Let \( p \) be a prime number. We define the set

\[ Q_p = \left\{ \frac{m}{p^r} \;\middle|\; m \in \mathbb{Z},\; r \in \mathbb{Z},\; r \ge 0 \right\}. \]

This means that \( Q_p \) consists of all rational numbers whose denominators are powers of \( p \). It is easy to verify that:

\( Q_p \) is a subgroup of the additive group \( (\mathbb{Q}, +) \), and
\( \mathbb{Z} \subseteq Q_p \).

Now, consider the quotient group

\[ G = Q_p / \mathbb{Z}. \]

Let \( H \) be any proper non-trivial subgroup of \( G \). By the Correspondence Theorem, there exists a unique subgroup \( K \) of \( Q_p \) such that:

\( \mathbb{Z} \subseteq K \subseteq Q_p \), and
\( H = K / \mathbb{Z} \).

Key Claim

We claim that:

\[ \frac{m}{p^r} \in K \quad \Longleftrightarrow \quad \frac{1}{p^r} \in K. \]

Proof

First, suppose \( \frac{1}{p^r} \in K \). Since \( K \) is a subgroup and hence closed under addition, it follows that:

\[ \frac{m}{p^r} = m \cdot \frac{1}{p^r} \in K \quad \text{for all integers } m. \]

Now, suppose \( \frac{m}{p^r} \in K \). Without loss of generality, assume \( (m, p^r) = 1 \) and \( r > 0 \). Then, there exist integers \( a, b \in \mathbb{Z} \) such that:

\[ am + bp^r = 1. \]

Dividing both sides by \( p^r \), we get:

\[ \frac{1}{p^r} = a \cdot \frac{m}{p^r} + b. \]

Since \( \frac{m}{p^r} \in K \) and \( b \in \mathbb{Z} \subseteq K \), it follows that \( \frac{1}{p^r} \in K \). This proves the claim.

Existence of a Largest Power

Since \( K \) is a proper subgroup of \( Q_p \), there must exist some positive integer \( t \) such that:

\[ \frac{1}{p^t} \notin K. \]

Then for any \( t' > t \), we can write:

\[ \frac{1}{p^t} = \frac{p^{t' - t}}{p^{t'}}. \]

Hence, \( \frac{1}{p^{t'}} \notin K \). Therefore, there exists a smallest positive integer \( n \) such that:

\[ \frac{1}{p^n} \in K \quad \text{but} \quad \frac{1}{p^{n+1}} \notin K. \]

This implies:

\[ \left\langle \frac{1}{p^n} \right\rangle \subseteq K. \]

Determination of Subgroup \( K \)

Let \( \frac{m}{p^r} \in K \) with \( (m, p) = 1 \). Then, by the earlier claim, \( \frac{1}{p^r} \in K \), which implies \( r \le n \).

Hence:

\[ \frac{m}{p^r} = \frac{m p^{n-r}}{p^n} \in \left\langle \frac{1}{p^n} \right\rangle. \]

Therefore, we conclude:

\[ K = \left\langle \frac{1}{p^n} \right\rangle. \]

Structure of Subgroups of \( G \)

Now,

\[ H = K / \mathbb{Z} = \left\langle \frac{1}{p^n} \right\rangle + \mathbb{Z} = \left\langle \frac{1}{p^n} + \mathbb{Z} \right\rangle. \]

Thus, every proper non-trivial subgroup of \( G \) is of the form:

\[ G_n = \left\langle \frac{1}{p^n} + \mathbb{Z} \right\rangle. \]

Also, the order of \( H \) is:

\[ |H| = p^n. \]

Final Conclusion

Since \( H \) was chosen arbitrarily, we conclude that:

Every proper subgroup of \( G \) is of the form \( G_n = \left\langle \frac{1}{p^n} + \mathbb{Z} \right\rangle \).
For each \( n \ge 0 \), the group \( G \) has exactly one subgroup of order \( p^n \).