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ON LEFT ALMOST SEMIGROUPS, RIGHT ALMOST SEMIGROUPS AND GROUPOIDS CONTAINING COMMUTATIVE ORTHODOX SEMIGROUPS AND
COMMUTATIVE GROUPS
Syed Aleem Shah, Nisar Ahmad, Aleem Syed, Nisar Shah, Kust Kohat
To cite this version:
Syed Aleem Shah, Nisar Ahmad, Aleem Syed, Nisar Shah, Kust Kohat. ON LEFT ALMOST SEMI- GROUPS, RIGHT ALMOST SEMIGROUPS AND GROUPOIDS CONTAINING COMMUTATIVE ORTHODOX SEMIGROUPS AND COMMUTATIVE GROUPS. 2020. �hal-02494598v2�
STUDY NEW RESULTS IN ALMOST SEMI- GROUPS AND MEDIALS
Syed Aleem Shah, Nisar Ahmad
[email protected] and [email protected]
Institute of Numerical Sciences, Kohat University of Science and Technology (KUST), Kohat, Pakistan
Corresponding Author Email: [email protected]
Abstract: In this article we constructed some new patterns of finite left almost semigroups, right almost semigroups and groupoids in such way that its idem- potent along with some non idempotent(s) form commutative groups and also some of its idempotents form orthodox semigroups. We solved open problem in [11] and constructed new theorems on medial, left double displacement semigroup (LDD-semigroup), right double displacement semigroup (RDD-Semigroup) and paramedial groupoids.
Key Words: Left Almost Semigroup; Right Almost Semigroup; Left Almost Monoid; Right Almost Monoid; Left Almost Group; Right Almost Group; Band;
Orthodox Semigroup; Medial; Regular; Left Regular; Right Regular; Unipotent;
Semilattice.
Preliminaries: In literature a groupoid S is called left almost semigroup (LA- Semigroup) if for all inputs a, b and c ∈ S the condition (ab)c = (cb)a holds.
In literature a groupoid S is called right almost semigroup (RA-Semigroup) if for all inputs a, b and c ∈ S the condition a(bc) = c(ba) holds. A groupoid S is medial if ∀ a, b, c and d lies in S, S satisfies medial law or bisymmetry law i.e.
(ab)(cd) = (ac)(bd). Every LA-Semigroup and RA-Semigroup satisfies medial law but the converse may not be true. A semigroup S is called left regular semigroup if for each input a ∈ S,∃ b ∈ S such that ba2 =a and similarly a semigroup S is called right regular semigroup if for each input a in S, ∃b in S such that a2b=a.
A semigroup S is called regular semigroup if for each element a ∈S, ∃b∈ S such that aba = a and bab = b. Semigroup S is called complete regular semigroup S is regular, left regular and right regular. Semigroup S is called E-semigroup if the subset of S containing its idempotents also form semigroup. A semigroup S is called orthodox semigroup if S is E-semigroup as well as regular semigroup. Band is such semigroup in which if each element is idempotent i.e. semigroup T is band if ∀ t ∈ T the condition t2 = t is satisfied. Each band is orthodox semigroup. A commutative band is called semilattice. A groupoid T is called locally associative groupoid if ∀ t ∈ T condition (a2)a = a(a2) is satisfied. If ∀ t ∈ groupoid T the condition t2 = t is satisfied then then T is locally associative groupoid. Every semigroup is locally associative but converse may not be true. LA-Semigroup S is
called LA-Monoid if ∃ left identity si in S such that for all sj in S the condition sisj =sj holds. LA-Semigroup S is called LA-Group if there exists left identity si
∈ S and inverse of each element exists. RA-Semigroup S is called RA-Monoid if there exists right identity si ∈ S such that ∀ sj ∈ S, sjsi =sj. RA-Semigroup S is called RA-Group if ∃right identity element si ∈ S and inverse of each element exists. A groupoid S is called paramedial if ∀ a, b, c and d ∈ S the condition (ab)(cd) = (db)(ca) holds. A groupoid S is called unipotent groupoid if ∃ only one idempotent s ∈ S i.e. s2 =s and for all t ∈S the condition t2 =s is satisfied and if the idempotent element is zero element of the groupoid T then T is called zeropotent groupoid. Zero semigroup is also zeropotent groupoid and commuta- tive semigroup however this is not regular neither right regular nor left regular semigroup. Klein-4 group is unipotent group. There are other groups of order 2n where n is positive integer andn≥2 and they are commutative and unipotent but they are not cyclic e.g. in mod24, w.r.t mod multiplication the subset of mod24 containing elements {1, 5, 7, 11, 13, 17, 19, 23} is unipotent group of order 8.
Introduction: D. Mclean [1]; gave concept of idempotent semigroups and ex- plained the conditions for which inputs of a semigroup become idempotents. N.
Kimura [2]; explained the idea of idempotent semigroups in his research article and explained the details about left regular, right regular and regular semigroups.
Kimura and Yamada [3]; discussed in detail about the left (right) normality of semigroup and proved that if semigroup S is left (right) normal then S is left (right) regular. Clifford and Preston [4]; elaborated the idea that “IfI∗ is an ideal of a semigroup S then I∗ and S/I∗ are regular (inverse) if and only if S is regular (inverse)” which they discussed in the corollary 7.37. They [4]; also proved the result that if R is some zero minimal right ideal of S then either rS =Rfor each r belongs toR\0 or elserS = 0 is satisfied withR={0, r}. T.E. Hall [5]; presented idea of such structures which are regular semigroups and their idempotents also form regular semigroup. T.E. Hall [5]; discussed the concept of commutativity, left normality, right normality and normality and achieved results. Kazim and Naseeruddin [6]; introduced the concept of left almost semigroup, right almost semigroup and Almost semigroup in 1971 and discussed a groupoid (S,∗) that if (a∗b)∗c= (c∗b)∗a for all a, b and c being inputs of groupoid S then S is left almost semigroup and ifa∗(b∗c) = c∗(b∗a) for all a, b and c being inputs of S then S is right almost semigroup and if S fulfils both the conditions then S is Almost semigroup. They [6]; proved that every commutative semigroup is left almost semigroup (LA-Semigroup) as well as right almost semigroup (RA-Semigroup).
