The Second Isomorphism Theorem for B-algebras 1 Introduction - Hikari

Applied Mathematical Sciences, Vol. 8, 2014, no. 38, 1865 - 1872 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.4291

The Second Isomorphism Theorem for B-algebras Joemar C. Endam and Jocelyn P. Vilela Department of Mathematics and Statistics College of Science and Mathematics MSU-Iligan Institute of Technology Tibanga, Iligan City, Philippines c 2014 Joemar C. Endam and Jocelyn P. Vilela. This is an open access article Copyright  distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper, some properties of B-homomorphism are provided and the Second Isomorphism Theorem for B-algebras is proved. Furthermore, a sufficient and necessary condition for the product HK of subalgebras to be a subalgebra is proved.

Mathematics Subject Classification: 08A35 Keywords: B-homomorphism, normal B-subalgebra, B-isomorphism theorem, commutative B-algebra, product of B-subalgebra

1

Introduction

The concept of B-algebras was introduced by J. Neggers and H.S. Kim [3]. A B-algebra (X; ∗, 0) is an algebra of type (2, 0), that is, a nonempty set X with a binary operation ∗ and a constant 0) satisfying the following axioms: (I) x ∗ x = 0, (II) x ∗ 0 = x, and (III) (x ∗ y) ∗ z = x ∗ (z ∗ (0 ∗ y)). In [1], the notions of a subalgebra and normality of B-algebras were introduced and some of their properties were established. A nonempty subset N of a B-algebra X is called a subalgebra of X if x ∗ y ∈ N for any x, y ∈ N. It is called normal in X if for any x ∗ y, a ∗ b ∈ N implies (x ∗ a) ∗ (y ∗ b) ∈ N. A normal subset of

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X is a subalgebra of X. The notion of B-homomorphism was also introduced and the First and Third Isomorphism Theorems for B-algebras were proved. In this paper, we provide some properties of B-homomorphism and prove a necessary and sufficient condition for HK to be a subalgebra and the Second Isomorphism Theorem for B-algebras.

2

Some Properties of B-homomorphism

Let (X; ∗, 0X ) and (Y ; ∗, 0Y ) be B-algebras. A mapping ϕ: X → Y is called a B-homomorphism from X into Y if ϕ(x ∗ y) = ϕ(x) ∗ ϕ(y) for any x, y ∈ X. A B-homomorphism ϕ is called a B-monomorphism, B-epimorphism, or Bisomorphism if ϕ is one-to-one, onto, or a bijection, respectively. An isomorphism ϕ: X → X is called a B-automorphism. Throughout this paper, X means a B-algebra (X; ∗, 0). Lemma 2.1 Let X be a B-algebra. i. If {Nα : α ∈ A} is any nonempty collection of subalgebras of X, then Nα is a subalgebra of X. α∈A

ii. If {Nα: α ∈ A} is any nonempty collection of normal subalgebras of X, Nα is a normal subalgebra of X. then α∈A

Proof : Let X be a B-algebra. i. Follows from the fact that a nonempty intersection of a system of subalgebras is a subalgebra. ii. Let {Nα : α ∈ A}  be any nonempty collection of normal subalgebras of X. By (i), Nα is a subalgebra of X. Suppose x ∗ y, a ∗ b ∈ α∈A



Nα . Then x ∗ y, a ∗ b ∈ Nα for all α ∈ A. Since each Nα is normal, α∈A  Nα (x ∗ a) ∗ (y ∗ b) ∈ Nα for all α ∈ A. Therefore, (x ∗ a) ∗ (y ∗ b) ∈ α∈A  and so Nα is a normal subalgebra of X.  α∈A

Since a subalgebra is also a B-algebra contained in a given B-algebra, the following remark easily follows. Remark 2.2 If N is a normal subalgebra of X, then N is normal for every subalgebra of X containing N.

