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

Please help me in this regard. Thank you very much.

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closed as off-topic by user98602, José Carlos Santos, Shaun, Saad, ancientmathematician Jan 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.

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  • $\begingroup$ You use Bolzano-Weierstrass theorem. Right? $\endgroup$ – Dbchatto67 Jan 1 at 6:03
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    $\begingroup$ But how do I prove that $O(2)$ is compact? $\endgroup$ – Dbchatto67 Jan 1 at 6:14
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    $\begingroup$ By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $\Bbb R^{2 \times 2}$ $\endgroup$ – Omnomnomnom Jan 1 at 6:18
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    $\begingroup$ So by using Heine-Borel theorem it follows $O(2)$ is compact. $\endgroup$ – Dbchatto67 Jan 1 at 7:32
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    $\begingroup$ @Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that. $\endgroup$ – user98602 Jan 1 at 8:44

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