Finite Multiplicative Subgroup of Field is Cyclic

Theorem
Let $\struct {F, +, \times}$ be a field.

Let $\struct {F^*, \times}$ denote the multiplicative group of $F$.

Let $C$ be a finite subgroup of $\struct {F^*, \times}$.

Then $C$ is cyclic.

Proof
We have that $\struct {F^*, \times}$ is an abelian group.

From Subgroup of Abelian Group is Abelian, $C$ is a finite abelian group.

From Fundamental Theorem of Finite Abelian Groups, $C$ is the internal group direct product of cyclic groups $H_1, H_2, \ldots, H_r$ whose orders are given as:
 * $\order {H_i} = {p_i}^{e_i}$

for some prime numbers $p_1, p_2, \ldots, p_r$ and positive integers $e_1, e_2, \ldots, e_r$.

From Internal and External Group Direct Products are Isomorphic:
 * $C \cong H_1 \times H_2 \times \cdots \times H_r$

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.

Hence $C$ is cyclic.