TITLE: Prove that the group $\mathrm{GL}(n, \mathbb{Z})$ is finitely generated
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: It is known, that for any Euclidean ring $R$ that $GL_n(R)$ is generated by the elementary matrices (proved via Gaussian elimination). For $\mathbb{Z}$ there are finitely many of them (as there are finitely many invertible elements)