# Discrete subgroups of $O(2)$ are all finite. [closed]

How to prove that discrete subgroups of the orthogonal group in dimension $$2$$ are all of finite order?

## closed as off-topic by user98602, José Carlos Santos, Shaun, Saad, ancientmathematicianJan 1 at 16:25

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It suffices to note that $$O(2)$$ is compact.
Suppose that $$G$$ is an infinite subgroup of $$O(2)$$. Then $$G$$ contains an infinite sequence $$(g_k)_{k \in \Bbb N}$$. Because $$O(2)$$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $$g_k \to g \in O(2)$$.
Verify that the induced topology on $$\{g_k\}_{k \in \Bbb N}$$ is not the discrete topology. In particular, note that any open subset of $$O(2)$$ containing $$g$$ must contain all but finitely many elements of the sequence.
• But how do I prove that $O(2)$ is compact? – Dbchatto67 Jan 1 at 6:14
• By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $\Bbb R^{2 \times 2}$ – Omnomnomnom Jan 1 at 6:18
• So by using Heine-Borel theorem it follows $O(2)$ is compact. – Dbchatto67 Jan 1 at 7:32