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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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Is $G$ a finite group? – user38268 Sep 21 '12 at 16:12
@BenjaLim: Yes. – Xodarap Sep 21 '12 at 16:28
up vote 6 down vote accepted

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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Why do we need inner products? $\rho$ is an injective homomorphism and so $G \cong \rho(G) \le \text{SL}_2(\mathbb{R})$. Or have I oversimplified matters? – Clive Newstead Sep 21 '12 at 16:34
@CliveN.: so $G$ is a subgroup of $SL_2(\mathbb{R})$ - and then? (the inner product was to show that $G$ is a subgroup of the circle $SO_2$) – user8268 Sep 21 '12 at 16:38
Oh of course! Sorry, I missed the magical transformation of L into O ;) – Clive Newstead Sep 21 '12 at 16:39

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.

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