Multiplicative group of a finite field is cyclic

Let $\mathbb{F}_q$ be a finite field with $q$ elements. Its nonzero elements form a group under multiplication,

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

which is an abelian group.

Theorem. The multiplicative group $\mathbb{F}_q^\times$ is cyclic. In particular,

|\mathbb{F}_q^\times| = q-1

and there exists an element $g\in\mathbb{F}_q^\times$ such that every nonzero element equals $g^k$ for some integer $k$ . Such a $g$ is often called a primitive element of $\mathbb{F}_q$ .

Relation to primitive elements

The term primitive element here means a generator of the multiplicative group. Such an element also generates $\mathbb F_q$ as a field over its prime subfield, but this cyclicity theorem is distinct from the primitive element theorem for finite separable field extensions.

Examples

$\mathbb{F}_5^\times=\{1,2,3,4\}$ has order $4$ and is cyclic: $2$ is a generator since $2^1=2,\quad 2^2=4,\quad 2^3=3,\quad 2^4=1 \pmod 5.$

$\mathbb{F}_4^\times$ has order $3$ , hence is cyclic of order $3$ . If $\alpha$ is a root of $x^2+x+1$ in $\mathbb{F}_4\cong \mathbb{F}_2[x]/(x^2+x+1)$ , then $\mathbb{F}_4^\times=\{1,\alpha,\alpha+1\}$ and $\alpha$ generates it.

$\mathbb{F}_8^\times$ has order $7$ , so it is cyclic of prime order $7$ . Thus every nonzero element other than $1$ is automatically a generator of $\mathbb{F}_8^\times$ .