While looking around for some abstract algebra problems online to play around with and solve, I found the following problem:
$\mathbb{C}_{p^{\infty}} = \{\exp^{\frac{2\pi i k}{p^n}} : k, n \in \mathbb{N}\}$. Prove that every proper subgroup of $\mathbb{C}_{p^{\infty}}$ is cyclic with $p^n$ elements, where $n = 1, 2, 3, 4, 5,\dots$
I found this question to be interesting although I am also a little stumped as to how to show it. What do you guys think?
Let $G \subset \mathbb{C}_{p^\infty}$ be a proper subgroup. Let $\zeta \in \mathbb{C}_{p^\infty} \setminus G$. Then there are $n \in \mathbb{N}$ and $k \in \mathbb{N}$ with $p \nmid k$ and $\zeta = \exp\left(\frac{2\pi ik}{p^n}\right)$.
Then $g^{p^{n-1}} = 1$ for all $g \in G$. For if $G$ contained an element $g_0$ of order $\geqslant p^n$, say of order $p^m$, then $g_1 = g_0^{p^{m-n}}$ would be a primitive $p^n$-th root of unity, hence $g_1 = \exp\left(\frac{2\pi i r}{p^n}\right)$ for some $r$ with $p \nmid r$. But then $g_1^{k\cdot r^{-1}} = \zeta \in G$, where $r^{-1}$ is the multiplicative inverse of $r$ modulo $p^n$.
So every proper subgroup is contained in the group of $p^n$-th roots of unity for some $n$, and that group is cyclic, hence all subgroups of it are cyclic.