Show that every proper subgroup of this group is finite. Let $G$ be the group of rational numbers in $[0,1)$ whose
denominator is a power of $2$:
\begin{align*}
G &= \{r/2^k : \text{$r \in \mathbb Z$, $0 \le r < 2^k$, $k = 0, 1, 
                     \ldots$} \} \\
   &=\{0, \frac12, \frac14, \frac34, \frac18, \frac38, 
               \frac58,\frac78, \frac1{16}, \ldots \}
\end{align*}
Addition in $G$ is modulo $1$. So $3/4 + 5/8 = 3/8$.
Show that every proper subgroup of $G$ is finite. 
I was planning to define $A_k = \{r/2^k : r = 0,1, \ldots, 2^k - 1\}$.
Then it is not hard to show that $A_k$ is a subgroup of $G$,
then, $A_k \subseteq A_{k+1}$, where $A_k$ is a cyclic group of order $2^k$,
and $G = \bigcup_{k=0}^\infty A_k$.
Since every $A_i$ is finite, I am done. 
Not sure if I'm overthinking too much, feel like this question is harder than a few lines. Am I missing something here?
Thanks in advance.

Hint: Let $H$ be a proper subgroup of $G$.  Let $\frac{2j+1}{2^k}\in G-H$.  Now argue that no reduced fraction in $G$ with denominator greater than or equal to $2^k$ is in $H$.  Thus all denominators in $H$ (when fractions are expressed in reduced form) are less than $2^k$.  How many such fractions are there?