Mushtaq and Yousaf [7]; extended the work of Kazim and Naseeruddin [6] in their research paper “On LA-semigroups” and developed the idea of locally associative LA-Semigroup T satisfying the condition a2 = a for all a in T. Q. Mushtaq [8];
did tremendous extension on his work and introduced the concept that on what conditions LA-Semigroup becomes commutative monoid and becomes group in his paper “Abelian groups defined by LA-semigroups”. N. Kehayopulu [9]; ex- plained the procedure of ordering of elements and by using specific multiplication and ordering of elements he constructed semigroups which are completely regular.
N. Kehayopulu [9]; also explained that “An ordered semigroup S is completely regular if and only if the bi-ideals of S are semiprime”. Extensions were made by Mushtaq and Kamran [10]; in the commutative LA-Semigroups and they proved
that LA-Semigroup S = S2 is commutative if and only if (ab)c = a(cb), ∀ a, b and c ∈ S. Ahmad et al. [11]; developed the idea of left double displacement semigroup (LDD-semigroup) “A groupoid U if ∀ a, b, c and d ∈ U the condition (ab)(cd) = (cb)(ad)” and right double displacement semigroup (RDD-semigroup)
“A groupoid V if∀a, b, c and d∈V the condition (ab)(cd) = (ad)(cb)”. Madadet al. [12]; constructed examples on idempotent LA-Semigroups known as LA-band or AG-band. Shah et al. [13] constructed new example on locally associative LA-Semigroup with zero element. J.L. Chrislock [14]; elaborated idea of medial semigroups. Cho et al. [15] explained and proved the conditions for which a groupoid X becomes paramedial groupoid. They [15]; also proved that if X is paramedial then X is medial if X is commutative, left(right) modular, unipotent and left(right) cancellative, unipotent and left(right) cancellative, idempotent or contains left(right) neutral i.e. identity element. We used definitions and concepts discussed in [1] to [16] and by using some conditions, we solved open problem in [11] and achieved results explained by following cases:
Case-1: LA-Semigroup With Left Identity
Madadet al. [12]; discussed LA-Group in such way that they assumed one element as idempotent element and in the column entries of the table one element precede pattern is followed and in the row one element succeed pattern is followed. Now we are giving its general formula that on non empty finite order set S with order m i.e if n(S) = m and if we define binary operation on S in such way that for all a and b ∈ S, ab is value of entry number c where c = bthentry−athentry+kthentry under the mod m then kth element is left identity of S and every element other that k is self inverse. This is LA-Group. For this we take set A1 = {a, b, c, d}
and A2 = {a, b, c, d, e, f} be two sets with binary operation ∗ defined on them elaborated in Tables 1 and 2.
Table 1: Table ForA1
* a b c d
a a b c d
b d a b c
c c d a b
d b c d a
Table 2: Table ForA2
* a b c d e f
a a b b d d f
b f a b c d e
c e f a b c d
d d e f a b c
e c d e f a b
f b c d e f a
Clearly from these tables A1 andA2 are not commutative, neither associative nor locally associative left invertive groupoids i.e. LA-Semigroups w.r.t binary op- eration. The subset of A1 that contains elements a and c and the subset of A2 that contains elements a and d form cyclic group and the absolute of the entry difference a and c is half of 4 and absolute of the entry difference a and d is half of 6. This is case of the unipotent LA-Semigroup.
Theorem 1: If binary operation on set S is defined in such way that for all a and b ∈ S, ab is value of entry number c where c = bthentry−athentry+kthentry under the mod m and n(S) = m then following conditions hold:
(a) kth element is left identity.
(b) if order of S i.e m is even then the subsetS1 ={si, sj}forms cyclic group and
|i−j|=|j−i|=m/2.
(c) If order of S i.e m is odd then subset of S that contains only left identity element forms groupoid and LA-Semigroup which is trivial.
(d) Every element sj is self inverse i.e. sjsj =sk.
Proof (a): Let binary operation on set S is defined in such way that for all a and b ∈ S, ab is value of entry number c where c = bthentry−athentry+kthentry under the mod m then ka= (a−k+k)modm so ka=a.
(b): Let si be the left identity of S then we are to show that there exists one other element sj such that they forms cyclic group when order of S is finite even number. By the pattern method si+1si =si−1 and si+2si =si−2 and stage comes when we have sjsi = sj because j = i +m/2 and sjsi = si−m/2 = sj. This is clear that when m is finite positive even number then in mod m the result a+m/2 = a−m/2 ∀ a ∈ Zm and this is already given that m is finite positive even number.
(c): Suppose if order of the groupoid S is finite and odd i.e. n(S) = m is fi- nite positive odd number and we take any arbitrary element si as left identity and the column entries in the table one element precede difference pattern and row elements follow one element succeed difference pattern then we are to show that the only subset of S which forms groupoid is the set that contains only left identity element si. m is positive odd number som/2 can not some entry because half of odd integer is not integer so there is some positive integer p such that pth term is middle term in the set {1, 2, 3, .... , p, .... , m} and p= (m+ 1)/2. Now to prove that S does not contain other elementsj that forms cyclic group with si. To prove this we have following three cases:
i: Let we take i < p and let q and r be some elements in S with q < p such that q = (m−g)/2 where g is finite positive odd numbers and let r > p then r = (m +h)/2 where h is also finite positive odd number but h > 1. By the pattern of preceding elements in the column and succeeding elements pattern in the row when n(S) = m is odd then sisp = sp but spsi 6= sp in mod m w.r.t the binary operation defined on S because (i+i−p) = (2i−p)6=pand by the same way sqsi 6= sq and srsi 6= sr in mod m w.r.t the binary operation defined on S because (i+i−q) = (2i−q)6=pand (i+i−r) = (2i−r)6=p.
(ii): If i > p then this is straightforward.
(iii): If i = p then si = sp is middle term and also left identity term of S al- ready proved in part-(a) then sisq =spsq = sq as well as sisr = spsr =sr but in mod m when m is finite positive odd number andith orpth element is idempotent as well as left identity then sqsp 6= sq and by the same way srsp 6= sr because (p+p−q) = (2p−q)6=q and (p+p−r) = (2p−r)6=r.
(d): We are to show that every element is self inverse i.e. ∀sj ∈S,sjsj =si. On Zm w.r.t binary operation defined byab=c where c is value of the entry number EN and EN = (bthentry−athentry+kthentry) sojj = (j−j+k)modm =k so value of the kth entry is sk that shows that each element is self inverse.
