Let $H$ be a subgroup of $\Bbb R^*$. If $\Bbb R^+ \subseteq H \subseteq \Bbb R^*$, prove that $H = \Bbb R^+$ or $H = \Bbb R^*$. 
Let $H$ be a subgroup of $\Bbb R^*$, the group of nonzero real numbers under multiplication. If $\Bbb R^+ \subseteq H \subseteq \Bbb R^*$, prove that $H = \Bbb R^+$ or $H = \Bbb R^*$.

I think that $\Bbb R^+$ is supposed to mean the set of non-negative reals? I am a bit confused on what do I need to prove here? If I assume that $H \ne \Bbb R^+$ and get that $H=\Bbb R^*$, then would that prove the result?
Suppose that $H \ne \Bbb R^+$, then I only have that there exists $h \in H$ such that $h \notin \Bbb R^+$?
 A: $\mathbb R^{+}$ is $(0,\infty)$. Suppose $\mathbb R^{+} \subset H$. (I write $\subset$ for proper subset). Then there exists $h \in H$ such that $h < 0$. Let $x < 0$. But then  $x=hy$ for some $y>0$. ($y=\frac  x  h$). Since $y >0$ it follows that $y \in H$. Since $H$ is  a subgroup we see that $x=yh \in H$. Thus $H$ contains  all negative numbers. By hypothesis it contains all positive numbers too, so $H=\mathbb R^{*}$.
A: $\Bbb{R}^+=\{a\in\Bbb{R}:a>0\}$.
$\Bbb{R}^+, \Bbb{R}^{\star}$ both are multiplicative group.
If $\Bbb{R}^+\subset H\subset \Bbb{R}^{\star}$ and $H\le \Bbb{R}^{\star}$ then $H=\Bbb{R}^+$ or $H=\Bbb{R}^{\star}$
Proof: Suppose $\Bbb{R}^+\subsetneq H$.
Then $\exists a\in H\setminus \Bbb{R}^+$ i.e $a\le 0$ and since $0\notin H$ , $a<0$.
Claim : $-1\in H$.
$a<0$ implies $-a\in \Bbb{R}^+$ and hence $-\frac{1}{a}\in \Bbb{R}^+$.
Hence $-\frac{1}{a}\in H$.
Since $H\le \Bbb{R}^{\star}$ , $a\cdot-\frac{1}{a}=-1\in H$
Now it's easy to show that $\langle \Bbb{R}^+, -1\rangle =\Bbb{R}^{\star}$
$(b<0$ , then $b=-b\cdot -1)$
Notations:
$\le $: subgroup.
$\langle S \rangle $: subgroup generated by $S$
A: Consider the canonical submersion $\pi:\Bbb R^*\to \Bbb R^*/\Bbb R^+=\Bbb Z_2$.  By the correspondence theorem, there's a bijection between the possible  subgroups $H$ such that $\Bbb R^+\le H\le\Bbb R^*$ and the subgroups of $\Bbb Z_2$.  Hence there are only two possibilities.
