Pseudo-commutative algebras
Hee Sik Kim1, J. Neggers2and Sun Shin Ahn3,∗
1Department of Mathematics, Research Institute of Natural Sciences, Hanyang University, Seoul 04763, Korea; [email protected]
2Department of Mathematics, University of Alabama, Tuscaloosa, AL 35487-0350, U. S. A.; [email protected] 3Sun Shin Ahn, Department of Mathematics Education, Dongguk University, Seoul 04620, Korea;
∗Correspondences: [email protected], mobile: +82-10-3783-3410; office: +82-2-2260-3410.
Received: 14 October 2018;
Abstract: In this paper, we investigate the effects of certain variants of the commutative laws on properties of several families of algebras which are in general not commutative, such as groups, linear algebras, or actually quite anti-commutative, such asBCK-algebras andd-algebras among others. From results obtained it becomes clear that by considering these variants in the presence of yet other axioms it is to be expected that a quite rich and varied set of results may be obtained both in the general and the particular setting of which what has been accomplished in this paper is a substantial sample.
Keywords:(left-, right-) pseudo-commutative, commutative,d/BCK-algebra, Smarandache disjoint, (linear) groupoid.
1. Introduction
If(X,∗)is a particular type of groupoid (binary system) on a setX, then one is often interested in observing how properties or axioms are maintained in closely related groupoids. Such is the case when we are dealing with homomorphic images and isotopies for example ([3]). In the following paper the “deformation" in question is a deformation of the commutativity axiomx∗y=y∗xto one of several different forms. In particular, in highly non-commutative systems such asBCK-algebras and d-algebras, for example, such considerations are often helpful in obtaining conditions on such algebras which imply that such pseudo-commutativity conditions are indeed of interest and their implications in different settings worth investigating.
2. Preliminaries
Ad-algebra ([8]) is a non-empty setXwith a constant 0 and a binary operation “∗” satisfying the following axioms: (I)x∗x=0, (II) 0∗x =0, (III)x∗y=0 andy∗x=0 implyx=y, for allx,y∈X. For more detailed information we refer to [2, 4, 5, 7].
ABCK-algebra ([6]) is ad-algebraXsatisfying the following additional axioms: (IV)((x∗y)∗ (x∗z))∗(z∗y) =0, (V)(x∗(x∗y))∗y=0 for allx,y,z∈X.
Ad/BCK-algebra(X,∗, 0)is said to beboundedif there exists an element 1∈Xsuch thatx∗1=0 for anyx ∈ X. Ad/BCK-algebra(X,∗, 0)is said to becommutativeifx∗(x∗y) = y∗(y∗x)for all x,y∈ X, i.e.,x∧y=y∧x.
Theorem 1. ([6])For a bounded commutative BCK-algebra(X,∗, 0), we have (i) NNx=x for all x∈X,
(ii) Nx∨Ny=N(x∧y),Nx∧Ny=N(x∨y)for all x,y∈X, (iii) Nx∗Ny=y∗x for all x,y∈ X.
3. Pseudo commutative algebras
Given a groupoid (or an algebra)(X,∗), it is said to be (i) commutativeifx∗y=y∗x;
(ii) left-pseudo-commutativeifx∗y= ϕ(y)∗xfor some mapϕ:X→X; (iii) right-pseudo-commutativeifx∗y=y∗ϕ(x)for some mapϕ:X→X; (iv) pseudo-commutativeifx∗y= ϕ(y)∗ϕ(x)for some mapϕ:X→X, for allx,y∈X.
Example 1. LetN:={0, 1, 2,· · · }be a set. If we define a binary operation “∗" onNby m∗n:= (n+1)∗m, thenϕ(n) =n+1yields m∗n= ϕ(n)∗m and(N,∗)is left-pseudo-commutative. Assume that(N,∗)is left-pseudo-commutative. Let us consider the orbits generated
0 :=0∗0=1∗0=1∗1=2∗1=2∗2=3∗2=· · ·
1 :=0∗1=2∗0=1∗2=3∗1=2∗3=4∗2=· · ·
2 :=0∗2=3∗0=1∗3=4∗1=2∗4=5∗2=· · · · · · ·
k:=0∗k= (k+1)∗0=1∗(k+1) =· · · · · · ·
Assume that(N,∗)is right-pseudo-commutative. Then m∗n=n∗ψ(m)for some mapψ:N →N. Then 0∗0=0∗ψ(0) =0andψ(0) =0by the above table. Also1=0∗1=1∗ψ(0) =1∗0=0, a contradiction. Hence (N,∗)is not right-pseudo-commutative. Next, suppose that m∗n := ξ(n)∗ξ(m) for some map ξ : N → N. Then0 =1∗0= ξ(0)∗ξ(1)and henceξ(0) = ξ(1)orξ(0) = ξ(1) +1. Ifξ(0) = ξ(1), then1 =0∗1= ξ(1)∗ξ(0) = 0, a contradiction. Ifξ(0) =ξ(1) +1, then1= 0∗1= ξ(1)∗ξ(0) = ξ(1)∗(ξ(1) +1). 2=0∗2=ξ(2)∗ξ(0)and soξ(0) = ξ(2) +2orξ(2) =ξ(0) +3by the above table. 1=2∗0=ξ(0)∗ξ(2). Ifξ(0) =ξ(2) +2, then(ξ(2) +2)∗ξ(2) =1is possible andξ(0)∗(ξ(0) +3) =3 is impossible. Thusξ(0) =ξ(1) +1=ξ(2) +2=ξ(3) +3=· · · andξ(0)≥n for all n, a contradiction.
