EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 3, 2019, 1248-1259 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On Companion B-algebras Lynnel D. Naingue1,∗, Jocelyn P. Vilela1 1 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University- Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. This study introduces the concept of companion B -algebra and establishes some of its properties. Also, this paper introduces the notions of �-subalgebra and �-ideal of a companion B - algebra and investigates their relationship. Furthermore, this study establishes some homomorphic properties of �-subalgebra and �-ideal. 2010 Mathematics Subject Classifications: 08C99, 08A05, 08A30 Key Words and Phrases: Companion B -algebra, �-subalgebra, �-ideal, companion B - homomorphism 1. Introduction Y. Imai and K. Iséki [7] first initiated the study of BCK -algebras in 1966. In the same year, K. Iséki [6] introduced another class of algebras, called BCI -algebras, which are generalizations of BCK -algebras. In 1999, J. Neggers and H. S. Kim [9], introduced the notion of d -algebra which is another generalization of BCK -algebra. In 2007, P. J. Allen, H. S. Kim and J. Neggers [3] developed the concept of companion d -algebra to demonstrate considerable parallelism with the theory of BCK -algebras. In 2002, J. Neggers and H. S. Kim [11] introduced and investigated another class of algebras called B -algebras and described it to have nice properties without being compli- cated. P. J. Allen, J. Neggers and H. S. Kim [2] proved that every group, under some conditions, determines a B -algebra. Also, M. Kondo and Y. B. Jun [8] proved the con- verse. This paper extends the study of B -algebras by defining the concept of companion op- eration and companion B -algebras and establishing some of its properties. This study also introduces the concepts of subalgebra and ideal of a companion B -algebra and determines some of its homomorphic properties. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3495 Email addresses: lynneldnaingue13@gmail.com (L.D. Naingue), jocelyn.vilela@g.msuiit.edu.ph (J.P. Vilela) http://www.ejpam.com 1248 c© 2019 EJPAM All rights reserved. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1249 2. Preliminaries Definition 2.1. [11] A B-algebra (X, ∗, 0) is a nonempty set X with a constant 0 and a binary operation “∗” satisfying the following axioms: for all x, y, z in X, (I) x ∗ x = 0, (II) x ∗ 0 = x, (III) (x ∗ y) ∗ z = x ∗ (z ∗ (0 ∗ y)). Example 2.2. The set of integers together with the usual subtraction and the constant 0 is a B -algebra. Theorem 2.3. [11] If (X, ∗, 0) is a B-algebra, then the following hold: for any x, y, z ∈ X, (a) (x ∗ y) ∗ (0 ∗ y) = x (b) y ∗ z = y ∗ (0 ∗ (0 ∗ z)) (c) x ∗ (y ∗ z) = (x ∗ (0 ∗ z)) ∗ y (d) x ∗ y = 0 implies x = y (e) 0 ∗ x = 0 ∗ y implies x = y (f) 0 ∗ (0 ∗ x) = x. Theorem 2.4. [13] If (X, ∗, 0) is a B-algebra, then the following hold: for any x, y, z ∈ X, 0 ∗ (x ∗ y) = y ∗ x. Definition 2.5. [11] A B -algebra (X, ∗, 0) is commutative if for any x, y ∈ X, x∗ (0∗y) = y ∗ (0 ∗ x). Theorem 2.6. [2] Let (X, ∗, 0) be a B-algebra. If x ◦ y = x ∗ (0 ∗ y) for all x, y ∈ X, then (X, ◦) is a group. Theorem 2.7. [11] Let (G, ◦) be a group with identity e. If we define x ∗ y = x ◦ y−1, then (G, ∗, e) is a B-algebra. Definition 2.8. [12] Let (X, ∗, 0) be a B -algebra. A nonempty subset H of X is called a B-subalgebra of X if x ∗ y ∈ H for any x, y ∈ H. Definition 2.9. [5] Let (X, ∗, 0) be a B -algebra. A nonempty subset I of X is called a B-ideal of X if 0 ∈ I and x ∗ y ∈ I and y ∈ I imply x ∈ I. Theorem 2.10. [1] Every subalgebra of a B-algebra X is an ideal. Definition 2.11. [10] Let (A, ∗A, 0A) and (B, ∗B, 0B) be B -algebras. The mapping φ : A → B is called a B-homomorphism if φ(x ∗A y) = φ(x) ∗B φ(y) for any x, y ∈ A. The kernel of f is defined as Kerf = {x ∈ A : φ(x) = 0B}. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1250 3. Basic Properties of Companion B-algebra Definition 3.1. Let (X, ∗, 0) be a B -algebra. A binary operation � on X is called a subcompanion operation of X if it satisfies for any x, y ∈ X, ((x� y) ∗ x) ∗ y = 0 (SC) A subcompanion operation � is a companion operation of X if for any x, y, z ∈ X, (z ∗ x) ∗ y = 0 implies z ∗ (x� y) = 0. (C) A companion B-algebra (X, ∗,�, 0) is a B -algebra (X, ∗, 0) with companion operation �. Example 3.2. Consider the B -algebra (X, ∗, 0) with ∗ defined below [11]. Define an operation � on X as follows: ∗ 0 1 2 3 4 5 0 0 2 1 3 4 5 1 1 0 2 4 5 3 2 2 1 0 5 3 4 3 3 4 5 0 2 1 4 4 5 3 1 0 2 5 5 3 4 2 1 0 � 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 0 5 3 4 2 2 0 1 4 5 3 3 3 4 5 0 1 2 4 4 5 3 2 0 1 5 5 3 4 1 2 0 By routine calculations, (X, ∗,�, 0) is a companion B -algebra. Example 3.3. Consider the B -algebra X = (Z,−, 0). Then for all x, y, z ∈ Z, ((x+ y)− x)− y = 0 and if (z − x)− y = 0, then z − (x+ y) = (z − x)− y = 0. Hence, the binary operation “+” is a companion operation of Z. Therefore, (Z,−,+, 0) is a companion B -algebra. Theorem 3.4. Let (X, ∗, 0) be a B-algebra. If X has a companion operation �, then it is unique. Proof : Assume that the binary operations �1 and �2 are companion operations on X. Then by (SC) applied on �1, for any x, y ∈ X, ((x�1 y) ∗ x) ∗ y = 0. By (C) applied on �2, (x �1 y) ∗ (x �2 y) = 0. Then by Theorem 2.3(d), x �1 y = x �2 y. Thus, �1 = �2 and the companion operation is unique. � Theorem 3.5. Let (X, ∗,�, 0) be a companion B-algebra. Let ? be a binary operation on X such that for all x, y, z ∈ X, (x ∗ y) ∗ z = x ∗ (y ? z). Then (X, ∗, ?, 0) is a companion B-algebra and ? is exactly the operation �. Proof : Suppose x, y, z ∈ X. By hypothesis and Definition 2.1(I), ((x ? y) ∗ x) ∗ y = (x ? y) ∗ (x ? y) = 0. Hence, ? is a subcompanion operation. Now, let (z ∗ x) ∗ y = 0. Then by hypothesis, z ∗ (x ? y) = (z ∗ x) ∗ y = 0. Thus, ? is a companion operation, which is unique by Theorem 3.4. Therefore, (X, ∗, ?, 