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.
Best Answer
The relationship between Z and Q is very different from the relationship between R and C.
Finite subgroups of GL(2,R) have a faithful real character of degree 2 whose irreducible components have Frobenius-Schur index 1, while finite subgroups of GL(2,C) have a faithful character of degree 2.
The group C4 × C4 for instance has no faithful real character of degree 2, so is isomorphic to a subgroup of GL(2,C) but not isomorphic to a subgroup of GL(2,R).
The smallest counterexample is the quaternion group of order 8.