Recently I have met an interesting problem $\rho$: $G \rightarrow SL(2,R)$ be a faithful representions of a finite group by real 2\times 2 matrices of determinant 1, then we can get this group is cylic.


what is more, how can we determine all finite groups wich have a faithful real two dimensional representation? I feel it have some connections with finite subgroup in $SL(2,R),SU\_{2},U\_{2}$


more generally, determine all finite groups which have a faithful real n dimensional representation? 


I do not konw how I takle with these problems



finite groups with faithful real two dimensional representation
Since it's not clear, I assume the question you're asking is the following. "Fix an $n\geq2$. Which finite groups have a faithful real $n$-dimensional representation? Equivalently, what are the finite subgroups of $\operatorname{GL}\_n \mathbb R$?" (If you just want to know which finite groups admit faithful real representations then the answer is easy: all of them.)


Unfortunately, it's hopelessly difficult to answer this question for general $n$. See the answers of Richard Borcherds and Geoff Robinson [here](https://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries) for an explanation.


For $n=2$, though, there is a simple answer. It turns out that a finite subgroup of $\operatorname{GL}\_2\mathbb R$ is either cyclic or dihedral. Here's a sketch of the argument (for a slightly expanded version, see Rees, *Notes on Geometry*). First note that such a subgroup $G$ can be conjugated into the orthogonal group $\operatorname{O}(2)$ (just average an inner product over $G$). Thus we may as well assume that $G \subset \operatorname{O}(2)$. Now consider the subgroup $H = G \cap \operatorname{SO}(2)$, which consists of rotations. It's not too hard to show that $H$ is in fact generated by one of these rotations -- namely the one whose rotation angle is the smallest. Thus if $H = G$, then $G$ is cyclic and we're done. Otherwise $[G : H] = 2$ and $G \backslash H \subset \operatorname{O}(2) \backslash \operatorname{SO}(2)$ consists of reflections. This forces $G$ to be dihedral.


There is also a relatively nice answer for $n=3$, but the list of possible subgroups this time is bigger...




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**Edit:** You can find a list of the finite subgroups of $\operatorname{O}(3)$ (hence of $\operatorname{GL}\_3 \mathbb R$) [here](http://www.gap-system.org/~john/MT4521/Lectures/L12.html). Let me briefly indicate how you get this list. As in the case of $\operatorname{GL}\_2 \mathbb R$, you start off by determining the finite subgroups of $\operatorname{SO}(3)$. These are given by Neil in his comment below. Next you use the fact that $\operatorname{O}(3) = \operatorname{SO}(3) \times \{\pm I\}$ together with [Goursat's lemma](http://en.wikipedia.org/wiki/Goursat%27s_lemma) to get the complete list.


