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Suppose there is a faithful representation $\rho:G\to SL_2(\mathbb{R})$. Prove that $G$ is cyclic.

I know there has to be something special about its representation being special (no pun intended) because e.g. the Klein 4 group has a non-special representation. Also it has to be important that it's in two dimensions, because $SO(3)$ contains non-cyclic groups.

Apart from that, I haven't really made any progress. Any hints?

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    @BenjaLim: Yes.2012-09-21

2 Answers 2

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I suppose that $G$ is finite (otherwise it doesn't seem true :)

Start with an inner product on $\mathbb{R}^2$ (e.g. with the standard one) and take its mean w.r.t. the action of $G$; you get a $G$-invariant inner product. $G$ can thus be seen as a finite subgroup of $SO_2(\mathbb{R})$, so it must be cyclic.

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    Oh of course! Sorry, I missed the magical transformation of L into O ;)2012-09-21
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If $G$ is a finite group, then the image of $\rho$ is in $SO_2$ up to conjugacy. Indeed, define

$A = \frac{1}{|G|} \sum_{g \in G} \,^t\rho(g) . \rho(g)$

Then you can check the matrix $A$ defines a symmetric definite positive bilinear form. If you write $A = \,^t B . B$, and $\rho'(g) = B \rho(g) B^{-1}$. Then $\rho' : G \to SL_2(\mathbb{R})$ has its image in $SO_2(\mathbb{R})$ (because $^t\rho(g) . A . \rho(g) = A$ for all $g \in G$). But $SO_2(\mathbb{R})$ is just the unit circle in $\mathbb{C}$ and the only finite subgroups of it are cyclic.