Kandasamy W. B. V. Smarandache Loops

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introduce a notation for S-subloops in L (m). When we say L (m) has subloops we
n n
assume n is an odd non-prime for otherwise it will have no subloops.
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Notation: Let L (m) be a loop, n an odd number which is not a prime for every t/n
n
there exist subloops of order k + 1 where k = n/t. So if t/n we denote the subloop by
Hi(t) ={e, i, i + t, & , i + (k t)t}, it is a S-subloop of L (m) for every i d" t.
n
Example 3.7.3: Now L9(8) " L9 the number which divides 9 is 3. So we have the
subloop
H1(t) = {e, 1, 4, 7}
H2(t) = {e, 2, 5, 8}
H3(t) = {e, 3, 6, 9}
The tables for them are as follows
Table for H1(t)
e 1 4 7
e e 1 4 7
1 1 e 7 4
4 4 7 e 1
7 7 4 1 e
Table for H2(t)
e 2 5 8
e e 2 5 8
2 2 e 8 5
5 5 8 e 2
8 8 5 2 e
Table for H3(t)
e 3 6 9
e e 3 6 9
3 3 e 9 6
6 6 9 e 3
9 9 6 3 e
Thus all subloops in L9(8) are abelian groups. So this is also a Smarandache
Hamiltonian loop as every subloop is an abelian group of order 4.
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Now we obtain a condition for the normalizers to be coincident in the loops L (m) "
n
L (n not a prime).
n
THEOREM 3.7.1: Let L (m) " L and Hi(t) be its S-subloop, then SN1(Hi(t)) =
n n
SN2(Hi(t)) if and only if (m2 m + 1, t) = (2m 1, t).
Proof: Let L (m) " L and Hi(t) be a S-subloop of L (m) first we show the first S-
n n n
normalizer SN1(Hi(t)) = Hi(k) where k = t/d and d = (2m 1, t) we use only simple
number theoretic arguments and the definition. SN1(Hi(t)) = {j " L (m)/ jHi(t) =
n
Hi(t)j} is the S-first normalizer of Hi(t). It is left for the reader to verify jHi(t) = Hi(t)j
if and only if (2m 1) (i j) a" 0(mod t) for j " Hi(t). More over for j " Hi(t) we
have jHi(t) = Hi(t)j.
So the possible values of j in L (m) for which jHi(t) = Hi(t)j are given by e, i, i + k, i
n
+ 2k, & , i + ((n/k) i) k. Thus SN1(Hi(t)) = Hi(k) using this fact we see k = t/d, d
= (2m 1, t). Now as we have worked for Smarandache first normalizer we can
work for Smarandache second normalizer and show that SN2(Hi(t)) = Hi(k) here k =
t/d and d = (m2 m + 1, t). This working is left as an exercise to the reader. Using
these two facts we can say SN1(Hi(t)) = SN2(Hi(t)) if and only if (m2 m+1, t) = ((2m
1), t). Hence the claim.
THEOREM 3.7.2: Let L be a S-loop which has no S-subloops. Then
1. S(N�) = N�
2. S(N�) = N�
3. S(N�) = N�
4. SN(L) = N(L)
5. SC(L) = C(L)
6. SZ(L) = Z(L)
Proof: The proof is obvious by the definitions of all these concepts and their
corresponding Smarandache definitions.
THEOREM 3.7.3: Let L (m) " L where n is a prime then SN(L (m)) = {e}.
n n n
Proof: Now we know when n is a prime. L (m) " L has no S-subloops but are S-
n n
loops. So we have N(L (m)) = {e} as we know SN(L (m)) = N(L (m)). Hence we
n n n
have SN� (L (m)) = N�(L (m)), SN�(L (m)) = N�(L (m)) SN�(L (m)) = N�(L (m)).
n n n n n n
Now N(L) = N� )" N� )" N�.
It is enough if we prove N�(L (m)) = {e} then it would imply N(L (m)) = {e}. So we
n n
shall prove N�(L (m)) = {e}. If x `" e " N�(L (m)), choose elements y, z " L (m)
n n n
such that z `" x, xy `" z and yz `" x. Now x " N�(L (m)) implies (x, y, z) = e which is
n
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possible only when x = z. However e " N�(L (m)) as (e, x, y) = e for all x, y, "
n
L (m). Hence the theorem.
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THEOREM 3.7.4: Let L (m) " L , n a prime, then S-Moufang centre of L (m) is
n n n
either {e} or L (m).
n
Proof: Using the fact SC(L (m)) = C(L (m)) and e " L (m) we see if e `" i "
n n n
C(L (m)) then i " j = j " i for all j " L (m). i " j = j " i implies (2m 1) (i j) a"
n n
0(mod n) choose j such that (|j i|, n) = 1 which implies (2m 1) a" zero(mod n).
Hence L (m) is commutative in which case C(L (m)) = SC(L (m)) = L (m). If L (m)
n n n n n
is non-commutative C(L (m)) = {e} = SC(L (m)).
n n
THEOREM 3.7.5: Let L (m) " L where n is a prime. Then NZ(L (m)) =
n n n
Z(L (m)) = {e}.
n
Proof: If n is a prime we have NZ(L (m)) = Z(L (m)). So it is enough if we prove
n n
Z(L (m)) = {e}.
n
We know SZ(L) = S(NCL) )" SC(L) = N(L) )" C(L) (if n is a prime) = {e} by our
earlier results.
PROBLEMS:
1. Find all S-Moufang centres for the loop L51(11).
2. Find all S-centres for the loop L55(13).
3. Find Z(L27(5)) and NZ(Z25(5)). Which of them is the larger subloop?
4. Find SN�, SN� and SN� for all S-subloops in L57(17).
5. For problem 4 find SN(L57(17)).
6. Find SN1 and SN2 for all subloops in SN(L57(29)).
7. If P and Q be subloops in a loop L such that P �" Q what can you say about
i. SN�(P) and SN�(Q).
ii. SZ(P) and SZ(Q)
iii. SN1(P) and SN1(Q) and
iv. SN2(P) and SN2(Q).
8. For all S-subloops in L35(9) find SN1,SN2, SZ, SC and SN.
9. For what S-subloops A in L. SN1 = SC = SN2 = SZ = SN?
10. Verify problem 9 in case of S-subloops in L15(8).
3.8. Smarandache mixed direct product of loops
In this section we define a new notion called Smarandache mixed direct product of
loops and prove these loops got as Smarandache mixed direct product are S-loops.
We define S-loop II. Using the definition of Smarandache mixed product we are able
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to get the Smarandache Cauchy theorem for S-loops. Further in this section we extend
two classical concepts, call them as Smarandache Lagrange criteria and Smarandache
Sylow criteria. These are substantiated by examples. In fact for each of these concepts
we have a class of loops, which satisfies it. Finally we give several problems to the
reader as exercise to solve. As solving of these problems alone can make the reader
get a deeper understanding of S-loops.
DEFINITION 3.8.1: Let L = L1 � Sn be the direct product of a loop and a group.
We call this the Smarandache mixed direct product of loops (S-mixed direct [ Pobierz całość w formacie PDF ]

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