What are the finite subgroups of $K^*$ if $K\in \{\Bbb Q,\Bbb R,\Bbb C\}$? What are the finite subgroups of $K^*$ if $K\in \{\Bbb Q,\Bbb R,\Bbb C\}$?
 A: Hint: If $G$ is a finite group of $K^*$, say $|G|=n$, then $x^n=1$ for all $x \in G$. Now think about the elements $x$ in  $\Bbb Q, \Bbb R$ or $\Bbb C$ that satisfy equations like $x^n=1$. 
A: In general, if $K$ is a field, then every finite subgroup of $K^*$ is cyclic. If it is order $n$, then it is therefore generated by an element of order $n$. Such elements are also known as primitive $n$ths roots of unity. If there is a primitive $n$th root of unity, then also for all divisors of $n$.
Since $\mathbb{C}$ is algebraically closed, it has a primitive $n$th root of unity for every $n$. So the question is only which of these are real or even rational. But they all have absolute value $1$, and $S^1 \cap \mathbb{R} = S^1 \cap \mathbb{Q} = \{\pm 1\}$.
Summary: For every $n \geq 1$ there is a unique finite subgroup of $\mathbb{C}^*$ of order $n$. For $\mathbb{R}^*$ and $\mathbb{Q}^*$ this is only the case for $n=1,2$.
A: I think as @copper.hat suggested, you are looking for the torsion subgroups of these groups. $$t(\mathbb Q^*,\cdot)=t(\mathbb R^*,\cdot)=\{\pm1\}$$
A: Of course all other answers are right and complete but I just want to point out the following not so difficult point

Let $G$ be an finitely generated abelian group. If for each $n\in\mathbb{N}$, there is at most $n$ elements in $G$ satisfying \begin{equation}n\cdot g=0,\end{equation}then $G$ is cyclic.

$n\cdot g$ means you add $n$ copies of $g$. This is not difficult once one is willing to apply the structure theorem for finite generated abelian groups.
In articular any finite subgroup $K^*$ of the multiplicative group of a field must be abelian and since $x^n=1$ has at most $n$ solutions, the fact above says $K^*$ is cyclic. That is, finite subgroups of multiplicative group of a field is cyclic.
Applying this to $\mathbb{Q},\mathbb{R},\mathbb{C}$ gives the desired answers.
