Algebras and Smarandache Types

2018·0.13 MB·English

by  Hee Sik Kim, J. Neggers, Sun Shin Ahn;

Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 5 November 2018 doi:10.20944/preprints201811.0117.v1
Article
Algebras and Smarandache Types
HeeSikKim1,J.Neggers2andSunShinAhn3,∗
1DepartmentofMathematics,ResearchInstituteofNaturalSciences,HanyangUniversity,Seoul04763,Korea;
[email protected]
2DepartmentofMathematics,UniversityofAlabama,Tuscaloosa,AL35487-0350,U.S.A.;[email protected]
3SunShinAhn,DepartmentofMathematicsEducation,DonggukUniversity,Seoul04620,Korea;
[email protected]
∗Correspondences:[email protected],mobile:[phone];office:[phone].
Abstract: Ifanalgebraoftype AcontainsasubalgebrawhichisalsoanalgebraoftypeB,thenitis
aSmarandacheB-type A-algebraprovidedthesubalgebraoftypeBcontainsatleasttwoelements.
ThisgeneralizesthenotionofSmarandachegroup,wherethegroupoforder≥2isasubsemigroup
ofasemigroup. Inthispaperweinvestigateanumberofsuchpairingsandwededuceanumberof
conclusionsasaconsequence.
Keywords: Smarandachealgebra,pointalgebra, p-derivedalgebra.
1. Introduction
Generally,inanyhumanfield,aSmarandachestructureonaset AmeansaweakstructureW
on AsuchthatthereexistsapropersubsetBof AwhichisembeddedwithastrongstructureS. In
[6],W.B.VasanthaKandasamystudiedtheconceptofSmarandachegroupoids,subgroupoids,ideals
of groupoids, semi-normal subgroupoids, Smarandache Bol groupoids and strong Bol groupoids
andobtainedmanyinterestingresultsaboutthem. Smarandacesemigroupsareveryimportantfor
thestudyofcongruences,anditwasstudiedbyR.Padilla([5]). Itwillbeveryinterestingtostudy
theSmarandachestructureinBCI-algebras. In[2],Y.B.JundiscussedtheSmarandachestructurein
BCI-algebras. Ifanalgebraoftype AcontainsasubalgebrawhichisalsoanalgebraoftypeB,thenit
isaSmarandacheB-type A-algebraprovidedthesubalgebraoftypeBcontainsatleasttwoelements.
ThisgeneralizesthenotionofSmarandachegroup,wherethegroupoforder≥2isasubsemigroup
ofasemigroup. Inthispaperweinvestigateanumberofsuchpairingsandwededuceanumberof
conclusionsasaconsequence.
2. Severalalgebras
Supposethat(X,∗)isabinarysystem,i.e.,X×X → X, (x,y) → x∗y,istheproductmapping.
If p ∈ Xisapointselectedasreferencepoint,then(X,∗,p)isapointalgebra.
Example1. Let(G,·)beagroup. Thentheidentityelementeisusuallytakentobethespecialpoint pselected,
writtenas(G,·,e). Wetakeallalgebrasdealtwithbelowtobepointedalgebras,withoutreallossofgenerality.
Often,forsimplicity’ssakeweshallwrite p = 0,notintending0tohavetheusualmeaning. Thus(G,·,e)
becomes (G,∗,0) unless it is important to distinguish the algebras (X,∗,0) which contains the subalgebra
(Y,∗)withitsowndistinguishedpoint ptoproduce(Y,∗,p).
Example2. (i)Let(X,∗,p)beasemigroupwithaselectedpoint pandlet(Y,∗,e)beasubsemigroupofX
withaselectedpointe. Then p (cid:54)= e.
(ii) Let (R,+,·) be a ring with identity. Then (R,+,0) is an abelian group. Let U be a group of units of
(R,+,0). Then(U,·,1)hasadifferentpole pfrom(R,+,0)
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Intermsoflistofaxiomstobeusedtodescribethevariousalgebratypeswenotethefollowing
sectionofaxioms:
(1) x∗x =0forallx ∈ X.
(2) x∗0= xforallx ∈ X.
(3) 0∗x = xforallx ∈ X.
(4) x∗y = y∗xforallx,y ∈ X.
(5) x∗y = y∗x =0⇔ x = yforallx,y ∈ X.
(6) x∗y = y∗x =0⇒ x = yforallx,y ∈ X.
(7) x∗y = y∗x ⇒ x = yforallx,y ∈ X.
