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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|0SN0MAEKYX]], and let $a$ be an integer with $p \\nmid a$.\nThen the [[multiplicative order|0SBRX5CZ5R]] of $a$ modulo $p$ divides $p - 1$.\n\n## Remarks\n\nThe proof: $a^{p-1} \\equiv 1 \\pmod p$ by [[Fermat's little theorem|0M25PRN3WQ]], and each exponent $m$ with $a^m \\equiv 1$ is a multiple of the order.\nThe order is equal to $p - 1$ exactly when $a$ is a [[primitive root|0KM2CJ80G7]] modulo $p$.\nFor example, modulo $13$, the order of $3$ is $3$ and the order of $2$ is $12$. Both divide $12$.\n\n## Explanations\n\n### L1 human\n\nThe powers of $a$ modulo $p$ go round a cycle. The length of the cycle always divides $p - 1$.\nModulo $11$, the powers of $5$ are $5, 3, 4, 9, 1$. The cycle has length $5$, and $5 \\mid 10$.\n",
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