Finite subgroups of the multiplicative group of a field are cyclic

Statement

Let $K$ be a field and let $G$ be a finite subgroup of the multiplicative group $K^\times$. Then $G$ is cyclic.

Remarks

The proof: for each $d$, the elements $g \in G$ with $g^d = 1$ are roots of $x^d - 1$. So there are at most $d$ of them, by the bound on the number of roots, and a finite group with this property is cyclic. For $K = \mathbb{C}$, the finite subgroups of $\mathbb{C}^\times$ are the groups of the $n$-th roots of unity. The statement fails for division rings: the quaternion group $\{\pm 1, \pm i, \pm j, \pm k\}$ is a finite subgroup of the nonzero quaternions that is not cyclic.

Explanations

L1 human

In a field, the equation $x^d = 1$ has at most $d$ solutions. This forces every finite group of nonzero elements to be generated by one element.