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  "body": "\n## Statement\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $G$ be a finite subgroup of the multiplicative group $K^\\times$.\nThen $G$ is cyclic.\n\n## Remarks\n\nThe proof: for each $d$, the elements $g \\in G$ with $g^d = 1$ are roots of $x^d - 1$.\nSo there are at most $d$ of them, by the [[bound on the number of roots|0KJVQV3E0F]], and a finite group with this property is cyclic.\nFor $K = \\mathbb{C}$, the finite subgroups of $\\mathbb{C}^\\times$ are the groups of the $n$-th roots of unity.\nThe statement fails for division rings: the quaternion group $\\{\\pm 1, \\pm i, \\pm j, \\pm k\\}$ is a finite subgroup of the nonzero [[quaternions|0GE4DVQKR8]] that is not cyclic.\n\n## Explanations\n\n### L1 human\n\nIn a field, the equation $x^d = 1$ has at most $d$ solutions.\nThis forces every finite group of nonzero elements to be generated by one element.\n",
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