A subgroup $G$ of $(\mathbb{R_{+}}, \cdot)$ such that $G \cap \left(\frac{1}{b}, b \right)$ is finite for any $b>1$ Let $(G, \cdot)$ be a nontrivial subgroup of $(\mathbb{R_{+}}, \cdot)$ such that $G \cap \left(\frac{1}{b}, b\right)$ is finite for any $b>1$. Show that $(G, \cdot)$ is isomorphic to $(\mathbb{Z},+)$.
I think that this problem boils down to proving that $G$ is a cyclic infinite group. However, I don't know how to use the given condition. I thought I could let $b\to \infty$, but I don't know how to proceed.
Solution following an idea from the comments : We get that $(G, \cdot) \cong (a\mathbb{Z},+)$ for some real $a$.
$a\mathbb{Z} \cap \left(\frac{1}{b}, b\right)$ must be a finite set for all $b>1$.
Assume that $a>1$. We have that $a\mathbb{Z} \cap \left(\frac{1}{a}, a\right)=\left(\frac{1}{a}, a\right)$, which is not a finite set, contradiction. The same reasoning works for $a<1$ and it follows that $a=1$.
EDIT: It doesn't work, we still have that $a\mathbb{Z} \cap \left(\frac{1}{a}, a\right)=\emptyset$, which is finite. 
 A: Proof. If $G$ has at least two elements, then there exists $g_1\in G$ such that $g_1>1$ (note that if $0<g<1$, then $g^{-1}>1$). Take $b=g_1+1$ and consider the set $G\cap (1,b)$, which is finite (by assumption) and nonempty ($\because g_1\in (1,b)$). Let $$1<g_0=\min(G\cap (1,b)).$$ Then one has the following 
Assertion. $G=\{g_0^k~|~k\in {\mathbb Z}\}=\{g_0^n~|~n\in {\mathbb N}\}\cup\{1\}\cup\{g_0^{-n}~|~n\in {\mathbb N}\}.$
Proof of assertion. Note that by construction, $$\nexists g\in G~{\rm such ~that ~}1<g<g_0.\qquad (1)$$ For every $g\in G,g>1,$ if $g\in \{g_0^n~|~n\in{\mathbb N}\},$ then by combining its inverses and the identity, the result is clear. Otherwise $$g_0^n<g<g_0^{n+1}~{\rm for ~some ~}n\geq 1.$$
$$\Rightarrow 1<gg_0^{-n}<g_0,$$ a contradiction to (1), since $gg_0^{-n}\in G.$ It follows that $G\cap (1,\infty)=\{g_0^n~|~n\in {\mathbb N}\}$ and $G=\{g_0^k~|~k\in{\mathbb Z}\}$, as required.
To complete the proof, the obvious isomorphism from $G$ to $\mathbb Z$ is given by $$\phi:G\rightarrow {\mathbb Z}$$ $$g_0^k\mapsto k.$$ QED
Remark. In fact $\phi(x)=\log_{g_0}(x),\forall x\in G.$
A: Assuming it isn't the trivial subgroup, here's an idea to get you started. If you want to prove a group is cyclic, you want to find a generator. Consider the following set
$$\left\{b\in\Bbb R_{>1}:\operatorname{card}\left(G\cap \left(\frac{1}{b},b\right)\right) = 1\right\}.$$
Since these sets are nested and $G$ has at least two elements, this set must be bounded and is clearly nonempty (why?). What could you say about its least upper bound? Can you take it from here? There's an extra step or two that needs to be taken to prove that $G$ is isomorphic to $\Bbb Z$ and does not merely just contain it as a subgroup.
A: Let's start with a nice way of restating the original claim:
Proposition 1: For any multiplicative subgroup $G$ of $\Bbb{R}_+$, if $G\cap\left(\frac{1}{b},b\right)$ is finite for every $b>1$, then $G$ is cyclic.
Note that any cyclic subgroup of $\Bbb{R}_+$ would have to either be trivial or isomorphic to $\Bbb{Z}$, since $\Bbb{R}_+$ contains no elements of finite order.
Next, I would like to suggest that even though the groups $(\Bbb{R},+)$ and $(\Bbb{R}_+,\cdot)$ are isomorphic, things will be clearer and easier to understand if we work with $(\Bbb{R},+)$. So I'm going to state an equivalent statement about $(\Bbb{R},+)$, and then I will show how to prove it. Also, henceforth I will refer to $(\Bbb{R},+)$ simply as $\Bbb{R}$. Our goal is to show the following:
Proposition 2: For any additive subgroup $G$ of $\Bbb{R}$, if $G\cap(-\epsilon,\epsilon)$ is finite for every $\epsilon>0$, then $G$ is cyclic.
Before we proceed, I want to point out that the above two propositions are equivalent. Let me briefly sketch how we could show this. First note that the map $\phi:\Bbb{R}\to\Bbb{R}_+$ with $\phi(x)=e^x$ for all $x\in\Bbb{R}$ is an isomorphism. For any multiplicative subgroup $H$ of $\Bbb{R}_+$, the pre-image $\phi^{-1}(H)$ will be an additive subgroup of $\Bbb{R}$. And for any interval $I$ of the form $\left(\frac{1}{b},b\right)$ for some $b>1$, the pre-image $\phi^{-1}(I)$ will be an interval of the form $(-\epsilon,\epsilon)$ for some $\epsilon>0$.
Proof of Proposition 2: Let $G$ be a subgroup of $\Bbb{R}$ and suppose that $G\cap(-\epsilon,\epsilon)$ is finite for every $\epsilon>0$. Our goal is to show that $G$ is cyclic. Of course if $G$ is trivial then $G$ is cyclic, so suppose $G$ is not trivial.
First, we want to find a positive element $b\in G$. Since $G$ is not trivial we can find a non-zero $a\in G$. If $a>0$ then let $b=a$. Otherwise let $b=-a$.
Now let $\epsilon=b+1$. Since $\epsilon>0$, we have that $G\cap(-\epsilon,\epsilon)$ is finite. We also know that $G\cap(-\epsilon,\epsilon)$ has at least one positive element, since $b\in G\cap(-\epsilon,\epsilon)$. Let $c$ be the smallest positive element of $G\cap(-\epsilon,\epsilon)$. We will complete the proof by showing that $G=c\Bbb{Z}$.
Note that $c\Bbb{Z}\subseteq G$ follows from the fact that $c\in G$. So it remains to show that $G\subseteq c\Bbb{Z}$. Let $g\in G$. Either $g>0$, $g=0$, or $g<0$. We'll handle each case separately.
Case 1: $g>0$. Since $c$ and $g$ are positive reals, it follows from the Archimedean property of $\Bbb{R}$ that $g<k\cdot c$ for some positive integer $k$. Let $m$ be the smallest positive integer with $g<m\cdot c$. So we have that
$$(m-1)\cdot c\le g<m\cdot c.$$
Suppose that the above inequalities are strict, so we have that
$$(m-1)\cdot c<g<m\cdot c.$$
In this case, let $d=g-(m-1)\cdot c$. It follows that $d\in G$ and $0<d<c$. But this is a contradiction, since $c$ was the smallest positive element of $G\cap(-\epsilon,\epsilon)$. It follows that $g=(m-1)\cdot c$. So $g\in c\Bbb{Z}$.
Case 2: If $g=0$, then $g\in c\Bbb{Z}$.
Case 3: If $g<0$, note that $-g\in G$ and $-g>0$. It follows from our work in Case 1 that $-g\in c\Bbb{Z}$. Hence $g\in c\Bbb{Z}$. $\;\Box$
