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  "body": "\n## Proof\n\n1. Let $F$ be a [[finite field|08RTDHX1SK]] of order $q$, and let $F^\\times$ be its [[group of units|05THZKR9AN]].\n2. In a field, every nonzero element is a unit. So $F^\\times = F \\setminus \\set{0}$ has $q - 1$ elements.\n3. So $F^\\times$ is a finite subgroup of the multiplicative group of the field $F$.\n4. By the [[theorem on finite subgroups of $K^\\times$|0CKGBEJX6B]], with $K = F$ and $G = F^\\times$, the group $F^\\times$ is cyclic.\n5. By step 2, it is cyclic of order $q - 1$.\n\n## Idea\n\nThe whole group $F^\\times$ is itself a finite subgroup of the multiplicative group of $F$. So the general theorem applies to it directly.\n",
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  "name": "Proof that the multiplicative group of a finite field is cyclic",
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