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

  • 6
    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ thanks for your answers, can you give me some materials about n=3,thanks $\endgroup$ – yaoxiao Oct 6 '11 at 5:26
  • 1
    $\begingroup$ 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. $\endgroup$ – Valerio Capraro Oct 6 '11 at 5:59
  • $\begingroup$ @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 $SO(3)$. $\endgroup$ – Neil Strickland Oct 6 '11 at 12:22
  • $\begingroup$ 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 $\operatorname{SO(3)}$. $\endgroup$ – Faisal Oct 6 '11 at 17:56
  • $\begingroup$ @Faisal: you are right, of course. I meant that every finite subgroup of $SL(3,\mathbb{R})$ is conjugate to a subgroup of $SO(3)$. $\endgroup$ – Neil Strickland Oct 6 '11 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.