Example 1.1: If A100 = {a1, a2, a3, a4, ..., a100} w.r.t binary operation ∗ defined onA100 inputa23 is idempotent input and column entries follow one element pre- ceding difference pattern and the row entries follow the one element succeeding difference pattern thena23 is left identity then the subsetA100C ={a23, a73}forms cyclic group and this is quite clear that|73−23|=|23−73|= 50.
Example 1.2: If A7 = {a1, a2, a3, ..., a7} and binary operation ∗ is defined on A7 in such way that a5 is idempotent element then the subset of A7 that is only groupoid is the set that contains a5 explained in Table 3.
Table 3: Table ForA7
∗ a1 a2 a3 a4 a5 a6 a7 a1 a5 a6 a7 a1 a2 a3 a4
a2 a4 a5 a6 a7 a1 a2 a3 a3 a3 a4 a5 a6 a7 a1 a2 a4 a2 a3 a4 a5 a6 a7 a1 a5 a1 a2 a3 a4 a5 a6 a7
a6 a7 a1 a2 a3 a4 a5 a6 a7 a6 a7 a1 a2 a3 a4 a5
Remarks in Case-1:
(a) In case of odd number of elements only left identity element si is both (left and right) regular.
(b) In Case of even number of elements only two elements left identity element si and the element sj that forms cyclic group are both left and right regular.
(c) In Case of even number of elements every element other than left identity si andsj that forms cyclic group is right regular w.r.t itself i.e. a2a =a and for each a,∃ b such that ba2 =a i.e. each a other than si and sj is left regular with other element b. When number of elements are odd then also this condition holds but in this case onlysi is both (left and right) regular with itself.
(d) Whensi is left identity then the elements a and b that satisfies the condition ba2 =a and ab2 =b also satisfies the condition that a∗si =b and b∗si =a.
Case-2: RA-Semigroup With Right Identity
We extended the work of Madad et al [12]; and assumed that S is non empty set of finite order m i.e. n(S) = m and if si is idempotent element and the column entries follows one element succeeding difference pattern and row elements follows one element preceding difference pattern then the idempotent elementsi becomes the right identity element and this is right almost semigroup (RA-Semigroup). We give the example 2.1 so readers can grasp the approach in effortless way.
Example 2.1: LetB1 = {b1, b2, b3, b4} and binary operation∗ is defined on B1 in such way that b1 is idempotent and we have the Table 4.
Table 4: Table For B1
* b1 b2 b3 b4 b1 b1 b4 b3 b2 b2 b2 b1 b4 b3 b3 b3 b4 b1 b4 b4 b4 b3 b2 b1
Pattern of Case-2: We take b1 as the idempotent and in the table in the row entries we take one element preceding difference pattern and in the column entries we take one element succeeding difference pattern sob1 becomes right identity and by the pattern S is RA-Group with right identityb1 and each element is self inverse i.e. ∀ bi in B1, bibi = bi. This is unipotent RA-Group as we defined unipotent groupoid in preliminaries that ∃ only one idempotent a in groupoid S and ∀ b ∈ S,b2 =a.
Generally if on any finite order groupoid S if ∀ a, b ∈ S, ab=c where c is value of entry number EN and EN = (athentry−bthentry+kthentry) under mod m then value of the kth entry is the right identity element and groupoid S becomes right almost group (RA-Group) with right identity k where each element is self inverse. To prove this result in general we have the following theorem:
Theorem 2: If binary operation ∗ on non empty set S of finite order is de- fined in such way that ∀ a and b ∈ S, a∗b = V(EN) where V(EN) means value of entry number N and N = (athentry−bthentry+kthentry) under the mod m where m is order of the set S then then following conditions hold:
(a) k is right identity element.
(b) if order of S i.e. n(S) = m where m is finite positive even then subset of S containing idempotent element k with only one other element h forms cyclic group and absolute of entry difference of (k - h) is half of m i.e |k−h|=|h−k|=m/2.
(c) If n(S) = m where m is finite positive odd number then the subset of S con- taining only right identity element k forms groupoid and right almost semigroup which is trivial case and@ h ∈ S such that {k, h}forms groupoid.
(d) ∀ a ∈ S the conditionaa =k holds i.e. S is unipotent RA-Semigroup.
Proof (a): Suppose S is finite set of order m and S ={s1, s2, s3, ...., sm} and on set S we define binary operation∗ in such way thatskbecomes the idempotent ele-
ment and∀siandsj ∈S,sisj =sawherea = (k+i−j)modm thenskis idempotent element as well as right identity. sksk = (kthentry +kthentry −kthentry)modm which is kth entry so sk is idempotent element. Now to show that sk is right identity so this is effortless to prove because ∀ si ∈ S, sisk = si because ik = (k+i−k)modm = (i)modm. So ik =i and this is proved generally that sk is right identity.
(b): When order of set S is finite positive even number i.e. n(S) = m where m ∈ M and M is finite subset of positive even numbers. We know that the bi- nary operation follows the pattern explained above in Pattern of Case-2 and we already proved in part(a) of Theorem-2 that sk is idempotent as well as right identity. Now we are to show that ∃ sh such that {sk, sh} forms cyclic group and |h−k| = |k−h| = m/2. From the pattern sksk+1 = sk−1, sksk+2 = sk−2, sksk+3 = sk−3 and this procedure goes until we have sksk+m/2 = sk−m/2. Now if
|h−k| = |k −h| = m/2, then h = m/2 + k = m/2−k. So we can say that sksk+m/2 =sk−m/2 becomes sksh =sh. So this is proved that sksh =sh. Now we are to show that shsh =sk so we can prove that {sk, sh} forms cyclic group. For this we have shsh =sc and c= (h−h+k)modm =k so shsh =sk. So subset of S say SC ={sk, sh}forms cyclic group.
(c): Straightforward.
(d): We are to show that each element is self inverse i.e. ∀ sa ∈ S the condi- tion sasa =sk is satisfied. Clearly sasa = sk because aa = (a−a+k)modm = k and sk is kth entry that proves that sasa =sk. So this is case of unipotent RA- Group whether order of RA-Semigroup S is finite positive even number or odd number.
Example 2.2: If B4 = {b1, b2, b3, ...b488} and binary operation is defined on B488 in such way that b175 is idempotent element and all elements follow the pat- tern explained in Case-2 then the subset of B488 that contains elements b175 and b419 forms cyclic group and clearly |419−175|=|175−419|= 488/2.
