EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 3, 2023, 1663-1674
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On B-commutators of B-algebras

Joel G. Adanza

Mathematics Department, Negros Oriental State University, Dumaguete City, Philippines

Abstract. In this paper, we investigate some properties of B-commutators of B-algebras. We also
characterize solvable B-algebras via B-commutators.

2020 Mathematics Subject Classifications: 08A05, 03G25
Key Words and Phrases: Solvable B-algebras, B-commutators, kth B-commutators

1. Introduction and Preliminaries

In 1966, Y. Imai and K. Iséki introduced the concept of BCK-algebras [14]. It is
known that BCK-algebras are inspired by some implicational logic. From then on, several
generalizations of BCK-algebras exist. In [15], K. Iséki introduced BCI-algebras and that
the class of BCK-algebras is a proper subclass of the class of BCI-algebras. In 1983, Q.P.
Hu and X. Li introduced a wide class of abstract algebras: BCH-algebras [13]. They have
shown that the class of BCI-algebras is a proper subclass of the class of BCH-algebras.
These algebras are of type (2, 0), that is, a nonempty set together with a binary operation
and a constant, satisfying some axioms. Up to this day, inspired by BCK/BCI/BCH-
algebras, there are more than twenty type (2, 0) algebras introduced and investigated.
One of these algebras is the concept of B-algebras.

In [21], J. Neggers and H.S. Kim introduced and established the notion of B-algebras.
A B-algebra is an algebra (X; ∗, 0) of type (2, 0) satisfying:
(I) x ∗ x = 0,
(II) x ∗ 0 = x,
(III) (x ∗ y) ∗ z = x ∗ (z ∗ (0 ∗ y)), for any x, y, z ∈ X.
X is said to be commutative if x ∗ (0 ∗ y) = y ∗ (0 ∗ x) for any x, y ∈ X. Let X be a
B-algebra. Recall that for any x, y, z ∈ X, we have the following properties:
(P1) 0 ∗ (0 ∗ x) = x [21],
(P2) x ∗ y = 0 ∗ (y ∗ x) [26],
(P3) x ∗ (y ∗ z) = (x ∗ (0 ∗ z)) ∗ y [21],
(P4) (x ∗ z) ∗ (y ∗ z) = x ∗ y [26].

We now present two examples of B-algebras, one is commutative and the other is
noncommutative.

DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4841

Email address: joel.adanza@norsu.edu.ph (J. Adanza)

https://www.ejpam.com 1663 © 2023 EJPAM All rights reserved.



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1664

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

∗ 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; ∗, 0) is a commutative B-algebra [10].

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

∗ 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

Then (X; ∗, 0) is a noncommutative B-algebra [20].

Throughout this paper, let X be a B-algebra (X; ∗, 0). In [20], a nonempty subset N
of X is called a subalgebra of X if x ∗ y ∈ N for any x, y ∈ N . A subalgebra N of X is
called normal in X if (x ∗ a) ∗ (y ∗ b) ∈ N for any x ∗ y, a ∗ b ∈ N . Let N be normal in X.
Define a relation ∼N on X by x ∼N y if and only if x ∗ y ∈ N , where x, y ∈ X. Then ∼N

is an equivalence relation on X. Denote the equivalence class containing x by xN , that
is, xN = {y ∈ X : x ∼N y}. Let X/N = {xN : x ∈ X}. The binary operation in X/N is
defined by xN ∗′ yN = (x ∗ y)N . The B-algebra X/N is called the quotient B-algebra of
X by N . In [1], xH = {x ∗ (0 ∗ h) : h ∈ H} and Hx = {h ∗ (0 ∗ x) : h ∈ H}, called the left
and right B-cosets of H in X, respectively. The subset HK [11] of X is given by HK =
{x ∈ X : x = h ∗ (0 ∗ k) for some h ∈ H, k ∈ K}.

