Multiplicative Group of a Finite Field Is Cyclic

Let $\mathbb{F}_q$ be a finite field with $q$ elements. Its multiplicative group of nonzero elements is

\mathbb{F}_q^\times=\mathbb{F}_q\setminus\{0\},

which is an abelian group under multiplication.

Theorem (cyclicity of $\mathbb{F}_q^\times$ ). The group $\mathbb{F}_q^\times$ is cyclic of order $q-1$ . Equivalently, there exists $\gamma\in\mathbb{F}_q^\times$ such that

\mathbb{F}_q^\times=\{\gamma^k:0\le k\le q-2\}.

Such a generator is often called a primitive element of $\mathbb{F}_q$ ; it is also a primitive $(q-1)$ st root of unity in the field.

Remarks

This statement is sometimes recorded as the cyclicity of the finite-field multiplicative group.

Examples

$\mathbb{F}_5^\times$ is cyclic of order $4$ . We have $\mathbb{F}_5^\times=\{1,2,3,4\}$ . The element $2$ generates: $2^1=2,\quad 2^2=4,\quad 2^3=3,\quad 2^4=1 \pmod 5,$ so $\mathbb{F}_5^\times=\langle 2\rangle$ .

$\mathbb{F}_7^\times$ is cyclic of order $6$ . Here $\mathbb{F}_7^\times=\{1,2,3,4,5,6\}$ and $3$ is a generator: $3,\,3^2=2,\,3^3=6,\,3^4=4,\,3^5=5,\,3^6=1 \pmod 7,$ so every nonzero element is a power of $3$ .

$\mathbb{F}_8^\times$ has prime order $7$ . Since $|\mathbb{F}_8^\times|=8-1=7$ is prime, every element of $\mathbb{F}_8^\times$ other than $1$ has order $7$ , hence is a generator. For instance, if $\mathbb{F}_8\cong \mathbb{F}_2[\alpha]/(f(\alpha))$ for some irreducible cubic $f$ over $\mathbb{F}_2$ (as in finite-field existence), then $\alpha\ne 1$ and therefore $\mathbb{F}_8^\times=\langle \alpha\rangle$ .