Every proper subgroup of $\mathbb{Z}(p^{\infty})$ is of the form $\left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$ for some $n>0$. Let $\displaystyle \mathbb{Z}(p^{\infty})=\bigcup_{n>0} \left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$. Show the following:
a) Every proper subgroup of $\mathbb{Z}(p^{\infty})$ is of the form $\displaystyle \left\langle \, \overline{\frac{1}{p^n}} \, \right\rangle$ for some $n>0$.
b) The multiplication map $\mathbb{Z}(p^{\infty}) \to \mathbb{Z}(p^{\infty})$ by $p^t$ is a surjective group homomorphism. Also find the kernel of this map.
I have noted that if $z \in \mathbb{Z}(p^{\infty})$, then $z=a/p^m$ for some $m>0$. I am not sure how to proceed.
 A: Note that $\bar{x}$ denotes the residue class of $x \in \Bbb Q$ modulo $\Bbb Z \unlhd \Bbb Q$.

Assume that $H \lneq \Bbb Z(p^\infty).$ For $t \ge 0$, define $\Bbb Z(p^t) = \left\langle \overline{\dfrac{1}{p^t}} \right\rangle.$ Note that
$$\Bbb Z(p^0) \subset \Bbb Z(p^1) \subset \Bbb Z(p^2) \subset \cdots$$
and
$$\Bbb Z(p^\infty) = \bigcup_{t \ge 0}\Bbb Z(p^t).$$
Note that $\Bbb Z(p^0)$ is the trivial group and hence $\Bbb Z(p^0) \le H.$ Now, since $H \lneq \Bbb Z(p^\infty),$ there is some $t \ge 0$ such that $\Bbb Z(p^{t+1}) \not\subseteq H$. Fix the smallest such $t$.
Thus, we have $\Bbb Z(p^t) \subset H$. We now show the reverse inclusion.
For the sake of contradiction, assume that there exists $x \in H \setminus\Bbb Z(p^t)$. Then, $$x = \overline{\dfrac{r}{p^m}}$$
for some $r \in \Bbb N,$ $m \ge 0$ with $\gcd(r, p) = 1.$ However, note that since $x \notin \Bbb Z(p^t)$, we have $m \ge t + 1$.
Now, by Bézout's identity, there exist integers $a, b$ such that $$ar + bp^m = 1 \quad\text{or}\quad a\frac{r}{p^m} + b = \frac{1}{p^m}.$$
Going modulo $\Bbb Z$ shows that $$\overline{\frac{1}{p^m}} = a\overline{\frac{r}{p^m}} \in \left\langle \overline{\frac{r}{p^m}} \right\rangle = \langle x \rangle \subset H.$$
But the above contradicts that $\Bbb Z(p^{t + 1}) \not\subset H$.
Thus, we get that $H = \Bbb Z(p^t)$.

An aside. The above shows that all the proper subgroups of $\Bbb Z(p^\infty)$ are finite. In particular, all descending chains of subgroups stabilise. This shows that $\Bbb Z(p^\infty)$ is an Artinian $\Bbb Z$-module. However, the chain $$\Bbb Z(p^0) \subsetneq \Bbb Z(p^1) \subsetneq \Bbb Z(p^2) \subsetneq \cdots$$ shows that it is not Noetherian.


Showing that the map is subjective should not be too difficult. Forgetting the residue classes for a moment, think about dividing by $p^t$ in $\Bbb Q$. Conclude from there.
Now, note that the kernel is not the whole group, since the map is subjective. Thus, the kernel is of the form $\Bbb Z(p^n)$. Now you can just check for the largest $n$ such that $\overline{\dfrac{1}{p^n}} \mapsto 0.$