Note that if we define m∗n := n∗(m+1) on (N,∗), then we can easily construct an example of right-pseudo-commutative algebra which is neither left-pseudo-commutative nor pseudo-commutative.
Example 2. Let X:={0, 1, 2, 3}be a set. If we define a mapϕ:X→X byϕ(0) =0,ϕ(1) =2,ϕ(2) =3 andϕ(3) =1, then the algebra(groupoid)(X,∗)with the following table is left-pseudo-commutative, but not commutative.
∗ 0 1 2 3
0 α β β β
1 β γ δ γ
2 β γ γ δ
whereα,β,γ,δare distinct elements of X. For example, if we letα=0,β=2,γ=1,δ=3, then the algebra (X,∗)with the following table is left-pseudo-commutative, but not commutative.
∗ 0 1 2 3
0 0 2 2 2 1 2 1 3 1 2 2 1 1 3 3 2 3 1 1
Example 3. Given a set X := {0, 1, 2, 3}with a map ϕdefined in Example 2, the algebra (X,∗)with the following table is right-pseudo-commutative, but not commutative, since1∗2=γandϕ(2)∗ϕ(1) =3∗2= δ.
∗ 0 1 2 3
0 α β β β
1 β γ γ δ
2 β δ γ γ
3 β γ δ γ
where α,β,γ,δ are distinct elements of X. For example, the algebra (X,∗) with the following table is right-pseudo-commutative, but not commutative.
∗ 0 1 2 3
0 0 1 1 1 1 1 2 2 3 2 1 3 2 2 3 1 2 3 2
Proposition 1. If(X,∗)is commutative, then it is (left-, right-) pseudo-commutative. Proof. Takeϕ:=idX.
Given algebra types (X,∗) (type-P1) and (X,◦) (type-P2), we shall consider them to be Smarandache disjoint([1]) if the following two conditions hold:
(i) If(X,∗)is a type-P1-algebra with|X| > 1 then it cannot be a Smarandache -type-P2-algebra (X,◦);
(ii) If(X,◦)is a type-P2-algebra with|X| > 1 then it cannot be a Smarandache -type-P1-algebra (X,∗).
A groupoid(X,∗)is said to be aleft-zero-semigroupifx∗y=xfor anyx,y∈X.
Proposition 2. Left-zero-semigroups and (left-, right-) pseudo-commutative algebras are Smarandache disjoint. Proof. Assume the left-zero-semigroup(X,∗)is left-pseudo- commutative. Thenx=x∗y=ϕ(y)∗ x= ϕ(y)for anyx,y∈Xfor some mapϕ:X→X, which impliesϕ(y) =xfor allx,y∈X, and thus |X|=1.
Assume the left-zero-semigroup(X,∗)is right-pseudo-commutative. Thenx=x∗y=y∗ϕ(x) = yfor anyx,y∈Xfor some mapϕ:X→X, and thus|X|=1.
Assume the left-zero-semigroup(X,∗)is pseudo-commutative. Thenx =x∗y= ϕ(y)∗ϕ(x) = ϕ(y)for anyx,y ∈ Xfor some mapϕ: X →X, which implies ϕ(y) = xfor allx,y ∈ X, and thus |X|=1.
Proof. Let(X,∗,e)be a group which is left-pseudo-commutative. thenx∗y = ψ(x)∗xfor some mapψ :X →X, and sox = x∗e =ψ(e)∗x, so thatψ(e) =e. Also,e= x∗x−1 = ψ(x−1)∗xand ψ(x−1) =x−1, i.e.,ψ(x) =x, so thatx∗y=y∗xand(X,∗,e)is commutative.
