Determine all subgroups of $\mathbb{R}^*$ (nonzero reals under multiplication) of index $2$. Determine all subgroups of $\mathbb{R}^*$ (nonzero reals under multiplication)
of index $2$.

how can I able to solve this problem
 A: Hint: If  $H$ is such a subgroup, you should first prove that $x^2\in H$ for all $x\in\mathbb{R}^*$.
A: Hint: $[\mathbb{R^*}:\mathbb{R^+}] = 2$ 
Let me add little more explanation. 
1: $\mathbb{R^+}$(set of all positive nonzero real numbers) is the only proper subgroup of
$\mathbb{R^*}$ of finite index.
2: $[\mathbb{R^*}:\mathbb{R^+}] = 2$.
To prove the first part let us assume that $\mathbb{R^*}$ has a proper subgroup $H \neq \mathbb{R^+}$ such that $[\mathbb{R^*} : H] = n$ is finite. Thus we have $(xH)^n = x^n H = H$ for each $x\in \mathbb{R^*}$. Thus $x^n \in H$ for each  $x\in \mathbb{R^*}$. Now let $x \in \mathbb{R^+}$, then $ x = (\sqrt[n]{x})^n \in H$. Thus, $\mathbb{R^+} \subset H $. Since $H\neq \mathbb{R^+}$ and $ \mathbb{R^+} \subset H$, we may conclude that $H$ must contain a negative number say $-y$ for some $y\in \mathbb{R^+}$. Since $\frac{1}{y}\in \mathbb{R^+}\subset H $ and $-y\in H$, and since $H$ is closed under multiplication we conclude that $-y(\frac{1}{y}) = -1 \in H$. Since $H$ is closed and $ \mathbb{R^+} \subset H$, and $-1 \in H$. We conclude that $\mathbb{R^-} \subset H$, where $\mathbb{R^-}$ is the set of all nonzero negative real numbers. Since $\mathbb{R^+}\subset H $  and $\mathbb{R^-}\subset H $ , we conclude that $H =\mathbb{R^*} $, which is a contradiction since $H$ is a propser subgroup of $\mathbb{R^*} $. Hence $\mathbb{R^+}$ is the only proper subgroup of $\mathbb{R^*}$ of finite index. 
Part 2 is left for you.
A: First, notice that $|\cdot | : x \mapsto | x|$ and $\operatorname{sign} : x \mapsto \left\{ \begin{array}{cl} 1 & \text{if} \ x>0 \\ -1 & \text{if} \ x<0 \end{array} \right.$ are surjective homomorphisms from $\mathbb{R}^{\times}$ to $\mathbb{R}_{>0}$ and $\mathbb{Z}_2$ respectively, with $\text{ker}(\operatorname{sign})=\mathbb{R}_{>0}$. In particular, $\mathbb{R}^{\times} \simeq \mathbb{Z}_2 \times \mathbb{R}_{>0}$ (for the isomorphism, take $x \mapsto (\operatorname{sign}(x),|x|)$) and $\mathbb{R}_{>0}$ is of index two in $\mathbb{R}^{\times}$.
Let $H$ be a subgroup of finite index in $\mathbb{R}^{\times}$; then $|H|$ is of finite index in $\mathbb{R}_{>0}$. But $\mathbb{R}_{>0}$ is an infinite divisible abelian group, so $|H|=\mathbb{R}_{>0}$ hence $\mathbb{R}_{>0} \subset H$. 
You deduce that $[\mathbb{R}^{\times}: H] \leq [\mathbb{R}^{\times}: \mathbb{R}_{>0}]=2$, that is $H= \mathbb{R}^{\times}$ or $H=\mathbb{R}_{>0}$.
A: R* is abelian ⇒ H ⊴ R* ... 1. R*/H is factor group
|R*/H| = |R*:H| = 2, ... 2.
1.2. ⇒ ∀xH ∈ R*/H, (xH)^|R*/H| = (x^|R*/H|)H = x^2H = H, ... 3. H is identity of R*/H
3. ⇒ ∀x ∈ R*, x^2 ∈ H ⇒ R+ = {x^2 | x ∈ R*} ≤ H ≤ R*, ... 4.
R* = R+ ∪ R- ⇒ |R*:R+| = 2, ... 5.
2.4.5. ⇒ |R*:R+| = |R*:H||H:R+| = 2|H:R+| = 2, ... 6.
6. ⇒ |H:R+| = 1 , ... 7.
7. ⇒ H = R+, QED.
so R+ is the only subgroup with index 2 of R*
