Step 1: Understanding the Concept:
This question tests fundamental theorems and properties of permutation groups, specifically the symmetric group \(S_n\) and the alternating group \(A_n\).
Step 2: Detailed Explanation:
Let's analyze each statement:
A. Every permutation of a finite set can be written as a cycle or a product of disjoint cycles.
This is the Fundamental Theorem of Permutation Groups. It states that any permutation has a unique decomposition into disjoint cycles (up to the order of the cycles). This statement is true.
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.
This is the standard method for calculating the order of a permutation. 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 a group of even permutation of n-symbol (\(n>1\)), then the order of \(A_n\) is n!.
This statement is false. The group of all permutations on n symbols is the symmetric group \(S_n\), and its order is \(|S_n| = n!\). The alternating group \(A_n\) is the subgroup of \(S_n\) containing all even permutations. For \(n \ge 2\), exactly half of the permutations are even and half are odd. Therefore, the order of \(A_n\) is \(|A_n| = \frac{n!}{2}\).
D. The pair of disjoint cycles commute.
This is a key property of disjoint cycles. If cycles \(\sigma\) and \(\tau\) are disjoint (meaning they move different sets of elements), then \(\sigma\tau = \tau\sigma\). This is because the action of \(\sigma\) does not affect the elements moved by \(\tau\), and vice versa. This statement is true.
Step 3: Final Answer:
Statements A, B, and D are true, while C is false. Therefore, the correct option is "A, B and D only".