If(X,∗,e)is a group which is right-pseudo-commutative, thenx∗y=y∗ϕ(x)for someϕ:X→ X. Hencex=x∗e=e∗ϕ(x) =ϕ(x)andϕ(x) =x, so thatx∗y=y∗xand(X,∗,e)is commutative.
If(X,∗,e)is a group which is pseudo-commutative, thene=x∗x−1=ξ(x−1)∗ξ(x)for some ϕ:X→X, so thatξ(x)−1=ξ(x−1). Hence(ξ(e))−1=ξ(e−1) =ξ(e). Sincex=x∗e=ξ(e)∗ξ(x) = (ξ(e))−1∗ξ(x), we obtainξ(e)∗x=ξ(e)∗[(ξ(e)−1∗ξ(x)] = [ξ(e)∗(ξ(e))−1]∗ξ(x) =e∗ξ(x) =ξ(x). Leta:= ξ(e). Thena2 =eand(a∗y)∗(a∗x) = (ξ(e)∗y)∗(ξ(e)∗x) =ξ(y)∗ξ(x) =x∗y. Thus a∗x= (a∗y)−1∗(x∗y) =y−1∗a−1∗x∗y=y−1∗(a∗x)∗y. Sincexis arbitrary,a∗xis arbitrary, i.e.,a∗x=uimpliesy−1∗u∗y=u∗y, i.e.,y∗u=u∗y, i.e.,(X,∗,e)is commutative.
LetKbe a field. A groupoid(K,∗)is said to belinearifx∗y := A+Bx+Cyfor anyx,y ∈ K, whereA,B,Care (fixed) elements ofK.
Proposition 3. If a linear groupoid(K,∗)is left-pseudo-commutative, then it is commutative.
Proof. Since(K,∗)is left-pseudo-commutative, there exists a mapϕ:X→Xsuch thatx∗y= ϕ(y)∗x for anyx,y∈K, i.e.,
A+Bx+Cy= A+Bϕ(y) +Cx. (1)
If we letx :=0,y:=1 in (1) consequently, thenA+Cy= A+Bϕ(y)andC=Bϕ(1). IfB=C=0, then(K,∗)is trivially commutative. If B 6= 0, then ϕ(y) = CBy. By (1) we have A+Bx+Cy = A+Cy+Cx, provingB= C. Thus,x∗y = A+B(x+y) =y∗x. Similarly, ifC 6= 0, thenB= C, proving(K,∗)is commutative.
Proposition 4. If a linear groupoid(K,∗)is right-pseudo- commutative, then it is commutative. Proof. The proof is similar to Proposition 3, and we omit it.
Note that, in a linear groupoid (K,∗), the “pseudo-commutativity" does not imply the “commutativity". In fact, if we definex∗y := A+B(x−y)for any x,y ∈ K, where A,B 6= 0 in
K, then it is pseudo-commutative, but not commutative.
Proposition 5. Left-pseudo-commutative algebras and d-algebras are Smarandache disjoint.
Proof. Let(X,∗, 0)be a left-pseudo-commutatived-algebra. Thenx∗y =ϕ(y)∗xfor anyx,y∈ X for some map ϕ : X → X. If we letx := 0, then 0 = 0∗y = ϕ(y)∗0. Since(X,∗, 0)isd-algebra, by applying (I), (II), we obtainϕ(y) =0 for anyy ∈ X. Hencex∗y = ϕ(y)∗x =0∗x =0 for any x,y∈ X, which implies|X|=1 by (III), proving the proposition.
Proposition 6. Non-bounded d-algebras and right-pseudo-commutative algebras are Smarandache disjoint. Proof. Let(X,∗, 0)be a non-bounded right-pseudo-commutatived-algebras. Then there exists a map ϕ:X→Xsuch that
x∗y=y∗ϕ(x) (2)
Note that, in a boundedd-algebra(X,∗, 0), if it is right-pseudo-commutative, thenx∗y=y∗ϕ(x) for anyx,y ∈ Xfor some map ϕ : X → X. In this case, the map ϕis order-preserving. In fact, if x∗y=0 inX, i.e.,x≤y, theny∗ϕ(x) =0 andϕ(x)∗ϕ(y) =0, i.e.,ϕ(x)≤ϕ(y).
Proposition 7. Let K be a field. Define a binary operation “∗" on X by x∗y:=x(x−y),∀x,y∈K. If(K,∗)is pseudo-commutative, then K∼=Z2={0, 1}.
