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and S-Lagranges criteria?
4. Will it be possible for all S-mixed direct product loops to satisfy S-Lagrange
criteria? Justify your answer.
5. Let L = L13(2) × G (where G = )#g / g7 = 1*# is the S-mixed direct product of
loops. Does L satisfy S-Lagranges criteria? Will L satisfy S-Sylow criteria?
6. Give an example of a S-loop II of order 26.
7. Give an example of a S-loop of order 25 which is not a S-loop II.
8. Can we say all prime order loops are not S-loops?
9. Is every S-mixed direct product of loop a S-loop II?
3.9. Smarandache cosets in loops
In this section we study and introduce the Smarandache coset representation in loops
when the loop has a subgroup i.e. only when L is a S-loop. This section defines
Smarandache right (left) coset of H in L. Several interesting examples are given like
cosets in a group, Smarandache cosets does not in general partition the loop or have
some nice representation. All these are explained by self-illustrative examples.
This problem leads to the definition of Smarandache right (left) coset equivalence
sets in a loop L related to a subgroup A of L. In chapter 5 several problems are given
for the interested reader to develop an interest and carry out research on the
Smarandache cosets of loops.
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DEFINITION 3.9.1: Let L be a S-loop. We define Smarandache right cosets (S-
right cosets) in L as follows:
Let A ‚" L be the subgroup of L and for m " L we have Am = {am/a " A}. Am is
called a S-right coset of A in L.
Similarly we can for any subgroup A of L define Smarandache left coset (S-left
coset) for some m " L as mA = {ma / a " A}. If the loop L is commutative then
only we have mA = Am. Even if L is S-commutative or S-strongly commutative
still we need not have Am = mA for all m " L.
Example 3.9.1: Let L5(2) be the non-commutative loop given by the following table:
e 1 2 3 4 5
"
e e 1 2 3 4 5
1 1 e 3 5 2 4
2 2 5 e 4 1 3
3 3 4 1 e 5 2
4 4 3 5 2 e 1
5 5 2 4 1 3 e
Let A = {e, 1} be the subgroup of the S-loop L5(2). The S-right coset of A is
A " 1 = {e, 1} A " 2 = {2, 3}
A " 3 = {3, 5} A " 4 = {4, 2}
A " 5 = {5, 4}
Thus the S-right cosets do not partition L5(2).
Consider the S-left cosets of A
1 " A = {e, 1} 2 " A = {5, 2}
3 " A = {3, 4} 4 " A = {4, 3}
5 " A = {5,2}
But the S-left cosets has partitioned L5(2). Thus we cannot make any comment about
the partition.
Example 3.9.2: Let L7(4) be the commutative loop given by the following table
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e 1 2 3 4 5 6 7
"
e e 1 2 3 4 5 6 7
1 1 e 5 2 6 3 7 4
2 2 5 e 6 3 7 4 1
3 3 2 6 e 7 4 1 5
4 4 6 3 7 e 1 5 2
5 5 3 7 4 1 e 2 6
6 6 7 4 1 5 2 e 3
7 7 4 1 5 2 6 3 e
Let A = {e, 5}, consider S-right coset of A.
A " 1 = {1, 3} A " 2 = {2, 7}
A " 3 = {3, 4} A " 4 = {4, 1}
A " 5 = {5, 1} A " 6 = {6, 2}
A " 7 = {7, 6}
Clearly A does not partition L5(2).
Take A = {e, 5} consider S-left coset of A
1 " A = {1, 3} 2 " A = {2, 7}
3 " A = {3, 4} 4 " A = {4, 1}
5 " A = {5, 1} 6 " A = {6, 2}
7 " A = {6, 7}
Clearly S-cosets do not get partitioned in this case. But mA = Am for all m " L7(4) as
L7(4) is a commutative loop.
Take B = {e, 4}
B " 1 = {1, 6} B " 2 = {2, 3}
B " 3 = {3, 7} B " 5 = {5, 1}
B " 6 = {6, 5} B " 7 = {7, 2}
This also does not partition L7(4). Clearly Bm = mB. Just we have seen for the new
class of loops; the S-right coset and S-left coset when the loop is non-commutative
and the loop is commutative.
In both these loops we do not have subgroups of other order. But one of the
important observations which is made by these problems are:
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We have a set of elements in L5(2) such that the set with a specified subgroup A gives
coset decomposition in a disjoint way which is a partition of L5(2). For example in the
loop L5(2) we have for the subgroup A, the elements {2, 5} and {3, 4} in L5(2) are
such that we get L5(2) = A *" A " 2 *" A " 5 i.e. {1. e} *" {2, 3} *" {4. 5} = L5(2)
which is a disjoint union. Further using the set of elements {3, 4} we see L5(2) = A *"
A " 3 *" A " 4 = {1, e} *" {3, 5} *" {4, 2}.
So unlike in a group we are able to divide the loop into equivalence classes not using
all elements but at the same time we cannot say the two sets partition in the same way.
This is possible only when the loop is a non-commutative one for the loop L7(4)
which is a commutative loop we are not able to get any such relation. So one more
relevant question would be should n of the loop L (m) be a prime number? We will
n
first illustrate examples before we try to answer these questions.
Example 3.9.3: Consider the non-commutative loop L7(3) given by the following
table:
e 1 2 3 4 5 6 7
"
e e 1 2 3 4 5 6 7
1 1 e 4 7 5 6 2 5
2 2 6 e 5 1 4 7 3
3 3 4 7 e 6 2 5 1
4 4 2 5 1 e 7 3 6
5 5 7 3 6 2 e 1 4
6 6 5 1 4 7 3 e 2
7 7 3 6 2 5 1 4 e
Let A1 = {1, e}, 2A1 = {6, 2}, 3A1 = {3, 4}, 4A1 = {4, 2}, 5A1 = {5, 7}, 6A1 = {6, 5}
and 7A1 = {7, 3}.
S-right coset decomposition by A1 is A1 = {1, e}, A12 = {2, 4}, A13 = {3,7}, A14 = {4,
3}, A15 = {5, 6}, A16 = {6, 2} and A17 = {7, 5}. The set {2, 3, 5} and {4, 6, 7} are
such that L7(3) = A1 *" {2, 4} *" {3, 7} *" {5,6} = A1 *" A12 *" A13 *" A15. Similarly
L7(3) = A1 *" A14 *" A16 *" A17 = {e, 1} *" {4, 3} *" {6, 2} *" {7, 5}.
S-left coset decomposition by A1 is A1= {1, e}, 2A1 = {6, 3}, 3A1 = {3, 4}, 4A1 = {4,
2}, 5A1 ={5, 7}, 6A1 = {6, 5} and 7A1 = {7, 3}. Here also the set {2, 3, 5} and {4, 6,
7} are such that L7(3) = A1 *" 2A1 *" 3A1 *" 5A1 = {e, 1} *" {2, 6} *" {3, 4} *" {5, 7}.
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L7(3) = A1 *" 4A1 *"6A1 *" 7A1 = {e, 1} *" {4, 2} *" {6, 5} *" {3, 7}. The chief thing
to be noticed is that for the same subgroup the sets associated with the S-left coset
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