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  "body": "\n## Proof\n\n1. Let $G$ be a group of order $p$, where $p$ is a [[prime number|0SN0MAEKYX]].\n2. Since $p \\ge 2$, $G$ contains an element $g$ that is not the identity $e$.\n3. [[The order of $g$ divides $p$|08V7JHS3QZ]]. Thus the order of $g$ is $1$ or $p$.\n4. Only the identity has order $1$, and $g \\ne e$. Thus $g$ has order $p$.\n5. Thus the [[cyclic subgroup|0YT586TKXF]] $\\langle g \\rangle$ has $p$ elements, so $\\langle g \\rangle = G$.\n6. Thus $G$ is [[cyclic|0B85Q41RP6]], with generator $g$.\n\n## Idea\n\nA prime has only two divisors. An element that is not the identity cannot have order one, so its order is the full order of the group.\n",
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