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!


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.

  • $\begingroup$ 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? $\endgroup$ – Peter Jan 12 '12 at 10:38
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    $\begingroup$ 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). $\endgroup$ – Derek Holt Jan 12 '12 at 15:01
  • $\begingroup$ 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 $GL_n$? $\endgroup$ – Leon Apr 29 '15 at 2:39
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    $\begingroup$ @Leon. Diagonal matrices are elementary matrices. $\endgroup$ – James Apr 29 '15 at 13:09

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

For any Euclidean domain R, 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 (https://www.math.ucla.edu/~rse/algebra_book.pdf).

  • $\begingroup$ 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 $\mathbb{Z}^n$??? $\endgroup$ – Peter Jan 13 '12 at 13:29

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