[ Pobierz całość w formacie PDF ]

a1 a b1 b2
ëø öø ëø öø
2
A = ìø
ìøa a ÷ø and B = ìø ÷ø
÷ø ìøb b4 ÷ø ,
íø 3 4 øø íø 3 øø
the operation 'o' defined on M2×2 is
a1 a2 b1 b2
ëø öø ëø öø
A o B = ìø ÷ø ìø ÷ø
o
ìøa a4 ÷ø ìøb b4 ÷ø
íø 3 øø íø 3 øø
a1b3 + a2b1(mod 2) a1b4 + a2b2(mod 2)
îø ùø
A o B =
ïøa b3 +a4b1 (mod2) a3b4 +a4b2 (mod 2)úø .
ðø 3 ûø
Clearly (M2×2, o) is a groupoid. It is left for the reader to verify ' o ' is non-associative on
M2×2. M2×2 is a SG for the set
ñø 0 0 1 0 üø
ëø öø ëø öø
A =
òøìø ÷ø,ìø ÷ø, oýø ,
ìø0 0 ÷ø ìø0 0÷ø
íø øø íø øø
óø þø
is a semigroup as
1 0 1 0 0 0
ëø öø ëø öø ëø öø
ìø ÷ø o ìø ÷ø = ìø ÷ø .
ìø0 0÷ø ìø0 0÷ø ìø0 0÷ø
íø øø íø øø íø øø
Now consider
ñø 0 0 1 0 0 1 1 1 üø
ëø öø ëø öø ëø öø ëø öø
A1 =
òøìø ÷ø,ìø ÷ø,ìø ÷ø,ìø ÷ø,'o'ýø,
ìø0 0 ÷ø ìø0 0÷ø ìø0 0÷ø ìø0 0÷ø
íø øø íø øø íø øø íø øø
óø þø
is a Smarandache subgroupoid. But A1 is non-commutative Smarandache subgroupoid for A1
contains the non-commutative semigroup S.
54
ñø 0 0 1 1 1 0 üø
ëø öø ëø öø ëø öø
S =
òøìø ÷ø,ìø ÷ø,ìø ÷ø,oýø
ìø0 0÷ø ìø0 0÷ø ìø0 0÷ø
íø øø íø øø íø øø
óø þø
such that
1 0 1 1 0 0
ëø öø ëø öø ëø öø
ìø ÷ø o ìø ÷ø = ìø ÷ø
ìø0 0÷ø ìø0 0÷ø ìø0 0÷ø
íø øø íø øø íø øø
and
1 1 1 0 1 0
ëø öø ëø öø ëø öø
ìø ÷ø ìø ÷ø = ìø ÷ø .
ìø0 0÷ø ìø0 0÷ø ìø0 0÷ø
íø øø íø øø íø øø
So (M2×2, o) is a Smarandache commutative groupoid but not a Smarandache inner
commutative groupoid.
PROBLEM 1: Is Z20 (3, 8) a SG? Does Z20 (3, 8) have semigroups of order 5?
PROBLEM 2: Find all the subsets, which are semigroups of the groupoid Z15 (7, 3).
PROBLEM 3: Can Z11 (7, 4) have subsets, which are commutative semigroups?
PROBLEM 4: Find Smarandache right ideals of Z15 (7, 8).
PROBLEM 5: Does Z15 (7, 8) have Smarandache left ideals and Smarandache ideals?
PROBLEM 6: Find all Smarandache subgroupoids of Z11 (3, 8).
PROBLEM 7: Find two Smarandache semiconjugate subgroupoids of Z24 (3, 11).
PROBLEM 8: Find two Smarandache conjugate subgroupoids of Z30 (5, 27).
PROBLEM 9: Does Z22 (9, 11) have Smarandache ideals?
PROBLEM 10: Find Smarandache seminormal groupoids of the groupoid Z22 (8, 13).
PROBLEM 11: Can Z22 (8, 15) have Smarandache normal groupoids?
PROBLEM 12: Find all the Smarandache seminormal groupoids which are Smarandache
semiconjugate in any groupoid G.
PROBLEM 13: Does there exist SGs in which all Smarandache normal subgroupoids are
a) Smarandache semiconjugate with each other
b) Smarandache conjugate with each other?
55
PROBLEM 14: Is Z18 (3, 14) Smarandache commutative? Justify.
PROBLEM 15: Can the groupoid Z19 (4, 11) be Smarandache inner commutative? Justify your
answer.
4.3 Identities in Smarandache Groupoids
In this section we introduce some identities in SGs called Bol identity and Moufang
identity, P-identity and alternative identity and illustrate them with examples.
DEFINITION: A SG (G, ") is said to be a Smarandache Moufang groupoid if there exist H ‚"
G such that H is a Smarandache subgroupoid satisfying the Moufang identity (xy)(zx) = (x
(yz)) x for all x, y, z in H.
DEFINITION: Let S be a SG. If every Smarandache subgroupoid H of S satisfies the Moufang
identity for all x, y, z in H then S is a Smarandache strong Moufang groupoid.
Example 4.3.1: Let G = Z10 (5, 6) be a groupoid given by the following table:
0 1 2 3 4 5 6 7 8 9
"
0 0 6 2 8 4 0 6 2 8 4
1 5 1 7 3 9 5 1 7 3 9
2 0 6 2 8 4 0 6 2 8 4
3 5 1 7 3 9 5 1 7 3 9
4 0 6 2 8 4 0 6 2 8 4
5 5 1 7 3 9 5 1 7 3 9
6 0 6 2 8 4 0 6 2 8 4
7 5 1 7 3 9 5 1 7 3 9
8 0 6 2 8 4 0 6 2 8 4
9 5 1 7 3 9 5 1 7 3 9
Clearly, Z10 (5, 6) is a SG as S = {2} is a semi-group. It can be verified Z10 (5, 6) is a
Smarandache strong Moufang groupoid as the Moufang identity (x " y) " (z " x) = [x " (y "
z)] " x is true for
(x " y) " (z " x) = (5x + 6y) " (5z + 6x)
= 5(5x + 6y) + 6(5z + 6x)
= x.
