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  "body": "\n## Proof\n\nWe compare $\\sigma\\tau(x)$ and $\\tau\\sigma(x)$ point by point. The argument works for any two permutations with disjoint sets of moved points.\n\n1. Let $\\sigma, \\tau \\in \\operatorname{Sym}(X)$ be [[cycles|0ZKPH2BD1T]], elements of the [[symmetric group|0RJMNEN3BJ]]. The product $\\sigma\\tau$ is the composition, so $\\sigma\\tau(x) = \\sigma(\\tau(x))$.\n2. Let $A = \\{x \\in X : \\sigma(x) \\ne x\\}$ and $B = \\{x \\in X : \\tau(x) \\ne x\\}$ be the sets of moved points. By hypothesis, $A$ and $B$ are [[disjoint|0W4RPA8QH3]].\n3. Let $x \\in A$, and put $y = \\sigma(x)$. If $\\sigma(y) = y$, then $\\sigma(y) = \\sigma(x)$, and $y = x$ since $\\sigma$ is [[injective|0PVG1CRAJN]]. This contradicts $x \\in A$.\n4. So $\\sigma(x) \\in A$ for each $x \\in A$.\n5. Let $x \\in A$. Then $x \\notin B$, so $\\tau(x) = x$ and $\\sigma(\\tau(x)) = \\sigma(x)$.\n6. By step 4, $\\sigma(x) \\in A$, so $\\sigma(x) \\notin B$ and $\\tau(\\sigma(x)) = \\sigma(x)$. With step 5, $\\sigma\\tau(x) = \\tau\\sigma(x)$.\n7. Let $x \\in B$. Exchanging the roles of $\\sigma$ and $\\tau$ in steps 3 to 6 gives $\\tau\\sigma(x) = \\tau(x) = \\sigma\\tau(x)$.\n8. Let $x \\notin A \\cup B$. Then $\\sigma(x) = x$ and $\\tau(x) = x$, so $\\sigma\\tau(x) = x = \\tau\\sigma(x)$.\n9. By steps 6 to 8, $\\sigma\\tau(x) = \\tau\\sigma(x)$ for every $x \\in X$. Hence $\\sigma\\tau = \\tau\\sigma$.\n\n## Idea\n\nA permutation sends its moved points to moved points, so it never sends a point into the other cycle's territory. On the points of $\\sigma$ only $\\sigma$ acts, on the points of $\\tau$ only $\\tau$ acts, and all other points stay put. So the order of the two factors does not matter.\n",
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