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  "body": "\n## Proof\n\nWe compute $\\sigma^m$ on the points of each cycle separately.\n\n1. Let $\\sigma \\in S_n$, the [[symmetric group|0RJMNEN3BJ]], and let $\\sigma = c_1 c_2 \\cdots c_r$ be its [[decomposition into disjoint cycles|0YWDAZPRPQ]], where $c_i$ has length $k_i$ and set of moved points $M_i$.\n2. Here $\\operatorname{lcm}(k_1, \\dots, k_r)$ is the least positive integer that each $k_i$ divides, as for the [[least common multiple|09S5SN35DJ]] of two integers. It exists, since $k_1 k_2 \\cdots k_r$ is such an integer.\n3. A [[cycle|0ZKPH2BD1T]] of length $1$ is the identity. Removing such a factor changes neither $\\sigma$ nor the lcm. So we may assume $k_i \\ge 2$, and then $|M_i| = k_i$.\n4. Let $c = (a_0\\ a_1\\ \\dots\\ a_{k-1})$ be a cycle of length $k$. By induction on $m \\ge 0$, $c^m(a_p) = a_q$, where $q$ is the remainder of $p + m$ on division by $k$. Also $c^m$ fixes every other point.\n5. Let $m \\ge 1$. If $k \\mid m$, step 4 gives $c^m(a_p) = a_p$ for each $p$, so $c^m$ is the identity.\n6. Conversely, if $c^m$ is the identity, then $c^m(a_0) = a_0$. By step 4 and since the $a_p$ are distinct, the remainder of $m$ on division by $k$ is $0$. So $k \\mid m$.\n7. Let $z \\in M_i$. If $c_i(c_i(z)) = c_i(z)$, then $c_i(z) = z$ since $c_i$ is [[injective|0PVG1CRAJN]]. This is false, so $c_i(z) \\in M_i$.\n8. For $j \\ne i$, the cycle $c_j$ fixes each point of $M_i$, since the sets of moved points are disjoint.\n9. In $c_1 \\cdots c_r(z)$ with $z \\in M_i$, the factors after $c_i$ fix $z$, then $c_i$ gives $c_i(z) \\in M_i$, and the factors before $c_i$ fix it. So $\\sigma(z) = c_i(z)$.\n10. A point outside all $M_i$ is fixed by each $c_j$, so it is fixed by $\\sigma$.\n11. By steps 7, 9 and 10 and induction on $m \\ge 0$, $\\sigma^m(z) = c_i^m(z)$ for $z \\in M_i$. Also $\\sigma^m$ fixes each point outside all $M_i$.\n12. Let $m \\ge 1$. By step 11, $\\sigma^m$ is the identity exactly when $c_i^m(z) = z$ for each $i$ and each $z \\in M_i$.\n13. Since $c_i^m$ fixes the points outside $M_i$, this holds exactly when each $c_i^m$ is the identity. By steps 5 and 6, this holds exactly when $k_i \\mid m$ for each $i$.\n14. The [[order|0JPYSD5E0B]] of $\\sigma$ is the least $m \\ge 1$ with $\\sigma^m$ the identity. By steps 2 and 13, it is $\\operatorname{lcm}(k_1, \\dots, k_r)$.\n15. If $r = 0$ after step 3, then $\\sigma$ is the identity, of order $1$, and the lcm of an empty list is $1$ by step 2.\n\n## Idea\n\nEach cycle of $\\sigma$ moves its own points, and on those points $\\sigma^m$ acts like the $m$-th power of that cycle. A cycle of length $k$ returns to the identity exactly at the multiples of $k$. So $\\sigma^m$ is the identity exactly when $m$ is a common multiple of all the cycle lengths, and the least such $m$ is the lcm.\n",
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