Regarding minimal elementary generators for $GL(n, \mathbb{Z})$ I have a result concerning the minimal number of elementary generators (and by this I mean generators which are elementary matrices) for $GL(3, \mathbb{Z})$. I'm currently working on extending the result to $GL(n, \mathbb{Z})$ and I'd like to know if someone already did this before I spend too much time on it. Any help would be greatly appreciated.
Thanks!
 A: The optimal number is $n$. Write $d_i$ the diagonal matrix with $-1$ at position $ii$ and $1$ at other diagonal places. For $i\neq j$, write $P_{ij}$ the transposition matrix $i\leftrightarrow j$, and $e_{ij}=I+E_{ij}$.
First, as you observed, $n$ can be achieved. Indeed, $\langle e_{12},e_{21},P_{1j},3\le j\le n\rangle$ generate for all $n\ge 3$ (for $n=2$ take $e_{12},P_{12}$ instead). Indeed, for $i,j\ge 3$ we have $P_{1i}e_{12}P_{1i}=e_{i2}$,  $P_{1j}e_{21}P_{1j}=e_{2j}$ and for such $i,j$ with $i\neq j$ we get $e_{ij}=[e_{i2},e_{2j}]$, and for a single $i$ we get $e_{1i}=[e_{12},e_{2i}]$ and similarly get $e_{2i},e_{i1},e_{i2}$. So the group they generate contains all $e_{ij}$, $1\le i\neq j\le n$, which generates $\mathrm{SL}_n(\mathbf{Z})$; since we also have $P_{13}$ of determinant $-1$, this generates $\mathrm{GL}_n(\mathbf{Z})$.
Conversely let us check that $n$ cannot be improved. Consider a set $S$ consisting of $k$ such matrices (or their powers, to allow all $e_{ij}^k=1+kE_{ij}$). Consider the graph with vertices $\{1,\dots,n\}$ and an edge from $i$ to $j$ whenever some matrix $P_{ij}$ or $e_{ij}$, $e_{ji}$ occurs; label this edge with the corresponding element of $S$. If this graph is not connected, clearly this does not generate (some nontrivial direct sum decomposition is preserved). If connected, after removing edges we get a tree, and which has one more vertex than edges. So, originally, the number $k'$ of edges (which is $\le k$) is $\ge n-1$. In case of equality $k'=n-1$, this is indeed a tree. If all edges are of the form $P_{ij}$, then the group generated by $S$ is finite. Otherwise, there is an edge $e_{ij}$. Consider the subtree "on the left", that is, all elements in $\{1,\dots,n\}$ which are closer to $i$ than $j$ in this tree. It forms a nonempty proper subset $I$. Up to a permutation of indices (possibly changing the value of $i$), we can suppose that $I$ is an initial segment $\{1,\dots,i\}$. This yields a nontrivial block decomposition preserved by $S$ (all but $e_{ij}$ are block-diagonal and $e_{ij}$ is block-upper triangular). Hence if $S$ generates, then $k\ge n$. (This also shows that if $S$ generates an infinite subgroup acting irreducibly on $\mathbf{Q}^n$, then $k\ge n$.)
A: @YCor, for $n = 3$ the answer is yes and was proven today by Bogdan. The proof, in full details, has 8 pages. The 3 elementary matrices that generate $GL_3 (Z)$ are:  $T_{12} (1) := I_3 + e_{12}$, $P_{12}$ and $P_{23}$, where the last two matrices are obtained from $I_3$ switching rows 1 and 2 and respectively 2 and 3. 
