# OnBE-Semigroups

Authors: Sun Shin Ahn; Young Hee Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

The notion of aBE-semigroup is introduced, and related properties are investigated. The concept of left (resp., right)
deductive systems of a BE-semigroup is also introduced.


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## Body

## 1. Introduction

Hu and Li, Iséki and Tanaka, respectively, introduced two classes of abstract algebras:BCK-algebras and BCI-algebras [1–3]. It is known that the class of BCK-algebras is a proper subclass of the class of BCI-algebras. In [1, 4] Hu and 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. We refer to [5] for general information on BCK-algebras. Neggers and Kim [6] introduced the notion of a d-algebra which is a generalization of BCK-algebras, and also they introduced the notion of a B-algebra [7, 8], that is, (I) x*x=0, (II) x*0=x, (III) (x*y)*z=x*(z*(0*y)), for any x,y,z∈X, which is equivalent to the idea of groups. Moreover, Jun et al. [9] introduced a new notion, called an BH-algebra, which is another generalization of BCH/BCI/BCK-algebras, that is, (I), (II), and (IV) x*y=0 and y*x=0 imply that x=y for any x,y∈X. Walendziak obtained other equivalent set of axioms for a B-algebra [10]. Kim et al. [11] introduced the notion of a (pre-) Coxeter algebra and showed that a Coxeter algebra is equivalent to an abelian group all of whose elements have order 2, that is, a Boolean group. C. B. Kim and H. S. Kim [12] introduced the notion of a BM-algebra which is a specialization of B-algebras. They proved that the class of BM-algebras is a proper subclass of B-algebras and also showed that a BM-algebra is equivalent to a 0-commutative B-algebra. In [13], H. S. Kim and Y. H. Kim introduced the notion of a BE-algebra as a generalization of a BCK-algebra. Using the notion of upper sets, they gave an equivalent condition of the filter in BE-algebras. In [14, 15], Ahn and So introduced the notion of ideals in BE-algebras and proved several characterizations of such ideals.In this paper, by combiningBE-algebras and semigroups, we introduce the notion of BE-semigroups. We define left (resp., right) deductive systems (LDS (resp., RDS) for short) of a BE-semigroup, and then we describe LDS generated by a nonempty subset in a BE-semigroup as a simple form.


## 2. Preliminaries

We recall some definitions and results discussed in [13].Definition 2.1 (see [13]).
An algebra(X;*,1) of type (2, 0) is called a BE-algebra if (BE1)
x*x=1 for all x∈X,(BE2)
x*1=1 for all x∈X,(BE3)
1*x=x for all x∈X,(BE4)
x*(y*z)=y*(x*z) for all x,y,z∈X (exchange).We introduce a relation “≤" on X by x≤y if and only if x*y=1.Proposition 2.2 (see [13]).
If(X;*,1) is a BE-algebra, then x*(y*x)=1 for any x,y∈X.Example 2.3 (see [13]).
LetX:={1,a,b,c,d,0} be a set with the following table:
(2.1)*1abcd011abcd0a11accdb111cccc1ab1abd11a11a0111111
Then (X;*,1) is a BE-algebra.Definition 2.4 (see [13]).
ABE-algebra (X;*,1) is said to be self-distributive if x*(y*z)=(x*y)*(x*z) for all x,y,z∈X.Example 2.5 (see [13]).
LetX:={1,a,b,c,d} be a set with the following table:
(2.2)*1abcd11abcda11bcdb1a1ccc11b1bd11111
Then it is easy to see that X is a self-distributive BE-algebra.Note that theBE-algebra in Example 2.3 is not self-distributive, since d*(a*0)=d*d=1, while (d*a)*(d*0)=1*a=a.Proposition 2.6.
LetX be a self-distributive BE-algebra. If x≤y, then z*x≤z*y and y*z≤x*z for any x,y,z∈X.Proof.
The proof is straightforward.