Other properties and characterizations of B-algebras can be found in some other pa-
pers ([2–7, 9, 10, 12, 16–19], [22, 23], [25, 26].) In particular, R. Soleimani [24] introduced
the notion of B-commutators of B-algebras. He also established some basic properties
of B-commutators. In [8], J.C. Endam and G.S. Dael introduced the notion of solv-
able B-algebras. In this paper, we established some basic properties of B-commutators
of B-algebras. These properties are used in characterizing solvable B-algebras via B-
commutators. As a result, we showed that a B-algebra X is solvable if and only if there
is positive integer m such that the mth B-commutator subalgebra X(m) is equal to {0}.

2. B-commutators

This section presents some identities satisfied by the B-commutators in B-algebras. We
recall first from [24] the definition of B-commutators. Let x, y ∈ X. The B-commutator
of x and y is given by



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1665

[x, y] = ((0 ∗ x) ∗ y) ∗ ((0 ∗ y) ∗ x).

The subalgebra of X generated by {[x, y] : x, y ∈ X} is called the derived B-algebra,
denoted by D(X).

Example 3. Let (X; ∗, 0) be the B-algebra in Example 2. We now compute for [x, y] for
all x, y ∈ X. These computations are used in the succeeding examples.

[0, 0] = 0 [1, 1] = 0 [2, 2] = 0 [3, 3] = 0 [4, 4] = 0 [5, 5] = 0

[0, 1] = 0 [1, 0] = 0 [2, 0] = 0 [3, 0] = 0 [4, 0] = 0 [5, 0] = 0

[0, 2] = 0 [1, 2] = 0 [2, 1] = 0 [3, 1] = 2 [4, 1] = 2 [5, 1] = 2

[0, 3] = 0 [1, 3] = 1 [2, 3] = 2 [3, 2] = 1 [4, 2] = 1 [5, 2] = 1

[0, 4] = 0 [1, 4] = 1 [2, 4] = 2 [3, 4] = 1 [4, 3] = 2 [5, 3] = 1

[0, 5] = 0 [1, 5] = 1 [2, 5] = 2 [3, 5] = 2 [4, 5] = 1 [5, 4] = 2

A map φ : X → Y is called a B-homomorphism [20] if φ(x ∗ y) = φ(x) ∗ φ(y) for any
x, y ∈ X.

Lemma 1. [24] Let φ : X → Y be a B-homomorphism and let x, y ∈ X. Then

i. [x, y] = 0 if and only if x ∗ (0 ∗ y) = y ∗ (0 ∗ x),

ii. φ([x, y]) = [φ(x), φ(y)],

iii. if φ is onto, then φ(D(X)) = D(φ(X)).

Lemma 2. Let x, y ∈ X. Then

i. [x, x] = [x, 0] = [0, x] = 0,

ii. 0 ∗ [x, y] = [y, x].

Proof. Clearly, (i) follows from Lemma 1(i) and (P1); (ii) follows from (P2).

Let x,w ∈ X. We define xw to be the element (0 ∗ w) ∗ ((0 ∗ w) ∗ x). For instance, let
X be the B-algebra in Example 1. Below are some sample computations to illustrate xw:

23 = (0 ∗ 3) ∗ ((0 ∗ 3) ∗ 2) = 3 ∗ (3 ∗ 2) = 3 ∗ 1 = 2
32 = (0 ∗ 2) ∗ ((0 ∗ 2) ∗ 3) = 2 ∗ (2 ∗ 3) = 2 ∗ 1 = 3
13 = (0 ∗ 3) ∗ ((0 ∗ 3) ∗ 1) = 3 ∗ (3 ∗ 1) = 3 ∗ 2 = 1
31 = (0 ∗ 1) ∗ ((0 ∗ 1) ∗ 3) = 1 ∗ (1 ∗ 3) = 1 ∗ 2 = 3
12 = (0 ∗ 2) ∗ ((0 ∗ 2) ∗ 1) = 2 ∗ (2 ∗ 1) = 2 ∗ 3 = 1
21 = (0 ∗ 1) ∗ ((0 ∗ 1) ∗ 2) = 1 ∗ (1 ∗ 2) = 1 ∗ 3 = 2

The following lemma presents the basic properties of xw.

