Subgroups of $R^*$ of finite index? Proof verification 
Subgroups of $G=R^*$ of finite index?

Kindly check my following proof: (Please point out the mistake if it is there)
Let $G=R^*$
Note: $R^*$ is the multiplicative group of non zero reals and $R^+$ is the multiplicative group of positive reals.
Let $H\leq G \ni\; [G:H] = n < \infty$. Also, $H$ is normal in $G$ because $G$ abelian so $G/H$ group makes sense with $H$ as identity.
Then, $\forall x\in G$ we have, $(xH)^n = x^nH = H \rightarrow x^n \in H$
Case $1$: $n$ is even
$\{x^n | x \in G\} = R^+\rightarrow R^+ \subseteq H \rightarrow R^+ \leq H$
But $[G:R^+] = 2$ and $R^+ \leq H \leq G$
so, $2 = [G:R^+] = [G:H][H:R^+]$
So the only possibilities are $[G:H] = 2$ and $[H:R^+] = 1 \rightarrow H = R^+$ or
$[G:H]= 1$ and $[H:R^+] = 2 \rightarrow H=G=R^*$
Case $2$: $n$ is odd
$\{x^n|x\in R^*\} = R^*,$ so $H=R^*$
So, in all cases, either $H=R^*$ or $H=R^+$
so $R^*$ and $R^+$ are the only subgroups of finite index. 
 A: Looks good! The only comment I can make is about notation: "$\forall x \in G \to (xH)^n$..." doesn't really make sense. You should write this as something like $\forall x \in G (x^n H = (xH)^n = H) \implies \forall x \in G (x^n \in H)$. Or, even better, in words: "for all $x \in G$, $x^n H = (xH)^n = H$, so $x^n \in H$".
A: It seems correct, but the symbology is very heavy and difficult to follow.
(1) For every $x\in R^*$, $x^2\in R^+$.
This is a well-known property of the reals.
(2) If $H$ is a subgroup of the abelian group $G$ and $[G:H]=n$ is finite, then $x^n\in H$, for every $x\in G$.
An easy properties of abelian groups.
Let $m$ be a positive integer. A multiplicatively written group $G$ is $m$-divisible if, for every $x\in G$, there exists $y\in G$ such that $y^m=x$.
(3) If $G$ is an $m$-divisible abelian group (with $m>1$), then it has no index $m$ subgroup.
Proof. Let $H$ be a subgroup of $G$ having index $m$. If $x\in G$, then there exists $y\in G$ with $y^m=x$. By (2), $y^m\in H$, so $x\in H$. Contradiction.
(4) The group $R^*$ is $m$-divisible for every odd $m$; the group $R^+$ is $m$-divisible for every $m$; $[R^*:R^+]=2$.
Standard properties of the reals (existence of $m$-th roots of positive reals).
Now we can do the required proof. Suppose $H$ is a finite index subgroup of $R^*$, with $[R^*:H]=n>1$. By (3) and (4), $n$ must be even.
Suppose $n$ is even. Let $x\in R^+$; then $x=y^n$, for some $y\in R^+$ by (4), so $x=y^n\in H$. Hence $R^+\subseteq H$. From the index formula
$$
2=[R^*:R^+]=[R^*:H][H:R^+]
$$
we get that either $[R^*:H]=1$ or $[H:R^+]=1$, so either $H=R^*$ or $H=R^+$.
