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

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

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.

• 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 $\mathbb{Z}^n$??? Commented Jan 13, 2012 at 13:29

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.

• Thanks. But there are only 3 Types of Nielsentransformations for a finite tuple $(u_1,...,u_n)$: T(1) replace some $u_i$ by ${u_i}^1$, (T2) replace some $u_i$ by $u_i u_j$ where $i\neq j$, (T3) delete some $u_i$ if $u_i=1$ But how can I find the row addition if $i\neq j$ is claimed. Whats up with the element in $\mathrm{GL}(n,\mathbb Z)$ which i get if I only multiply the i-th row with some integer of $\mathbb Z$. What's the preimage of this one? Or is this no Element of $\mathrm{GL}(n,\mathbb Z)$ since $\mathbb Z$ is no field? Commented Jan 12, 2012 at 10:38
• Since elements of ${\rm GL}(n,{\mathbb Z})$ need to be invertible, you are only allowed to multiply rows by units of ${\mathbb Z}$ - that is by 1 or $-1$. The preimage of this has type (T1). Commented Jan 12, 2012 at 15:01
• How can $[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}]$ be obtained from $[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]$ by sequence of elementary row operations? Don't you also need diagonal matrices to generate $GL_n$?
– Leo
Commented Apr 29, 2015 at 2:39
• @Leon. Diagonal matrices are elementary matrices. Commented Apr 29, 2015 at 13:09