Lemma 3. Let x, y, w ∈ X. Then the following properties hold:

i. 0 ∗ xw = (0 ∗ x)w,



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1666

ii. (x ∗ y)w = 0 ∗ (y ∗ x)w,

iii. (0 ∗ x)x = 0 ∗ x,

iv. xx = x,

v. x0∗x = x,

vi. x ∗ yx = (0 ∗ y) ∗ (0 ∗ x),

vii. xy = x ∗ [y, x],

viii. x0∗y = y ∗ (y ∗ x),

ix. [xy, 0 ∗ y] = [y, x].

Proof. Let x, y, w ∈ X.
i. By (P2) and (III), we have

0 ∗ xw = 0 ∗ ((0 ∗ w) ∗ ((0 ∗ w) ∗ x))
= ((0 ∗ w) ∗ x) ∗ (0 ∗ w)
= (0 ∗ w) ∗ ((0 ∗ w) ∗ (0 ∗ x))
= (0 ∗ x)w.

ii. By (i), we have (x ∗ y)w = (0 ∗ (y ∗ x))w = 0 ∗ (y ∗ x)w.
iii. By (I) and (II), we have

(0 ∗ x)x = (0 ∗ x) ∗ ((0 ∗ x) ∗ (0 ∗ x))
= (0 ∗ x) ∗ 0
= 0 ∗ x.

iv. By P3, (I), and P1, we have

xx = (0 ∗ x) ∗ ((0 ∗ x) ∗ x)
= ((0 ∗ x) ∗ (0 ∗ x)) ∗ (0 ∗ x)
= 0 ∗ (0 ∗ x)
= x.

v. By P1, (I), and (II), we have

x0∗x = (0 ∗ (0 ∗ x)) ∗ ((0 ∗ (0 ∗ x)) ∗ x)
= x ∗ (x ∗ x)
= x ∗ 0
= x.



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1667

vi. By P3, P2, (III), and (I), we have

x ∗ yx = x ∗ ((0 ∗ x) ∗ ((0 ∗ x) ∗ y))
= (x ∗ (0 ∗ ((0 ∗ x) ∗ y))) ∗ (0 ∗ x)
= (x ∗ (y ∗ (0 ∗ x))) ∗ (0 ∗ x)
= ((x ∗ x) ∗ b) ∗ (0 ∗ x)
= (0 ∗ y) ∗ (0 ∗ x).

vii. By P3, P2, (III), and (I), we have

x ∗ [y, x] = x ∗ (((0 ∗ y) ∗ x) ∗ ((0 ∗ x) ∗ y))
= (x ∗ (0 ∗ ((0 ∗ x) ∗ y))) ∗ ((0 ∗ y) ∗ x)
= (x ∗ (y ∗ (0 ∗ x))) ∗ ((0 ∗ y) ∗ x)
= ((x ∗ x) ∗ y) ∗ ((0 ∗ y) ∗ x)
= (0 ∗ y) ∗ ((0 ∗ y) ∗ x)
= xy.

viii. This follows from P1.
ix. By P1, (vi), P4, (vii), P2, and (II), we get

[xy, 0 ∗ y] = ((0 ∗ xy) ∗ (0 ∗ y)) ∗ ((0 ∗ (0 ∗ y)) ∗ xy)
= ((0 ∗ xy) ∗ (0 ∗ y)) ∗ (y ∗ xy)
= ((0 ∗ xy) ∗ (0 ∗ y)) ∗ ((0 ∗ x) ∗ (0 ∗ y))
= (0 ∗ xy) ∗ (0 ∗ x)
= (0 ∗ (x ∗ [y, x])) ∗ (0 ∗ x)
= ([y, x] ∗ x) ∗ (0 ∗ x)
= [y, x] ∗ 0
= [y, x].

The following lemma is used to prove the succeeding theorems.

