---
tag: 0CKGBEJX6B
kind: statement
role: theorem
name: Finite subgroups of the multiplicative group of a field are cyclic
msc: ["12E20"]
uses: [05THZKR9AN, 0KJVQV3E0F]
sources:
  - url: https://en.wikipedia.org/wiki/Finite_field
    locator: Multiplicative structure
---

## Statement

Let $K$ be a [[field|0QA8CHH8MG]] 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|0KJVQV3E0F]], 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|0GE4DVQKR8]] 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.