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I've read the following question:

Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)

and it made me wonder. It's easy to see that $\operatorname{SL}_2(\mathbb{Z})=\operatorname{Sp}_2(\mathbb{Z})$. So does it remain true that $\operatorname{Sp}_{2n}(\mathbb{Z})$ has only finitely many finite subgroups, if $n$ is general? If not, can we still say something about the possible orders of its finite subgroups? (For example, must they be divisible by only finitely many primes?)

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    $\begingroup$ Yes, there's a bound on the order that depends on $n$. The idea is to show that most congruence subgroups are torsion-free. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 20, 2016 at 7:43
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    $\begingroup$ For any Chevalley group $G$ (e.g., ${\rm{Sp}}_{2n}$) there is a systematic approach to getting a uniform upper bound (in terms of the root system) on the size of finite subgroups, even working inside $G(k)$ for a fixed number field $k$. See Theorem 5 in section 5.4 of part II of college-de-france.fr/media/jean-pierre-serre/… (where $t=\ell-1$ for $\ell$ unramified in $k$). The conjugacy aspects seem to be much more subtle (especially if you only work with $\mathbf{Z}$-points); see Theorem 8 in section 6.6 of part II for a sample. $\endgroup$
    – nfdc23
    Commented Mar 20, 2016 at 7:51
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    $\begingroup$ It has infinitely many finite subgroups. Probably you have another question in mind, namely: does it have finitely isomorphism type of subgroups (yes, because it's virtually torsion-free, and you have bounds on the possible orders using the embedding into $\mathrm{SL}_{2n}(\mathbb{Z})$), or, more interesting, does it have finitely many conjugacy classes of finite subgroups. The latter is still true, but less obviously and I'm not sure of a reference (and unlike the previous question, it does not boil down to $\mathrm{SL}_{2n}(\mathbb{Z})$). $\endgroup$
    – YCor
    Commented Mar 20, 2016 at 9:15
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    $\begingroup$ As mentioned in MO106338, Zassenhaus proved the finiteness result ( following results of Blichfeldt,Schur, Jordan etc), and there is a proof in the 1962 edition of Curtis and Reiner. $\endgroup$
    – Geoff Robinson
    Commented Mar 20, 2016 at 10:00
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    $\begingroup$ @GeoffRobinson: just to make your statement more precise: the link is mathoverflow.net/questions/106338, and it refers to the finiteness of the number of conjugacy classes of finite subgroups in $\mathrm{GL}_m(\mathbb{Z})$. But this does not imply that all its subgroups also have finitely many conjugacy classes of finite subgroups; it's probably false in general, while it's certainly true for $\mathrm{Sp}_{2n}(\mathbb{Z})$. $\endgroup$
    – YCor
    Commented Mar 20, 2016 at 10:23

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This seems to be done by Markus Kirschner (see these 2011 talk notes).

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  • $\begingroup$ The problem (of the classification, not just the finiteness result) is indeed tackled these slides. It provides some kind of procedure; he obtains classification of maximal finite subgroups up to dimension $2n=22$. $\endgroup$
    – YCor
    Commented Mar 20, 2016 at 18:24

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