.e., the class of B-algebras, which is related to several classed of algebras of interest such as BCH/BCI/BCK-algebras and which seems to have rather nice properties without being excessively complicated otherwise. Furthermore, a digraph on algebras defined below demonstrates a rather interesting connection between B-algebras and groups. J. R. Cho and H. S. Kim[1] discussed further relations between B-algebras and other classed of algebras, such as quasigroups. J. Neggers and H. S. Kim [10] introduced the notion of normality in B-algebras and obtained a fundamental theorem of B-homomorphism for B-algebras.

In this paper we introduce the notion of medial B-algebras, and we obtain a fundamental theorem of B-homomorphism for B-algebras.

 

2. Preliminaries

In this section, we introduce some notions and results which have also been discussed in [1,9]. A B-algebra is a non-empty set X with a constant 0 and a binary operation "∗" satisfying the following axioms:

for all x, y, z in X.

Example 2.1. Let X := {0, 1, 2} be a set with the following table:

Then (X; ∗, 0) is a B-algebra.

Example 2.2 ([9]). Let X be the set of all real numbers except for a negative integer −n. Define a binary operation ∗ on X by

Then (X; ∗, 0) is a B-algebra.

Example 2.3. Let X := {0, 1, 2, 3, 4, 5} be a set with the following table:

Then (X; ∗, 0) is a B-algebra (see[10]).

Example 2.4 ([9]). Let F < x, y, z > be the free group on three elements. Define u ∗ v := vuv−2. Thus u ∗ u = e and u ∗ e = u. Also e ∗ u = u−1. Now, given a, b, c, ∈ F < x, y, z >, let

Let N(∗) be the normal subgroup of F < x, y, z > generated by the elements w(a, b, c). Let G = F < x, y, z > /N(∗). On G define the operation "·" as usual and define

It follows that (uN(∗)) ∗ (uN(∗)) = eN(∗), (uN(∗)) ∗ (eN(∗)) = uN(∗) and

Hence (G; ∗, eN(∗)) is a B-algebra.

Lemma 2.5 ([9]). If (X; ∗, 0) is a B-algebra, then y ∗ z = y ∗ (0 ∗ (0 ∗ z)) for any y, z ∈ X.

Proposition 2.6 ([9]). If (X; ∗; 0) is a B-algebra, then

for any x, y, z ∈ X.

Lemma 2.7 ([1]). Let (X; ∗, 0) be a B-algebra. Then we have the following statements.

Let (X; ∗, 0X) and (Y ; •, 0Y) be B-algebras. A mapping φ : X → Y is called a B-homomorphism[10] if φ(x ∗ y) = φ(x) • φ(y) for any x, y ∈ X.

Example 2.8 ([10]). Let X := {0, 1, 2, 3} be a set with the following table:

Then (X; ∗, 0) is a B-algebra[1]. If we define φ(0) = 0, φ(1) = 3, φ(2) = 3 and φ(3) = 0, then φ : X → Y is a B-homomorphism.

A B-homomorphism φ : X → Y is called a B-isomorphism[10] if φ is a bijection, and denote it by X ≅ Y. Note that if φ : X → Y is a B-isomorphism then φ−1 : Y → X is also a B-isomorphism. If we define φ(0) = 0, φ(1) = 2, φ(2) = 1 and φ(3) = 3 in Example 2.8, then φ : X → Y is a B-isomorphism. Let φ : X → Y be a B-homomorphism. Then the subset {x ∈ X | φ(x) = 0Y} of X is called the kernel of the B-homomorphism φ, and denote it by Kerφ

Definition 2.9 ([10]). Let (X; ∗, 0) be a B-algebra. A non-empty subset N of X is called a subalgebra of X if x ∗ y ∈ N, for any x, y ∈ N.

In Example 2.8, N1 := {0, 3} is a subalgebra of X, while N2 := {0, 1} is not a subalgebra of X, since 0 ∗ 1 = 2 ∉ N2. Note that any subalgebra of a B-algebra is also a B-algebra.

Theorem 2.10 ([10]). Let (X; ∗, 0) be a B-algebra and Then the following are equivalent:

Note that any kernel of a B-homomorphism is a subalgebra of X.

 

3. Medial B-algebras

Let (X; ∗, 0) be a B-algebra and let N be a subalgebra of X. The set X(resp., N) is said to be medial if (x ∗ n1) ∗ (y ∗ n2) = (x ∗ y) ∗ (n1 ∗ n2) for any x, y, n1, n2 ∈ X(resp., for any x, y, n1, n2 ∈ N).