Example 2.3: IfB5 ={b1, b2, b3, b4, b5}and∗ is defined onB5 in such way thatb3
is idempotent element and all elements follows the same pattern and binary oper- ation explained in Case-2 pattern then the subset ofB5 that forms groupoid is set that contains only b3 and no other element forms cyclic group or even groupoid with b3 explained in Table 5.
Table 5: Table For B5
* b1 b2 b3 b4 b5
b1 b3 b2 b1 b5 b4 b2 b4 b3 b2 b1 b5 b3 b5 b4 b3 b2 b1 b4 b1 b5 b4 b3 b2
b5 b2 b1 b5 b4 b3
Matlab Code For Modulo n example in Case-1 and Case-2: We con- structed example in modulo n and made matlab code for LA-Group and for RA- Group. On Zn if binary operation ∗ is defined in such way that ∀ a and b ∈ Zn if a∗b = (b−a+k)modn then Zn is LA-Group w.r.t binary operation ∗ and k is left identity and if . is defined on Zn in such way that ∀ a and b ∈ Zn if a.b= (a−b+k)modn then Zn is RA-Group w.r.t binary operation .and k is right identity.
Following is Matlab Code For LA-Group:
function [] LA 1 () S = 1:40;
IE = 15;
Table = zeros(length(S));
for i = 1:length(S) a = S(i);
for k = 1:length(S)
AA = mod(IE +(S(k)-S(i)),length(S));
if AA==0
Table(i,k) = length(S);
else
Table(i,k) = AA;
end end end
Table = [[‘B’,sym(S)];[S’,Table]];
disp(Table) end
Following is Matlab Code For RA-Group:
function [] RA 1 () S = 1:40;
IE = 11;
Table = zeros(length(S));
for i = 1:length(S) a = S(i);
for k = 1:length(S)
AA = mod(IE +(S(i)-S(k)),length(S));
if AA==0
Table(i,k) = length(S);
else
Table(i,k) = AA;
end end end
Table = [[‘B’,sym(S)];[S’,Table]];
disp(Table) end
Extensions in Case-1 and Case-2: The idea elaborated in cases 1 and 2 is ex- tended to groupoids in which number whether numbers of elements in set are even or odd we get cyclic groups of order more than two. We take setA4 ={1,2,3,4}
and A5 = {1,2,3,4,5} and binary operation ∗, . and o are defined on A4 and A5 explained in Tables 6 and 7.
Table 6: Tables for A4
∗ 1 2 3 4 . 1 2 3 4 o 1 2 3 4
1 1 2 3 4 1 1 2 3 1 1 1 2 3 4
2 2 3 1 1 2 2 3 1 1 2 2 3 1 1
3 3 1 2 1 3 3 1 2 1 3 3 1 2 1
4 1 1 1 1 4 4 1 1 1 4 4 1 1 1
Table 7: Tables for A5
∗ 1 2 3 4 5 . 1 2 3 4 5 o 1 2 3 4 5
1 1 2 3 4 5 1 1 2 3 4 1 1 1 2 3 4 5
2 2 3 4 1 1 2 2 3 4 1 1 2 2 3 4 1 1
3 3 4 1 2 1 3 3 4 1 2 1 3 3 4 1 2 1
4 4 1 2 3 1 4 4 1 2 3 1 4 4 1 2 3 1
5 1 1 1 1 1 5 5 1 1 1 1 5 5 1 1 1 1
Discussion on Extensions of Case-1 and Case-2:
Clearly from tables Table 6 and Table 7 the magmas A4 and A5 are not com- mutative, neither associative nor locally associative groupoids and also A4 and A5 does not satisfy left invertive law, right invertive law and medial law. With respect to binary operation∗ A4 andA5 are groupoids with left identity 1 in which cyclic groups{1, 2, 3}and {1, 2, 3, 4}are contained respectively. With respect to binary operation . A4 and A5 are groupoids with right identity 1 in which again cyclic groups{1, 2, 3}and{1, 2, 3, 4}respectively are contained. With respect to binary operation o A4 is groupoid with identity 1 in which cyclic groups {1, 2, 3}
and {1, 4} are contained and A5 is also groupoid with identity 1 in which cyclic groups {1, 2, 3, 5} and {1, 5} are contained.
Next we will show that on what conditions groupoids becomes commutative semigroup but first we have some already proved results from [6] to [13] which we will use to prove our theorems which are following:
Proved Results-1 In [6] To [13]:
If groupoid S holds
(a) left invertive law and associative law then S is commutative semigroup as well as RA-Semigroup.
(b) right invertive law and associative law then S is commutative semigroup as well as LA-Semigroup.
(c) associative law and commutative law then S is LA-semigroup as well as RA- Semigroup.
(d) left invertive law and commutative law then S is commutative semigroup as well as RA-Semigroup.
(e) right invertive law and commutative law then S is commutative semigroup as well as LA-Semigroup.
Proved Results-2 In [6] To [13]:
If groupoid S is
(a) LA-Semigroup with right identity then S is commutative monoid.
(b) RA-Semigroup with left identity then S is commutative monoid.
(c) Non commutative group, non commutative monoid and non commutative semi- group then S can neither be LA-Semigroup nor RA-Semigroup.
(d) Non commutative group, non commutative monoid then S can not hold medial law. (This will be proved in Theorem 21)
(e) Non commutative semigroup then S may be medial or may not be medial.
Next we proved the following theorems:
Theorem 3: If S is LA-semigroup and satisfies condition (ab)c= (ac)b then S is commutative semigroup.
Proof: S is LA-Semigroup so ∀ a, b and c ∈ S the condition (ab)c= (cb)a holds and also this is given that (ab)c = (ac)b. Let we have a, b, c and d ∈ S then we have following steps:
Method-1:
Let we have x and y ∈S and we suppose x=aband y=cdso xy= (ab)(cd) and we do the following steps:
(ab)(cd) = (ab)f = (f b)a= ((cd)b)a. Let cd=f
So ((cd)b)a = ((cd)a)b = ((ca)d)b= ((ca)b)d. By using the condition (ab)c= (ac)b So ((ca)b)d= ((ba)c)d= (dc)(ba). By using left invertive law
Now (cd)(ab) = (cd)g = (gd)c= ((ab)d)c. Letab=g
So ((ab)d)c = ((db)a)c = ((da)b)c = ((da)c)b. By using left invertive law and condition (ab)c= (ac)b
So ((da)c)b = ((ca)d)b = ((ca)b)d = ((ba)c)d. By using left invertive law and condition (ab)c= (ac)b
So ((ba)c)d= (dc)(ba). By using left invertive law
Hence xy= (ab)(cd) = (dc)(ba) and yx= (cd)(ab) = (dc)(ba). So xy=yx.
