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  "body": "\n## Proof\n\n1. Let $K$ be a [[field|0QA8CHH8MG]], and let $G$ be a finite subgroup of the [[group of units|05THZKR9AN]] $K^\\times$. Let $n = \\abs{G}$.\n2. Multiplication in $K$ is commutative, so $G$ is an [[abelian group|071DDFVG4J]].\n3. Every $g \\in G$ satisfies [[$g^n = 1$|0ZEPNS9BXK]]. So some positive integer $m$ has $g^m = 1$ for all $g \\in G$.\n4. Let $e$ be the [[exponent|0FCCTFA275]] of $G$: the least positive integer with $g^e = 1$ for all $g \\in G$. By step 3, $e \\le n$.\n5. Every element of $G$ is a root in $K$ of the polynomial $x^e - 1 \\in K[x]$, which is nonzero of degree $e$.\n6. This polynomial has [[at most $e$ roots in $K$|0KJVQV3E0F]]. The $n$ elements of $G$ are distinct roots, so $n \\le e$.\n7. By steps 4 and 6, the exponent $e$ equals $\\abs{G} = n$.\n8. A finite abelian group whose exponent equals its order [[is cyclic|0MC0AQZSHE]]. So $G$ is cyclic.\n\n## Idea\n\nIn a field, the equation $x^e = 1$ has at most $e$ solutions. So the exponent of a finite subgroup of $K^\\times$ cannot be smaller than its order. For an abelian group, exponent equal to order means that one element has the full order.\n",
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