---
tag: 04XGKK8WQW
kind: statement
role: theorem
name: Existence of primitive roots modulo a prime
msc: [11A07]
uses: [0KM2CJ80G7, 0SN0MAEKYX]
sources:
  - url: https://en.wikipedia.org/wiki/Primitive_root_modulo_n
    locator: Definition
---

## Statement

Let $p$ be a [[prime number|0SN0MAEKYX]].
Then there is a [[primitive root|0KM2CJ80G7]] modulo $p$.
That is, the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic of order $p - 1$.

## Remarks

A proof: a polynomial of degree $d$ has at most $d$ roots in the field $\mathbb{Z}/p\mathbb{Z}$. So for each $d \mid p - 1$, at most $\varphi(d)$ elements have order $d$. The [[sum of the totient over the divisors|09JXE6XMCC]] then forces exactly $\varphi(d)$ elements of each order $d$.
In particular, there are exactly $\varphi(p - 1)$ primitive roots modulo $p$ in $\{1, \dots, p - 1\}$.
More generally, each finite subgroup of the multiplicative group of a field is cyclic.

## Explanations

### L1 human

For each prime $p$, some number $g$ has powers that run through all of $1, 2, \dots, p - 1$ modulo $p$.
For $p = 11$, the number $2$ works: its powers are $2, 4, 8, 5, 10, 9, 7, 3, 6, 1$.