(8) 0∗x =0forallx ∈ X.
(9) (x∗y)∗z = (x∗z)∗yforallx,y,z ∈ X.
(10) (x∗y)∗z = x∗(z∗y)forallx,y,z ∈ X.
(11) (x∗y)∗z = x∗(z∗(0∗y))forallx,y,z ∈ X.
(12) (x∗y)∗z = (x∗z)∗(y∗z)forallx,y,z ∈ X.
(13) (x∗y)∗(0∗y) = xforallx,y ∈ X.
(14) x∗(y∗z) = (x∗y)∗zforallx,y,z ∈ X.
(15) (x∗(x∗y))∗y =0forallx,y,z ∈ X.
(16) ((x∗y)∗(x∗z))∗(x∗y) =0forallx,y,z ∈ X.
(17) ∀x ∈ X∃y ∈ Xwithx∗y =0.
(18) ∀x ∈ X∃y ∈ Xwithy∗x =0.
Analgebra(X,∗,0)iscalledagroupifitsatisfies(2),(3),(14),(17),and(18). Analgebra(X,∗)is
calledasemigroupifitsatisfies(14). Analgebra(X,∗,0)iscalledasemigroupwithidentityifitsatisfies
(2),(3), and (14). An algebra (X,∗,0) is called a B-algebra if it satisfies (1),(2), and (11). An algebra
(X,∗,0)iscalledaBG-algebraifitsatisfies(1),(2),and(13). Analgebra(X,∗,0)iscalledaBH-algebra
if it satisfies (1), (2), and (6). An algebra (X,∗,0) is called a Q-algebra if it satisfies (1), (2), and (9).
Analgebra(X,∗,0)iscalledaQˆ-algebraifitsatisfies(1),(2),and(10). Analgebra(X,∗,0)iscalled
ad-algebraifitsatisfies(1),(5),and(8). Analgebra(X,∗,0)iscalleda BCK-algebraifitsatisfies(1),
(5),(8),(15),and(16). Analgebra(X,∗,0)iscalleda gBCK-algebraifitsatisfies(1),(2),(9)and(12).
Analgebra(X,∗,0)iscalledanabeliangroupifitsatisfies(2),(3),(4),(14),(17),and(18). Analgebra
(X,∗,0)iscalledanabeliansemigroupifitsatisfies(4)and(14).
3. Smarandachetypes
Let (X,∗) be a binary algebra. Then (X,∗) is a Smarandache-type P-algebra if it contains a
subalgebra (Y,∗), where Y is non-trivial, i.e, |Y| ≥ 2, or Y contains at least two distinct elements,
and (Y,∗) is itself of type P. Thus we have Smarandache-type semigroup (the type P-algebra is a
semigroup),Smarandache-typegroups(thetype P-algebraisagroup),Smarandache-typeabelian
groups(thetypeP-algebraisanabeliangroup). Ifanalgebraoftype Acontainsasubalgebrawhichis
alsoanalgebraoftypeB,thenitisaaSmarandacheB-type A-algebraprovidedthesubalgebraoftype
Bcontainsatleasttwoelements.
Theorem1. If(X,∗,0)isanabeliand-algebra,thenitisatrivialalgebra.
Proof. Supposethat(X,∗,0)isanabeliand-algebra. By(8),wehave0=0∗xforallx ∈ X. SinceXis
abelian,0∗x = x∗0. Hence0∗x = x∗0=0.By(5),weobtainx =0. HenceX = {0}.
Theorem2. If(X,∗,0)isasemigroupandd-algebra,thenitisatrivialalgebra.
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Proof. Foranyx ∈ X,wehave
x∗0=x∗(0∗(x∗0))
=(x∗0)∗(x∗0) =0.
By(8),wehave0∗x =0. By(5),wehavex =0. HenceX = {0}.
Definition1. Let (X,∗,p) beapointedalgebra. Define x·y := x∗(p∗y), ∀x,y ∈ X. Thenthealgebra
(X,·,p)iscalleda p-derivedalgebra.
Example3. (i)Let(X,∗,e)beagroup. Thenthee-derivedalgebra(X,·,e)has x·y = x∗(e∗y) = x∗y,
and(X,∗,e) = (X,·,e).
(ii)If(X,∗,0)isad-algebra,thenx·y = x∗(0∗y) = x∗0.
(iii)If(X,∗,0)isaBCK-algebra,thenx·y = x∗(0∗y) = x∗0= x. Hence(X,·,0)isaleftsemigroupwith
aselectedpoint0.
