TITLE: Why is the multiplicative group of a finite field cyclic?
QUESTION [24 upvotes]: Why is the multiplicative group $(K\smallsetminus\{0\},\cdot)$ of a finite field $(K,+,\cdot)$ always cyclic?

REPLY [10 votes]: Let $G$ be the multiplicative group of a finite field, $n$ its order.
Let $x$ be an element of $G$, $d$ its order.
Let $H$ be the subgroup of $G$ generated by $x$.
If $G = H$, we are done.
Suppose otherwise.
There exists $y \in G - H$.
Let $m$ be the order of $y$, $l$ = lcm$(d, m)$.
Suppose $d = l$.
Then $m\mid d$.
Hence $y^d = 1$.
This contradicts the fact that the number of solutions of the equation $X^d = 1$ is less than or equal to $d$.
Hence $d \lt l$.
By the answer to this question, there exists an element $z$ of order $l$ in $G$.
We can repeat this process until we find a generator of $G$.
This completes the proof.