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A nice result about $GL(n,\mathbb{Z})$ is that it has finitely many finite subgroups upto isomorphism; and also any finite subgroup of $GL(n,\mathbb{Q})$ is conjugate to a subgroup of $GL(n,\mathbb{Z})$.

Next, I would like to ask natural question, what can be said about finite subgroups of $GL(n,\mathbb{R})$, $GL(n,\mathbb{C})$; at least for $n=2$ . Does every finite group can be embedded in $GL(2,\mathbb{R})$? (I couldn't find some reference for this.)

Only thing I convinced about $GL(2\mathbb{R})$ is that it contains elements of every order and hence cyclic groups of all finite order.

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    Can you perhaps give a reference to the "finitely many subgroups" result for $GL(n, \mathbb{Z})$? Or even a list of the subgroups of $GL(2, \mathbb{Z})$?2011-08-19
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    Finite subgroups of $GL(n,R)$ (resp. $GL(n,C)$)are conjugate to a subgroup of $O(n,R)$ (resp. $U(n,C)$) (it is a classical result). For $n=2$ note that $O(2,R) = S^1 \rtimes \{ -1,1 \}$then you can deduce of finites subgroups of $GL(2)$ up to conjugacy class.2011-08-19

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