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  "body": "\n## Statement\n\nLet $p$ be a prime number and let $G$ be a [[group|G5S1M9]] of order $p$.\nThen $G$ is [[cyclic|0KVGH0J5B3]], and each element of $G$ other than the identity generates $G$.\n\n## Remarks\n\nIf $g \\neq e$, then the order of $\\langle g \\rangle$ is greater than $1$ and divides $p$ by [[Lagrange's theorem|3C53TD]], so it is $p$.\nThus each group of order $p$ is isomorphic to $\\mathbb{Z}/p\\mathbb{Z}$.\nA group of prime order has no subgroups except $\\{e\\}$ and itself.\n\n## Explanations\n\n### L1 human\n\nWhen the size of a group is a prime, there is no room for proper subgroups other than the trivial one.\nAny element other than the identity generates a subgroup larger than $1$, so it generates everything.\nThus, for example, each group with $7$ elements is a clock with $7$ hours.\n\n### L2 agent\n\nA prime $p$ has only the divisors $1$ and $p$. With Lagrange's theorem, this leaves a group of order $p$ no room for subgroups other than $\\{e\\}$ and the whole group.\n\nThe proof is short:\n\n1. Let $g \\in G$ with $g \\ne e$.\n2. The order of $g$ divides $p$, so it is $1$ or $p$.\n3. Only $e$ has order $1$. Thus $\\operatorname{ord}(g) = p$.\n4. Thus $\\langle g \\rangle$ has $p$ elements, and it is all of $G$.\n\nA worked example with $p = 5$ and $G = \\mathbb{Z}/5\\mathbb{Z}$:\n\n1. Take $g = 2$. Its multiples are $2, 4, 1, 3, 0$.\n2. All five elements occur, so $2$ generates $G$.\n3. The same is true for $1$, $3$ and $4$, as the result states.\n\nA consequence is that, up to isomorphism, there is exactly one group of each prime order $p$, namely $\\mathbb{Z}/p\\mathbb{Z}$. Thus the groups of order $2$, $3$, $5$ and $7$ are all cyclic and abelian.\n\nThe hypothesis that $p$ is prime is necessary. The Klein four-group has order $4$, and it is not cyclic. The group $S_3$ has order $6$, and it is not even abelian.\n\nA common error is to think that the result also gives a generator for groups of prime power order. A group of order $p^2$ is abelian, but it need not be cyclic, as $\\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$ shows.\n",
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        "adapter": "Math2.Match.prime_card",
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    "name": "Groups of prime order are cyclic",
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        "url": "https://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)"
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  "name": "Groups of prime order are cyclic",
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