## 3.BE-Semigroups

Definition 3.1.
An algebraic system(X;⊙,*,1) is called a BE-semigroup if it satisfies the following:  (i)
(X;⊙) is a semigroup,(ii)
(X;*,1) is a BE-algebra,(iii)
the operation “⊙” is distributive (on both sides) over the operation “*”.Example 3.2.
(1) Define two operations “⊙” and “*” on a set X:={1,a,b,c} as follows:
(3.1)⊙1abc11111a1111b1111c1abc*1abc11abca11bcb1a1cc1111
It is easy to see that (X;⊙,*,1) is a BE-semigroup.
(2) Define two binary operations “⊙” and “*” on a set A:={1,a,b,c} as follows:(3.2)⊙1abc11111a1111b111bc11bc*1abc11abca11bcb1a1cc1111
It is easy to show that (A;⊙,*,1) is a BE-semigroup.Proposition 3.3.
Let(X;⊙,*,1) be a BE-semigroup. Then  (i)
(∀x∈X)(1⊙x=x⊙1=1),(ii)
(∀x,y,z∈X)(x≤y⇒x⊙z≤y⊙z, z⊙x≤z⊙y).Proof.
(i) For allx∈X, we have that 1⊙x=(1*1)⊙x=(1⊙x)*(1⊙x)=1 and x⊙1=x⊙(1*1)=(x⊙1)*(x⊙1)=1.
(ii) Letx,y,z∈X be such that x≤y. Then(3.3)(x⊙z)*(y⊙z)=(x*y)⊙z=1⊙z=1,(z⊙x)*(z⊙y)=z⊙(x*y)=z⊙1=1.
Hence x⊙z≤y⊙z and z⊙x≤z⊙y.Definition 3.4.
An elementa(≠1) in a BE-semigroup (X;⊙,*,1) is said to be a left (resp., right) unit divisor if
(3.4)(∃b(≠1)∈X)(a⊙b=1(resp.,b⊙a=1)).
A unit divisor is an element of X which is both a left and a right unit divisors.Theorem 3.5.
Let(X;⊙,*,1) be a BE-semigroup. If it satisfies the left  (resp., right ) cancellation law for the operation ⊙, that is,
(3.5)(∀x(≠1),y,z∈A)(x⊙y=y⊙z(resp.,y⊙x=z⊙x)⟹y=z),
then X contains no left  (resp., right) unit divisors.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 by Proposition 3.3(i), which implies y=1. Similarly it holds for the right case. Hence there is no left (resp., right) unit divisors in X.Now we consider the converse of Theorem3.5.Theorem 3.6.
Let(X;⊙,*,1) be a BE-semigroup in which there are no left (resp., right ) unit divisors. Then it satisfies the left (resp., right) cancellation law for the operation ⊙.Proof.
Letx,y,z∈X be such that x⊙y=x⊙z and x≠1. Then
(3.6)x⊙(y*z)=(x⊙y)*(x⊙z)=1,x⊙(z*y)=(x⊙z)*(x⊙y)=1.
Since X has no left unit divisor, it follows that y*z=1=z*y so that y=z. The argument is the same for the right case.Definition 3.7.
Let(X;⊙,*,1) be a BE-semigroup. A nonempty subset D of X is called a left (resp., right) deductive system (LDS (resp., RDS), for short) if it satisfies  (ds1)
X⊙D⊆D (resp., (D⊙X⊆D)),(ds2)
(∀a∈D)((∀x∈X)(a*x∈D⇒x∈D).Example 3.8.
LetX:={x,y,z,1} be a set with the following Cayley tables:
(3.7)⊙1xyz11111x1x11y11yzz11zy*1xyz11xyzx11yzy111zz1111
It is easy to show that (X;⊙,*,1) is a BE-semigroup. We know that D:={1,x} is an LDS of X, but E:={1,y} is not an LDS of X, since z⊙y=z∉E and/or y*x=1∈E, y∈E but x∉E.Let(X;*,1) be a BE-algebra, and let a,b∈X. Then the set (3.8)A(a,b):={x∈X∣a*(b*x)=1}
is nonempty, since 1,a,b∈A(a,b).Proposition 3.9.
IfD is an LDS of a BE-semigroup (X;⊙,*,1), then
(3.9)(∀a,b∈D)(A(a,b)⊆D).Proof.
Letx∈A(a,b) where a,b∈D. Then a*(b*x)=1∈D and so x∈D by (ds2). Therefore A(a,b)⊆D.Theorem 3.10.
Let{Di} be an arbitrary collection of LDSs of a BE-semigroup (X;⊙,*,1), where i ranges over some index set I. Then ∩i∈IDi is  also an LDS of A.Proof.