Method-2:
S is LA-Semigroup and also holds condition (ab)c= (ac)b ∀a, b and c ∈ S. Then (ab)c = (cb)a = (ca)b and also we can say that (ca)b = (ba)c = (bc)a. So ∀ a, b and c ∈ S, (ab)c= (ac)b= (ca)b= (ba)c= (bc)a= (cb)a.
Hence we can also use these conditions and prove that So ∀ a, b, c and d ∈ S, (ab)(cd) = (cd)(ba).
(ab)(cd) = (ab)y
So (ab)y= (yb)a= ((cd)b)a= ((cb)d)a = ((cb)a)d. By using conditions So ((cb)a)d= ((ab)c)d= ((ab)d)c. By using conditions
So xy = (ab)(cd) = ((ab)d)c = (xd)c = (cd)x = (cd)(ab) = yx. By using left invertive law
Theorem 4: If a groupoid S satisfies the conditions (ab)c = (ac)b and (ab)c = (ba)c ∀ a, b and c ∈ S then S satisfies left invertive law i.e. (ab)c = (cb)a and S is also commutative semigroup.
Proof: Using the given conditions we have the following steps:
(ab)c = (ac)b = (ca)b = (cb)a. By using given conditions this is proved that S is LA-semigroup. Now we do the following steps:
So (cb)a = (bc)a = (ba)c. By using conditions this is proved that ∀ a, b and c ∈ S, S satisfies conditions (ab)c = (ba)c = (bc)a. We can use these conditions and this is effortless to prove that S is commutative semigroup.
We can also use Theorem-3 that if S is LA-semigroup and satisfies condition (ab)c= (ac)b then S is commutative semigroup.
Theorem 5: If S is LA-Semigroup and satisfies the condition a(bc) = (cb)a for all a, b and c ∈ S then S is commutative semigroup.
Proof: Given that ∀ a, b and c ∈ S, (ab)c= (cb)a and also a(bc) = (cb)a. So if we compare the right side of the given conditions then (ab)c=a(bc). So S satis- fies associative property i.e. S is LA-Semigroup and semigroup. So from Proved Results 1 S is commutative semigroup as well as RA-Semigroup.
Theorem 6: Is T is RA-semigroup and satisfies the condition (ab)c = c(ba)
∀ a, b and c ∈ T then T is commutative semigroup.
Proof: Given that ∀ a, b and c ∈ T, a(bc) = c(ba) and (ab)c = c(ba) so by comparing these two results we have a(bc) = (ab)c.
Hence S is associative. So From Proved Results 1 S is commutative semigroup as well as LA-Semigroup.
Theorem 7: If ∀ a, b and c ∈ T, a(bc) = a(cb) and a(bc) = b(ac) then T is commutative semigroup.
Proof: Using the given conditions we have a(bc) =a(cb) = c(ab) =c(ba). So S is RA-Semigroup and also satisfies propertya(bc) =c(ab).
So c(ba) =b(ca) =b(ac). So by applying all these conditions we do the following procedure to prove that S is commutative semigroup:
(ab)(cd) = g(cd) = d(cg) = d(c(ab)). Let g = ab and f = cd because binary operation is well defined so if a and b∈ T then ab also∈ T so we suppose ab=g and cd=f thus we have the following steps:
d(c(ab)) =d(a(cb)) = a(d(cb) =a(b(cd) a(b(cd)) =a(bf) = f(ab) = (cd)(ab)
Theorem 8: If T is RA-semigroup and ∀ a, b and c ∈ T holds condition a(bc) = (ba)cthen T is commutative semigroup.
Proof: Given that ∀ a, b and c ∈ T, a(bc) = c(ba) and a(bc) = (ba)c. So c(ba) = (bc)a. So ∀ a, b and c ∈ T a(bc) = (ba)c= (bc)a. So if we suppose that if g = ab and f = cd and gf = (ab)(cd) and we use the conditions and do the following procedure:
(ab)(cd) =f(cd) = d(cf)
d(cf) = (cd)f =gf. By using a(bc) = (ba)c Also we can do (cd)(ab) =f(ab) = b(af) b(a(cd) = (ab)(cd) By using a(bc) = (ba)c
Theorem 9: If a groupoid T satisfies conditions (ab)c = (ba)c and (ab)c = (bc)a then T is commutative semigroup.
Proof: Given that ∀a, b and c ∈ T, (ab)c= (ba)cand (ab)c= (bc)a then we use these conditions and do the following steps:
(ab)c = (ba)c = (ac)b = (ca)b = (cb)a. So T holds left invertive law as well as T holds the condition (ab)c = (ac)b so by using theorem-3 this is effortless to prove that T is commutative semigroup.
Theorem 10: If a groupoid T satisfies conditions a(bc) = (ba)c and (ab)c = (bc)a then T is commutative semigroup.
Proof: ∀ a, b and c ∈ T, a(bc) = (ba)c = (ac)b by using the condition (ab)c = (bc)a. So if (ab)c= (bc)a and (ab)c= (ac)b then ∀ a, b and c ∈ S, (ac)b = (cb)a, hence T is LA-Semigroup and by using Theorems 3 and 4 this is effortless to prove that T is commutative semigroup as well as RA-Semigroup.
Open Problem 1: In Case-1 we study finite LA-Semigroup in which when order of LA-Group is finite even number then cyclic group is contained. We left this to researchers as an open problem to find the finite LA-Semigroup T of odd order in which where m is some finite positive odd number and subsetT1 of LA-Semigroup T is cyclic group of order 2 and when order of T is finite even number then@ any subset T1 of T that contains cyclic group of two elements.
Open Problem 2: In Case-2 we study finite RA-Semigroup in which when order of RA-Semigroup is finite even number then cyclic group is contained. We left this to researchers as an open problem to find the finite RA-Semigroup T of odd order in which where m is some finite positive odd number and subset T1 of RA- Semigroup T is cyclic group of order 2 and when order of T is finite even number then @ any subset T1 of T that contains cyclic group of two elements.