Lemma 4. Let a, b, c ∈ X. Then the following properties hold:

i. ((a ∗ b) ∗ a) ∗ (a ∗ (a ∗ c)) = a ∗ (a ∗ ((0 ∗ b) ∗ c)),

ii. (a ∗ (a ∗ b)) ∗ (a ∗ (a ∗ c)) = a ∗ (a ∗ (b ∗ c)),

iii. [((a ∗ b) ∗ ((0 ∗ b) ∗ (0 ∗ a))) ∗ c] ∗ (c ∗ (c ∗ b)) = (a ∗ b) ∗ (c ∗ (0 ∗ a)).

Proof. i. By (III), P2, P4, and P3, we have

((a ∗ b) ∗ a) ∗ (a ∗ (a ∗ c)) = (a ∗ b) ∗ [(a ∗ (a ∗ c)) ∗ (0 ∗ a)]



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1668

= a ∗ [((a ∗ (a ∗ c)) ∗ (0 ∗ a)) ∗ (0 ∗ b)]
= a ∗ [(a ∗ ((0 ∗ a) ∗ (0 ∗ (a ∗ c)))) ∗ (0 ∗ b)]
= a ∗ [(a ∗ ((0 ∗ a) ∗ (c ∗ a))) ∗ (0 ∗ b)]
= a ∗ [(a ∗ (0 ∗ c)) ∗ (0 ∗ b)]
= a ∗ (a ∗ ((0 ∗ b) ∗ c)).

ii. By (III), P2, and P4, we have

(a ∗ (a ∗ b)) ∗ (a ∗ (a ∗ c)) = a ∗ [(a ∗ (a ∗ c)) ∗ (0 ∗ (a ∗ b))]
= a ∗ ((a ∗ (a ∗ c)) ∗ (b ∗ a))
= a ∗ [a ∗ ((b ∗ a) ∗ (0 ∗ (a ∗ c)))]
= a ∗ (a ∗ ((b ∗ a) ∗ (c ∗ a)))
= a ∗ (a ∗ (b ∗ c)).

iii. By (III), P2, P4, P3, (I), and (II), we have
[((a ∗ b) ∗ ((0 ∗ b) ∗ (0 ∗ a))) ∗ c] ∗ (c ∗ (c ∗ b))

= [(a ∗ b) ∗ (c ∗ (0 ∗ ((0 ∗ b) ∗ (0 ∗ a))))] ∗ (c ∗ (c ∗ b))
= [(a ∗ b) ∗ (c ∗ ((0 ∗ a) ∗ (0 ∗ b)))] ∗ (c ∗ (c ∗ b))
= (a ∗ b) ∗ [(c ∗ (c ∗ b)) ∗ (0 ∗ (c ∗ ∗((0 ∗ a) ∗ (0 ∗ b))))]
= (a ∗ b) ∗ [(c ∗ (c ∗ b)) ∗ (((0 ∗ a) ∗ (0 ∗ b)) ∗ c)]
= (a ∗ b) ∗ [c ∗ ((((0 ∗ a) ∗ (0 ∗ b)) ∗ c) ∗ (0 ∗ (c ∗ b)))]
= (a ∗ b) ∗ [c ∗ ((((0 ∗ a) ∗ (0 ∗ b)) ∗ c) ∗ (b ∗ c))]
= (a ∗ b) ∗ [c ∗ (((0 ∗ a) ∗ (0 ∗ b)) ∗ b)]
= (a ∗ b) ∗ [c ∗ ((0 ∗ a) ∗ (b ∗ b))]
= (a ∗ b) ∗ (c ∗ ((0 ∗ a) ∗ 0))
= (a ∗ b) ∗ (c ∗ (0 ∗ a)).

Theorem 1. Let w, x, y ∈ X. Then [x, y]w = [xw, yw].