Example 3.1. The B-algebra in Example 2.8, is medial. The B-algebra in Example 2.3, is not medial, since (5∗2)∗(4∗3) = 4∗1 = 5 ≠ 3 = 1∗5 = (5∗4)∗(2∗3).

J. Neggers and H. S. Kim[10] introduced the notion of a normal subalgebra in B-algebras. A nonempty subset N of X is said to be normal (or normal subalgebra) of X if (x ∗ a) ∗ (y ∗ b) ∈ N for any x ∗ a, y ∗ b ∈ N.

Example 3.2. The subalgebra N1 = {0, 3} is both a normal and a medial subalgebra of X in Example 2.8, while the subalgebra N2 = {0, 3} in Example 2.3 is medial, but not normal.

Example 3.3. Let X := {0, 1, 2, 3} be a set with the following table:

Then (X; ∗, 0) is a B-algebra and the subalgebra N3 = {0, 2} is a medial subal-gebra of X.

Let (X; ∗, 0) be a B-algebra and let N be a subalgebra of X. Define a relation ∼N on X by x ∼N y if and only if x ∗ N = y ∗ N, where x, y ∈ X. Then it is easy to show that ∼N is an equivalence relation on X. Assume X is medial (or N is a medial subalgebra of X). If x ∼N y and a ∼N b, where x, y, a, b ∈ N, then x ∗ N = y ∗ N and a ∗ N = b ∗ N and hence x = y ∗ n1, a = b ∗ n2 for some n1, n2 ∈ N. Hence x ∗ a = (y ∗ n1) ∗ (b ∗ n2) = (y ∗ b) ∗ (n1 ∗ n2) ∈ (y ∗ b) ∗ N, since X(resp., N) is medial. For any (x ∗ a) ∗ n3 ∈ (x ∗ a) ∗ N, we have

Hence (x ∗ a) ∗ N ⊆ (y ∗ b) ∗ N. Similarly, we obtain (y ∗ b) ∗ N ⊆ (x ∗ a) ∗ N. This means that x ∗ a ∼N y ∗ b, i.e., ∼N is a congruence relation on X. Denote the equivalence class containing x by [x]N, i.e., [x]N = {y ∈ X | x ∼N y} and let X/N := {[x]N | x ∈ X}. We show that X/N is a B-algebra.

Theorem 3.4. Let X be a medial B-algebra and let N be a subalgebra of X. Then X/N is a medial B-algebra with N = [0]N.

Proof. If we define [x]N ∗ [y]N := [x∗y]N then the operation "∗" is well-defined, since ∼N is a congruence relation on X. We claim that [0]N = N. If x ∈ [0]N, then x ∗ N = 0 ∗ N, and hence by (II) x = x ∗ 0 ∈ x ∗ N = 0 ∗ N, i.e., x = 0 ∗ n for some n ∈ N. Since N is a subalgebra and 0 ∈ N, x = 0 ∗ n ∈ N. Hence [0]N ⊆ N.

For any x ∈ N, since N is subalgebra of X, 0 ∗ x ∈ N, say n1 = 0 ∗ x. By applying Lemma 2.7-(iii), x = 0 ∗ (0 ∗ x) ∈ 0 ∗ N. We show that x ∗ N = 0 ∗ N. For any x ∗ n ∈ x ∗ N,

Hence x ∗ N ⊆ 0 ∗ N. If y ∈ 0 ∗ N, then y = 0 ∗ n2 for some n2 ∈ N. Hence y = 0 ∗ n2 = (x ∗ x) ∗ n2 = x ∗ (n2 ∗ (0 ∗ x)). Since x ∈ N, by Theorem 2.10, n2 ∗ (0 ∗ x) ∈ N. Hence y ∈ x ∗ N, i.e., 0 ∗ N ⊆ x ∗ N. Thus x ∗ N = 0 ∗ N, i.e., x ∼N 0. Hence x ∈ [0]N, proving N ⊆ [0]N. Checking three axioms and mediality is trivial and we omit the proof.

Theorem 3.4 can be replaced by the following statement:

Theorem 3.4′. Let X be a B-algebra and N be a medial subalgebra of X. Then X/N is a medial B-algebra with N = [0]N.

The B-algebra X/N discussed in Theorems 3.4 and 3.4′ is called the quotient B-algebra of X by N.

Proposition 3.5. Let N be a medial subalgebra of the B-algebra (X; ∗, 0). Then the mapping γ : X → X/N, given by γ(x) := [x]N, is a surjective B-homomorphism, and Kerγ = N.

