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  text: '[summary: In a symmetric group, if we are applying a collection of permutations which are each disjoint cycles, we get the same result no matter the order in which we perform the cycles.]\n\nConsider two [49f cycles] $(a_1 a_2 \\dots a_k)$ and $(b_1 b_2 \\dots b_m)$ in the [-497] $S_n$, where all the $a_i, b_j$ are distinct.\n\nThen it is the case that the following two elements of $S_n$ are equal:\n\n- $\\sigma$, which is obtained by first performing the permutation notated by $(a_1 a_2 \\dots a_k)$ and then by performing the permutation notated by $(b_1 b_2 \\dots b_m)$\n- $\\tau$, which is obtained by first performing the permutation notated by $(b_1 b_2 \\dots b_m)$ and then by performing the permutation notated by $(a_1 a_2 \\dots a_k)$\n\nIndeed, $\\sigma(a_i) = (b_1 b_2 \\dots b_m)[(a_1 a_2 \\dots a_k)(a_i)] = (b_1 b_2 \\dots b_m)(a_{i+1}) = a_{i+1}$ (taking $a_{k+1}$ to be $a_1$), while $\\tau(a_i) = (a_1 a_2 \\dots a_k)[(b_1 b_2 \\dots b_m)(a_i)] = (a_1 a_2 \\dots a_k)(a_i) = a_{i+1}$, so they agree on elements of $(a_1 a_2 \\dots a_k)$.\nSimilarly they agree on elements of $(b_1 b_2 \\dots b_m)$; and they both do not move anything which is not an $a_i$ or a $b_j$.\nHence they are the same permutation: they act in the same way on all elements of $\\{1,2,\\dots, n\\}$.\n\nThis reasoning generalises to more than two disjoint cycles, to show that disjoint cycles commute.\n',
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