# finite groups with faithful real two dimensional representation

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

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The image of a finite group will be contained in a maximal compact subgroup. The maximal compact subgroup of $SL_2({\mathbb R})$ is $SO(2)$, the circle group. Finite subgroups of the circle are cyclic. If we replace $SL_2$ with $GL_2$, the maximal compact subgroup is $O(2)$. Finite subgroups are either cyclic or dihedral. In higher dimensions, we are looking for finite subgroups of $O(n)$. Life gets complicated. See the answers to this MO question: mathoverflow.net/questions/17072/the-finite-subgroups-of-sun –  B R Oct 6 '11 at 5:03

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 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...

Edit: You can find a list of the finite subgroups of $\operatorname{O}(3)$ (hence of $\operatorname{GL}_3 \mathbb R$) here. 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 to get the complete list.

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thanks for your answers， can you give me some materials about n=3，thanks –  yaoxiao Oct 6 '11 at 5:26
Just a comment that is always good to keep in mind. It is true that every finite group embeds into some $GL(n,\mathbb R)$, but this is no longer true for finitely generated groups: Malcev's theorem states that f.g. subgroups of $GL(n,\mathbb R)$ are residually finite. –  Valerio Capraro Oct 6 '11 at 5:59
@yaoxiao: Finite subgroups of $SO(3)$ are cyclic or dihedral or isomorphic to $S_4$ or $A_4$ or $A_5$. This is well-known and proved in many places, including Section 11 of these notes: shef.ac.uk/nps/courses/groups/notes/groups.pdf. More generally, if $G$ is a finite subgroup of $GL(3,\mathbb{R})$ then it will be conjugate to a subgroup of $SO(3)$. –  Neil Strickland Oct 6 '11 at 12:22
Neil, why is your last sentence true? A finite subgroup of $\operatorname{GL_3 \mathbb R}$ is certainly conjugate to a subgroup of $\operatorname{O}(3)$ but not necessarily to a subgroup of $\operatorname{SO(3)}$. –  Faisal Oct 6 '11 at 17:56
@Faisal: you are right, of course. I meant that every finite subgroup of $SL(3,\mathbb{R})$ is conjugate to a subgroup of $SO(3)$. –  Neil Strickland Oct 6 '11 at 21:35
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