Proof. By P2, we have

[xw, yw] = ((0 ∗ xw) ∗ yw) ∗ ((0 ∗ yw) ∗ xw)
= [(0 ∗ ((0 ∗ w) ∗ ((0 ∗ w) ∗ x))) ∗ yw] ∗ [(0 ∗ ((0 ∗ w) ∗ ((0 ∗ w) ∗ y))) ∗ xw]
= [(((0 ∗ w) ∗ x) ∗ (0 ∗ w)) ∗ yw]︸ ︷︷ ︸

(1)

∗ [(((0 ∗ w) ∗ y) ∗ (0 ∗ w)) ∗ xw]︸ ︷︷ ︸
(2)

We first consider (1), by Lemma 4(i) [with a = 0 ∗ w, b = x, c = y], we have

(((0 ∗ w) ∗ x) ∗ (0 ∗ w)) ∗ yw = (((0 ∗ w) ∗ x) ∗ (0 ∗ w)) ∗ ((0 ∗ w) ∗ ((0 ∗ w) ∗ y))
= (0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ x) ∗ y)).

Similarly for (2), by Lemma 4(i) [with a = 0 ∗ w, b = y, c = x], we have

(((0 ∗ w) ∗ y) ∗ (0 ∗ w)) ∗ xw = (((0 ∗ w) ∗ y) ∗ (0 ∗ w)) ∗ ((0 ∗ w) ∗ ((0 ∗ w) ∗ x))



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1669

= (0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ y) ∗ x)).

Thus,

[xw, yw] = [(((0 ∗ w) ∗ x) ∗ (0 ∗ w)) ∗ yw]︸ ︷︷ ︸
(1)

∗ [(((0 ∗ w) ∗ y) ∗ (0 ∗ w)) ∗ xw]︸ ︷︷ ︸
(2)

= [(0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ x) ∗ y))] ∗ [(0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ y) ∗ x))].

Applying Lemma 4(ii) [with a = 0 ∗ w, b = (0 ∗ x) ∗ y, c = (0 ∗ y) ∗ x], we have

[xw, yw] = [(0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ x) ∗ y))] ∗ [(0 ∗ w) ∗ ((0 ∗ w) ∗ ((0 ∗ y) ∗ x))]
= (0 ∗ w) ∗ [(0 ∗ w) ∗ (((0 ∗ x) ∗ y) ∗ ((0 ∗ y) ∗ x))]
= (0 ∗ w) ∗ ((0 ∗ w) ∗ [x, y])
= [x, y]w.

Theorem 2. Let w, x, y, z ∈ X. Then [x ∗ (0 ∗ y), z] = [x, z]y ∗ [z, y].

Proof. By (III) and P2, we have

[x, z]y ∗ [z, y] = ((0 ∗ y) ∗ ((0 ∗ y) ∗ [x, z])) ∗ [z, y]
= (0 ∗ y) ∗ ([z, y] ∗ (0 ∗ ((0 ∗ y) ∗ [x, z])))
= (0 ∗ y) ∗ ([z, y] ∗ ([x, z] ∗ (0 ∗ y)))
= (0 ∗ y) ∗ [(((0 ∗ z) ∗ y) ∗ ((0 ∗ y) ∗ z)) ∗ ((((0 ∗ x) ∗ z) ∗ ((0 ∗ z) ∗ x)) ∗ (0 ∗ y)))]

For simplicity, we write x′ = 0 ∗x, y′ = 0 ∗ y, z′ = 0 ∗ z. Thus, by (III), P1, and P2, we get

[x, z]y ∗ [z, y] = y′ ∗ [((z′ ∗ y) ∗ (y′ ∗ z)) ∗ (((x′ ∗ z) ∗ (z′ ∗ x)) ∗ y′)]
= y′ ∗ [(z′ ∗ ((y′ ∗ z) ∗ y′)) ∗ (((x′ ∗ z) ∗ (z′ ∗ x)) ∗ y′)]
= y′ ∗ [z′ ∗ ((((x′ ∗ z) ∗ (z′ ∗ x)) ∗ y′) ∗ (y′ ∗ (y′ ∗ z)))]