Case-3: Locally Associative LA-Semigroup With Zero Element
Shah et al. [13] constructed new example on locally associative LA-Semigroup
with zero element in which they took set set S = {0,1,2} and binary operation . was defined in such way in the table 1.0 = 0 and all other entries were sub- stituted by 2, so 2 is zero element of the structure (S, .). Also we are using the concept of zero semigroup A in which binary operation is defined in such way that any two elements are operated and we get zero element of that structure A or in simple words a semigroup in which the we get only zero element in the cayley table. But in Case-3 we do some changes and there are two elements a1 and a2 which are not zero elements but they disturbs the commutative and associative properties of the whole structure in which either a1a2 = a2 or a1a2 = a1 or we have the other situation in which a2a1 =a1 ora2a1 =a2. We extend this concept to such LA-Semigroup C in such way that we can have two, three, four or more idempotents in which one idempotent is the zero element of the LA-Semigroup C and the subsetC1 of C having these idempotent members of C forms commutative orthodox semigroup and C1 is also the ideal of C. We take C4 = {c1, c2, c3, c4}, C5 ={c1, c2, c3, c4, c5} and C6 ={c1, c2, c3, c4, c5, c6} and binary operation η (eta) is defined on C1, C2 and C3 elaborated in Table-8.
Table 8: Table-8 For C4, C5 and C6
c1 c2 c3 c4 c1 c2 c3 c4 c5 c1 c2 c3 c4 c5 c6 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c1 c2 c1 c1 c1 c1 c2 c1 c1 c1 c1 c1 c2 c1 c1 c1 c1 c1 c1
c3 c1 c2 c1 c1 c3 c1 c2 c1 c1 c1 c3 c1 c2 c1 c1 c1 c1 c4 c1 c1 c1 c4 c4 c1 c1 c1 c4 c1 c4 c1 c1 c1 c4 c1 c1 c5 c1 c1 c1 c1 c5 c5 c1 c1 c1 c1 c5 c1 c6 c1 c1 c1 c1 c1 c6
Discussion On Examples 3.1 To 3.3 elaborated In Tables 8:
Clearly from the tables Table-8 the magmas(groupoids)C4,C5 andC6 are neither commutative nor associative however they are locally associative LA-Semigroups with zero element c1. The subset of C4 say C4A = {c1, c4} is commutative or- thodox semigroup and ideal of C4. The set C5A = {c1, c4, c5} is commutative orthodox semigroup and ideal of C5. Similarly this is also clear from the table that C6A={c1, c4, c5, c6} is commutative orthodox semigroup and ideal of C6. Important Results In Case-3:
(3a) This is such locally associative LA-Semigroup with zero element.
(3b) Zero element with other idempotent(s) forms commutative orthodox semi- group as well as ideal of LA-Semigroup C.
(3c) Zero element with two non idempotent members that violates commutative and associative property only form LA-Semigroup.
(3d) Each commutative orthodox semigroup which is also ideal of LA-Semigroup structure must contain zero element.
(3e) The product of two idealsCIandCJ is{0}whenCI ={0, c1i, c2i, c3i, ...., cni}and CJ ={0, c1j, c2j, cj3, ...., cnj} where I = {c1i, ci2, c3i, ...., cni} and J = {c1j, c2j, c3j, ...., cnj} are disjoint sets. Here we are not taking power of elements, here this means i1, i2, ..., in and j1, j2, ...., jn. We can simply write CI = {0, i1, i2, i3, ...., in} and
CJ ={0, j1, j2, j3, ...., jn}.
(3f) LA-Semigroup C has many proper ideals but this has no zero minimal ideal and the trivial zero minimal ideal is {0}.
(3h) Case-3 is the case of LA-Semigroup which is the not regular, neither left not right regular.
POSSIBLE PRACTICAL APPLICATIONS OF CASE-3: A useful prac- tical application of Case-3 can be for setting some password for some bank locker or some safe if by clicking any two digits entries give specific same number but the last two entries give different number e.g. if password is to close and open some safe or vault is 55557 then 1.1, 2.1, 3.1 and 4.1 execute 5 but 6.7 executes 7 and the lock can be closed or opened. For securing the important documents in some safe or vault the pattern discussed in Case-3 can be useful tool. Also from 0to9 we can select two or more than idempotents digits and password can be set for securing digital locks of safes or vaults. Case-3 can be used in cryptography procedures. The more complicated will be the binary operation the more safe and secure will be information.
First Extension in Case-3: We extend this idea of Case-3 that we can have LA- Semigroup in which non cyclic but commutative group (Klein-4 group) is contained if we just have setC7 ={c1, c2, c3, c4, c5, c6, c7}andC8 ={c1, c2, c3, c4, c5, c6, c7, c8} and binary operation . is defined on C7 and C8 in such way that C7 and C8 are LA-Semigroups and first four elements forms Klein-4 group; see Table 9.
Table 9: Table For First Extension In Case-3
. c1 c2 c3 c4 c5 c6 c7 . c1 c2 c3 c4 c5 c6 c7 c8 c1 c1 c2 c3 c4 c5 c5 c5 c1 c1 c2 c3 c4 c5 c5 c5 c5
c2 c2 c1 c4 c4 c5 c5 c5 c2 c2 c1 c4 c4 c5 c5 c5 c5 c3 c3 c4 c1 c2 c5 c5 c5 c3 c3 c4 c1 c2 c5 c5 c5 c5 c4 c4 c3 c2 c1 c5 c5 c5 c4 c4 c3 c2 c1 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5 c5
c6 c5 c5 c5 c5 c5 c5 c5 c6 c5 c5 c5 c5 c5 c5 c5 c5 c7 c5 c5 c5 c5 c5 c6 c5 c7 c5 c5 c5 c5 c5 c6 c5 c5 c8 c5 c5 c5 c5 c5 c5 c5 c8
Second Extension in Case-3: This idea can be extended to infinite set on which binary operation can be defined is in such way that the set having infinite idempotents in which only one element is zero element and the set containing all the elements that are idempotents forms commutative orthodox semigroup e.g. If we take set of non negative integers (say W) set of whole numbers and binary operation is defined on W in such way that 0 becomes zero element in W and the subset of W sayW1that includes inputs 1, 2, 3 and 4 forms Klein-4 group. The set W2 = {8,10,12,14, ...} is set of idempotents while the set W3 ={7,9,11,13, ....}
is set in which every element is self zero divisor and the product of each element in W3 with each element of W gives 0. Let W4 = {5,6} and 5.5 = 6.6 = 6.5 = 0 but 5.6 = 6. Thus W w.r.t binary operation . is infinite locally associative LA-
Semigroup with zero element 0 and this contains W −1 which Klein-4 group and the subset of W say WA={0,5,6}is locally associative LA-Semigroup with zero element 0. The subset of W say WB ={0,8,10,12,14, ....} is commutative ortho- dox semigroup andWB is also ideal of W. This is example in Case-3 in which first four elements do not form Klein-4 group.
