What is the easiest way to generate $\mathrm{GL}(n,\mathbb Z)$?

Question

I'm searching for a way to generate the group $\mathrm{GL}(n,\mathbb Z)$. Does anyone have an idea? The intention of my question is that I am searching for an easy proof of the existence of the epimorphism:

$\Phi:\mathrm{Aut}(F_n )\to \mathrm{Aut}(F_n/[F_n,F_n])=\mathrm{Aut}(\mathbb {Z}^n)=\mathrm{GL}(n,\mathbb {Z})$

I know that $\Phi$ is an canonical homomorphism since the commutator subgroup is characteristic in $F_n$. So I need some nice generators of $\mathrm{GL}(n,\mathbb {Z})$ to find their preimages in $\mathrm {Aut}(F_n)$ to prove the surjectivity of $\Phi$.

Thanks for help!

Answer 1

From linear algebra we know that every invertible matrix can be obtained from the identity matrix by a sequence of elementary row operations. The corresponding elementary matrices therefore generate the general linear group. Now you can find pre-images in $\mathrm {Aut}(F_{n})$ by considering Nielsen transformations, which match up fairly directly with the elementary matrices.

Answer 2

James has correctly identified the elementary matrices as generating $\operatorname{GL}(n,\mathbb{Z})$. I'd like to address the why - since the relevant fact that $\mathbb{Z}$ is a Euclidean domain is not quite captured in the "from linear algebra" remark in the previous answer.

For any Euclidean domain $R$, $\operatorname{GL}(n,R)$ is generated by elementary matrices. This follows from the proof of Smith normal form. But which proof? Not the usual one that works for an arbitrary principal ideal domain, but a more algorithmic version valid for Euclidean domains only. A presentation is in Chapter 108 ("Smith Normal Form over a Euclidean Ring"; the numbering might shift) of Richard Elman's notes.