$\DeclareMathOperator{\gl}{GL}\DeclareMathOperator{\sl}{SL}$From de la Harpe's book "Topics in Geometric Group Theory" I learnt that $\gl(n,\mathbb{Z})$ is generated by the matrices $$s\_1 = \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{bmatrix}$$ $$s\_2 = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ $$s\_3 = \begin{bmatrix} 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ $$s\_4 = \begin{bmatrix} -1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ whenever $n \geq 2$. It can be shown that $s\_1, s\_3, s\_4$ suffice (a result by Hua and Reiner). The matrices $s\_1$ and $s\_3$ generate $\gl(n,\mathbb{Z})$ when $n$ is even and $\sl(n,\mathbb{Z})$ when $n$ is odd. But what about the odd case?



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> For odd $n$, is $\gl(n,\mathbb{Z})$ generated by two elements?
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Minimal number of generators for $GL(n,\mathbb{Z})$
Assuming the assertions your claim are true, it follows that $\mathrm{GL}\_n(\mathbf{Z})$ is generated by 2 elements for all $n$; for $n$ odd (your missing case), the matrices $-s\_1,s\_3$ indeed generate $\mathrm{GL}\_n(\mathbf{Z})$.


This follows from the following three facts:


1) Let $p$ be prime and $C\_p=\langle c\rangle$ the cyclic group of order $p$. If $G$ is an arbitrary group with $\mathrm{Hom}(G,C\_p)=0$, and if $G$ is generated by $k\ge 1$ elements, then so is $G\times C\_p$. Proof: if $g\_1,\dots,g\_k$ generate $G$, then $(g\_1,c),(g\_2,1)\dots,(g\_k,1)$ generate $G\times C\_p$. Indeed they generate a group projecting onto both factors and hence if by contradiction it's a proper subgroup, this subgroup defines the graph of an isomorphism between some quotient of $G$ with $C\_p$, contradiction.


2) for all $n\neq 2$, $\mathrm{SL}\_n(\mathbf{Z})$ is perfect and in particular $\mathrm{Hom}(\mathrm{SL}\_n(\mathbf{Z}),C\_2)=0$. Proof: this is obvious for $n\le 1$ and for $n\ge 3$ it's generated by the elementary matrices $e\_{ij}$, $i\neq j$, each of which being a commutator, namely $e\_{ij}=[e\_{ik},e\_{kj}]$ for some $k\notin\{i,j\}$.


3) If $n$ is odd then $\mathrm{GL}\_n(\mathbf{Z})\simeq\mathrm{SL}\_n(\mathbf{Z})\times C\_2$, where $C\_2$ corresponds to $\{\pm I\_n\}$.