0) is a companion B-algebra. � Example 3.6. Let X = {0, 1, 2, 3} be a set with the following table of operations: L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1251 ∗ 0 1 2 3 0 0 3 2 1 1 1 0 3 2 2 2 1 0 3 3 3 2 1 0 � 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 Then (X, ∗, 0) is a B -algebra [2] and by routine calculations, (X, ∗,�, 0) is a companion B -algebra. If x = 1 and y = 3, then ((1∗3)∗1)∗3 = 2 6= 0. Hence, ∗ is not a subcompanion operation and so not a companion operation. Remark 3.7. If (X, ∗, 0) is a B-algebra, then (X, ∗, ∗, 0) is not always a companion B- algebra. In Example 3.6, the condition x ∗ y = y ∗ (0 ∗ x) does not hold. Example 3.8. Consider the Klein B -algebra K4 = {0, 1, 2, 3} with the following table of operation [4]: ∗ 0 1 2 3 0 0 1 2 3 1 1 0 3 2 2 2 3 0 1 3 3 2 1 0 Then x ∗ y = y ∗ (0 ∗ x) for any x, y ∈ K4 and (K4, ∗, ∗, 0) is a companion B -algebra. The observation in Example 3.8 is generalized in the next theorem. Theorem 3.9. Let (X, ∗, 0) be a B-algebra. X satisfies x∗y = y ∗ (0∗x) for any x, y ∈ X if and only if (X, ∗, ∗, 0) is a companion B-algebra. Proof : Suppose x∗y = y ∗ (0∗x). By Definition 2.1(III), assumption and Definition 2.1(I), ((x ∗ y) ∗ x) ∗ y = (x ∗ y) ∗ (y ∗ (0 ∗ x)) = (x ∗ y) ∗ (x ∗ y) = 0. Suppose (z ∗ x) ∗ y = 0. By Definition 2.1(III), z ∗ (y ∗ (0 ∗ x)) = 0 and by assumption, z ∗ (x ∗ y) = 0. Therefore, (X, ∗, ∗, 0) is a companion B-algebra. Conversely, suppose (X, ∗, ∗, 0) is a companion B -algebra. By Definition 3.1, (X, ∗, 0) is a B -algebra. Let x, y ∈ X. Then by (SC), ((x ∗ y) ∗ x) ∗ y = 0. By Definition 2.1(III), (x ∗ y) ∗ (y ∗ (0 ∗ x)) = 0. So, x ∗ y = y ∗ (0 ∗ x) by Theorem 2.3(d). � Lemma 3.10. Let (X, ∗,�, 0) be a companion B-algebra. Then for any x, y, z ∈ X, the following hold: (a) 0� y = y and y � 0 = y; (b) x� y = y ∗ (0 ∗ x); (c) if x ∗ z = y, then x = z � y; (d) � is associative in X; (e) x = (x� y)� (0 ∗ y); (f) if (X, ∗, 0) is commutative, then x� y = x ∗ (0 ∗ y). L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1252 Proof : Let (X, ∗,�, 0) be a companion B -algebra and x, y, z ∈ X. (a) In (SC), take x = 0, that is, 0 = ((0� y) ∗ 0) ∗ y = (0� y) ∗ y. By Theorem 2.3(d), 0�y = y. Now, take x = y and y = 0 in (SC). Then, 0 = ((y�0)∗y)∗0 = (y�0)∗y. Hence, by Theorem 2.3(d), y � 0 = y. (b) By (SC), ((x� y) ∗ x) ∗ y = 0. So, by Definition 2.1(III), (x� y) ∗ (y ∗ (0 ∗ x)) = 0. Thus, by Theorem 2.3(d), x� y = y ∗ (0 ∗ x). (c) If x ∗ z = y, then (x ∗ z) ∗ y = y ∗ y = 0. By (C), x ∗ (z� y) = 0. Hence, by Theorem 2.3(d), x = z � y. (d) By Lemma 3.10(b), Definition 2.1(III) and Theorem 2.3(c), we have (x� y)� z = z ∗ (0 ∗ (x� y)) = z ∗ (0 ∗ (y ∗ (0 ∗ x))) = z ∗ ((0 ∗ x) ∗ y) = (z ∗ (0 ∗ y)) ∗ (0 ∗ x) = (y � z) ∗ (0 ∗ x) = x� (y � z). Thus, the companion operation � is associative. (e) Note that by Theorem 2.3(f), Definitions 2.1(I), 2.1(III), Theorems 2.3(b) and 2.4, and Lemma 3.10(b), x = 0 ∗ (0 ∗ x) = ((0 ∗ y) ∗ (0 ∗ y)) ∗ (0 ∗ x) = (0 ∗ y) ∗ ((0 ∗ x) ∗ (0 ∗ (0 ∗ y))) = (0 ∗ y) ∗ ((0 ∗ x) ∗ y) = (0 ∗ y) ∗ (0 ∗ (y ∗ (0 ∗ x))) = (0 ∗ y) ∗ (0 ∗ (x� y)) = (x� y)� (0 ∗ y). (f) Suppose (X, ∗, 0) is commutative. By Lemma 3.10(b) and Definition 2.5, x � y = y ∗ (0 ∗ x) = x ∗ (0 ∗ y). � Notice that in Example 3.6, X is commutative and 1 � 1 = 2 6= 0. Hence, we have found x = 1 ∈ X such that x � x 6= 0. Also, 1 6= 3 = 0 ∗ 1. Thus, we have the following remark. Remark 3.11. If (X, ∗,�, 0) is a companion B-algebra, then (X,�, 0) is not necessarily a B-algebra. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1253 Proposition 3.12. Suppose (X, ∗,�, 0) is a companion B-algebra. If (X, ∗, 0) is a com- mutative B-algebra and x = 0 ∗ x for any x ∈ X, then (X,�, 0) is a B-algebra. Proof : Suppose (X, ∗, 0) is a commutative B -algebra and x, y ∈ X. By Lemma 3.10(b), Definition 2.5 and by assumption, x � y = y ∗ (0 ∗ x) = x ∗ (0 ∗ y) = x ∗ y. Hence, (X,�, 0) = (X, ∗, 0) is a B -algebra. � Example 3.13. Consider the companion B -algebra in Example 3.2 and consider the following table of operation: ⊗ 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 0 4 5 3 2 2 0 1 5 3 4 3 3 4 4 0 2 1 4 4 3 5 1 0 2 5 5 4 3 2 1 0 Applying Theorem 2.6, we conclude that (X,⊗, 0) is the group where x⊗y = x∗(0∗y). Note that � 6= ⊗ since 1� 5 = 4 6= 3 = 1⊗ 5. Thus, by definition of ⊗, x� y 6= x⊗ y = x ∗ (0 ∗ y). Hence, we cannot apply Theorem 2.6 to immediately conclude that (X,�, 0) is a group. However, the following theorem says so. Theorem 3.14. Let (X, ∗,�, 0) be a companion B-algebra. Then (X,�, 0) is a group. Proof : Note that X 6= ∅ since 0 ∈ X. By Lemma 3.10(d) the companion operation � is associative. Note that by Lemma 3.10(a), 0 acts as the �-identity element in (X,�, 0). Find y such that x� y = 0 and y � x = 0. Suppose x� y = 0. Then by Lemma 3.10(b), y ∗ (0 ∗ x) = 0. So, by Theorem 2.3(d), y = 0 ∗ x. Also, suppose y � x = 0. By Lemma 3.10(b), x ∗ (0 ∗ y) = 0. Then by Theorem 2.3(f), (0 ∗ (0 ∗x)) ∗ (0 ∗ y) = 0 and by Theorem 2.3(d), 0 ∗ (0 ∗ x) = 0 ∗ y. Hence, by Theorem 2.3(e), y = 0 ∗ x. Thus, we have found x−1 = y = 0 ∗ x in (X,�, 0). Therefore, (X,�, 0) is a group. � Remark 3.15. For any x ∈ X, x−1 = 0∗x is called the inverse of x in the group (X,�, 0). Theorem 3.16. Let (G, ◦) be a group with identity e. Then G determines a companion B-algebra (G, ∗,⊗, e) where x ∗ y = x ◦ y−1 and x⊗ y = y ∗ x−1. Proof : Let (G, ◦) be a group with identity e and x, y ∈ G. Define two binary operations ∗ and ⊗ by x ∗ y = x ◦ y−1 and x⊗ y = y ∗ x−1. By Theorem 2.7, (G, ∗, e) is a B -algebra. Observe that ((x⊗ y) ∗ x) ∗ y = ((x⊗ y) ◦ x−1) ∗ y = ((x⊗ y) ◦ x−1) ◦ y−1 = (x⊗ y) ◦ (x−1 ◦ y−1) = (x⊗ y) ◦ (y ◦ x)−1 L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1254 = (y ∗ x−1) ◦ (y ◦ x)−1 = (y ◦ (x−1)−1) ◦ (y ◦ x)−1 = (y ◦ x) ◦ (y ◦ x)−1 = e. Hence, ⊗ is a subcompanion operation on G. Suppose (z ∗ x) ∗ y = e. Then z ◦ (y ◦ x)−1 = z ◦ (x−1 ◦ y−1) = (z ◦ x−1) ◦ y−1 = (z ∗ x) ◦ y−1 = (z ∗ x) ∗ y = e. Observe that z ∗ (x⊗ y) = z ∗ (y ∗ x−1) = z ∗ (y ◦ (x−1)−1) = z ∗ (y ◦ x) = z ◦ (y ◦ x)−1 = e. Hence, ⊗ is a companion operation on G. Thus, (G, ∗,⊗, e) is a companion B -algebra. � Consider the B -algebra given in Example 3.2. Note that X is not commutative since there exist x = 3 and y = 4 such that 3∗(0∗4) = 2 6= 1 = 4∗(0∗3). Define x◦y = x∗(0∗y). If x = 3 and y = 2, then ((x◦y)∗x)∗y = 2 6= 0. Hence, ◦ is not a subcompanion operation. Remark 3.17. If (X, ∗, 0) is a B-algebra, then (X, ∗, ◦, 0) is not necessarily a companion B-algebra where the operation ◦ is defined by x ◦ y = x ∗ (0 ∗ y). Example 3.18. Let X = {0, 1, 2} be a set with the following table of operations, where x ◦ y = x ∗ (0 ∗ y): ∗ 0 1 2 0 0 2 1 1 1 0 2 2 2 1 0 ◦ 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1 By routine calculations, (X, ∗, 0) is a commutative B -algebra and (X, ∗, ◦, 0) is a compan- ion B -algebra. Theorem 3.19. If (X, ∗, 0) is a commutative B-algebra, then (X, ∗, ◦, 0) is a companion B-algebra where x ◦ y = x ∗ (0 ∗ y). Proof : Let (X, ∗, 0) be a commutative B -algebra and x, y, z ∈ X. Define the operation ◦ by x ◦ y = x ∗ (0 ∗ y). Note that by Definition 2.1(III), the definition of ◦, Definitions 2.5 and 2.1(I), we have ((x ◦ y) ∗ x) ∗ y = (x ◦ y) ∗ (y ∗ (0 ∗ x)) = (x ∗ (0 ∗ y)) ∗ (y ∗ (0 ∗ x)) = (y ∗ (0 ∗ x)) ∗ (y ∗ (0 ∗ x)) = 0. Now, suppose (z ∗ x) ∗ y = 0. Then by the definition of ◦, Definition 2.5 and Definition 2.1(III), z ∗ (x ◦ y) = z ∗ (x ∗ (0 ∗ y)) = z ∗ (y ∗ (0 ∗ x)) = (z ∗ x) ∗ y = 0. Hence, ◦ is a companion operation. Therefore, (X, ∗, ◦, 0) is a companion B -algebra. � L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1255 4. On �-subalgebras Definition 4.1. Let (X, ∗,�, 0) be a companion B -algebra and I be a nonempty subset of X. Then I is called a �-subalgebra if x� y ∈ I for any x, y,∈ I. Example 4.2. In Example 3.2, the set I1 = {0, 1, 2} is a �-subalgebra of X, while I2 = {3, 4, 5} is not a �-subalgebra since 3� 4 = 1 /∈ I2. Theorem 4.3. Let (X, ∗,�, 0) be a companion B-algebra. If I is a B-ideal of X, then I is a �-subalgebra of X. Proof : Let (X, ∗,�, 0) be a companion B -algebra and I be a B -ideal of X. Then I 6= ∅. Let x, y ∈ I. By (SC), ((x � y) ∗ x) ∗ y = 0 ∈ I. Since I is a B -ideal of X and y ∈ I, (x � y) ∗ x ∈ I by Definition 2.9. Furthermore, since x ∈ I, x � y ∈ I. Therefore, I is a �-subalgebra of X. � The converse of Theorem 4.3 need not be true in general. In the companion B -algebra (Z,−,+, 0) in Example 3.3, I = Z+ is a �-subalgebra since for all x, y ∈ I, x + y ∈ I. However, 0 /∈ I, thus, I is not a B -ideal. Hence, we have the following remark. Remark 4.4. If I is a �-subalgebra of a companion B-algebra (X, ∗,�, 0), then I is not necessarily a B-ideal. Let (Z,−,+, 0) be the companion B -algebra given in Example 3.3. Then I = Z+ is a +-subalgebra . Note that I1 = Z+ ∪ {0} is a B -ideal since 0 ∈ I1. Now, let x− y ∈ I1 and y ∈ I1. Then x− y ≥ 0 and y ≥ 0. So x ≥ 0 and x ∈ I1. Theorem 4.5. Let (X, ∗,�, 0) be a companion B-algebra. Suppose I is a �-subalgebra and 0 ∈ I. Then I is a B-ideal. Proof : Suppose I is a �-subalgebra of X and 0 ∈ I. Let u ∗ v ∈ I and v ∈ I. Then by Theorem 2.3(a) and Lemma 3.10(b), u = (u ∗ v) ∗ (0 ∗ v) = v� (u ∗ v) ∈ I. Therefore, I is a B -ideal. � The following result follows from Theorem 4.3 and Theorem 2.10. Corollary 4.6. Let (X, ∗,�, 0) be a companion B-algebra. If S is a B-subalgebra of X, then S is a �-subalgebra of X. Consider again the companion B -algebra (Z,−,+, 0) and +-subalgebra I = Z+. Notice that 3−5 = −2 /∈ I. Hence, I is not a B -subalgebra. Thus, we have the following remark. Remark 4.7. A �-subalgebra of X is not necessarily a B-subalgebra. Example 4.8. Consider Example 3.2 and �-subalgebra I = {0, 1, 2}. It is easy to see that I is a B -subalgebra and 0 ∗ a ∈ I, for any a ∈ I. Theorem 4.9. Let (X, ∗,�, 0) be a companion B-algebra. Suppose I is a �-subalgebra and 0 ∗ a ∈ I, for any a ∈ I. Then I is a B-subalgebra. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1256 Proof : Suppose I is a �-subalgebra and 0 ∗ a ∈ I, for any a ∈ I. Let x, y ∈ I. Then 0 ∗ y ∈ I. By Theorem 2.3(b) and Lemma 3.10(b), x ∗ y = x ∗ (0 ∗ (0 ∗ y)) = (0 ∗ y)�x ∈ I. Thus, I is a B -subalgebra. � Consider again the companion B -algebra (Z,−,+, 0) and +-subalgebra I = Z+. Take a = 2 and b = 3 ∈ I. Then b−1 = 0 − b = −3 and a + b−1 = −1 /∈ I. Hence, I is not a subgroup of the group (Z,+, 0). So, we have the following remark. Remark 4.10. If I is a �-subalgebra, then I is not necessarily a subgroup. Consider the companion B -algebra (Z,−,+, 0), H1 = Z+ and H2 = Z−. Then H1 and H2 are +-subalgebras. However, H1 ∩H2 = ∅ and hence, not a +-subalgebra. Thus, we have the following remark. Remark 4.11. The intersection of �-subalgebras need not be a �-subalgebra. The proof of the following theorem is straightforward. Theorem 4.12. Let {Ik : k ∈ K} be a nonempty collection of �-subalgebras of a com- panion B-algebra. If I = ⋂ k∈K Ik 6= ∅, then I is a �-subalgebra. Consider Example 3.2. Take A = {0, 3} and B = {0, 4}. Then A and B are �- subalgebras. However, A ∪ B = {0, 3, 4} is not a �-subalgebra since 3 � 4 = 1 /∈ A ∪ B. Hence, we have the following remark. Remark 4.13. The union of �-subalgebras need not be a �-subalgebra. 5. On �-ideals Definition 5.1. Let (X, ∗,�, 0) be a companion B -algebra. A nonempty subset I of X is called a �-ideal if it satisfies: for any x, y ∈ X, (i) 0 ∈ I and (ii) x� y ∈ I and y ∈ I imply x ∈ I. Example 5.2. In Example 3.2, {0, 3} is a �-ideal of X. But, I = {0, 1} is not a �-ideal since 2� 1 = 0 ∈ I and 1 ∈ I but 2 /∈ I. Lemma 5.3. Let (X, ∗,�, 0) be a companion B-algebra and let I be a �-ideal. If x ∈ I, then x−1 = 0 ∗ x ∈ I. Proof : By Remark 3.15, x−1 = 0 ∗ x is the inverse of x. Thus, (0 ∗ x)� x = 0 ∈ I. Since x ∈ I and I is a �-ideal, then 0 ∗ x ∈ I. � Theorem 5.4. Let (X, ∗,�, 0) be a companion B-algebra. If I is a �-ideal of X, then I is a �-subalgebra. