IIT JAM Exam > IIT JAM Questions > In the permutation group Sn(n 5) , if H is t...
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In the permutation group Sn (n ≥ 5) , if H is the smallest subgroup containing all the 3-cycles, then which one of the following is TRUE?
- a)order of H is 2
- b)Index of H in Sn is 2
- c)H is ablian
- d)H = Sn
Correct answer is option 'B'. Can you explain this answer?
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In the permutation group Sn(n 5) , if H is the smallest subgroup cont...
Explanation:
In the permutation group Sn(n=5), the smallest subgroup containing all the 3-cycles is H. We need to find which of the given options is true for H.
Order of H:
The order of H is the number of elements in the subgroup H. Since H contains all the 3-cycles, the number of 3-cycles in Sn(n=5) is (5 choose 3) = 10. Each 3-cycle contains 3 elements, so the total number of elements in H is 3 x 10 = 30. Therefore, the order of H is 30.
Index of H in Sn(n=5):
The index of H in Sn(n=5) is the number of left cosets of H in Sn(n=5). By Lagrange's theorem, the index of H in Sn(n=5) is [Sn(n=5) : H] = n! / |H|. In this case, n=5 and |H|=30, so the index of H in Sn(n=5) is [Sn(n=5) : H] = 5! / 30 = 20. Therefore, option B is true.
H is abelian:
H is not necessarily abelian. The subgroup generated by all the 3-cycles is not abelian in general. For example, the product of two 3-cycles is a 5-cycle, which is not a 3-cycle, so the product of two elements in H may not be in H.
H = Sn:
H is not equal to Sn(n=5) because H is a proper subgroup of Sn(n=5) and does not contain all the permutations of Sn(n=5).
In the permutation group Sn(n=5), the smallest subgroup containing all the 3-cycles is H. We need to find which of the given options is true for H.
Order of H:
The order of H is the number of elements in the subgroup H. Since H contains all the 3-cycles, the number of 3-cycles in Sn(n=5) is (5 choose 3) = 10. Each 3-cycle contains 3 elements, so the total number of elements in H is 3 x 10 = 30. Therefore, the order of H is 30.
Index of H in Sn(n=5):
The index of H in Sn(n=5) is the number of left cosets of H in Sn(n=5). By Lagrange's theorem, the index of H in Sn(n=5) is [Sn(n=5) : H] = n! / |H|. In this case, n=5 and |H|=30, so the index of H in Sn(n=5) is [Sn(n=5) : H] = 5! / 30 = 20. Therefore, option B is true.
H is abelian:
H is not necessarily abelian. The subgroup generated by all the 3-cycles is not abelian in general. For example, the product of two 3-cycles is a 5-cycle, which is not a 3-cycle, so the product of two elements in H may not be in H.
H = Sn:
H is not equal to Sn(n=5) because H is a proper subgroup of Sn(n=5) and does not contain all the permutations of Sn(n=5).
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In the permutation group Sn(n 5) , if H is the smallest subgroup containing all the 3-cycles, then which one of the following is TRUE? a)order of H is 2b)Index of H in Snis 2c)H is abliand)H =SnCorrect answer is option 'B'. Can you explain this answer?
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