The proper subgroups of $A:=\left\lbrace m/p^n : m\in \mathbb{Z}; n \geq 0 \right\rbrace$ Let $p$ be a fixed prime integer. Consider the subgroup
\begin{align}
A=\left\lbrace  \frac{m}{p^n} : m\in \mathbb{Z}; n \geq 0       \right\rbrace
\end{align}
of the additive group $\mathbb{Q}$. Clearly, $A$ contains $\mathbb{Z}$.
For each $i \geq 0$, set $A_i= \lbrace m/p^i : m\in \mathbb{Z} \rbrace$. Clearly, $A_i$ is a proper subgroup of $A$ containing $\mathbb{Z}$.
I claim that the only subgroups of $A$ containing $\mathbb{Z}$ are $A$ itself and $A_i$ ($i\geq 0$). How can I prove that any proper subgroup of $A$ containing $\mathbb{Z}$ must be of the form $A_k$ for some $k\geq 0$?.
Thanks in advance.
 A: A special case of this question was asked relatively recent here. Since nobody answered this 6 months old question even though the answer is in the comments I will answer it now.
Throughout $V$ denotes a subgroup of $A$. We need some lemmas.
Lemma 1: If $v\in V$, then $m\cdot v\in V$ for all $m\in \mathbb{Z}$.
Proof: The proof is basic group theory and induction, so I'll leave it as an exercise.
Lemma 2: Let $n\in \mathbb{N}$. If $\frac{m}{p^n}\in V$ for some $m$ coprime to $p$, then $\frac{1}{p^n}+t\in V$ for some $t\in \mathbb{Z}$. Moreover, if $V$ contains $\mathbb{Z}$ then $\frac{1}{p^n}\in V$.
Proof: Since $m$ and and $p^n$ are coprime, we can find $s,t\in \mathbb{Z}$ so that $s\cdot m - t\cdot p^n = 1$. By Lemma 1, $\frac{1}{p^n}+t = s\cdot \frac{m}{p^n} \in V$. If in addition $V$ contains $\mathbb{Z}$, then $-t\in V$ and so $\frac{1}{p^n}\in V$.
Now we prove the question in hand.
Assume $V$ contains $\mathbb{Z}$. We have three cases.

*

*Case 1: The set $\{n\in \mathbb{N} : \frac{1}{p^n}\in V\}$ is empty.

In that case $V=\mathbb{Z}$. The inclusion $\mathbb{Z}\subseteq V$ is given by assumption. For the other inclusion, let $v\in V$ and $v=\frac{m}{p^n}$ where $m$ is coprime to $p$ (this can always be done as if $m$ is not coprime to $p$ one can write $m=m'p^r$ for some $m'$ coprime to $p$ and $r\in\mathbb{Z}$, then one has $v=\frac{m'}{p^{n-r}}$).
Our goal is to show that $n=0$. If by contradiction it doesn't then, by Lemma 2 we have that $\frac{1}{p^n}\in V$. But this contradicts the assumption.

*

*Case 2: The set $\{n\in \mathbb{N} : \frac{1}{p^n}\in V\}$ is finite.

In that case I claim that $V=A_i$, where $i = \max \{n\in \mathbb{N} : \frac{1}{p^n}\in V\}$. Indeed, let $x\in A_i$, then $x= \frac{m}{p^i}$ for some $m\in \mathbb{N}$. Since $\frac{1}{p^i}\in V$, the claim follows by Lemma 1.
Now to prove the other inclusion, let $v=\frac{m}{p^n}\in V$ with $m$ coprime to $p$. By Lemma 2, we have that $\frac{1}{p^n}\in V$. Therefore $n\leq i$ by definition of $i$. But then $v = \frac{m\cdot p^{i-n}}{p^i}\in A_i$.

*

*Case 3: The set $\{n\in \mathbb{N} : \frac{1}{p^n}\in V\}$ is infinite.

In that case I claim that $V=A$. Let $x=m\cdot \frac{1}{p^n}\in A$ and choose $l>n$ for which $\frac{1}{p^l}\in V$. By Lemma $1$ we have $x=m\cdot p^{l-n} \cdot \frac{1}{p^l}\in V$ which completes the proof.