Now [x " (y " z)] " x = (x " [5y + 6z]) " x
= [5x + 6(5y + 6z)] " x
= (5x + 6z) " x
= 5(5x + 6z) + 6x
56
= 5x + 6x
= x.
Since x, y, z are arbitrary elements from Z10 (5,6) we have this groupoid to be a
Smarandache strong Moufang groupoid. Since every Smarandache, subgroupoid of G will
also satisfy the Moufang identity.
In view of this example, we give the following nice theorem.
THEOREM 4.3.1: Let S be a SG if the Moufang identity (x " y) " (z " x) = (x " (y " z)) " x is
true for all x, y, z " S then S is a Smarandache strong Moufang groupoid.
Proof: If (x " y) " (z " x) = (x " (y " z)) " x for all x, y, z " S then clearly we see every
Smarandache subgroupoid of S satisfies the Moufang identity hence S is a Smarandache
strong Moufang groupoid.
Now we see an example of a groupoid, which is only a Smarandache Moufang
groupoid, and not a Smarandache strong Moufang groupoid.
Example 4.3.2: Let G = Z12 (3, 9) be a groupoid given by the following table:
0 1 2 3 4 5 6 7 8 9 10 11
"
0 0 9 6 3 0 9 6 3 0 9 6 3
1 3 0 9 6 3 0 9 6 3 0 9 6
2 6 3 0 9 6 3 0 9 6 3 0 9
3 9 6 3 0 9 6 3 0 9 6 3 0
4 0 9 6 3 0 9 6 3 0 9 6 3
5 3 0 9 6 3 0 9 6 3 0 9 6
6 6 3 0 9 6 3 0 9 6 3 0 9
7 9 6 3 0 9 6 3 0 9 6 3 0
8 0 9 6 3 0 9 6 3 0 9 6 3
9 3 0 9 6 3 0 9 6 3 0 9 6
10 6 3 0 9 6 3 0 9 6 3 0 9
11 9 6 3 0 9 6 3 0 9 6 3 0
Clearly, A1 = {0, 4, 8} and A2 = {0, 3, 6, 9} are Smarandache subgroupoids of G as
P1 = {0, 4} ‚" A1 is a semigroup and P2 = {0, 6} ‚" A2 is a semigroup of G. A2 does not satisfy
Moufang identity but A1 satisfies the Moufang identity. So G is only a Smarandache
Moufang groupoid but not a Smarandache strong Moufang groupoid.
THEOREM 4.3.2: Every Smarandache strong Moufang groupoid is a Smarandache Moufang
groupoid and not conversely.
Proof: By the very definition of the Smarandache strong Moufang groupoid and the
Smarandache Moufang groupoid it is clear that every Smarandache strong Moufang groupoid
is a Smarandache Moufang groupoid.
57
To prove the converse, consider the example 4.3.2 clearly that groupoid is proved to
be Smarandache Moufang groupoid but it is not Smarandache strong Moufang groupoid for
x, y, z " Z12 (3, 9) we see (x " y) " (z " x) `" (x " (y " z)) " x for
(x " y) " (z " x) = 6x + 3y + 3z (mod 12)
Now (x " (y " z)) " x = 6x + 9y + 3z (mod 12).
Thus, Z12 (3, 9) satisfies the Moufang identity for the subgroupoid {0, 6}.
Example 4.3.3: Let G = Z12 (3, 4) be a groupoid given by the following table:
0 1 2 3 4 5 6 7 8 9 10 11
"
0 0 4 8 0 4 8 0 4 8 0 4 8
1 3 7 11 3 7 11 3 7 11 3 7 11
2 6 10 2 6 10 2 6 10 2 6 10 2
3 9 1 5 9 1 5 9 1 5 9 1 5
4 0 4 8 0 4 8 0 4 8 0 4 8
5 3 7 11 3 7 11 3 7 11 3 7 11
6 6 10 2 6 10 2 6 10 2 6 10 2
7 9 1 5 9 1 5 9 1 5 9 1 5
8 0 4 8 0 4 8 0 4 8 0 4 8
9 3 7 11 3 7 11 3 7 11 3 7 11
10 6 10 2 6 10 2 6 10 2 6 10 2
11 9 1 5 9 1 5 9 1 5 9 1 5
This groupoid is a SG as it contains semigroups viz. {2}, {4} and {10}. Now Z12 (3,
4) is a Smarandache strong Bol groupoid for ((x " y) " z) " y = x " [(y " z) " y] for all x, y, z
" Z12 (3, 4).
Consider,
((x " y) "z) " y = [(3x + 4y) " z] " y
= [9x + 12y + 4z] " y
= [9x + 4z] " y
= 27x + 12z + 4y
= 3x + 4y.
Now
x " [(y " z) " y] = x " [(3y + 4z) " y]
= x " [9y + 12z + 4y]
= x " y
= 3x + 4y.
Thus Z12 (3, 4) is a Smarandache strong Bol groupoid as every triple satisfies the Bol
identity so every subgroupoid of Z12 (3, 4) which is a Smarandache subgroupoid of G will
satisfy the Bol identity.
58
DEFINITION: Let G be a groupoid, G is said to be a Smarandache Bol groupoid if G has a [ Pobierz całość w formacie PDF ]