Subgroups of $\Bbb R^{\times}$ 
*

*What are the open subgroups of $\Bbb R^{\times}$ ?

*What are the closed subgroups of $\Bbb R^{\times}$ ?

*What are the finite index subgroups of $\Bbb R^{\times}$ ?


My thoughts:


*

*Since $\Bbb R_{>0}$ is a connected Lie group, any open neighborhood of $1$ generates it as a group. So only $\Bbb  R_{>0}$ and $\Bbb  R^{\times}$ are the open subgroups of $\Bbb R^{\times}$. Is it correct?

*Here I don't really know. 

*If $H \subset \Bbb R_{>0}$ has finite index in $\Bbb R^{\times}$ then it is $\Bbb R_{>0}$ by this answer – because $ \Bbb R_{>0}$ is divisible. If $H \subset \Bbb R^{\times}$ is a subgroup of finite index not contained in $\Bbb R_{>0}$, do we have $H = \Bbb R^{\times}$ ? Then we would have only 2 subgroups of finite index.
 A: Since the OP's reasoning in 1. and 3. is sound, I will only deal with 2. 
The key observation is that most the closed subgroup of $\mathbb{R}^\times$ generated by only two elements is at least $\mathbb{R}_{>0}$ unless something very special happens. 
Lemma. Let $x, y \in \mathbb{R}_{>0}$ and put $A = \{x^n y^m : n,m \in \mathbb{Z}\}$. Then either (1) $A = \{z^n : n \in \mathbb{Z} \}$ for some $z \in \mathbb{R}_{>0}$, or (2) $A$ is dense in $\mathbb{R}_{>0}$. 
Proof. Apply Kronecker's theorem to $\log(x)/\log(y)$. 
Suppose that $G < \mathbb{R}^\times$ is a closed subgroup, and let $H = G \cap \mathbb{R}_{>0}$. Since the square of any element of $G$ is in $H$, we have $G = H$ or $G = H \cup xH$ where $x \in \mathbb{R}_{<0}$ and $x^2 \in H$. So, it's enough to characterise all possible $H$.
If $H$ is the trivial group, we are done, so suppose there is some $x \in H \setminus \{1\}$. Likewise, if $H = \mathbb{R}_{>0}$ we are done, so suppose that $H$ is not dense in $\mathbb{R}_{>0}$. In particular, there are only finitely many positive integers $n$ such that $x^{1/n} \in H$; let $t = x^{1/n}$ where $n$ is largest possible. For any $y \in H$ it follows from the above Lemma that $y$ and $t$ are both integer powers of some $z \in H$, but by the choice of $t$ this is only possible if $z = t$ and so $y$ is a power of $t$. Since $y$ was arbitrary, $H = \{t^n : n \in \mathbb{Z} \}$ (the opposite inclusion is clear). 
Conversely, for any $t > 0$, $H = \{t^n : n \in \mathbb{Z} \}$ is a closed subgroup of $\mathbb{R}_{>0}$ for any $t \geq 0$. Hence, the closed subgroups of $\mathbb{R}^\times$ are: $\mathbb{R}^\times$, $\mathbb{R}_{>0}$, $\{t^n : n \in \mathbb{Z} \}$ for $t > 0$, $\{\pm t^n : n \in \mathbb{Z} \}$ for $t > 0$, and $\{(-t)^n : n \in \mathbb{Z} \}$ for $t > 0$.
Edit: The reasoning can be shortened if we assume we already know the closed subgroups of $(\mathbb{R},+)$, but since the problems are very similar I avoid that assumption :-)