Third Extension in Case-3: This is not difficult to construct a LA-Semigroup with zero element that contains cyclic group, Klein-4 group if we see Table-10 and Table-11.
Table 10: Third Extensions A in Case-3
. 1 2 3 4 5 6 7 8 9 10 . 1 2 3 4 5 6 7 8 9 10 11
1 1 2 3 4 5 5 5 5 5 5 1 1 2 3 4 5 5 5 5 5 5 5
2 2 1 4 4 5 5 5 5 5 5 2 2 1 4 4 5 5 5 5 5 5 5
3 3 4 1 2 5 5 5 5 5 5 3 3 4 1 2 5 5 5 5 5 5 5
4 4 3 2 1 5 5 5 5 5 5 4 4 3 2 1 5 5 5 5 5 5 5
5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
6 5 5 5 5 5 5 5 5 5 5 6 5 5 5 5 5 6 5 5 5 5 5
7 5 5 5 5 5 6 5 5 5 5 7 5 5 5 5 5 5 5 8 5 5 5
8 5 5 5 5 5 5 5 8 9 10 8 5 5 5 5 5 5 5 5 5 5 5
9 5 5 5 5 5 5 5 9 10 8 9 5 5 5 5 5 5 5 5 5 5 5
10 5 5 5 5 5 5 5 10 1 9 10 5 5 5 5 5 5 5 5 5 5 5
11 5 5 5 5 5 5 5 5 5 5 5
Table 11: Third Extensions B in Case-3
. 1 2 3 4 5 6 7 8 9 10 11
1 1 2 3 4 5 5 5 5 5 5 5
2 2 1 4 4 5 5 5 5 5 5 5
3 3 4 1 2 5 5 5 5 5 5 5
4 4 3 2 1 5 5 5 5 5 5 5
5 5 5 5 5 5 5 5 5 5 5 5
6 5 5 5 5 5 5 5 5 5 5 5
7 5 5 5 5 5 6 5 5 5 5 5
8 5 5 5 5 5 5 5 8 9 10 11
9 5 5 5 5 5 5 5 9 8 11 10
10 5 5 5 5 5 5 5 10 11 8 9
11 5 5 5 5 5 5 5 11 10 9 8
Scheme In Case-3: We select two elements which are neither left nor right zero and also not idempotents say c1 and c2 and used this condition that either c1c2 =c2 and c2c1 = co i.e. zero element or c1c2 =c1 and c2c1 =co. This clearly
makes the groupoid neither commutative nor associative but this groupoid will be locally associative LA-Semigroup.
Fourth Extension in Case-3: We construct LA-Semigroup containing cyclic group and commutative orthodox semigroup which is locally associative and con- tain zero element elaborated in Table 12.
Table 12: Table of Fourth Extension In Case-3
. 1 2 3 4 5 6 7 8 . 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 5 5 5 1 1 2 3 4 5 5 5 5 5
2 2 3 4 1 5 5 5 5 2 2 3 4 1 5 5 5 5 5
3 3 4 1 2 5 5 5 5 3 3 4 1 2 5 5 5 5 5
4 4 1 2 3 5 5 5 5 4 4 1 2 3 5 5 5 5 5
5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
7 5 5 5 5 5 6 5 5 5 5 5 5 5 5 5 5 5 5
8 5 5 5 5 5 5 5 8 5 5 5 5 5 5 5 5 5 5
5 5 5 5 5 5 5 5 5 5
Fifth Extension of Case-3: We extended this idea to infinite as well as un- countable set. We take set of real numbers R and we define binary operation . on R in such way that the subset of R sayR1 ={1,2,3,4} forms Klein-4 group with 1 as the identity element. The element 0 is zero element in the structure R and let we have 6.5 = 5 but 5.6 = 0 and 5.5 = 0 and 5.5 = 6.6 = 0 and all other members of R other than these elements are idempotents i.e. aa =a andbb=bbut ab= 0.
This is infinite LA-Semigroup which is not regular, neither left nor right regular LA-Semigroup. R does not satisfy commutative, associative, regular, left regular and right regular properties.
Sixth Extension of Case-3: If we have set of real numbers R and we define binary operation . on R in such way that first four positive integers form Klein-4 group with identity element 1 and 5.6 = 6 and 6.5 = 1 and all other real numbers when are operated with each other and we get 1. This is locally associative LA- Semigroup which is without left (right) identity and which has no zero, neither left nor right zero element.
Seventh Extension of Case-3: If we have set of real numbers R and we define binary operation . on R in such way that first four positive integers form Klein-4 group with identity element 1 and 1.5 = 4 and 5.1 = 5 and 2.5 = 5.2 = 3.5 = 5.3 = 4.5 = 5.4 = 5.5 = 1 and 6.7 = 7 but 7.6 = 1. All other elements that are rational are idempotents suppose ∀ p, q ∈ Q (set of rational numbers) but p.q = q.p = 1 when p neq q and when elements are irrational i.e. ∀ s, t ∈ Q’
(set of irrational numbers) then ss = tt = st = ts = 1. This is groupoid which is not commutative, neither associative nor locally associative and this is neither LA-Semigroup nor RA-Semigroup and this is also not medial.
Similarly we can extend all these cases to LA-Semigroup containing cyclic groups of order 4, 5, 6 and more elements and containing semilattice of order 4, 5, 6 and more elements.
Case-4: RA-Monoid Contains Cyclic Group, Orthodox Semigroup and RA-Group:
We constructed the RA-Semigroup D with binary operation . defined on Z5 ex- plained in Table 13.
Table 13: Table For Z5 In Case-4
. 0 1 2 3 4
0 0 0 0 4 3
1 0 1 2 4 3
2 0 2 1 4 3
3 3 3 3 0 4
4 4 4 4 3 0
Clearly from the Table this is RA-Monoid with right identity 1. The Subset of Z5 say Z5S = {0,1} is semilattice. The subset Z5R1 = {0,3,4} is unipotent RA-Group. ClearlyZ5C ={1,2} is cyclic group. This is further extended and we have groupoidZ10w.r.t binary operation .that contains cyclic groups, semilattice and unipotent RA-Group and the unipotent RA-Group also contains cyclic group;
see Table 14.
