Finite Subgroups of GL(n,R) 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.

I might as well mention Jordan's theorem: There is a function $f: \mathbb{N} \to \mathbb{N}$ such that whenever $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C}),$ there is Abelian $A \lhd G$ with that $[G:A] \leq f(n).$ Explicit bounds were given by Blichfeldt and Frobenius.
To give some general background for the statement regarding only finitely many isomorphism types of subgroups of ${\rm GL}(n,\mathbb{Z})$, let me mention another classical result (perhaps of Schur, though I am not certain). Let $K$ be a finite (degree) extension of the rationals, and suppose that $K$ contains $m$ roots of unity.
Then any finite subgroup of ${\rm GL}(n,K)$ has order less than $(2n)^{m^n}.$