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For any x, y, z ∈ X, we have: (P1) 0 ∗ (0 ∗ x) = x [3], (P2) x ∗ y = 0 ∗ (y ∗ x) [4], and (P3) x ∗ (y ∗ z) = (x ∗ (0 ∗ z)) ∗ y [3]. Theorem 2.3 [1] Let ∅ =  N ⊆ X. Then N is a subalgebra of X if and only if x ∗ (0 ∗ y), 0 ∗ y ∈ N for any x, y ∈ N. Theorem 2.4 [2] Let N be a subalgebra of X. Then N is normal in X if and only if x ∗ (x ∗ y) ∈ N for any x ∈ X, y ∈ N. Definition 2.5 Let H, K be subalgebras of X. Define the subset HK of X to be the set HK = {x ∈ X : x = h ∗ (0 ∗ k) for some h ∈ H, k ∈ K}. Example 2.6 Let X = {0, 1, 2, 3, 4, 5} be a set with the following table: ∗ 0 1 2 3 4 5

0 0 1 2 3 4 5

1 2 0 1 4 5 3

2 1 2 0 5 3 4

3 3 4 5 0 1 2

4 4 5 3 2 0 1

5 5 3 4 1 2 0

Then (X; ∗, 0) is a B-algebra [3]. Let H = {0, 3} and K = {0, 4}. Clearly, H and K are subalgebras of X. However, we see that HK = {0, 2, 3, 4} is not a subalgebra of X since 4 ∗ 3 = 1 ∈ / HK. In general, HK need not be a subalgebra. In the succeeding results, a necessary and sufficient condition for HK to be a subalgebra will be proved. First, we consider the following lemma. Lemma 2.7 Let H and K be subalgebras of X. Then i. H ⊆ HK, KH and K ⊆ HK, KH, ii. HH = H, iii. H ⊆ K implies HK = KH = K. Proof : Let H and K be subalgebras of X. If h ∈ H, then h = h∗0 = h∗(0∗0) ∈ HK by (I) and (II). Also, h = 0 ∗ (0 ∗ h) ∈ KH by (P1). Thus, H ⊆ HK, KH. Similarly, K ⊆ HK, KH.(ii) and (iii) are easy.  The following theorem gives the necessary and sufficient condition for HK to be a subalgebra.

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Theorem 2.8 Let H and K be subalgebras of X. Then HK is a subalgebra of X if and only if HK = KH. Proof : Suppose HK is a subalgebra of X. Let x ∈ KH. Then x = k ∗ (0 ∗ h) for some k ∈ K, h ∈ H. By Lemma 2.7(i), k, h ∈ HK. Since HK is a subalgebra, 0 ∗ h ∈ HK by Theorem 2.3 and so x = k ∗ (0 ∗ h) ∈ HK. Thus, KH ⊆ HK. On the other hand, let x ∈ HK. Then 0 ∗ x ∈ HK. Hence, 0 ∗ x = h ∗ (0 ∗ k) for some h ∈ H, k ∈ K. Since H and K are subalgebras, 0 ∗ h ∈ H and 0 ∗ k ∈ K. Thus, by (P1) and (P2), we have x = 0 ∗ (0 ∗ x) = 0 ∗ (h ∗ (0 ∗ k)) = (0 ∗ k) ∗ h = (0 ∗ k) ∗ (0 ∗ (0 ∗ h)) ∈ KH. Hence, HK ⊆ KH. Therefore, HK = KH. Conversely, suppose that HK = KH and let x, y ∈ HK. Then x = h1 ∗ (0 ∗ k1 ), y = h2 ∗ (0 ∗ k2 ) for some h1 , h2 ∈ H, k1 , k2 ∈ K. Now, since (0 ∗ k2 ) ∗ h2 = (0 ∗ k2 ) ∗ (0 ∗ (0 ∗ h2 )) ∈ KH = HK, (0 ∗ k2 ) ∗ h2 = h3 ∗ (0 ∗ k3 ) for some h3 ∈ H, k3 ∈ K. Similarly, k1 ∗ (0 ∗ h3 ) = h4 ∗ (0 ∗ k4 ) for some h4 ∈ H, k4 ∈ K. Thus, by (P1), (P2), and (III), we have x ∗ y = (h1 ∗ (0 ∗ k1 )) ∗ (h2 ∗ (0 ∗ k2 )) = (h1 ∗ (0 ∗ k1 )) ∗ [0 ∗ ((0 ∗ k2 ) ∗ h2 )] = (h1 ∗ (0 ∗ k1 )) ∗ [0 ∗ (h3 ∗ (0 ∗ k3 ))] = (h1 ∗ (0 ∗ k1 )) ∗ ((0 ∗ k3 ) ∗ h3 ) = h1 ∗ [((0 ∗ k3 ) ∗ h3 ) ∗ (0 ∗ (0 ∗ k1 ))] = h1 ∗ [((0 ∗ k3 ) ∗ h3 ) ∗ k1 ] = h1 ∗ [(0 ∗ k3 ) ∗ (k1 ∗ (0 ∗ h3 ))] = h1 ∗ [(0 ∗ k3 ) ∗ (h4 ∗ (0 ∗ k4 ))] = h1 ∗ [((0 ∗ k3 ) ∗ k4 ) ∗ h4 ] = h1 ∗ [((0 ∗ k3 ) ∗ k4 ) ∗ (0 ∗ (0 ∗ h4 ))] = (h1 ∗ (0 ∗ h4 )) ∗ ((0 ∗ k3 ) ∗ k4 ) = (h1 ∗ (0 ∗ h4 )) ∗ [0 ∗ (k4 ∗ (0 ∗ k3 ))] ∈ HK. Therefore, HK is a subalgebra of X.  In [3], X is commutative if x ∗ (0 ∗ y) = y ∗ (0 ∗ x) for any x, y ∈ X. Corollary 2.9 If H and K are subalgebras of a commutative B-algebra X, then HK is a subalgebra of X. Proof : Since X is commutative, HK = KH. By Theorem 2.8, HK is a subalgebra of X.  Lemma 2.10 If N and K are subalgebras of X with K normal in X, then N ∩ K is a normal subalgebra of N. Proof : Since N ∩ K ⊆ N, N ∩ K is a subalgebra of N by Lemma 2.1(i). Let x ∈ N and y ∈ N ∩ K. Since K is normal of X, y ∈ K, x ∈ N ⊆ X, we have