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1257 Proof : Let x, y ∈ I. Note that by Lemma 5.3, 0 ∗ y ∈ I. Observe that by Lemma 3.10(b), Theorems 2.4, 2.3(c), Definition 2.1(I) and Theorem 2.3(f), (x� y)� (0 ∗ y) = (y ∗ (0 ∗ x))� (0 ∗ y) = (0 ∗ y) ∗ (0 ∗ (y ∗ (0 ∗ x))) = (0 ∗ y) ∗ ((0 ∗ x) ∗ y) = ((0 ∗ y) ∗ (0 ∗ y)) ∗ (0 ∗ x) = 0 ∗ (0 ∗ x) = x. Since x ∈ I, 0 ∗ y ∈ I and I is a �-ideal, x� y ∈ I. Therefore, I is a �-subalgebra. � The converse of Theorem 5.4 need not be true in general. Note that I = Z+ is a �-subalgebra of (Z,−,+, 0) since for all x, y ∈ I, x + y ∈ I. However, 0 /∈ I. Hence, I is not a �-ideal. Thus, we have the following remark. Remark 5.5. If I is a �-subalgebra, then I is not necessarily a �-ideal. Example 5.6. Consider Example 3.6 and �-subalgebra I = {0, 2}. Observe that 0 ∗ 0 = 0 ∈ I and 0 ∗ 2 = 2 ∈ I, so, 0 ∗ a ∈ I, for any a ∈ I. It is clear that I is also a �-ideal. Theorem 5.7. Let (X, ∗,�, 0) be a companion B-algebra. Suppose I is a �-subalgebra of X and 0 ∗ a ∈ I for any a ∈ I. Then I is a �-ideal. Proof : Suppose I is a �-subalgebra and 0∗a ∈ I for any a ∈ I. Let x ∈ I. Then 0∗x ∈ I. Since I is �-subalgebra, 0 = x � (0 ∗ x) ∈ I. Now, suppose u � v ∈ I and v ∈ I. Then 0 ∗ v ∈ I. By Lemma 3.10(e), u = (u � v) � (0 ∗ v). Since I is a �-subalgebra, u ∈ I. Therefore, I is a �-ideal. � Theorem 5.8. Let (G, ∗,�, 0) be a companion B-algebra. A nonempty subset I of G is a �-ideal of G if and only if I is a subgroup of the group (G,�, 0). Proof : Let I be a �-ideal and a, b ∈ I. By Lemma 5.3, b−1 = 0 ∗ b ∈ I. Because I is also a �-subalgebra by Theorem 5.4, a� b−1 ∈ I. Hence, I is a subgroup. Conversely, suppose I is a subgroup of the group (G,�, 0) and a, b ∈ I. Then a � b−1 ∈ I. Note that a � a−1 = 0. So, 0 ∈ I. Suppose x � y ∈ I and y ∈ I. Then by Lemma 3.10(e), x = (x� y)� (0∗ y) = (x� y)� y−1 ∈ I. Thus, I is a �-ideal. � The following corollary follows from Theorem 5.8 and 5.4. Corollary 5.9. Let (G, ∗,�, 0) be a companion B-algebra. If I is a subgroup of the group (G,�, 0), then I is a �-subalgebra. The following corollary follows from Theorem 5.8. Corollary 5.10. Let {Ik : k ∈ K} be a nonempty collection of �-ideals of a companion B-algebra. If I = ⋂ k∈K Ik 6= ∅, then I is a �-ideal. L.D. Naingue, J.P. Vilela / Eur. J. Pure Appl. Math, 12 (3) (2019), 1248-1259 1258 Observe that in Example 3.2, I1 = {0, 3} and I2 = {0, 4} are �-ideals. But their union, I = I1 ∪ I2 = {0, 3, 4} is not a �-ideal because 1� 4 = 3 ∈ I and 4 ∈ I but 1 /∈ I. Thus, we have the following remark. Remark 5.11. The union of �-ideals need not be a �-ideal. 