every proper compact subgroup of the circle group is finite

Question

let $\mathbb T$ be the circle group and $H \subsetneq \mathbb T$ be a proper, compact subgroup. i'm trying to show that $H$ must be finite using only topological arguments, but i'm having trouble doing so.

so i want to show that $1$ is isolated in $H$, as then $H$ is discrete and compact, hence finite. but what if $1$ happens to be a limit point of $H$? does anyone have an idea?

background. i'm trying to give a high level argument for why $\langle x \rangle \subseteq \mathbb T$ must be dense for all infinite order elements $x \in \mathbb T$. i'm aware of low level analysis arguments for this, but i'm not interested in them.

Answer 1

Consider the map$$\begin{array}{rccc}\Pi\colon&\Bbb R&\longrightarrow&\Bbb T\\&x&\mapsto&\bigl(\cos(x),\sin(x)\bigr),\end{array}$$which is a surjective continuous group homomorphism. If $G$ is a subgroup of $\Bbb T$, $\Pi^{-1}(G)$ is a subgroup of $(\Bbb R,+)$. But every subgroup of $(\Bbb R,+)$ is either dense or of the form $\Bbb Z\lambda$, for some $\lambda>0$. If $G$ is compact, then it is closed, and therefore $\Pi^{-1}(G)$ is closed too. If it happened to be dense, then it would have to be equal to $\Bbb R$. But then $G$ would be equal to $S^1$, and it is being assumed that that's not the case. So, $\Pi^{-1}(G)=\Bbb Z\lambda$, for some $\lambda\in\Bbb R$. Can you take it from here?

Answer 2

If $1$ is not isolated, then there is a sequence of elements $g_i \in H\setminus\{1\}$ converging to $1$. That means $G_i = \langle n g_i \mid n \in \mathbb{Z} \rangle \subset H$ because $H$ is a subgroup. Now just show that $\bigcup_{i \geq 1} G_i$ is dense in $S^1$.