---
tag: 0EBZ3G455G
kind: statement
role: corollary
name: The multiplicative group of a finite field is cyclic
msc: ["12E20"]
uses: [0CKGBEJX6B, 08RTDHX1SK]
sources:
  - url: https://en.wikipedia.org/wiki/Finite_field
    locator: Multiplicative structure
---

## Statement

Let $F$ be a [[finite field|08RTDHX1SK]] of order $q$.
Then the multiplicative group $F^\times$ is cyclic of order $q - 1$.

## Remarks

This is the case $G = F^\times$ of the [[theorem on finite subgroups|0CKGBEJX6B]].
A generator of $F^\times$ is a [[primitive element|0YAA9CJQXF]] of $F$.
For $F = \mathbb{Z}/p\mathbb{Z}$, the statement says that there is a primitive root modulo $p$.
For example, $3$ generates $(\mathbb{Z}/7\mathbb{Z})^\times$, with powers $3, 2, 6, 4, 5, 1$.

## Explanations

### L1 human

All the nonzero elements of a finite field are powers of one element.
So multiplication in a finite field is like addition of exponents modulo $q - 1$.