Like every other group also $\mbox{GL}(n,\mathbb{Z})$ acts on the set of all its subgroups, by conjugation: if $\phi \in \mbox{GL}(n,\mathbb{Z})$, then $\phi$ acts by $H \mapsto \phi H \phi^{-1}$, where $H \leq \mbox{GL}(n,\mathbb{Z})$.
A theorem by Jordan (and later Zassenhaus) implies that the number of conjugacy classes of finite subgroups of $\mbox{GL}(n,\mathbb{Z})$ is finite.
About this fact I have a few questions:
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Is the actual number of conjugacy classes known for each $n$? Maybe at least asymptotically?
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Is it known which of these classes are particularly big for every $n$?(Edit: Doesn't make sense in this context, see comments.)
Best Answer
A rough estimate can be obtained by comparison with the number of isomorphism classes of (symmorphic) space groups (see also Agol's comments): The map $Q \mapsto Q \ltimes\mathbb{Z}^n$ induces a bijection between the conjugacy classes $C_n$ of finite subgroups of $GL_n(\mathbb{Z})$ and the isomorphism classes of symmorphic space groups. A proof thereof can be found in my answer to this question:
Subgroups of the Euclidean group as semidirect products
In particular, $|C_n| = 73,\; 710,\; 6079,\; 85311$ for $n=3,4,5,6$ respectively:
http://www.math.ru.nl/~souvi/papers/acta03.pdf
The number of isomorphism classes of (all) space groups has been estimated (cf. Remark 5.5) in
http://www.unige.ch/math/folks/bucher/Affine/pdfAffine/BuserBieberbach.pdf
and yields the upper bound $|C_n| \le e^{\displaystyle e^{4n^2}}$.
Actually it is conjectured by Schwarzenberger in a 1974 paper that the number of isomorphism classes of space groups is asymptotically $O(2^{\displaystyle n^2})$. But I don't know if this has been proved in the meanwhile.