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x ∗ (x ∗ y) ∈ K by Theorem 2.4. Since N is a subalgebra and x, y ∈ N, we have x ∗ (x ∗ y) ∈ N. Thus, x ∗ (x ∗ y) ∈ N ∩ K. By Theorem 2.4, N ∩ K is normal of N.  Lemma 2.11 If N and K are subalgebras of X with K normal in X, then NK is a subalgebra of X. Proof : Let x, y ∈ NK. Then x = n1 ∗ (0 ∗ k1 ) and y = n2 ∗ (0 ∗ k2 ) for some n1 , n2 ∈ N and k1 , k2 ∈ K. Since N and K are subalgebras of X, n1 ∗ n2 ∈ N and k1 ∗ k2 ∈ K. By Theorem 2.4, n2 ∗ (n2 ∗ (k1 ∗ k2 )) ∈ K. Therefore, by (I), (II), (III), and (P1), we have x ∗ y = (n1 ∗ (0 ∗ k1 )) ∗ (n2 ∗ (0 ∗ k2 )) = n1 ∗ [(n2 ∗ (0 ∗ k2 )) ∗ (0 ∗ (0 ∗ k1 ))] = n1 ∗ [(n2 ∗ (0 ∗ k2 )) ∗ k1 ] = [n1 ∗ ((0 ∗ n2 ) ∗ (0 ∗ n2 ))] ∗ [(n2 ∗ (0 ∗ k2 )) ∗ k1 ] = ((n1 ∗ n2 ) ∗ (0 ∗ n2 )) ∗ [(n2 ∗ (0 ∗ k2 )) ∗ k1 ] = (n1 ∗ n2 ) ∗ {[(n2 ∗ (0 ∗ k2 )) ∗ k1 ] ∗ (0 ∗ (0 ∗ n2 ))} = (n1 ∗ n2 ) ∗ {[(n2 ∗ (0 ∗ k2 )) ∗ k1 ] ∗ n2 } = (n1 ∗ n2 ) ∗ {[n2 ∗ (k1 ∗ (0 ∗ (0 ∗ k2 )))] ∗ n2 } = (n1 ∗ n2 ) ∗ [(n2 ∗ (k1 ∗ k2 )) ∗ n2 ] = (n1 ∗ n2 ) ∗ {0 ∗ [n2 ∗ (n2 ∗ (k1 ∗ k2 ))]} ∈ NK. Therefore, NK is a subalgebra of X.  Lemma 2.12 If N and K are normal subalgebras of X, then NK = KN is a normal subalgebra of X. Proof : By Lemma 2.11 and Theorem 2.8, NK = KN is a subalgebra of X. Let x ∈ X and y ∈ NK. Then y = n ∗ (0 ∗ k) for some n ∈ N and k ∈ K. Therefore, by (I), (II), (III), (P1), (P2), and (P3), we have x ∗ (x ∗ y) = x ∗ [x ∗ (n ∗ (0 ∗ k))] = x ∗ ((x ∗ k) ∗ n) = x ∗ [((0 ∗ (0 ∗ x)) ∗ k) ∗ (0 ∗ (0 ∗ n))] = x ∗ [(0 ∗ (k ∗ x)) ∗ (0 ∗ (0 ∗ n))] = (x ∗ (0 ∗ n)) ∗ (0 ∗ (k ∗ x)) = [(x ∗ (0 ∗ n)) ∗ ((0 ∗ x) ∗ (0 ∗ x))] ∗ (0 ∗ (k ∗ x)) = [((x ∗ (0 ∗ n)) ∗ x) ∗ (0 ∗ x)] ∗ (0 ∗ (k ∗ x)) = [(x ∗ (0 ∗ n)) ∗ x] ∗ [(0 ∗ (k ∗ x)) ∗ (0 ∗ (0 ∗ x))] = [(x ∗ (0 ∗ n)) ∗ x] ∗ [(0 ∗ (k ∗ x)) ∗ x] = [x ∗ (x ∗ (0 ∗ (0 ∗ n)))] ∗ [0 ∗ (x ∗ (0 ∗ (k ∗ x)))] = (x ∗ (x ∗ n)) ∗ [0 ∗ (x ∗ (x ∗ k))] ∈ NK. By Theorem 2.4, NK is normal in X.  Lemma 2.13 If ϕ: X → Y and ψ: Y → Z are B-homomorphisms, then ψ ◦ ϕ: X → Z is also a B-homomorphism.

