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Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.
For n > 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), then
For n > 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), then
- a)H = Sn
- b)H is abelian
- c)the index of H in Sn is 2
- d)H is cyclic
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Let Sn be the group of all permutations on the set {1, 2, ..., n} unde...
**Explanation:**
To prove that option 'A' is the correct answer, we need to show that H = Sn, i.e., H is equal to the group of all permutations on the set {1, 2, ..., n}.
**Proof:**
We are given that H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n).
Let σ be any permutation in Sn. We need to show that σ is in H.
Consider the permutation σ(1, 2)σ^(-1). This permutation takes 1 to σ(1) and 2 to σ(2). In other words, it swaps the positions of 1 and σ(1), and also swaps the positions of 2 and σ(2).
Now, consider the permutation σ(1, 2, ..., n)σ^(-1). This permutation takes 1 to σ(1), 2 to σ(2), and so on, up to n to σ(n). In other words, it cycles the positions of 1, 2, ..., n according to the permutation σ.
Since H is the smallest subgroup containing (1, 2) and (1, 2, ..., n), it must contain the permutations σ(1, 2)σ^(-1) and σ(1, 2, ..., n)σ^(-1).
Now, we can write σ as a product of these two permutations:
σ = (σ(1, 2)σ^(-1))(σ(1, 2, ..., n)σ^(-1))
Note that (σ(1, 2)σ^(-1))(σ(1, 2, ..., n)σ^(-1)) is a composition of two permutations, and therefore, it is also a permutation. This means that σ is in H.
Since σ was an arbitrary permutation in Sn, this shows that every permutation in Sn is in H. Therefore, H = Sn.
Hence, the correct answer is option 'A'.
To prove that option 'A' is the correct answer, we need to show that H = Sn, i.e., H is equal to the group of all permutations on the set {1, 2, ..., n}.
**Proof:**
We are given that H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n).
Let σ be any permutation in Sn. We need to show that σ is in H.
Consider the permutation σ(1, 2)σ^(-1). This permutation takes 1 to σ(1) and 2 to σ(2). In other words, it swaps the positions of 1 and σ(1), and also swaps the positions of 2 and σ(2).
Now, consider the permutation σ(1, 2, ..., n)σ^(-1). This permutation takes 1 to σ(1), 2 to σ(2), and so on, up to n to σ(n). In other words, it cycles the positions of 1, 2, ..., n according to the permutation σ.
Since H is the smallest subgroup containing (1, 2) and (1, 2, ..., n), it must contain the permutations σ(1, 2)σ^(-1) and σ(1, 2, ..., n)σ^(-1).
Now, we can write σ as a product of these two permutations:
σ = (σ(1, 2)σ^(-1))(σ(1, 2, ..., n)σ^(-1))
Note that (σ(1, 2)σ^(-1))(σ(1, 2, ..., n)σ^(-1)) is a composition of two permutations, and therefore, it is also a permutation. This means that σ is in H.
Since σ was an arbitrary permutation in Sn, this shows that every permutation in Sn is in H. Therefore, H = Sn.
Hence, the correct answer is option 'A'.
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Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer?
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Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer?.
Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Sn be the group of all permutations on the set {1, 2, ..., n} under the composition of mappings.For n 2, if H is the smallest subgroup of Sn containing the transposition (1, 2) and the cycle (1, 2, ..., n), thena)H = Snb)H is abelianc)the index of H in Sn is 2d)H is cyclicCorrect answer is option 'A'. Can you explain this answer?.
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