A finite group is isomorphic to a subgroup of $GL_n(\mathbb Z)$

Question

I'm trying to prove that any finite group $G$ of order $n\ge 2$ is isomorphic to a subgroup of $GL_n(\mathbb Z)$.

My thoughts:

By Cayley, $G$ is isomorphic to a subgroup of $S_n$. Now I need to establish somehow a connection between $S_n$ and $GL_n(\mathbb Z)$. In the case of $n$ small, say $n=2$, $S_3$ acts on $\{e_1,e_2,e_1+e_2\}$, and this action is faithful, which gives an injective homomorphism $GL_2(\mathbb Z)\to S_3$. But a) I guess I need a homomorphism to the other direction (in this case this is an isomorphism, but for larger $n$ it is not I believe) and b) I don't know how to generalize this for larger $n$.

Answer 1

Any finite group embeds into symmetric group (Cayley's theorem), and symmetric group embeds into $GL_n(Z)$ through permutation matrices.

Answer 2

Why embed $S_3$ into $GL_2$? An easier way is to embed $\phi:S_3 \to GL_3$, where the function for any permutation $\sigma$ gives rise to the map

$\phi(\sigma)(x_1,x_2,x_3)=(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)}).$

This is a homomorphism, and it has a matrix representation, which can be determined by studying how $\phi(\sigma)$ acts on a basis $e_1,e_2,e_3$. You can do this for any $n \in \mathbb N$.