Applying Lemma 4(iii) [with a = x′, b = z, c = y′], P2, and (III), we get

[x, z]y ∗ [z, y] = y′ ∗ [z′ ∗ ((x′ ∗ z) ∗ (y′ ∗ x))]
= y′ ∗ [z′ ∗ ((x′ ∗ z) ∗ (0 ∗ (x ∗ y′)))]
= y′ ∗ [(z′ ∗ (x ∗ y′)) ∗ (x′ ∗ z)]
= y′ ∗ [(z′ ∗ (x ∗ y′)) ∗ (0 ∗ (z ∗ x′))]
= (y′ ∗ (z ∗ x′)) ∗ (z′ ∗ (x ∗ y′))
= ((0 ∗ y) ∗ (z ∗ (0 ∗ x))) ∗ ((0 ∗ z) ∗ (x ∗ (0 ∗ y)))
= (((0 ∗ y) ∗ x) ∗ z) ∗ ((0 ∗ z) ∗ (x ∗ (0 ∗ y)))
= ((0 ∗ (x ∗ (0 ∗ y))) ∗ z) ∗ ((0 ∗ z) ∗ (x ∗ (0 ∗ y)))
= [x ∗ (0 ∗ y), z].



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1670

Corollary 1. Let x, y, z ∈ X. Then [x, y ∗ (0 ∗ z)] = [x, z] ∗ [y, x]z.

Proof. By Lemma 2(ii), Theorem 2, and P2, we get

[x, y ∗ (0 ∗ z)] = 0 ∗ [y ∗ (0 ∗ z), x]
= 0 ∗ ([y, x]z ∗ [x, z])
= [x, z] ∗ [y, x]z.

Theorem 3. Let x, y ∈ X. Then [x, 0 ∗ y] = [y, x]0∗y.

Proof. By Theorem 1, Lemma 3(v, vi), P1, P4, Lemma 3(viii), P2, and P3, we get

[y, x]0∗y = [y0∗y, x0∗y]

= [y, x0∗y]

= ((0 ∗ y) ∗ x0∗y) ∗ ((0 ∗ x0∗y) ∗ y)
= ((0 ∗ x) ∗ (0 ∗ (0 ∗ y))) ∗ ((0 ∗ x0∗y) ∗ y)
= ((0 ∗ x) ∗ y) ∗ ((0 ∗ x0∗y) ∗ y)
= (0 ∗ x) ∗ (0 ∗ x0∗y)
= (0 ∗ x) ∗ (0 ∗ (y ∗ (y ∗ x)))
= (0 ∗ x) ∗ ((y ∗ x) ∗ y)
= ((0 ∗ x) ∗ (0 ∗ y)) ∗ (y ∗ x)
= ((0 ∗ x) ∗ (0 ∗ y)) ∗ ((0 ∗ (0 ∗ y)) ∗ x)
= [x, 0 ∗ y].

Corollary 2. Let x, y ∈ X. Then [0 ∗ x, y] = [y, x]0∗x.

Proof. By Lemma 2(ii), Theorem 3, and Lemma 3(i), we get

[0 ∗ x, y] = 0 ∗ [y, 0 ∗ x]
= 0 ∗ [x, y]0∗x

= (0 ∗ [x, y])0∗x

= [y, x]0∗x.



J. Adanza / Eur. J. Pure Appl. Math, 16 (3) (2023), 1663-1674 1671

3. kth B-commutators

We recall first the concept of solvable B-algebras [8]. Let

X = H0 ⊇ H1 ⊇ H2 ⊇ · · · ⊇ Hn = {0}

be a series of subalgebras of X. The series is called a subnormal B-series if each Hi is
normal in Hi−1. The series is called a normal B-series if each Hi is normal in X. Since
{0} is normal in X, every B-algebra has a normal B-series. If X has a subnormal B-series
X = H0 ⊇ H1 ⊇ H2 ⊇ · · · ⊇ Hn−1 ⊇ Hn = {0} such that Hi/Hi+1 is commutative,
i = 0, 1, . . . , n − 1, then we say that X is solvable. Such a subnormal B-series is called a
solvable B-series for X.

For simplicity, we write the derived B-algebra D(X) as X ′.

