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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 pete for this nice answer.I must have used my brain little bit before asking this one.2011-03-04
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    No Pete, I have still one doubt. When $[G:H] = 2$, you are claiming that $G = D_n$. Now to show this, it is enough to show that our $G$ contains a reflection of the form $\begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}$ after some conjugation. Since our field is real we cannot claim that any matrix outside rotation group is diagonalizable.2011-03-04
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    @Anjan: every matrix of order $2$ over a field of characteristic different from $2$ is diagonalizable.2011-03-04
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    I agree with you. But that diagonal matrix may be $-I$ also. You need a reflection. I guess you are arguing that since $G$ has order even it must have an element of order 2. Why that element is reflection.2011-03-04
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    Sorry, I think I got my extension the wrong way round. The result is still true, but it is less obvious that the extension splits as a semidirect product. Off the top my head, one can use for instance the Cartan-Dieudonne Theorem, which implies here that any element of $O(n) \setminus SO(n)$ is a reflection: see e.g. Theorem 32 http://math.uga.edu/~pete/quadraticforms.pdf. I have to run now; I will touch up my answer later.2011-03-04
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    @Pete: any element $g \in G \setminus H$ when composed by a reflection is a rotation. So it looks like \begin{pmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{pmatrix}. Its square is identity and therefore it is diagonalizable to reflection. Cartan–Dieudonné theorem says $g$ is a odd number of reflection as its determinant is -1. Please tell me if my argument is correct.2011-03-05
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    @Anjan: well, the way I'm thinking of it: in this very special case, Cartan–Dieudonné says that any element of $O_2(\mathbb{R})$ is either a reflection [in the general form of this theorem, reflection means "orthogonal hyperplane reflection", so here it means linear reflection] or a product of two reflections. Since a product of k reflections has determinant $(-1)^k$, in dimension $2$ an element of negative determinant must be a linear reflection.2011-03-05
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    Thanks for this interpretation2011-03-06
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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 $O(2,\mathbb R)$.