Finite Multiplicative Subgroup of Field is Cyclic

Theorem
Let $(F,\cdot,+)$ be a field.

Let $C$ be a finite subgroup of $(F^\times,\cdot)$.

Then $C$ is cyclic.

Proof
Write $|C| = p_1^{e_1} \cdots p_r^{e_r}$ with $p_i$, $i = 1,\ldots,r$ distinct primes (using the Fundamental Theorem of Arithmetic).

By the Fundamental Theorem of Finite Abelian Groups $C$ is a direct sum of subgroups $H_1,\ldots,H_r$ of orders $p_1^{e_1},\ldots, p_r^{e_r}$ respectively.

Let $p_i$ be a prime dividing $|C|$.

Then by $p$-group is cyclic $H_i$ has a generator $c_i$ of order $e_i$.

Finally by Group Direct Product of Cyclic Groups, since $\operatorname{gcd}(p_i^{e_i},p_j^{e_j}) = 1$ when $i \neq j$ we have that $C = H_1 \oplus \cdots \oplus H_r$ is cyclic.