Timeline for Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)
Current License: CC BY-SA 2.5
10 events
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| Jun 7, 2010 at 14:00 | comment | added | Skip | To translate what Venselaar writes ("is not the best possible"): Serre shows that the Minkowski-style upper bound is optimal at every odd prime p. That is, if Minkowski says finite subgroups have order <= M, then for every prime p dividing M, there is a finite subgroup H such that the same power of p divides M and |H|. | |
| Jun 7, 2010 at 13:25 | comment | added | Pete L. Clark | For those who are interested: I wrote up the arguments from Serre's Lie Algebras and Lie Groups on finite groups of $\operatorname{GL}_n(\mathbb{Q})$ here: math.uga.edu/~pete/8410Chapter9.pdf | |
| Jun 7, 2010 at 11:53 | comment | added | Roland Bacher | Indeed. Nice argument! | |
| Jun 7, 2010 at 11:15 | comment | added | Keivan Karai | @Bacher: Serre considers the finite group G as a subgroup of $SL(n,Z_p)$ where $Z_p$ is the ring of $p$-adic integers. Now, it is easy to see that the kernel of the reduction map from $SL(n, Z_p)$ to $SL(n, F_p)$ is torsion-free. This implies that $G$ maps injectively into $SL(n, F_p)$ | |
| Jun 7, 2010 at 10:47 | history | edited | Roland Bacher | CC BY-SA 2.5 |
Added a second part: (Starting from: I intended ...)
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| Jun 7, 2010 at 9:46 | comment | added | Jan Jitse Venselaar | For $GL_3(\mathbb{Z})$ and $SL_3(\mathbb{Z})$ you could look at "On the finite subgroups of GL(3,Z)" by Tahara in Nagoya Mathematical Journal Vol 41. Apparently, Serre's upper bound on the cardinalities of finite subgroups is actually due to Minkowski, and is not the best possible. There is a sharper bound by Minkowski ("Zur theorie der positiven quadratischen formen") which is the best possible if $n=3$. | |
| Jun 7, 2010 at 9:31 | comment | added | Torsten Ekedahl | Indeed, for $p>2$ the reduction is injective and for $p=2$ the kernel consists of elements of order $2$ for a general $\mathrm{GL}_n(\mathbb Z)$. Hence, in the case of $\mathrm{SL}_2(\mathbb Z)$ we get that the kernel has at most order $2$ and order of a finite subgroup is a divisor of $2\cdot|\mathrm{SL}_2(\mathbb Z)/2|=12$. If we restrict ourselves to commutative subgroups we get a divisor of $4$ or $6$ (as the order of elements of $\mathrm{SL}_2(\mathbb Z)/2$ are $1$, $2$ or $3$) but that also follows from the fact that these are the only $d$ with $\varphi(d)\leq 2$. | |
| Jun 7, 2010 at 8:32 | comment | added | Roland Bacher | No idea. I remember having seen a proof in a lecture by Serre giving an upper bound for cardinalities of finite subgroups in $SL_n(\mathbb Z)$ based on the fact that the canonical morphism into $SL_n(\mathbb F_p)$ is in fact injective when restricted to finite subgroups even for quite small primes (I believe even $p=3$ works.) The argument (which I forgot) was very elementary. | |
| Jun 7, 2010 at 8:17 | comment | added | Robin Chapman | Does this generalize to $SL_n(\mathbb{Z})$? | |
| Jun 7, 2010 at 7:43 | history | answered | Roland Bacher | CC BY-SA 2.5 |