any proper subgroup is cyclic We need to prove that any proper subgroup of $C_{p^\infty}$ is cyclic of order $p^n$ for some $n$ ($p$ is prime).
 Consider the chain

$0\subset C_p \subset C_{p^2}\subset \dots$

which has $C_{p^\infty}$ as its union. Let $H$ be a proper subgroup of $C_{p^\infty}$. $g$ is in some $C_{p^n}$, but not in $H$. $|g|=p^n$. Consider $h$ in $H$. $|h|<|g|=p^n$. 
Then I got stuck here and don't know how to proceed further.
 A: It is not hard to see that $C_{p^{\infty}}=\{\frac{a}{p^n}+\mathbb Z\,|a\in\mathbb Z,n\in \mathbb N \}$. Let $H$ be proper subgroup of $C_{p^{\infty}}$. First, we claim that if $\frac{a}{p^n}+\mathbb Z\in H$ and $(a,p)=1$, then  $\frac{1}{p^n}+\mathbb Z\in H$.
There exits $r,s\in\mathbb Z$ such that $ra+sp^n=1$. We can conclude that $\frac{1-sp^n}{p^n}+\mathbb Z\in H$ and so $\frac{1}{p^n}-s+\mathbb Z\in H$ and the claim is proved. Now, we show that there exist $n\in \mathbb N$ such that $H=\langle\frac{1}{p^n}+\mathbb Z\rangle$. Assume that $n$ is the largest number such that $\frac{1}{p^n}+\mathbb Z$. Note that if $n$ doesnt exist, then we can infer that $H=C_{p^{\infty}}$. It is obvious that $\langle\frac{1}{p^n}+\mathbb Z\rangle\subseteq H$. Suppose that $\frac{b}{p^m}+\mathbb Z\in H \setminus\langle\frac{1}{p^n}+\mathbb Z\rangle$ and so $\frac{1}{p^m}+\mathbb Z\in H$. Since $n$ is largest number, $m<n$. Therefore, we have $\frac{1}{p^m}+\mathbb Z\in \langle\frac{1}{p^n}+\mathbb Z\rangle$ as, $\frac{1}{p^m}+\mathbb Z=p^{n-m}(\frac{1}{p^n}+\mathbb Z)$. It is contradiction.
