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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 $GL(2, \mathbb{Z})$? – user1729 Aug 19 '11 at 11:14
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. – user10676 Aug 19 '11 at 18:19

The relationship between Z and Q is very different from the relationship between R and C.

Not every finite subgroup of GL(2,C) can be embedded in GL(2,R).

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.

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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 $(2n)^{m^n}.$

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A branch of algebra called representation theory gives answers to this kind of questions.

Jack Schmidt gave you examples of groups without injective homomorphisms to GL(2,R). Possibly the simplest (well, the first that occurred to me :-) example of a group that has no isomorphic copies of it inside GL(2,C) is the alternating group $A_5$ . Its irreducible representations have dimension 1,4,5,3 and 3. The 1-dimensional representation maps all the elements of $A_5$ to the identity matrix. The existence of 3-dimensional representations means that there exists a monomorphism from $A_5$ to GL(3,C).

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fwiw, 2×2×2 is the smallest group not embedding into GL(2,C). – Jack Schmidt Aug 19 '11 at 10:48
@Jack: Thanks (and oops!). For some reason I was thinking about irreducible representations only :-) – Jyrki Lahtonen Aug 19 '11 at 11:02

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