Let $F$ be a finite field. Then the multiplicative group $(F \setminus \{0\}, \cdot)$ is cyclic

Question

Let $F$ be a finite field. I'm trying to prove

The multiplicative group $(F \setminus \{0\}, \cdot)$ is cyclic.

Then I figure out that it's sufficient to prove

Different multiplicative subgroups of $F$ have different orders.

I would like to prove it by myself, but fail after one week of trying. Could you please leave me some minimal hints so that I can go on?

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Update: I've just figured out that it's sufficient to prove

For any divisor $d$ of $|F \setminus \{0\}|$, there is at least one multiplicative cyclic subgroup of order $d$.

Answer 1

HINT

Let $F^\times = F\setminus\{0\}.$
Let $l = \operatorname{lcm}\{\operatorname{order}(x) \mid x \in F^\times\}.$
Let $n = |F^\times|.$

1. Consider the polynomial $x^l - 1$. How many roots does this have in $F$? What can you conclude?

$n \le l.$

2. Consider an element $g \in F^\times$ such that $\operatorname{order}(g) = l.$ (Why does this exist?)
   What can you conclude?

$l \mid n$ and hence, $n = l.$ Thus, $g$ is an element of order $n$ and therefore, must generate $F^\times$.