{
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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]] with identity element $e$, and let $g \\in G$ be an element of finite [[order|0WE2NHZJ7B]] $m$.\nLet $n$ be an integer.\nThen $g^n = e$ if and only if $m$ divides $n$.\n\n## Remarks\n\nThe proof divides $n$ by $m$ with remainder: $n = qm + r$ with $0 \\leq r \u003c m$, so that $g^n = (g^m)^q g^r = g^r$.\nBy the minimality of $m$, the power $g^r$ is $e$ only when $r = 0$.\nA consequence is that $g^a = g^b$ if and only if $a \\equiv b \\pmod m$.\n\n## Explanations\n\n### L1 human\n\nAn element of order $m$ comes back to the identity exactly after $m$, $2m$, $3m$ steps, and so on.\nIt is never at the identity after a number of steps that is not a multiple of $m$.\nFor example, a rotation by $90$ degrees is the identity after $8$ steps, but not after $6$ steps.\n",
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        "locator": "Order and structure",
        "url": "https://en.wikipedia.org/wiki/Order_(group_theory)"
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  "name": "g to the n equals 1 iff the order divides n",
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