Definition 1. Set X(1) = X ′ and define inductively X(k+1) = X(k)′, the B-commutator
subalgebra of X(k), k > 0. For any positive integer k, X(k) is called the kth B-commutator
subalgebra of X.

By Lemma 1, a B-algebra X is commutative if and only if X ′ = {0}.

Example 4. Let (X; ∗, 0) be the noncommutative B-algebra in Example 2. Then from
the computations in Example 3, we see that X ′ = {0, 1, 2} and X(2) = {0, 1, 2}′ = {0}.
Thus, X(k) = {0} for all k ≥ 2.

The following theorem characterizes solvable B-algebra.

Theorem 4. X is solvable if and only if there is positive integer m such that X(m) = {0}.

Proof. Suppose that X is solvable. Then X has a solvable series, say,

X = H0 ⊇ H1 ⊇ H2 ⊇ · · · ⊇ Hn−1 ⊇ Hn = {0}.

Since Hi+1 is normal in Hi and Hi/Hi+1 is commutative, H ′
i ⊆ Hi+1 by [24, Theorem

4.14]. Hence,

H1 ⊇ H ′
0 = X(1), H2 ⊇ H ′

1 ⊇ X(2), . . .,{0} = Hn ⊇ H ′
n−1 ⊇ X(n).

Thus, X(n) = {0}.
Conversely, suppose that X(m) = {0}. The series X ⊇ X(1) ⊇ · · · ⊇ X(m−1) = {0} is a
solvable B-series. Thus, X is solvable.

Proposition 1. Let H ̸= {0} be a subalgebra of a solvable B-algebra X. Then H ′ ̸= H.

Proof. Suppose H ′ = H. Then H(2) = (H ′)′ = H ′ = H ̸= {0}. By induction,
H(n) = H ̸= {0} for any positive integer n. By [8, Theorem 12], H is solvable. Thus, by
Theorem 4, there exists a positive integer n such that H(n) = {0}, a contradiction. Hence,
H ′ ̸= H.



REFERENCES 1672

Theorem 5. A finite B-algebra X is solvable if and only if H ′ ̸= H for any subalgebra
H ̸= {0} of X.

Proof. Let X be a finite B-algebra. Suppose that X is solvable. By Proposition
1, H ′ ̸= H for any subalgebra H ̸= {0} of X. Conversely, suppose that H ′ ̸= H for
any subalgebra H ̸= {0} of X. Then X ̸= X ′. Thus, X ′ ⊂ X. If X(n) ̸= {0}, then
X(n) ̸= X(n+1), that is X(n+1) ⊂ X(n). Hence, we have the following strictly descending
series of subalgebras:

X ⊃ X ′ ⊃ · · · ⊃ X(n) ⊃ X(n+1) ⊃ · · · .

Since X is finite and H ′ ̸= H for any subalgebra H ̸= {0} of X, there exists a positive
integer n such that X(n) = {0}. Hence, X is solvable.

Example 5. Let (X; ∗, 0) be the noncommutative B-algebra in Example 2. The nontrivial
subalgebras of X are the following: H1 = {0, 3}, H2 = {0, 4}, H3 = {0, 5}, H4 = {0, 1, 2}.
Clearly, from the computations in Example 3, we get H ′

1 = {0} ≠ H1, H
′
2 = {0} ≠ H2,

H ′
3 = {0} ≠ H3, and H ′

4 = {0} ≠ H4. In Example 4, X ′ = {0, 1, 2} ≠ X. Hence, H ′ ̸= H
for any subalgebra H ̸= {0} of X. Therefore, by Theorem 5, X is solvable, which confirms
the result in [8, Example 11].

4. Conclusion

We established some basic properties of B-commutators of B-algebras. These prop-
erties are used in characterizing solvable B-algebras via B-commutators. As a result, we
showed that a B-algebra X is solvable if and only if there is positive integer m such that
the mth B-commutator subalgebra X(m) is equal to {0}.

Acknowledgements

The author would like to thank the referees for the comments and suggestions which
were incorporated into this revised version.

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