Question:medium

Which of the following statements are true for group of permutations?
A. Every permutation of a finite set can be written as a cycle or a product of disjoint cycles
B. The order of a permutation of a finite set written in a disjoint cycle form is the least common multiple of the lengths of the cycles
C. If \(A_n\) is a group of even permutation of n-symbol (\(n>1\)), then the order of \(A_n\) is n!
D. The pair of disjoint cycles commute

Show Hint

Memorize the key facts about \(S_n\) and \(A_n\):
Any permutation is a product of disjoint cycles.
Order of a permutation is the LCM of its disjoint cycle lengths.
Disjoint cycles always commute.
\(|S_n| = n!\)
\(|A_n| = n!/2\) for \(n \ge 2\). These four facts answer the vast majority of basic questions about permutation groups.
Updated On: Feb 10, 2026
  • A, B and D only
  • A, B and C only
  • A, B, C and D
  • B, C and D only
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Core Idea:
This question assesses knowledge of permutation groups, specifically \(S_n\) and \(A_n\).

Step 2: Breakdown of Statements:
Let's examine each statement:
A. Every permutation of a finite set can be expressed as a cycle or a product of separate cycles. This is the Fundamental Theorem of Permutation Groups. It guarantees a unique decomposition into disjoint cycles (ignoring cycle order). This statement is true.
B. The order of a permutation (in disjoint cycle form) is the least common multiple of its cycle lengths. This is how we calculate the permutation's order. For example, the order of \((1 2 3)(4 5)\) in \(S_5\) is lcm(3, 2) = 6. This statement is true.
C. If \(A_n\) is the group of even permutations of n symbols (\(n>1\)), then the order of \(A_n\) is n!. This statement is false. The symmetric group \(S_n\) has order \(|S_n| = n!\). The alternating group \(A_n\) (even permutations) has order \(|A_n| = \frac{n!}{2}\) for \(n \ge 2\).
D. Disjoint cycles commute. This is a key property. If cycles \(\sigma\) and \(\tau\) are disjoint, then \(\sigma\tau = \tau\sigma\). This is because their actions don't overlap. This statement is true.
Step 3: Conclusion:
Statements A, B, and D are correct, while C is incorrect. The solution is "A, B and D only".
Was this answer helpful?
0