Abstract Algebra: Permutation Groups Quiz

1.

PLEASE READ CAREFULLY.1.There are 5 Multi-SELECT Questions, in which one or MORE options are correct.2. There is no NEGATIVE marking.

2.

Let S_n be the symmetric group on a finite set with n symbols. Then which of the following is(are) TRUE?

A.

Sn contains a subgroup of order m, where 1 ≤ m ≤ n.
B.

Sn contains a subgroup of order (n -1)!.
C.

Sn is non-Abelian for all n ∈ N.
D.

There exists a cycle of order m > 1, which belong to Z(Sn).

Correct Answer(s)
A. Sn contains a subgroup of order m, where 1 ≤ m ≤ n.
B. Sn contains a subgroup of order (n -1)!.

3.

Which of the following is(are) CORRECT?

A.

There exist three transpositions whose product is the identity e of the symmetric group Sn.
B.

The order of the product of two transpositions is either 2 or 3.
C.

If α and β are two disjoint permutations such that αβ = e, then α = β = e.
D.

Every permutation in Sn can be expressed as a product of disjoint cycles.

Correct Answer(s)
B. The order of the product of two transpositions is either 2 or 3.
C. If α and β are two disjoint permutations such that αβ = e, then α = β = e.
D. Every permutation in Sn can be expressed as a product of disjoint cycles.

Explanation
The order of the product of two transpositions is either 2 or 3 because a transposition is a permutation that swaps two elements, so when you multiply two transpositions, it results in either swapping the same two elements twice (order 2) or swapping three elements cyclically (order 3).

If α and β are two disjoint permutations such that αβ = e, then α = β = e because the identity permutation e is the only permutation that satisfies the property of being able to cancel out any other permutation when multiplied.

Every permutation in Sn can be expressed as a product of disjoint cycles because a cycle is a permutation that cyclically permutes a set of elements, and any permutation can be decomposed into a product of disjoint cycles.

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4.

Which among the following are TRUE?

A.

(1 2)(1 3)(1 4)(2 5) is an even permutation.
B.

An odd permutation is even.
C.

(1 2 3 . . . n) is an odd permutation, if n is odd.
D.

There exists a subgroup of Sn with index 2.

Correct Answer(s)
A. (1 2)(1 3)(1 4)(2 5) is an even permutation.
B. An odd permutation is even.
D. There exists a subgroup of Sn with index 2.

Explanation
The given answer is correct.

1. (1 2)(1 3)(1 4)(2 5) is an even permutation because it can be expressed as the product of an even number of transpositions.

2. An odd permutation is even. This is a known property of permutations. The parity (even or odd) of a permutation is determined by the number of transpositions required to express it. An odd permutation is one that requires an odd number of transpositions, which means it is also an even permutation.

3. (1 2 3 . . . n) is an odd permutation if n is odd. This is also a known property of permutations. A cyclic permutation of length n is odd if n is odd and even if n is even.

4. There exists a subgroup of Sn with index 2. This is true because the alternating group, denoted by An, is a subgroup of Sn and has index 2. The alternating group consists of all even permutations in Sn.

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5.

Which of the following is(are) CORRECT? 1. There exists a permutation α in S_n,such that α(1 2)α^-1 = (1 2 3).2. There exists NO permutation α in S_n, such that α(1 2 3)α^-1 = (4 5 6).3. If α∈ S_nand |α| = m, then for any permutation we have | βαβ^-1| = m.4. A cycle which is conjugate with (1 2 3 . . . n) is an n-cycle.

A.

Option1
B.

Option2
C.

Option3
D.

Option4

Correct Answer(s)
C. Option3
D. Option4

Explanation
Option 3 states that for any permutation α in Sn with |α| = m, the order of the permutation βαβ^-1 is also m. This is a correct statement because the order of a conjugate permutation is equal to the order of the original permutation.

Option 4 states that a cycle which is conjugate with (1 2 3 ... n) is an n-cycle. This is also a correct statement because conjugating a cycle with (1 2 3 ... n) does not change the cycle structure, so it remains an n-cycle.

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6.

Which of the following is(are) correct?

A.

A subgroup H in S4, generated by (1 2 3) and (1 2) is of order 6.
B.

The largest possible order of elements of the alternating group A5 is 10.
C.

Each element in the alternating group A4 can be written as a product of 3-cycles,
D.

There are exactly n!/2 odd permutations in Sn.

Correct Answer(s)
A. A subgroup H in S4, generated by (1 2 3) and (1 2) is of order 6.
C. Each element in the alternating group A4 can be written as a product of 3-cycles,
D. There are exactly n!/2 odd permutations in Sn.

Explanation
The given answer is correct because a subgroup generated by two elements in S4 has the order equal to the least common multiple of the orders of the generating elements. In this case, the order of (1 2 3) is 3 and the order of (1 2) is 2. The least common multiple of 3 and 2 is 6, so the subgroup H has an order of 6. Additionally, each element in the alternating group A4 can be written as a product of 3-cycles, and there are exactly n!/2 odd permutations in Sn.

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Abstract Algebra: Permutation Groups Quiz

PLEASE READ CAREFULLY.1.There are 5 Multi-SELECT Questions, in which one or MORE options are correct.2. There is no NEGATIVE marking.

Let Sn be the symmetric group on a finite set with n symbols. Then which of the following is(are) TRUE?

Which of the following is(are) CORRECT?

Which among the following are TRUE?

Which of the following is(are) correct?

Let S_n be the symmetric group on a finite set with n symbols. Then which of the following is(are) TRUE?