Proof. The mapping γ is obviously surjective. For all x, y ∈ X, γ(x ∗ y) = [x ∗ y]N = [x]N ∗ [y]N = γ(x) ∗ γ(y). Hence γ is a B-homomorphism. We claim that {x ∈ X | [x]N = [0]N} = N. For any n ∈ N, we show that n ∗ N = 0 ∗ N. If n1 ∈ N, by Lemma 2.7-(iii), n ∗ n1 = (0 ∗ (0 ∗ n)) ∗ n1 = 0 ∗ (n1 ∗ (0 ∗ (0 ∗ n))) = 0 ∗ (n1 ∗ n) ∈ 0 ∗ N, i.e., n ∗ N ⊆ 0 ∗ N. For any 0 ∗ n2 ∈ 0 ∗ N, 0 ∗ n2 = (n ∗ n) ∗ n2 = n ∗ (n2 ∗ (0 ∗ n)) ∈ n ∗ N, i.e., 0 ∗ N ⊆ n ∗ N. This proves 0 ∗ N = n ∗ N, i.e., [n]N = [0]N. If [x]N = [0]N, then x ∗ N = 0 ∗ N, i.e., x = 0 ∗ n1 for some n1 ∈ N. Since N is a subalgebra of X, x = 0 ∗ n1 ∈ N. Hence

proving the proposition.

The mapping γ discussed in Proposition 3.5 is called the natural(or canonical) B-homomorphism of X onto X/N.

Proposition 3.6. Let X be a medial B-algebra. If φ : X → Y is a B-homomorphism, then the kernel Kerφ is a medial subalgebra of X.

Proof. Straightforward.

By Theorem 3.4 and Proposition 3.6, if φ : X → Y is a B-homomorphism, then X/Kerφ is a B-algebra.

A B-algebra (X; ∗, 0) is said to be commutative[9] if a ∗ (0 ∗ b) = b ∗ (0 ∗ a) for any a, b ∈ X. The B-algebra in Example 2.1 is commutative, while the B-algebra in Example 2.3 is not commutative, since 3 ∗ (0 ∗ 4) = 2 ≠ 1 = 4 ∗ (0 ∗ 3).

Theorem 3.7. Let X be a commutative medial B-algebra and let φ : X → Y be a B-homomorphism. Then X/Kerφ ≅ Imφ. In particular, if φ is surjective, then X/Kerφ ≅ Y.

Proof. Let K := Kerφ. If we define Ψ : X/K → Imφ by Ψ([x]K) := φ(x), then Ψ is well-defined. In fact, suppose that [x]K = [y]K. Then x ∼K y and x ∗ K = y ∗ K, i.e., x = y ∗ k1, y = x ∗ k2 for some k1, k2 ∈ K. Hence φ(x) = φ(y ∗ k1) = φ(y) ∗ φ(k1) = φ(y) ∗ 0 = φ(y), i.e., Ψ([x]K) = Ψ([y]K). Suppose that Ψ([x]K) = Ψ([y]K), where [x]K, [y]K ∈ X/K. Then φ(x) = φ(y). If α ∈ [x]K, then α ∼K x and α∗K = x∗K. This means that α = x∗k1, x = α∗k2 for some k1, k2 ∈ K. Hence φ(α) = φ(x ∗ k1) = φ(x) ∗ φ(k1) = φ(x) = φ(y), which implies φ(α ∗ y) = φ(α) ∗ φ(y) = 0. Hence α ∗ y ∈ Kerφ = K, i.e., α ∗ y = k3 for some k3 ∈ K. Similarly, φ(y) ∗ φ(α) = 0 implies y ∗ α = k4 for some k4 ∈ K. Sice X is commutative,

For any α ∗ k4 ∈ α ∗ K, α ∗ k = (y ∗ k4) ∗ k = y ∗ (k ∗ (0 ∗ k4)) ∈ y ∗ K. Hence α ∗ K ⊆ y ∗ K. Conversely, we have

proving y ∗K ⊆ α∗K. Hence α∗K = y ∗K, i.e., α ∼K y. This proves α ∈ [y]K. Similarly, [y]K ⊆ [x]K. Thus [x]K = [y]K, proving that Ψ is injective. Obviously Ψ is surjective. Since Ψ([x]K ∗ [y]K) = Ψ([x ∗ y]K) = φ(x ∗ y) = φ(x) ∗ φ(y) = Ψ([x]K) ∗ Ψ([y]K), Ψ is a B-homomorphism. Hence X/Kerφ ≅ Imφ.

Example 3.8. In Example 2.8, since K = Kerφ = {0, 3}, we have [0]K = {0, 3} and [1]K = {x ∈ X | x ∗ 1 ∈ K} = {1, 2}. Hence X/Kerφ = {[0]K, [1]K} and X/Kerφ ≅ Imφ by defining Ψ([0]K) = φ(0) and Ψ([1]K) = φ(1).