12
$\begingroup$

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!

2 Answers 2

7

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.

  • 2
    @Leon. Diagonal matrices *are* elementary matrices.2015-04-29
6

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.

  • 0
    Thank you! I have one more question: Derek Holt says, that we are only allowed to multiply rows by units of $\mathbb{Z}$. I know that this has to be right, because of the determinantfunction, which has to be equal to 1 or -1 for an invertible Matrix with integers in $\mathbb{Z}$. But how do we get an invertibe Matrix with some integers elements of the set $\mathbb{Z}\backslash {1,-1}$. I think we can't get this with elemetentary matrices. If we could, how? Or do we only find a Matrix generated by elementary matrices, such that these two Matrices operate in the same way on $\mathbb{Z}^n$???2012-01-13