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  "body": "\n## Statement\n\nLet $G$ be a [[cyclic group|0KVGH0J5B3]] of finite order $n$, generated by $g$.\nFor each positive divisor $d$ of $n$, $G$ has exactly one [[subgroup|0X1JXT47GA]] of order $d$, namely $\\langle g^{n/d} \\rangle$.\nEach subgroup of $G$ has this form for some positive divisor $d$ of $n$.\n\n## Remarks\n\nThus the number of subgroups of $G$ is the number of positive divisors of $n$.\nFor $d, d'$ dividing $n$, the subgroup of order $d$ is contained in the subgroup of order $d'$ exactly when $d$ divides $d'$.\nFor example, $\\mathbb{Z}/12\\mathbb{Z}$ has six subgroups, of the orders $1, 2, 3, 4, 6, 12$.\n\n## Explanations\n\n### L1 human\n\nIn a cyclic group, the subgroups are as simple as possible: there is one for each divisor of the order.\nIn the clock group $\\mathbb{Z}/12\\mathbb{Z}$, the subgroup of order $4$ is $\\{0, 3, 6, 9\\}$, generated by $12/4 = 3$.\nThe order of each subgroup divides $12$, as [[Lagrange's theorem|3C53TD]] says, and here every divisor occurs.\n",
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