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text: 'Cauchy's theorem states that if $G$ is a finite [-3gd] and $p$ is a [4mf prime] dividing $|G|$ the [3gg order] of $G$, then $G$ has a subgroup of order $p$. Such a subgroup is necessarily [47y cyclic] ([4jh proof]).\n\n# Proof\n\nThe proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.\n\nLet $$X = \\{ (x_1, x_2, \\dots, x_p) : x_1 x_2 \\dots x_p = e \\}$$ the collection of $p$-[-tuple]s of elements of the group such that the group operation applied to the tuple yields the identity.\nObserve that $X$ is not empty, because it contains the tuple $(e, e, \\dots, e)$.\n\nNow, the cyclic group $C_p$ of order $p$ [3t9 acts] on $X$ as follows: $$(h, (x_1, \\dots, x_p)) \\mapsto (x_2, x_3, \\dots, x_p, x_1)$$ where $h$ is the generator of $C_p$.\nSo a general element $h^i$ acts on $X$ by sending $(x_1, \\dots, x_p)$ to $(x_{i+1}, x_{i+2} , \\dots, x_p, x_1, \\dots, x_i)$.\n\nThis is indeed a group action (exercise).\n\n%%hidden(Show solution):\n\n- It certainly outputs elements of $X$, because if $x_1 x_2 \\dots x_p = e$, then $$x_{i+1} x_{i+2} \\dots x_p x_1 \\dots x_i = (x_1 \\dots x_i)^{-1} (x_1 \\dots x_p) (x_1 \\dots x_i) = (x_1 \\dots x_i)^{-1} e (x_1 \\dots x_i) = e$$\n- The identity acts trivially on the set: since rotating a tuple round by $0$ places is the same as not permuting it at all.\n- $(h^i h^j)(x_1, x_2, \\dots, x_p) = h^i(h^j(x_1, x_2, \\dots, x_p))$ because the left-hand side has performed $h^{i+j}$ which rotates by $i+j$ places, while the right-hand side has rotated by first $j$ and then $i$ places and hence $i+j$ in total.\n%%\n\nNow, fix $\\bar{x} = (x_1, \\dots, x_p) \\in X$.\n\nBy the [4l8 Orbit-Stabiliser theorem], the [4v8 orbit] $\\mathrm{Orb}_{C_p}(\\bar{x})$ of $\\bar{x}$ divides $|C_p| = p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\\bar{x} \\in X$.\n\nNow, what is the size of the set $X$?\n%%hidden(Show solution):\nIt is $|G|^{p-1}$.\n\nIndeed, a single $p$-tuple in $X$ is specified precisely by its first $p$ elements; then the final element is constrained to be $x_p = (x_1 \\dots x_{p-1})^{-1}$.\n%%\n\nAlso, the orbits of $C_p$ acting on $X$ partition $X$ ([4mg proof]).\nSince $p$ divides $|G|$, we must have $p$ dividing $|G|^{p-1} = |X|$.\nTherefore since $|\\mathrm{Orb}_{C_p}((e, e, \\dots, e))| = 1$, there must be at least $p-1$ other orbits of size $1$, because each orbit has size $p$ or $1$: if we had fewer than $p-1$ other orbits of size $1$, then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$, and all the remaining orbits would have to be of size $p$, contradicting that $p \\mid |X|$.\n[todo: picture of class equation]\n\nHence there is indeed another orbit of size $1$; say it is the singleton $\\{ \\bar{x} \\}$ where $\\bar{x} = (x_1, \\dots, x_p)$.\n\nNow $C_p$ acts by cycling $\\bar{x}$ round, and we know that doing so does not change $\\bar{x}$, so it must be the case that all the $x_i$ are equal; hence $(x, x, \\dots, x) \\in X$ and so $x^p = e$ by definition of $X$.',
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