6. On Companion-B-homomorphisms Definition 6.1. Let (X, ∗X ,�X , 0X) and (Y, ∗Y ,�Y , 0Y ) be companion B -algebras. A map f : X → Y is called a companion-B-homomorphism if for any a, b ∈ X, f(a ∗X b) = f(a) ∗Y f(b) and f(a�X b) = f(a)�Y f(b). Example 6.2. Let m ∈ Z be fixed. The function f : Z→ Z defined by f(x) = mx, x ∈ Z, is a companion-B -homomorphism. Remark 6.3. A companion B-homomorphism is a B-homomorphism and a group homo- morphism. Example 6.4. Consider the companion B -algebra (X, ∗1,�1, 0) in Example 3.6 and (Y, ∗2,�2, 0) in Example 3.8 where �2 = ∗2. Let f : X → Y and f(x) = { 0, if x = 0, 2, 3, if x = 1, 3. Then f is a companion-B -homomorphism. Theorem 6.5. Suppose f : X → Y is a companion B-homomorphism. Then Kerf is a �-subalgebra of X. Proof : Note that by Remark 6.3, Kerf is a subgroup of X. Thus, by Corollary 5.9, Kerf is also a �-subalgebra. � The proof of the following theorem is straightforward. Theorem 6.6. Suppose f : X → Y is a companion B-homomorphism. If I is a �- subalgebra of X, then f(I) is a �-subalgebra of Y . Theorem 6.7. Suppose f : X → Y is a companion B-epimorphism and B is a �- subalgebra of Y . Then f−1(B) is a �-subalgebra of X. Proof : Let B ⊆ Y be a �-subalgebra of Y . Since B 6= ∅ and f is onto, there exist a ∈ B and x ∈ X such that f(x) = a. Hence, x ∈ f−1(B). So, f−1(B) 6= ∅. Note that f−1(B) = {a ∈ X : f(a) ∈ B} ⊆ X. Now, let x, y ∈ f−1(B). Then f(x), f(y) ∈ B. Because B is a �-subalgebra, f(x � y) = f(x) � f(y) ∈ B. Hence, x � y ∈ f−1(B). Therefore, f−1(B) is a �-subalgebra of X. � By Theorem 5.8, a �-ideal is equivalent to a subgroup of (X,�). Thus, the following corollary holds: Corollary 6.8. Suppose f : X → Y is a companion B-homomorphism. REFERENCES 1259 (i) If I is a �-ideal of X, then f(I) is a �-ideal of Y . (i) If B ⊆ Y is a �-ideal of Y , then f−1(B) is a �-ideal of X. (iii) Kerf is a �-ideal of X. References [1] H.K. Abdullah and A.A. Atshan. Complete ideal and n-ideal of B -algebra. Applied Mathematical Sciences, 11(35):1705–1713, 2017. [2] P.J. Allen, H.S. Kim, and J. Neggers. B -algebras and groups. Scientiae Mathematicae Japonicae Online, 9:159–165, 2003. [3] P.J. Allen, H.S. Kim, and J. Neggers. Companion d -algebras. Mathematica Slovaca, 57(2):93–106, 2007. [4] J.C. Endam and R.C. Teves. Some properties of cyclic B -algebras. International Mathematical Forum, 11(8):387–394, 2016. [5] E. Fitria, S. Gemawati, and Kartini. Prime ideals in B -algebras. International Journal of Algebra, 11(7):301–309, 2017. [6] K. Iséki. On BCI -algebras. Math. Seminar Notes, 8:125–130, 1980. [7] K. Iséki and Tanaka. An introduction to theory of BCK -algebras. Math. Japonica, 23:1–26, 1978. [8] M. Kondo and Y.B. Jun. The class of B -algebras coincides with the class of groups. Scientiae Mathematicae Japonicae Online, 2:175–177, 2002. [9] J. Neggers and H.S. Kim. On d -algebras. Mathematica Slovaca, 49(1):19–26, 1999. [10] J. Neggers and H.S. Kim. A fundalmental theorem of B -homomorphism for B - algebras. Int.Math.J., 2(3):207–214, 2002. [11] J. Neggers and H.S. Kim. On B -algebras. Math. Vesnik, 54:21–29, 2002. [12] A. Walendziak. A note on normal subalgebras in B -algebras. Scientiae Mathematicae Japonicae Online, pages 49–53, 2005. [13] A. Walendziak. Some axiomizations of B -algebras. Mathematica Slovaca, 56(3):301– 306, 2006.