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?

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    $\begingroup$ Not every subgroup of $\mathrm{GL}(n,\mathbb Z)$ is finite. $\endgroup$ – Matt Samuel Mar 17 '16 at 22:23
  • $\begingroup$ Well...then where should I to start? $\endgroup$ – Math yocoo Mar 17 '16 at 22:30
  • $\begingroup$ Maybe I should find the generator of it? But I cannot either find a good way to generate GL(n,Z). $\endgroup$ – Math yocoo Mar 17 '16 at 22:36
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    $\begingroup$ See math.stackexchange.com/questions/98308/… and mathoverflow.net/questions/181366/…. $\endgroup$ – lhf Mar 18 '16 at 1:11
  • $\begingroup$ Wow! Thanks Ihf! This actually helps a lot! $\endgroup$ – Math yocoo Mar 18 '16 at 1:14

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