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Assuming compatibility of functions so that composition is defined, the following corollary easily follows. Corollary 2.14 The composition of B-monomorphisms is a B-monomorphism, the composition of B-epimorphisms is an B-epimorphism, the composition of B-isomorphisms is a B-isomorphism, and the composition of B-automorphisms is a B-automorphism. Clearly, the identity mapping idX : X → X of a B-algebra X is an automorphism of X. If ϕ: X → Y is a B-homomorphism from X into Y , then ϕ(0X ) = 0Y and ϕ(0X ∗x) = 0Y ∗ϕ(x) for all x ∈ X. Given a B-homomorphism, the following lemma also holds. Lemma 2.15 Let ϕ: X → Y be a B-homomorphism from X into Y . i. If N is a subalgebra of X, then ϕ(N ) is a subalgebra of Y . Moreover, if N is commutative, then ϕ(N ) is commutative. ii. If K is a subalgebra of Y , then ϕ−1 (K) is a subalgebra of X containing Ker ϕ. iii. If N is a normal subalgebra of X and ϕ is onto, then ϕ(N ) is a normal subalgebra of Y . iv. If K is a normal subalgebra of Y , then ϕ−1 (K) is a normal subalgebra of X. Proof : Let ϕ: X → Y be a B-homomorphism. i. Suppose N is commutative. If x, y ∈ ϕ(N ), then there exist a, b ∈ N such that ϕ(a) = x and ϕ(b) = y. Since N is commutative, x ∗ (0Y ∗ y) = ϕ(a) ∗ (ϕ(0X ) ∗ ϕ(b)) = ϕ(a ∗ (0X ∗ b)) = ϕ(b ∗ (0X ∗ a)) = ϕ(b) ∗ (ϕ(0X ) ∗ ϕ(a)) = y ∗ (0Y ∗ x). Therefore, ϕ(N ) is commutative. iii. Let N is a normal subalgebra of X. By (i), ϕ(N ) is a subalgebra of Y . Let x ∈ Y and y ∈ ϕ(N ). Then y = ϕ(n) for some n ∈ N. Since ϕ is onto, there exists a ∈ X such that ϕ(a) = x. Since N is normal in X, a ∗ (a ∗ n) ∈ N. Thus, x ∗ (x ∗ y) = ϕ(a) ∗ (ϕ(a) ∗ ϕ(n)) = ϕ(a ∗ (a ∗ n)) ∈ ϕ(N ). Therefore, ϕ(N ) is normal in Y . iv. Let K be a normal subalgebra of Y . By (ii), ϕ−1 (K) is a subalgebra of X. Let x ∈ X and y ∈ ϕ−1 (K). Then ϕ(x) ∈ Y and ϕ(y) ∈ K. Since K is normal of Y , ϕ(x ∗ (x ∗ y)) = ϕ(x) ∗ (ϕ(x) ∗ ϕ(y)) ∈ K. Thus, x ∗ (x ∗ y) ∈ ϕ−1 (K) and so ϕ−1 (K) is normal in X. 

