On 𝐵𝐸-Semigroups

Abstract

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

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 combining 𝐵𝐸-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]). Let 𝑋∶={1,𝑎,𝑏,𝑐,𝑑,0} be a set with the following table: ∗1𝑎𝑏𝑐𝑑01𝑎1𝑎𝑏𝑐𝑑0𝑏11𝑎𝑐𝑐𝑑𝑐111𝑐𝑐𝑐𝑑1𝑎𝑏1𝑎𝑏011𝑎11𝑎111111 (2.1) Then (𝑋;∗,1) is a 𝐵𝐸-algebra.

Definition 2.4 (see [13]). A 𝐵𝐸-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]). Let 𝑋∶={1,𝑎,𝑏,𝑐,𝑑} be a set with the following table: ∗1𝑎𝑏𝑐𝑑1𝑎1𝑎𝑏𝑐𝑑𝑏11𝑏𝑐𝑑𝑐1𝑎1𝑐𝑐𝑑11𝑏1𝑏11111 (2.2) Then it is easy to see that 𝑋 is a self-distributive 𝐵𝐸-algebra.

Note that the 𝐵𝐸-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. Let 𝑋 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: ⊙1𝑎𝑏𝑐1𝑎1111𝑏1111𝑐1111∗1𝑎𝑏𝑐1𝑎𝑏𝑐1𝑎1𝑎𝑏𝑐𝑏11𝑏𝑐𝑐1𝑎1𝑐1111 (3.1) 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:⊙1𝑎𝑏𝑐1𝑎1111𝑏1111𝑐111𝑏∗11𝑏𝑐1𝑎𝑏𝑐1𝑎1𝑎𝑏𝑐𝑏11𝑏𝑐𝑐1𝑎1𝑐1111 (3.2) 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 all 𝑥∈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) Let 𝑥,𝑦,𝑧∈𝑋 be such that 𝑥≤𝑦. Then()∗=()∗𝑥⊙𝑧𝑦⊙𝑧𝑥∗𝑦⊙𝑧=1⊙𝑧=1,𝑧⊙𝑥𝑧⊙𝑦=𝑧⊙𝑥∗𝑦=𝑧⊙1=1. (3.3) Hence 𝑥⊙𝑧≤𝑦⊙𝑧 and 𝑧⊙𝑥≤𝑧⊙𝑦.

Definition 3.4. An element 𝑎(≠1) in a 𝐵𝐸-semigroup (𝑋;⊙,∗,1) is said to be a left (resp., right) unit divisor if (∃𝑏(≠1)∈𝑋)𝑎⊙𝑏=1resp.,𝑏⊙𝑎=1. (3.4) 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, ∀𝑥(≠1),𝑦,𝑧∈𝐴𝑥⊙𝑦=𝑦⊙𝑧resp.,𝑦⊙𝑥=𝑧⊙𝑥⟹𝑦=𝑧, (3.5) 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 Theorem 3.5.

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. Let 𝑥,𝑦,𝑧∈𝑋 be such that 𝑥⊙𝑦=𝑥⊙𝑧 and 𝑥≠1. Then =∗()=()∗𝑥⊙𝑦∗𝑧𝑥⊙𝑦𝑥⊙𝑧=1,𝑥⊙𝑧∗𝑦𝑥⊙𝑧𝑥⊙𝑦=1. (3.6) 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. Let 𝑋∶={𝑥,𝑦,𝑧,1} be a set with the following Cayley tables: ⊙1𝑥𝑦𝑧1𝑥1111𝑦1𝑥11𝑧11𝑦𝑧∗11𝑧𝑦1𝑥𝑦𝑧1𝑥1𝑥𝑦𝑧𝑦11𝑦𝑧𝑧111𝑧1111 (3.7) 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 𝐴(𝑎,𝑏)∶={𝑥∈𝑋∣𝑎∗(𝑏∗𝑥)=1} (3.8) is nonempty, since 1,𝑎,𝑏∈𝐴(𝑎,𝑏).

Proposition 3.9. If 𝐷 is an LDS of a 𝐵𝐸-semigroup (𝑋;⊙,∗,1), then (∀𝑎,𝑏∈𝐷)(𝐴(𝑎,𝑏)⊆𝐷). (3.9)

Proof. Let 𝑥∈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 {𝐷_𝑖} be an arbitrary collection of LDSs of a 𝐵𝐸-semigroup (𝑋;⊙,∗,1), where 𝑖 ranges over some index set 𝐼. Then ∩_𝑖∈𝐼𝐷_𝑖 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 by 𝐷, 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).

A 𝐵𝐸-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∣𝑦_𝑛∗(⋯∗(𝑦₁∗𝑎)⋯)=1 for some 𝑦₁,…,𝑦_𝑛∈𝐷}.