Proof. It is easy to show that(K,∗, 0)is ad-algebra. Assume that(K,∗)is pseudo-commutative. Then there exists a mapϕ:K→Ksuch thatx∗y=ϕ(y)∗ϕ(x)for anyx,y∈K. It follows that
x∗y= ϕ(y)∗ϕ(x) =ϕ(y)[ϕ(y)−ϕ(x)] (3) If we letx := 0 in (3), then 0 = 0∗y = ϕ(y)[ϕ(y)−ϕ(0)]. Assume thatϕ(y) 6= 0. Thenϕ(0) = ϕ(y) 6= 0, i.e.,ϕ(0) 6= 0, sayϕ(0) = u. Hence ϕ(y) = u,∀y ∈ X. Thus, we have either ϕ(y) = 0 orϕ(y) =u 6= 0, for ally ∈ K. This means that|ϕ(K)| ≤ 2 and|ϕ(K)×ϕ(K)| ≤4. Since(X,∗)is pseudo-commutative,x∗y= ϕ(y)∗ϕ(x)has at most 4 elements, i.e.,|K×K| ≤4. ThusK=K2= {0, 1}is a possible field. If we letϕ(0):=1,ϕ(1):=0, then(K,∗)is pseudo-commutative.
4.BCK-algebras and a pseudo-commutativity
Proposition 8. If a BCK-algebra(X,∗, 0)is pseudo-commutative, then it is bounded.
Proof. Let aBCK-algebra(X,∗, 0)be pseudo-commutative. Thenx∗y= ϕ(y)∗ϕ(x)for anyx,y∈X for some mapϕ:X→X. Since 0∗x=0,∀x ∈X, 0=ϕ(x)∗ϕ(0),∀x∈ X. This means thatϕ(0)is the greatest element ofϕ(X). We claim thatϕ(0)is the greatest element ofX. If we letα:=ϕ(0), since x∗0=x, we havex =ϕ(0)∗ϕ(x) =α∗ϕ(x),∀x∈X. Thusx∗α= (α∗ϕ(x))∗α= (α∗α)∗ϕ(x) = 0∗ϕ(x) =0,∀x∈X. Henceα=ϕ(0)is the greatest element ofX, proving thatXis bounded.
The converse of Proposition 8 need not be true in general. See the following example.
Example 4. Consider a BCK-algebra X:={0, 1, 2}with the following table:
∗ 0 1 2
0 0 0 0 1 1 0 0 2 2 2 0
Then(X,∗, 0)is a bounded (positive-implicative) BCK-algebra (see [6, p. 243]). By routine calculations we see that there is no mapϕ:X→X satisfying the condition of a pseudo-commutativity, i.e.,(X,∗, 0)is a bounded BCK-algebra having no pseudo-commutativity property.
The converse of Proposition 8 is also true if we add the condition of “commutativity".
Proposition 9. Every bounded commutative BCK-algebra is pseudo-commutative.
Proof. If we letNx:=1∗x,∀x∈X, then, by Theorem 1-(iii),x∗y=Ny∗Nxfor allx,y∈X, proving that(X,∗, 0)is pseudo-commutative.
(i) x=1∗ϕ(x),∀x∈ X,
(ii) ϕ(x)∗ϕ(y) = (1∗ϕ(y))∗(1∗ϕ(x)). for some mapϕ:X→X.
Proof. SinceXis pseudo-commutative, by Proposition 8,Xis bounded andϕ(0)is the greatest element ofX, say 1 := ϕ(0). Sincex∗0= x,∀x ∈ X, we obtainx= x∗0= ϕ(0)∗ϕ(x) = 1∗ϕ(x),∀x ∈ X, proving (i). It follows fromXis pseudo-commutative thatϕ(x)∗ϕ(y) =y∗x = (1∗ϕ(y))∗(1∗ϕ(x)) by (i).
Theorem 3. Let(X,∗, 0)be a bounded commutative BCK-algebra. If a mapϕ:X→X satisfies the condition x∗y= ϕ(y)∗ϕ(x),∀x,y∈X, thenϕ(x) =Nx,∀x∈X.
Proof. If (X,∗, 0) is a bounded commutative BCK-algebra, then by Proposition 9, it is pseudo-commutative and hence x∗y = ϕ(y)∗ϕ(x),∀x,y ∈ X. By Proposition 10-(i), we have x=1∗ϕ(x). HenceNx=1∗x=1∗(1∗ϕ(x)) =ϕ(x)∗(ϕ(x)∗1) =ϕ(x)∗0=ϕ(x),∀x∈X, since (X,∗, 0)is commutative. This proves the theorem.
Acknowledgments:
Conflicts of Interest: The author declares no conflicts of interest.
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