# 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.

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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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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).