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,

which is an abelian group .

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

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 $\mathbb{F}_q$.

This cyclicity is frequently paired with the primitive element theorem : it provides explicit generators for many finite field extensions , especially in the finite-field setting.

Examples

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

   $$2^1=2,\quad 2^2=4,\quad 2^3=3,\quad 2^4=1 \pmod 5.$$

2. $\mathbb{F}_4^\times$ 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.

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