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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|0SN0MAEKYX]].\nThen there is a [[primitive root|0KM2CJ80G7]] modulo $p$.\nThat is, the multiplicative group $(\\mathbb{Z}/p\\mathbb{Z})^\\times$ is cyclic of order $p - 1$.\n\n## Remarks\n\nA proof: a polynomial of degree $d$ has at most $d$ roots in the field $\\mathbb{Z}/p\\mathbb{Z}$. So for each $d \\mid p - 1$, at most $\\varphi(d)$ elements have order $d$. The [[sum of the totient over the divisors|09JXE6XMCC]] then forces exactly $\\varphi(d)$ elements of each order $d$.\nIn particular, there are exactly $\\varphi(p - 1)$ primitive roots modulo $p$ in $\\{1, \\dots, p - 1\\}$.\nMore generally, each finite subgroup of the multiplicative group of a field is cyclic.\n\n## Explanations\n\n### L1 human\n\nFor each prime $p$, some number $g$ has powers that run through all of $1, 2, \\dots, p - 1$ modulo $p$.\nFor $p = 11$, the number $2$ works: its powers are $2, 4, 8, 5, 10, 9, 7, 3, 6, 1$.\n",
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