Discussion
Let \(\sigma\) be a permutation in \(S_{n}\). Let \(\sigma = \sigma_{1}, \sigma_{2} \cdots \sigma_{k}\), where \(\sigma_{1}, \sigma_{2}, \ldots, \sigma_{k}\) are disjoint cycles of lengths \(l_{1}, l_{2}, \ldots, l_{k}\) respectively. Then the order of \(\sigma\) is equal to :
Answer: D
The order is \(lcm(l_{1}, \ldots, l_{k})\).
Disjoint cycles commute, so \(\sigma^{m} = \sigma_{1}^{m}\sigma_{2}^{m}\cdots\sigma_{k}^{m}\), and since they move different points, \(\sigma^{m}\) is the identity only if each \(\sigma_{i}^{m}\) is. A cycle of length \(l_{i}\) has order \(l_{i}\), so \(m\) must be a multiple of every \(l_{i}\); the least such \(m\) is the lcm.
For example, \((1\,2)(3\,4\,5)\) has order 6, which is not the sum 5 or the gcd 1; and \((1\,2)(3\,4)\) has order 2, not the product 4.
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