The proof is straightforward.Let(X;⊙,*,1) be a BE-semigroup. For any subset D of X, the intersection of all LDSs (resp., RDSs) of X containing D is called the LDSs (resp., RDSs) generated byD, and is denoted by 〈D〉l (resp., 〈D〉r). It is clear that if D and E are subsets of a BE-semigroup (X;⊙,*,1) satisfying D⊆E, then 〈D〉l⊆〈E〉l (resp., 〈D〉r⊆〈E〉r), and if D is an LDS (resp., RDS) of X, then 〈D〉l=D (resp., 〈D〉r=D).ABE-semigroup (X;⊙,*,1) is said to be self-distributive if (X;*,1) is a self-distributive BE-algebra.Theorem 3.11.
Let(X;⊙,*,1) be a self-distributive BE-semigroup and let D be a nonempty subset of X such that A⊙D⊆D. Then 〈D〉l:={a∈X∣yn*(⋯*(y1*a)⋯)=1 for some y1,…,yn∈D}.Proof.
Denote(3.10)B:={a∈X∣yn*(⋯*(y1*a)⋯)=1forsomey1,…,yn∈D}.
Let a∈X and b∈B. Then there exist y1,…,yn∈D such that yn*(⋯*(y1*b)⋯)=1. It follows that
(3.11)1=x⊙1=x⊙(yn*(⋯*(y1*b)⋯))=(x⊙yn)*(⋯*((x⊙y1)*(x⊙b))⋯).
Since x⊙yi∈D for i=1,…,n, we have that x⊙b∈B. Let x,a∈X be such that a*x∈B and a∈B. Then there exist y1,…,yn,z1,…,zm∈D such that
(3.12)yn*(⋯*(y1*(a*x))⋯)=1,(3.13)zm*(⋯*(z1*a)⋯)=1.
Using (BE4), it follows from (3.12) that a*(yn*(⋯*(y1*x)⋯))=1, that is, a≤yn*(⋯*(y1*x)⋯), and so from (3.13) and Proposition 2.6 it follows that
(3.14)1=zm*(⋯*(z1*a)⋯)≤zm*(⋯*(z1*(yn*(⋯*(y1*x)⋯)))⋯).
Thus zm*(⋯*(z1*(yn*(⋯*(y1*x)⋯)))⋯)=1, which implies x∈B. Therefore B is an LDS of X. Obviously D⊆B. Let G be an LDS containing D. To show B⊆G, let a be any element of B. Then there exist y1,…,yn∈D such that yn*(⋯*(y1*a)⋯)=1. It follows from (ds2) that a∈G so that B⊆G. Consequently, we have that 〈D〉l=B.In the following example, we know that the union of any LDSs (resp., RDSs)D and E may not be an LDS (resp., RDS) of a self-distributive BE-semigroup (X;·,*,1).Example 3.12.
LetX:={1,a,b,c,d} be a set with the following Cayley tables:
(3.15)⊙1abcd111111a11111b11111c11111d1111d*1abcd11abcda11bbdb1a1adc1111dd11bb1
It is easy to check that (X;⊙,*,1) is a self-distributive BE-semigroup. We know that D:={1,a} and E:={1,b} are LDSs of X, but D∪E={1,a,b} is not an LDS of X, since b*c=a∈D∪E, c∉D∪E.Theorem 3.13.
LetD and E be LDSs of a self-distributive BE-semigroup (X;·,*,1). Then
(3.16)〈D∪E〉l:={a∈X∣x*(y*a)=1forsomex∈D,y∈E}.Proof.
Denote(3.17)K:={a∈X∣x*(y*a)=1forsomex∈D,y∈E}.
Obviously, K⊆〈D∪E〉l. Let b∈〈D∪E〉l. Then there exist y1,…,yn∈D∪E such that yn*(⋯*(y1*b)⋯)=1 by Theorem 3.11. If yi∈D (resp., E) for all i=1,…,n, then b∈D (resp., E). Hence b∈K since b*(1*b)=1 (resp., 1*(b*b)=1). If some of y1,…,yn belong to D and others belong to E, then we may assume that y1,…,yk∈D and yk+1,…,yn∈E for 1≤k<n, without loss of generality. Let p=yk*(⋯*(y1*b)⋯). Then
(3.18)yn*(⋯*(yk+1*p)⋯)=yn*(⋯*(yk+1*(yk*(⋯*(y1*b)⋯)))⋯)=1,
and so p∈E. Now let q=p*b=(yk*(⋯*(y1*b)⋯))*b. Then
(3.19)yk*(⋯*(y1*q)⋯)=yk*(⋯*(y1*((yk*(⋯*(y1*b)⋯))*b))⋯)=(yk*(⋯*(y1*b)⋯))*(yk*(⋯*(y1*b)⋯))=1,
which implies that q∈D. Since p*(q*b)=q*(p*b)=q*q=1, it follows that b∈K so that 〈D∪E〉l⊆K. This completes the proof.



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