# Prove that the group $\mathrm{GL}(n, \mathbb{Z})$ is finitely generated

Knowing that for $n \geq 2$, $\mathrm{GL}(n, \mathbb{Z}) = \big\{ A \in \mathrm{M}_{n,n}(\mathbb{Z}) \mid \det(A) \in \{ 1, −1 \} \big\}$ is a group with respect to matrix multiplication, prove that for every integer $n \geq 2$ the group $\mathrm{GL}(n, \mathbb{Z})$ is finitely generated.

If I prove that $\mathrm{GL}(n, \mathbb{Z})$ has finite subgroups does that mean it has a finite set of generators so that it is finitely generated?

• Not every subgroup of $\mathrm{GL}(n,\mathbb Z)$ is finite. – Matt Samuel Mar 17 '16 at 22:23
• Well...then where should I to start? – Math yocoo Mar 17 '16 at 22:30
• Maybe I should find the generator of it? But I cannot either find a good way to generate GL(n,Z). – Math yocoo Mar 17 '16 at 22:36
• – lhf Mar 18 '16 at 1:11
• Wow! Thanks Ihf! This actually helps a lot! – Math yocoo Mar 18 '16 at 1:14