Structure of subgroups of $A = \{ m2^n \mid m,n\in \mathbb{Z} \}$ Let $A = \{ m2^n \mid m,n\in \mathbb{Z} \} \le (\mathbb{Q}, +)$. I am looking to find how subgroups $H\le A$ work. My suspect is that they have some kind of "boundedness" condition on the denominator. I'll explain in more detail what I mean.
Given $H < A$ proper, does there exist a $d\in \mathbb{Z}$ such that $\forall m2^n \in H$ we have $d\le n$?

What I have noted is that there exist obvious subgroups that are of the form
$$H_i = \{ m2^i \mid m\in \mathbb{Z} \} \cong 2^i\mathbb{Z}$$
with $H_i \subset H_{i-1}$ for all $i\in \mathbb{Z}$. These subgroups look like they might be the only ones of $A$, in which case we would have a positive answer. If you have counterexamples or some insight, it would be greatly appreciated.

UPDATE: As suggested in the comments of an answer, a generalization of the possible subgroups would be $m2^n \mathbb{Z}$ with $m,n\in \mathbb{Z}$.
 A: That is wrong.
The group $3\mathbb{Z}$ is a subgroup of $A$.
Edit: To answer the updated question
My proof to The proper subgroups of $A:=\left\lbrace m/p^n : m\in \mathbb{Z}; n \geq 0 \right\rbrace$ implies that if $H$ contains $\mathbb{Z}$, then it takes the form $2^{-i}\mathbb{Z}$ for some $i\geq 0$ (or $H=A$).
Now take $H$ an arbitrary subgroup. Then $H\oplus \mathbb{Z}$ contains $\mathbb{Z}$ and so isomorphic to $2^{-i}\mathbb{Z}$, or to $A$ which we will now informally denote by $2^{-\infty}\mathbb{Z}$.
$H\cap \mathbb{Z}$ is a subgroup of $\mathbb{Z}$ and therefore isomorphic to some $m\mathbb{Z}$ where $m\in\mathbb{N}$.
We now prove that $H=m2^{-i}\mathbb{Z}$, where $i$ is potentially infinity.
To see that $H\subseteq m2^{-i}\mathbb{Z}$, let $h\in H$. The obvious inclusion $H\subseteq H\oplus \mathbb{Z}$ implies that $h\in 2^{-i}\mathbb{Z}$, write $h=\frac{n}{2^i}$ for some $n$ coprime to $2$ (if $i$ is infinity then, write $h=\frac{n}{2^l}$ for any $l$). Indeed, since $h\in H$, we also have that $n\in H$, since $H\cap \mathbb{Z} = m\mathbb{Z}$ we deduce that $n$ is a multiple of $m$. Hence, $h\in m\cdot 2^{-i}\mathbb{Z}$.
Now we prove the other direction. Let $x\in m\cdot 2^{-i}\mathbb{Z}$, we need to prove that $x\in H$. Now write $x = m\cdot t$ where $t\in 2^{-i}\mathbb{Z} = H\oplus \mathbb{Z}$. We can therefore find some $h\in H$ and $z\in \mathbb{Z}$ so that $t=h+z$. Hence, $x=mh + mz$. Clearly $mh\in H$ and since $z\in\mathbb{Z}$ and $H\cap\mathbb{Z} = m\mathbb{Z}$ we deduce that $mz\in H$. Hence, $x\in H$ and the claim follows.
