Finite Multiplicative Subgroup of Field is Cyclic

Theorem
Let $\left({F, +, \times}\right)$ be a field.

Let $\left({F^*, \times}\right)$ denote the multiplicative group of $F$.

Let $C$ be a finite subgroup of $\left({F^*, \times}\right)$.

Then $C$ is cyclic.

Proof
From the Fundamental Theorem of Arithmetic:
 * $\left\vert{C}\right\vert = p_1^{e_1} \cdots p_r^{e_r}$

where $p_i$, $i = 1, \ldots, r$ are distinct prime numbers.

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 number which divides $\left\vert{C}\right\vert$.

Then by Prime Group is Cyclic $H_i$ has a generator $c_i$ of order $e_i$.

We have that $\operatorname{gcd} \left({p_i^{e_i}, p_j^{e_j} }\right) = 1$ when $i \ne j$.

So by Group Direct Product of Cyclic Groups, $C = H_1 \oplus \cdots \oplus H_r$ is cyclic.