Proof. Denote 𝐵∶=𝑎∈𝑋∣𝑦_𝑛∗⋯∗𝑦₁∗𝑎⋯=1forsome𝑦₁,…,𝑦_𝑛∈𝐷. (3.10) Let 𝑎∈𝑋 and 𝑏∈𝐵. Then there exist 𝑦₁,…,𝑦_𝑛∈𝐷 such that 𝑦_𝑛∗(⋯∗(𝑦₁∗𝑏)⋯)=1. It follows that 𝑦1=𝑥⊙1=𝑥⊙_𝑛∗⋯∗𝑦₁⋯=∗𝑏𝑥⊙𝑦_𝑛∗⋯∗𝑥⊙𝑦₁∗(𝑥⊙𝑏)⋯. (3.11) Since 𝑥⊙𝑦_𝑖∈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,…,𝑦_𝑛,𝑧₁,…,𝑧_𝑚∈𝐷 such that 𝑦_𝑛∗⋯∗𝑦₁∗(𝑎∗𝑥)⋯=1,𝑧_𝑚∗⋯∗𝑧₁∗𝑎⋯=1. (3.12)(3.13) Using (BE4), it follows from (3.12) that 𝑎∗(𝑦_𝑛∗(⋯∗(𝑦₁∗𝑥)⋯))=1, that is, 𝑎≤𝑦_𝑛∗(⋯∗(𝑦₁∗𝑥)⋯), and so from (3.13) and Proposition 2.6 it follows that 1=𝑧_𝑚∗⋯∗𝑧₁∗𝑎⋯≤𝑧_𝑚∗⋯∗𝑧₁∗𝑦_𝑛∗⋯∗𝑦₁∗𝑥⋯⋯. (3.14) Thus 𝑧_𝑚∗(⋯∗(𝑧₁∗(𝑦_𝑛∗(⋯∗(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,…,𝑦_𝑛∈𝐷 such that 𝑦_𝑛∗(⋯∗(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. Let 𝑋∶={1,𝑎,𝑏,𝑐,𝑑} be a set with the following Cayley tables: ⊙1𝑎𝑏𝑐𝑑1𝑎11111𝑏11111𝑐11111𝑑11111∗1111𝑑1𝑎𝑏𝑐𝑑1𝑎1𝑎𝑏𝑐𝑑𝑏11𝑏𝑏𝑑𝑐1𝑎1𝑎𝑑𝑑1111𝑑11𝑏𝑏1 (3.15) 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. Let 𝐷 and 𝐸 be LDSs of a self-distributive 𝐵𝐸-semigroup (𝑋;⋅,∗,1). Then ⟨𝐷∪𝐸⟩_𝑙∶=𝑎∈𝑋∣𝑥∗𝑦∗𝑎=1forsome𝑥∈𝐷,𝑦∈𝐸. (3.16)

Proof. Denote 𝐾∶=𝑎∈𝑋∣𝑥∗𝑦∗𝑎=1forsome𝑥∈𝐷,𝑦∈𝐸. (3.17) Obviously, 𝐾⊆⟨𝐷∪𝐸⟩_𝑙. Let 𝑏∈⟨𝐷∪𝐸⟩_𝑙. Then there exist 𝑦₁,…,𝑦_𝑛∈𝐷∪𝐸 such that 𝑦_𝑛∗(⋯∗(𝑦₁∗𝑏)⋯)=1 by Theorem 3.11. If 𝑦_𝑖∈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,…,𝑦_𝑛 belong to 𝐷 and others belong to 𝐸, then we may assume that 𝑦₁,…,𝑦_𝑘∈𝐷 and 𝑦_𝑘+1,…,𝑦_𝑛∈𝐸 for 1≤𝑘<𝑛, without loss of generality. Let 𝑝=𝑦_𝑘∗(⋯∗(𝑦₁∗𝑏)⋯). Then 𝑦_𝑛∗⋯∗𝑦_𝑘+1∗𝑝⋯=𝑦_𝑛∗⋯∗𝑦_𝑘+1∗𝑦_𝑘∗⋯∗𝑦₁∗𝑏⋯⋯=1, (3.18) and so 𝑝∈𝐸. Now let 𝑞=𝑝∗𝑏=(𝑦_𝑘∗(⋯∗(𝑦₁∗𝑏)⋯))∗𝑏. Then 𝑦_𝑘∗⋯∗𝑦₁∗𝑞⋯=𝑦_𝑘∗⋯∗𝑦₁∗𝑦_𝑘∗⋯∗𝑦₁⋯⋯=𝑦∗𝑏∗𝑏_𝑘∗⋯∗𝑦₁∗𝑏⋯∗𝑦_𝑘∗⋯∗𝑦₁∗𝑏⋯=1, (3.19) 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.