(iv)If(X,∗,p)isaleftsemi-groupwithaselectedpoint p,thenx·y = x∗(p∗y) = x∗p = x,i.e.,x·y = x
and(X,∗,p) = (X,·,p).
Definition2. AB-algebraiscalledaSmarandachegroup-typeB-algebraifitcontainsasubalgebra(Y,∗,e)
whichisagroup,whereYisnon-trivial,|Y| ≥2.
Noticethate∗e = 0inthegroup,whilee∗e = 0inthe B-algebra,sothate = 0,i.e.,thepoles
coincidenaturally. Moreoverx∗x =0= eforallx ∈ X. Henceeveryelementinthegrouphasorder
2andthus (Y,∗,e = 0) isaBooleangroup. HenceSmarandachegroup-type B-algebraisequalto
SmarandacheBooleangroup-typeB-algebra.
Example4. Considerthecaseofagroup-typeQ-algebra. Let(X,∗,0)beaQ-algebraandthesubalgebrawhich
isagroupis(Y,∗,e). Thene∗e = einthegroupande∗e =0intheQ-algebraisaverycommonsituationand
itfollowsthatx∗x =0= e,whenceeveryelementinthegrouphasorder2. HenceSmarandachegroup-type
Q-algebrasisSmarandacheBooleangroup-typeQ-algebra.
Theorem3. Let(X,∗,0)beagroupandlet(Y,∗,0)beasubalgebra(notnecessaryasubgroup)whichisa
B-algebra. ThenSmarandacheB-algebra-typegroupsareSmarandacheBooleangroup-typegroup.
Proof. Since(X,∗,0)isagroup,0∗e = e = e∗e. Bythecancellationlawin(X,∗,0),wehavee =0.
Now x ∈ Y has x∗x = eandthus x = x−1. (Y,∗,e),sothat(Y,∗,e)has x∗y−1 = x∗y ∈ Y forall
x,y ∈ Y. HenceY isasubgroupof X. Butalsoforall x ∈ Y andhenceY isaBooleangroup. Thus
SmarandacheB-algebra-typegroupsareSmarandacheBooleangroup-typegroup.
Corollary1. SmarandachQ-algebra-typegroupsisSmarandacheBooleangroup-typegroups.
Analgebra(X,∗,0)isaBQ-algebraifitsatisfies(1),(2),(11),and(9).
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Theorem4. Let(X,∗,0)beBQ-algebra. Let(X,∗,·)beanalgebrawherex·y := x∗(0∗y). Then(X,·,0)
isanabeliangroup.
Proof. Since(X,∗,0)isaB-algebra,(X,·,0)isagroup.Now(x∗(0∗y))∗(0∗z) = (x∗(0∗z))∗(0∗y)
by(9). Hence (x·y)·z = (x·z)·y = x·(z·y) since (X,∗,0) isagroup. Put x := 0in (x·y)·z =
(x·z)·y. Then(0·y)·z =0·(z·y). Since0·y =0∗(0∗y) = y,wehavey·z = z·y,i.e.,(X,·,0)isa
commutativegroup.
Theorem 5. Let (X,·,0) be an abelian group. Define x∗y=x·y−1 for any x,y ∈ X. Then (X,∗,0) is a
BQ-algebra.
Proof. Foranyx ∈ X,x∗x = x·x−1 =0. Thus(1)holds. By(2),wehavex∗0= x·0−1 = x·0= x.
Hence (2) holds. Using (X,·,0) is a commutative group, we obtain (x∗y)∗z = (x·y−1)·z−1 =
(x·0−1)y−1 = (x∗z)∗y.(9)holds. Since(X,·,0)isanabeliangroup,wehave
x∗(z∗(0∗y)) =x·(z·(0·y−1)−1)−1
=x·[z·((y−1)−1·0−1)]−1
=[z·(y·0)]−1
=x·[z·y]−1
=x·(y−1·z−1).
But(x∗y)∗z = (x·y−1)··z−1.Henceweobtain(x∗y)∗z = x∗(z∗(0∗y)).(11)holds.Thus(X,∗,0)
isaBQ-algebra.
Theinterestingfacttonoteisthatweareabletotakeadvantageoftherelationshipx·y = x∗(0∗y)
tounderstandbetterwhatthemeaningoftheclassof BQ-algebrais. Othersuchquestionsaround
inthis settingaswell asothers. E.q., whatclassof B-algebrascorrespondstothe classofsolvable
groups? Canitbeconsideredtobeoftheform: BV-algebrascorrespondstosolvablegroupswhere
“V-algebras"issomenicelyidentifiableclass,astheclassforBQ-algebras?
