Let $p$ be a prime integer. Let $C(p^n)$ be the cyclic group of order $p^n$ ($C(p^n)=\{z \in C\mid z^{p^n}=1\}$ and let $C(p^\infty)$ be the union of the groups $C(p^n)$, i.e., $C(p^\infty)=\{z\in C\mid z^{p^n}=1 \text{ for some positive }n=n(z)\}$.
Prove:
a) $C(p^\infty)$ is a group under multiplication
b) $C(p^\infty)$ is not cyclic
c) Every proper subgroup is a proper cyclic group (one of the groups $C(p^n)$)
We say $C_p\subset C_{p^2}\subset C_{p^3}\subset\cdots$ is a chain of subsets in order to interpret the union. An important fact about this construction is that every element of $C(p^\infty)=\bigcup C(p^n)$ is in one of the $C(p^n)$s.
Prove that every two elements in $C(p^\infty)$ are contained in some $C(p^n)$. (In fact, every finite subset is contained in one the cyclic groups.) Use this to argue $C(p^\infty)$ is indeed a group.
Suppose $C(p^\infty)$ is generated by $x$. In view of the aforementioned "important fact" what can you conclude? Now this will tell you $C(p^\infty)$ isn't cyclic.
Considering general subgroups requires slightly more insight. Every element of $C(p^\infty)$ can be "tagged" by the cyclic subgroup it generates. Think about the $n$'s in $\{n:\langle x\rangle=C(p^n),x\in H\}$.