I was trying the following problem from Berkeley problems in mathematics.

Given $R$ is an integral domain and $G$ is a finite subgroup of $R^{*}$. Show that $G$ is cyclic

Turns out, we simply consider the field of quotients. And the following statement is true.

If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$, then $G$ is cyclic.

We consider the field because then we can use the fact that $F[x]$ is a PID and hence a UFD. I liked this proof in MSE, I will share it here.

Let $G$ a finite group with $n$ elements. If for every $d \mid n$, $\# \{x \in G \mid x^d = 1 \} \leq d$, then $G$ is cyclic.
Fix $d \mid n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d \neq \varnothing$, so there exists $y \in G_d$; it is clear that $\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$. But the subgroup $\langle y \rangle$ has cardinality $d$, so from the hypothesis we have that $\langle y \rangle = \{ x \in G \mid x^d = 1 \}$. Therefore $G_d$ is the set of generators of the cyclic group $\langle y \rangle$ of order $d$, so $\# G_d = \phi(d)$. We have proved that $G_d$ is empty or has cardinality $\phi(d)$, for every $d \mid n$. So we have: $$\begin{align} n &= \# G\\ & = \sum_{d \mid n} \# G_d \\ &\leq \sum_{d \mid n} \phi(d) \\ &= n. \end{align}$$ Therefore $\# G_d = \phi(d)$ for every $d \vert n$. In particular $G_n \neq \varnothing$. This proves that $G$ is cyclic. QED
If $G$ is a finite subgroup of the multiplicative group of a field, then $G$ satisfies the hypothesis because the polynomial $x^d - 1$ has $d$ roots at most.