The second isomorphism theorem for B-algebras

3

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The Second Isomorphism Theorem for Balgebras

In [3], if N is a normal subalgebra of X, then (X/N; ∗, [0]N ) is a B-algebra, where X/N = {[x]N : x ∈ X} and ∗ is defined by [x]N ∗ [y]N = [x ∗ y]N . For x ∈ X, [x]N is the equivalence class containing x, that is, [x]N = {y ∈ X : x ∼N y}, where x ∼N y if and only if x ∗ y ∈ N for any x, y ∈ X. The algebra X/N is called the quotient B-algebra of X by N. The following theorem is labeled as the First Isomorphism Theorem for B-algebras and can be found in [1]. Theorem 3.1 Let ϕ: X → Y be a B-homomorphism from X into Y . Then X/Ker ϕ ∼ = Imϕ. In particular, if ϕ is surjective, then X/ Ker ϕ ∼ = Y. The following theorem is labeled as the Third Isomorphism Theorem for B-algebras and can be found in [1]. Theorem 3.2 Let N and K be normal subalgebras of X, and let K ⊆ N. Then X/N ∼ = (X/K)/(N/K). Lemma 3.3 If K and N are normal subalgebras of X such that K ⊆ N, then N/K is a normal subalgebra of X/K. Proof : By Remark 2.2, N/K is well-defined. Note that X/K and N/K are B-algebras. Since K ⊆ N, N/K ⊆ X/K and so N/K is a subalgebra of X/K. Let [x]K ∗ [y]K , [a]K ∗ [b]K ∈ N/K. Since [x ∗ y]K = [x]K ∗ [y]K ∈ N/K, x ∗ y ∈ N. Similarly, a ∗ b ∈ N. Since N is normal, (x ∗ a) ∗ (y ∗ b) ∈ N and so ([x]K ∗ [a]K ) ∗ ([y]K ∗ [b]K ) = [x ∗ a]K ∗ [y ∗ b]K = [(x ∗ a) ∗ (y ∗ b)]K ∈ N/K. Therefore, N/K is a normal subalgebra of X/K.  This means that (X/K)/(N/K) in Theorem 3.2 is, indeed, well-defined. The following proves the Second Isomorphism Theorem for B-algebras. Let N and K be subalgebras of X with K normal in X. By Lemma 2.10, N ∩ K is a normal subalgebra of N. Thus, N/(N ∩ K) is well-defined. By Lemma 2.11 and Theorem 2.8, NK = KN is a subalgebra of X. By Lemma 2.7(i) and Remark 2.2, K is normal in NK. Hence, NK/K is well-defined. Theorem 3.4 If N and K are subalgebras of X with K normal in X, then N/(N ∩ K) ∼ = NK/K.

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Proof : Define ϕ: N → NK/K by ϕ(n) = [n]K for all n ∈ N. Since N ⊆ NK, ϕ(n) = [n]K ∈ NK/K for all n ∈ N. Let m, n ∈ N such that m = n. Then ϕ(m) = [m]K = [n]K = ϕ(n). Thus, ϕ is well-defined. Let [x]K ∈ NK/K.Then x = n ∗ (0 ∗ k) for some n ∈ N, k ∈ K. It follows that ϕ is onto since [x]K = [n ∗ (0 ∗ k)]K = [n]K ∗ [0 ∗ k]K = [n]K = ϕ(n). Also, ϕ is a B-homomorphism since ϕ(a ∗ b) = [a ∗ b]K = [a]K ∗ [b]K = ϕ(a) ∗ ϕ(b). Hence, by Theorem 3.1, N/Ker ϕ ∼ = NK/K. But Ker ϕ = {n ∈ N: ϕ(n) = K} = {n ∈ N: [n]K = [0]K } = {n ∈ N: n ∈ K} = N ∩ K. Therefore, N/(N ∩ K) ∼  = NK/K. Acknowledgements. This research is funded by the Department of Science and Technology (DOST) through Accelerated Science and Technology Human Resource Development Program (ASTHRDP).

References [1] J. Neggers and H.S. Kim, A fundamental theorem of B-homomorphism for B-algebras, Int. Math. J., 2(3)(2002), 207-214. [2] A. Walendziak, A note on normal subalgebras in B-algebras, Scientiae Mathematicae Japonicae Online, (2005), 49-53. [3] J. Neggers and H.S. Kim, On B-algebras, Mat. Vesnik, 54(2002), 21-29. [4] A. Walendziak, Some Axiomatizations of B-algebras, Math. Slovaca, 56(3)(2006), 301-306. Received: March 20, 2014