17
$\begingroup$

$\DeclareMathOperator{\gl}{GL}\DeclareMathOperator{\sl}{SL}$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?

For odd $n$, is $\gl(n,\mathbb{Z})$ generated by two elements?

$\endgroup$
17
$\begingroup$

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 $\{\pm I_n\}$.

$\endgroup$
  • $\begingroup$ This is very convincing, thanks a lot. $\endgroup$ – eins6180 Sep 20 '14 at 21:55
6
$\begingroup$

Though it's a good exercise to work things out directly (as YCor does), the ideas here are all fairly old and can be extracted without much difficulty from the literature as Dietrich's reference indicates. An old paper of Hua-Reiner, for instance, is freely available online here. To supplement this with more details about generators, there is a discussion over more general rings than $\mathbb{Z}$ (but specialized frequently to $\mathbb{Z}$) in the first five sections of Chapter VII in the textbook by M. Newman, Integral Matrices, Academic Press, 1972.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.