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  "body": "\n## Statement\n\nLet $G$ be a finite [[group|G5S1M9]] and let $g \\in G$.\nThen the [[order|0WE2NHZJ7B]] of $g$ divides the [[order|0B85S2V2DR]] $|G|$ of $G$.\n\n## Remarks\n\nThe order of $g$ is the order of the cyclic subgroup $\\langle g \\rangle$, [[by this fact|031M8HX7DA]], and [[Lagrange's theorem|3C53TD]] applies to $\\langle g \\rangle$.\nFor example, in a group of order $10$, each element has order $1, 2, 5$ or $10$.\n\n## Explanations\n\n### L1 human\n\nAn element generates a subgroup with as many elements as its order.\nThe size of a subgroup always divides the size of the group.\nSo the order of an element divides the order of the group.\n\n### L2 agent\n\nThe order of an element $g$ is the smallest $n \\ge 1$ with $g^n = e$. The result says that this number divides the order of the group.\n\nThe proof reduces the result to Lagrange's theorem:\n\n1. The powers of $g$ form the cyclic subgroup $\\langle g \\rangle$.\n2. This subgroup has exactly $\\operatorname{ord}(g)$ elements, namely $e, g, \\dots, g^{n-1}$ with $n = \\operatorname{ord}(g)$.\n3. By Lagrange's theorem, $|\\langle g \\rangle|$ divides $|G|$.\n\nA worked example in the group of units of $\\mathbb{Z}/7\\mathbb{Z}$, which has $6$ elements:\n\n1. Take $g = 2$. Then $2^1 = 2$, $2^2 = 4$ and $2^3 = 8 = 1$.\n2. Thus $\\operatorname{ord}(2) = 3$, and $3$ divides $6$.\n3. Take $g = 3$. The powers are $3, 2, 6, 4, 5, 1$, so $\\operatorname{ord}(3) = 6$.\n\nA useful consequence is that $g^{|G|} = e$ for each $g \\in G$. Applied to the units of $\\mathbb{Z}/p\\mathbb{Z}$, this gives Fermat's little theorem: $a^{p-1} \\equiv 1 \\pmod p$ when $p$ does not divide $a$.\n\nA common error is to think that each divisor of $|G|$ is the order of some element. The Klein four-group has order $4$, but it has no element of order $4$. Cauchy's theorem gives a partial converse: each prime divisor of $|G|$ is the order of some element.\n",
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