Does the group $\mathbb{R}^{\times} / \mathbb{Q}^{\times}$ have a subgroup of order 5? 
Does the group $\mathbb{R}^{\times} / \mathbb{Q}^{\times}$ have a subgroup of order 5?

I don't know how I should approach this problem. Could you give me some hints on how to solve this? A more general explanation on how to check if a group $G$ has a subgroup $H$ of order $n$ would be more appreciated. Thank you!
 A: Hint: Is there a non-zero real number $\alpha\in\mathbb{R}$ that is irrational, but whose 5th power is rational?
A: Since subgroups of prime order are cyclic, the question is equivalent to: Is there an element of $\mathbb{R}^{\times}/\mathbb{Q}^{\times}$ of order $5$? Equivalently, is there a non-zero real number $r$ with $r^5 \in \mathbb{Q}$, but $r \notin \mathbb{Q}$? Well, of course, for example $r=\sqrt[5]{2}$.
A: Regarding your general question, if $G$ is finite, then Lagrange's theorem gives a necessary condition for the existence of a subgroup $H$ of order $n$, namely that $n$ divides the order of $G$. However, this is not sufficient in general. For supersolvable groups it is sufficient. 
If $G$ is an infinite group, then the existence of a subgroup $H$ of order $n$ means that there exists an element $g\in G$ of finite order $n$. Here there are many possibilities. The group $(\mathbb{Z},+)$ has no subgroup $H$ of oder $n>1$. On the other hand there are infinite groups, even finitely-generated ones, where every non-trivial subgroup is finite, e.g., the Tarski-monster, or Grigorchuk group.
