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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:

  1. Is the actual number of conjugacy classes known for each $n$? Maybe at least asymptotically?

  2. Is it known which of these classes are particularly big for every $n$? (Edit: Doesn't make sense in this context, see comments.)

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    $\begingroup$ I am confused by question (2). These conjugacy classes are infinite, so how are you measuring their size? $\endgroup$
    – David E Speyer
    Sep 4, 2012 at 14:19
  • $\begingroup$ If I counted right $GL_1$ has $2$ and $GL_2$ has $9$. The number should increase fairly rapidly with $n$. $\endgroup$
    – Will Sawin
    Sep 4, 2012 at 15:29
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    $\begingroup$ The largest finite subgroup is the orthogonal group of order $2^nn!$, at least for all $n > 10$. These questions have been discussed on MO before, but I do not believe that any asymptotic results on the number of classes are known. $\endgroup$
    – Derek Holt
    Sep 4, 2012 at 19:14
  • $\begingroup$ @David Speyer: You're right, it doesn't make sense to speak about big conjugacy classes. Since I was thinking about space group and their arithmetic equivalence for the last few days, things got mixed up in my head. Thanks for pointing that out. $\endgroup$
    – Gregor Samsa
    Sep 4, 2012 at 20:42
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    $\begingroup$ I think the number of Bieberbach groups gives an upper bound. For each finite subgroup $G< GL_n(\mathbb{Z})$, one obtains a Bieberbach group $G \ltimes \mathbb{Z}^n$. Moreover, if two Bieberbach groups are obtained this way, then by Bieberbach's theorem, they are affinely equivalent. Since the $\mathbb{Z}^n$ subgroup is a maximal subgroup of translations, this means the affine map must send $\mathbb{Z}^n$ to $\mathbb{Z}^n$, so it lies in $GL_n(\mathbb{Z})$. However, this will be an overcount, since there are Bieberbach groups which are not of this form (e.g. Klein bottle group), ie not split. $\endgroup$
    – Ian Agol
    Sep 5, 2012 at 12:42

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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.

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  • $\begingroup$ All this is very interesting, thank you. $\endgroup$
    – Gregor Samsa
    Sep 12, 2012 at 12:31

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