Can you show any finite subgroup of $SL\_2(R)$ is cyclic without using an invariant form?



Finite Subgroups of $SL_2(R)$
Here's one approach with no geometry/integration, using finite groups. First prove it for a (nontrivial) finite $p$-group. If $p$ is odd, it contains a central element of order $p$, whose centralizer is $\mathbf{C}$-diagonalizable, hence embeddable as a $p$-subgroup of the nonzero complexes; this has to be cyclic. For $p=2$, it contains $-I$, and you need to proceed with the same argument in $\text{PSL}(2,\mathbf{R})$, so your group stands as an extension with order 2 kernel and cyclic quotient (of order some power of 2), with only one element of order 2: lifting a generator from the quotient we obtain a group which is either cyclic or direct product of cyclic with $\{I,-I\}$; since there's only one element of order 2, we deduce that that such a group is cyclic. So all Sylow subgroups are cyclic: this was the easy part.


Now a group whose Sylow subgroups are cyclic is called a [Z-group](http://en.wikipedia.org/wiki/Z-group) and these are known, apparently by Zassenhaus in the 30's, that such a group is metacyclic. I don't know how hard is this result, but it is certainly not relying on the Classification, nor any deep part of it.


So let's go ahead with a (nontrivial) metacyclic group $G$. It contains a nontrivial normal cyclic subgroup $H$, with cyclic quotient. If $H$ has order 2, the lifting argument above shows that the group is either cyclic or direct product of a cyclic group of order 2 and another cyclic group, and in this case, the latter has to be of odd order, so the whole group is cyclic anyway. Otherwise, $H$ has order $\ge 3$, and hence preserves exactly two lines in $\mathbf{R}^2$. So the normalizer $N$ of $H$ in $\text{GL}\_2(\mathbf{C})$ preserves this unordered pair of lines. Let $Z$ be the centralizer of $H$ in $\text{GL}\_2(\mathbf{C})$, which is abelian and has index 2 in $N$. Let us check that any two elements $g,h$ in $G\cap (N-Z)$ commute. Observe that all elements in $N-Z$ are square roots of $-I$. It follows that $g$ conjugates $gh$ to its inverse $h^{-1}g^{-1}$ (which is equal to $-hg^{-1}=g^2hg^{-1}$). Now if $x$ is a element in $\text{SL}\_2(\mathbf{R})$ of finite order $\ge 3$, its centralizer in $\text{GL}\_2(\mathbf{R})$ is contained in $\text{SL}\_2(\mathbf{R})$, and it is conjugate to its inverse by some element not in $\text{SL}\_2(\mathbf{R})$; it follows that $x$ is *not* conjugate to its inverse in $\text{SL}\_2(\mathbf{R})$. Thus $gh$ has order at most 2 and $g$ and $h$ commute. Thus either $G\subset Z$ and $G$ is commutative, or $G$ is generated by $G\smallsetminus Z$ which is a commutative subset and hence $G$ is commutative anyway. So $G$ is abelian and we're done.


Edit: Here's a way to have a self-contained approach, avoiding the use of Zassenhaus's theorem. 


Taking a minimal counterexample, we get a finite group all of whose maximal subgroups are cyclic. 


Assume that a finite group $G$ is not abelian and has all its maximal subgroups abelian; let us show that $G$ is abelian-by-cyclic. It amounts to proving that one of its maximal subgroups is abelian. Its center cannot be maximal (this holds in any group!). So any maximal subgroup contains a non-central element. Since the maximal subgroup is abelian, it is actually the centralizer of any of its non-central elements. Besides, any maximal subgroup contains the center $Z$ of $G$ (otherwise $G$ would be abelian). It follows that $M\cap N=Z$ for any two distinct maximal subgroups.
Let $(M\_i)\_{1\le i\le k}$ representatives of conjugacy classes of maximal subgroups. Let $r\_i$ be the index of the normalizer of $M\_i$, so that $M\_i$ has exactly $r\_i$ conjugates. Since the maximal subgroups minus $Z$ partition $G\smallsetminus Z$, and setting $z=|Z|$ and $m\_i=|M\_i|$, $g=|G|$, we get 
$z+\sum\_i (m\_i-z)r\_i=g$. Assume by contradiction that for all $i$, $M\_i$ is not normal; thus $m\_ir\_i=g$ for all $i$, and we get $(k-1)g=z(-1+\sum\_i r\_i)$. If $M\_i$ is the centralizer of some $x\_i$, then $r\_i$ is the cardinal of the conjugacy class of $x\_i$. These conjugacy classes are disjoint modulo $Z$, hence $\sum r\_i\le g/z$, hence $k<2$. So $k=1$ and the above equation then reads $z(-1+r\_1)=0$, so $r\_1=1$, a contradiction, so $G$ is abelian-by-cyclic. 


Turning back to our minimal counterexample, it is thus metacyclic. The first and third part of the previous argument, which do not rely on the Zassenhaus theorem, show that any metacyclic subgroup of $\text{SL}\_2(\mathbf{R})$ is cyclic.


