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  "body": "\n## Statement\n\nLet $G$ be a [[cyclic group|0KVGH0J5B3]] of finite order $n$, and let $d$ be a positive divisor of $n$.\nThen $G$ has exactly $\\varphi(d)$ elements of [[order|0WE2NHZJ7B]] $d$, where $\\varphi$ is Euler's totient function.\n\n## Remarks\n\nHere $\\varphi(d)$ is the number of integers $k$ with $1 \\leq k \\leq d$ and $\\gcd(k, d) = 1$.\nThe elements of order $d$ are the generators of the unique subgroup of order $d$.\nCounting all elements by their orders gives the identity $\\sum_{d \\mid n} \\varphi(d) = n$.\nIn particular, $G$ has exactly $\\varphi(n)$ generators.\n\n## Explanations\n\n### L1 human\n\nIn $\\mathbb{Z}/6\\mathbb{Z}$, the element $0$ has order $1$ and the element $3$ has order $2$. The elements $2$ and $4$ have order $3$, and $1$ and $5$ have order $6$.\nThese counts are $1, 1, 2, 2$, which are $\\varphi(1), \\varphi(2), \\varphi(3), \\varphi(6)$.\nThe totient function counts the elements of each order.\n",
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