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
As C is a finite abelian group, due to the Fundamental Theorem of Finite Abelian Groups, $C$ is the Internal Group Direct Product of Cyclic Groups $H_1, \ldots, H_r$ with orders $p_1^{e_1}, \ldots, p_r^{e_r}$ respectively.

As Internal and External Group Direct Products are Isomorphic,


 * $C \cong H_1 \times \ldots \times H_r$

where $\cong$ denotes (group) isomorphism.

As Powers of Coprime Numbers are Coprime, the orders of $H_1, \ldots, H_r$ are coprime.

By Group Direct Product of Cyclic Groups, $H_1 \times \ldots \times H_r$ is cyclic, so $C$ is cyclic.