Table 14: Table-14 ForZ10 In Case-4
. 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 9 8 7
1 0 1 2 3 4 5 6 9 8 7
2 0 2 3 4 5 6 1 0 0 0
3 0 3 4 5 6 1 2 0 0 0
4 0 4 5 6 1 2 3 0 0 0
5 0 5 6 1 2 3 4 0 0 0
6 0 6 1 2 3 4 5 0 0 0
7 7 7 0 0 0 0 0 0 9 8
8 8 8 0 0 0 0 0 9 0 8
9 9 9 0 0 0 0 0 8 7 0
Clearly from the table subset ofZ10say Z10R1 ={0,7,8,9}is unipotent RA-group and the subset of Z10 say Z10R2 ={0,8} is cyclic group which is subset of Z10R1. 1 is right identity ∀ x ∈ Z10. Z10C1 = {1,2,3,4,5,6} is cyclic group. Similarly Z10C2 = {1,3,5} and Z10C2 = {1,4} are also cyclic groups and they both are subsets of Z10C1. Z10S ={0,1} is semilattice.
Case-5: Finite Commutative Groupoid Containing Semilattices and Cyclic Groups
Let we take a set S1 = {a, b, c}, S2 = {a, f, g} and S3 = {a, b, c, d, e} and define the binary operation o on S1, S2 and S3 elaborated in Table 15.
Table 15: Table-15 For Case-5
o a b c o a f g o a b c d e
a a b c a a f g a a b c d e
b b b a f f f g b b b c a a
c c a a g g g f c c c b a a
d d a a d e
e e a a e d
Clearly S1, S2 and S3 are commutative groupoids but do not hold associative, medial, ldd and rdd laws. We have examples of each case if we suppose S1 is set that contains three distinct sets SSZ = {0}, Z+ = {1,2,3,4, ....} and Z− ={−1,−2,−3, ...} and we define binary operation ∗ with the following prop- erties:
(a) ∀ a and b∈ X if a∗b=b∗a =a+b =b+a and a∗b ∈ X then X∗X =X.
(b) ∀ a ∈ X and ∈ b ∈ Y if a∗b and b∗a ∈ both X and Y in different cases then XY =Y X =SSZ.
(c) ∀ a ∈ X and ∈ b∈ Y if a∗b and b∗a ∈X then XY =Y X =X and if a∗b and b∗a ∈Y then XY =Y X =Y.
Clearly SSZ∗Z+ = Z∗+SSZ = Z+ and SSZ∗Z− = Z∗−SSZ = Z−. Also we know Z+∗Z+ = Z+, Z− ∗Z− = Z− and Z∗+Z− as well as Z∗+Z− ∈ both Z+ and Z− because when a ∈ Z+ and b ∈ Z− and |a| > |b| then a∗b = b∗a = c and c > 0 and when |a| < |b| then a∗b = b∗a = d and d < 0. In the case when
|a| = |b| then a∗b =b∗a = 0. Clearly the set S1 that contains three disjoint sets of positive integers, negative integers and the singleton set containing only zero element is commutative idempotent groupoid that contains two semilatices say S1A={SSZ, Z+} and S1B ={SSZ, Z−}.
The example for S2 is also the examples in the set such that 2 contains three sets ZZS = {0}, E+ (set of positive even numbers) and O+ (set of positive odd numbers) and this is the set such that this is commutative monoid that contains cyclic group S2A={E+, O+} and the sum of two even numbers is even numbers, the sum of two odd numbers is also even number and the sum of an even number with odd number is odd number. The behaviour ofS2 can easily be understood if we see that in mod6 w.r.t mod multiplication the subset of Z6 say Z6A={1,2,4}
is commutative monoid that is also commutative orthodox semigroup and the sub- set ofZ6 say Z6B ={2,4} w.r.t mod6 multiplication is cyclic group.
Clearly from the table S3 is only commutative groupoid that does not satisfies associative, left invertive, right invertive, ldd, rdd and medial properties but S3 contains two commutative orthodox semigroups and two cyclic groups. Again we have example in sets if we take setsSSZ ={0},E+, O+, E− and O−.
CASE-6: NON COMMUTATIVE GROUPOID WITH LEFT ZERO ELEMENT THAT CONTAINS SEMILATTICE:
Let we take set T1 = {0} then power set of T1 is P(T1) = {Θ, T1} and if we define binary operation . on P(T) by ∀ A and B ∈ P(T), AB = (AUB)cUA where
U means union and c is for compliment symbol then clearly ΘΘ = Θ, ΘT1 = T1, T1Θ = T1 and T1T1 = T1 but when we have the set T2 = {0,1} then the power set of T2 is P(T2) = {Θ,{0},{1}, T2} and in this case w.r.t the binary opera- tion AB = (AUB)cUA, P(T2) is only groupoid that contains semilattice. We can substitute sets in P(T2) by substituting empty set by O, a = {0}, b = {1} and T2 ={0,1} in Table-16.
Table 16: Table In Case-6
. Θ T1 . Θ a b T2
Θ Θ T1 Θ Θ b b T2
T1 T1 T1 a T2 T2 a a b T2 b T2 b T2 T2 T2 T2 T2
Similarly we can take set T3 = {0,1,2} and T4 = {0,1,2,3} and by taking their power set and defining binary operation.onP(T3) andP(T4) byAB = (AUB)cUA where A and B are the sets and members of power set of sets T3 and T4 this is quite easy that the power set is non commutative groupoid that always contains semilattice of two elements which are empty set Θ and the setTk. where k is some finite positive integer.
This idea can be extended to infinite set TI and surely the power set P(TI) will also be infinite set and we have infinite subsets of setTI so power set also contains infinite members and in this case also TI which is member in its power set is left zero element of its power set P(TI) and the subset ofP(TI) say P(TA) ={Θ, TI} is semilattice.
Case-7 Groupoid Contains Cyclic Group And Non Commutative Or- thodox Semigroup:
First we prove a general theorem of this case “Case-7” which is stated by the following statement:
Theorem 11: If U is groupoid where order of U is finite positive even num- ber i.e. n(U) = m and m is finite positive even number and binary operation ˜b is defined on U in such way thatui is idempotent element and in the table of (U,˜b)
