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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]] and let $g \\in G$.\nIf $g$ has finite [[order|0WE2NHZJ7B]] $n$, then $\\langle g \\rangle = \\{e, g, g^2, \\ldots, g^{n-1}\\}$, these $n$ elements are distinct, and thus $|\\langle g \\rangle| = n$.\nIf $g$ has infinite order, then the powers $g^k$ for $k \\in \\mathbb{Z}$ are distinct, and $\\langle g \\rangle$ is infinite.\n\n## Remarks\n\nHere $\\langle g \\rangle$ is the [[cyclic subgroup generated by $g$|08ABT85339]], and $|\\langle g \\rangle|$ is its [[order|0B85S2V2DR]].\nThe proof uses [[the divisibility criterion for powers|0F5QC101Q7]]: $g^a = g^b$ exactly when $n$ divides $a - b$.\nWith [[Lagrange's theorem|3C53TD]], this fact shows that the order of an element of a finite group divides the order of the group.\n\n## Explanations\n\n### L1 human\n\nThe two meanings of the word \"order\" agree.\nThe order of an element is the number of distinct powers that it has, and these powers form the subgroup that it generates.\nFor example, a rotation by $120$ degrees has order $3$, and its powers are the three rotations by $0$, $120$ and $240$ degrees.\n",
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