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text: 'The [-497] $S_n$ contains elements which are made up from [4cn transpositions] ([4cp proof]).\nIt is a fact that if $\\sigma \\in S_n$ can be made by multiplying together an even number of transpositions, then it cannot be made by multiplying an odd number of transpositions, and vice versa.\n\n%%%knows-requisite([47y]):\nEquivalently, there is a [-47t] from $S_n$ to the [-47y] $C_2 = \\{0,1\\}$, taking the value $0$ on those permutations which are made from an even number of permutations and $1$ on those which are made from an odd number.\n%%%\n\n# Proof\n\nLet $c(\\sigma)$ be the number of cycles in the [49f disjoint cycle decomposition] of $\\sigma \\in S_n$, including singletons.\nFor example, $c$ applied to the identity yields $n$, while $c((12)) = n-1$ because $(12)$ is shorthand for $(12)(3)(4)\\dots(n-1)(n)$.\nWe claim that multiplying $\\sigma$ by a transposition either increases $c(\\sigma)$ by $1$, or decreases it by $1$.\n\nIndeed, let $\\tau = (kl)$.\nEither $k, l$ lie in the same cycle in $\\sigma$, or they lie in different ones.\n\n- If they lie in the same cycle, then $$\\sigma = \\alpha (k a_1 a_2 \\dots a_r l a_s \\dots a_t) \\beta$$ where $\\alpha, \\beta$ are disjoint from the central cycle (and [49g hence commute] with $(kl)$).\nThen $\\sigma (kl) = \\alpha (k a_s \\dots a_t)(l a_1 \\dots a_r) \\beta$, so we have broken one cycle into two.\n- If they lie in different cycles, then $$\\sigma = \\alpha (k a_1 a_2 \\dots a_r)(l b_1 \\dots b_s) \\beta$$ where again $\\alpha, \\beta$ are disjoint from $(kl)$.\nThen $\\sigma (kl) = \\alpha (k b_1 b_2 \\dots b_s l a_1 \\dots a_r) \\beta$, so we have joined two cycles into one.\n\nTherefore $c$ takes even values if there are evenly many transpositions in $\\sigma$, and odd values if there are odd-many transpositions in $\\sigma$.\n\nMore formally, let $\\sigma = \\alpha_1 \\dots \\alpha_a = \\beta_1 \\dots \\beta_b$, where $\\alpha_i, \\beta_j$ are transpositions.\n%%%knows-requisite([modular_arithmetic]):\n(The following paragraph is more succinctly expressed as: "$c(\\sigma) \\equiv n+a \\pmod{2}$ and also $\\equiv n+b \\pmod{2}$, so $a \\equiv b \\pmod{2}$.")\n%%%\nThen $c(\\sigma)$ flips odd-to-even or even-to-odd for each integer $1, 2, \\dots, a$; it also flips odd-to-even or even-to-odd for each integer $1, 2, \\dots, b$.\nTherefore $a$ and $b$ must be of the same [even_odd_parity parity].\n',
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