Theorem6. If(X,∗,0)isaSmarandacheB-algebra-typeQ-algebra,then(X,·,p)isaSmarandacheabelian
group-type0-derivedQ-algebra,wherex·y := x∗(0∗y)foranyx,y ∈ X.
Proof. Supposethat(X,∗,0)isaQ-algebraandlet(Y,∗,e)beaB-algebraandasubalgebraofX. For
any x ∈ Y, x∗x = e and so x∗x = 0, so that e = 0 in any case. Hence (Y,∗,0) is a BQ-algebra.
Therefore if (X,·,0) is the 0-derived algebra obtained from a Q-algebra, then (Y,·,0) is a abelian
subgroup. Thusif(X,·,0)isaSmarandacheB-algebra-typeQ-algebra,then(Y,·,0)isaSmarandache
abeliangrouptype0-derivedQ-algebra.
Corollary 2. (X,∗,0) is a Smarandache Q-algebra-type B-algebra, then (X,·,0) is a Smarandach abelian
group-type0-derived. i.e.,aSmarandacheabeliangrouptypegroup.
Proof. ItissimilartoTheorem6.
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Proposition1. EveryB-algebraisaSmarandacheQ-algebra-typeB-algebra.
Proof. Let (X,·,0) be a group and let x ∈ X. Then < x > the group generated by x is an abelian
group, | < x > | ≥ 2 if x (cid:54)= 0. Hence if (X,∗,0) is obtained by x∗y = x·y−1, then (X,∗,0) is a
B-algebra,while(< x >,∗,0)isaBQ-algebra. Thusif|x| ≥2,theneveryB-algebraisaSmarandache
Q-algebra-typeB-algebra.
Theorem7. Anynon-triviald-algebracannotbeaSmarandachegrouptyped-algebra.
Proof. Let(X,∗,0)bead-algebraandlet(Y,∗,e)beagroupwhichisasubalgebra. Thene∗e = e =0
andx = e∗x = 0∗x = 0foranyy ∈ Y. Hence|Y| = 1. Thusanynon-triviald-algebracannotbea
Smarandachegroup-typed-algebra.
Theorem8. Anynon-trivialgBCK-algebracannotbeaSmarandachegroup-typegBCK-algebra.
Proof. Let (X,∗,0) be a gBCK-algebra and let (Y,∗,e) be a group which is a subalgebra. Then
e = e∗e =0andx∗x =0= eforanyx ∈Y. Henceo(x) =2foranyx ∈Y. Therefore(Y,∗,e =0)is
trivialoraBooleangroup. Also,(x∗y)∗z = (x∗z)∗(y∗z)foranyz ∈ Xandthefactthat(Y,∗,e)
isagroup,requiresz =0and|Y| =1. Thusanynon-trivialgBCK-algebracannotbeaSmarandache
group-typegBCK-algebra.
Theorem9. Anynon-trivialgroupcannotbeaSmarandached-algebra-typegroup.
Proof. Let(X,∗,e)beagroupandlet(Y,∗,0)bead-algebrawhichisasubalgebra. Foranyx ∈ Y,
x∗x = 0,0∗x = 0. Bycancellationlawinthegroup, x = 0forany x ∈ Y,i.e.,|Y| = 1. Henceany
non-trivialgroupcannotbeaSmarandached-algebra-typegroup.
Theorem10. AnygroupcannotgroupbeaSmarandachegBCK-algebratypegroup.
Proof. Let(X,∗,e)beagrouplet(Y,∗,0)bea gBCK-algebrawhichisasubalgebra. Foranyx ∈ Y,
x∗0= x = x∗eandsoe =0bycancellationlawinthegroup. Nowconsidertheaxiom(x∗y)∗z =
(x∗z)∗(y∗z). Since x∗x = 0 = e,Y consistsofelementsoforder2. Thusforany x ∈ Y, x = x−1
meansYisnotonlyasubalgebrabutalsoasubgroupwhichisabelian.Hencey = z∗y.Bycancellation
lawandy = e∗y,weobtainz = e =0,foranyz ∈Y,i.e.,|Y| =1. Thusanygroupcannotgroupbea
SmarandachegBCK-algebratypegroup.
Acknowledgments:
ConflictsofInterest: Theauthordeclaresnoconflictsofinterest.
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