Timeline for Finite Subgroups of $SL_2(R)$

8 events

when toggle format	what		by	license	comment
Sep 26, 2012 at 2:53	vote	accept	i. m. soloveichik
Sep 26, 2012 at 0:44	comment	added	i. m. soloveichik		So I think your argument might work over any field (of $char\ne 0$) so that the roots of unity in the field is just $\pm 1$. N'est pas?
Sep 25, 2012 at 19:07	vote	accept	i. m. soloveichik
Sep 26, 2012 at 2:53
Sep 25, 2012 at 15:08	comment	added	YCor		you're right, they're order 4 (if you diagonalize $H$ over complex, it's antidiagonal matrices of determinant 1). I edited accordingly.
Sep 25, 2012 at 15:06	history	edited	YCor	CC BY-SA 3.0	Corrected order 2 issue.
Sep 25, 2012 at 14:25	comment	added	John Shareshian		I am confused by the claim in the third paragraph of the original argument that all elements of $N \setminus Z$ have order two. Consider the cyclic subgroup $H$ of $GL_2({\mathbb C})$ generated by a diagonal matrix whose diagonal entries are the two primitive cube roots of $1$. Let $x$ be a $2 \times 2$ matrix whose diagonal elements are zero and whose off diagonal elements are $1$ and $-1$. Then $x$ has order four and normalizes but does not centralize $H$. Maybe I am missing something. If I am, could you finish instead by noting that $SL_2({\mathbf R})$ has one element of order two?
Sep 25, 2012 at 7:17	history	edited	YCor	CC BY-SA 3.0	added 1796 characters in body
Sep 23, 2012 at 21:34	history	answered	YCor	CC BY-SA 3.0