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I know that all finite subgroups of $\operatorname{SL}_2(\mathbb R)$ are cyclic by standard averaging argument. They are all conjugate to some finite subgroup of $\operatorname{SO}_2(\mathbb R)$ and therefore cyclic. My question is how to classify all finite subgroups of $\operatorname{GL}_2(\mathbb R)$.

Thanking you.

2 Answers 2

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The same averaging argument you and rvk describe gives you that any finite subgroup $G$ of $\operatorname{GL}_2(\mathbb{R})$ is conjugate to a subgroup of $O_2(\mathbb{R})$. (For that matter, this is also shown in $\S 1.3$ of these notes. The rest of the notes treat complex and $p$-adic analogues, with applications to classifying finite subgroups of $\operatorname{GL}_n(\mathbb{Q})$.)

Without loss of generality, we may as well assume that $G \subset O_2(\mathbb{R})$. Let $H = G \cap SO_2(\mathbb{R})$. Since $[O_2(\mathbb{R}):SO_2(\mathbb{R})] = 2$, we have $[G:H] = 1$ or $2$. If $G = H$ then $G$ itself is a finite subgroup of $SO_2(\mathbb{R})$, hence cyclic. If $[G:H] = 2$, let $g \in G \setminus H$. Then $\det(g) = -1$, so by the Cartan-Dieudonné Theorem $g$ is a linear reflection and thus $G = \langle H, g \rangle$ is a dihedral group $D_n$.

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    Thanks for this interpretation – 2011-03-06
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Consider a finite subgroup $F$. It acts on $\mathbb R^2$. We can define a $F$-invariant positive definite inner product by Weyl's averaging trick. Then an isomomorphic image of $F$ sits inside $O(2,\mathbb R)$.