0

$SL_2(\bf F_3)$ is generated by $A=\begin{bmatrix} 1\ 1 \\ 0\ 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1\ 0 \\ 1\ 1 \end{bmatrix}$.

I think $GL_2(\bf F_3)$ is generated minimally by $\begin{bmatrix}1 & 0\\ 0 & 2 \end{bmatrix}$ and $\begin{bmatrix}2 & 2\\1 & 0\end{bmatrix}.$

I am interested in generating these three elements: $$\begin{bmatrix} 0\ 2 \\ 1\ 1 \end{bmatrix}, \begin{bmatrix} 1\ 0 \\ 1\ 1 \end{bmatrix}, \begin{bmatrix} 1\ 2 \\ 1\ 1 \end{bmatrix}$$ each have absolute value of determinant $1$. How can I use the generators to generate these elements? Is there a systematic procedure?

Turbo
  • 6,319
  • 1
    $SL_2$ has index $2$ here, so $GL_2$ is generated by any set of generators of $SL_2$ together with any matrix of determinant $-1$. A nice one to take is $\left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right]$. – Qiaochu Yuan Sep 29 '22 at 00:14
  • Thank you. So \begin{bmatrix}1&0\0&2\end{bmatrix} suffices as additional generator. So do we need three generators? – Turbo Sep 29 '22 at 00:15
  • 1
    @Turbo you have to be able to generate the entire special linear group then add one with a negative determinant. Note that there is a homomorphism from $GL_2(F_3) \rightarrow {-1,1}$ which has two cosets that explains why this is sufficient. – CyclotomicField Sep 29 '22 at 00:18
  • 2
    Groupprops claims that two generators suffices (https://groupprops.subwiki.org/wiki/General_linear_group:GL(2,3) ) but doesn't give a presentation with two generators (that I can find, anyway). Maybe GAP can do the calculation, I don't know. – Qiaochu Yuan Sep 29 '22 at 00:19
  • GAP says $\begin{bmatrix}2 & 0 \ 0 & 1\end{bmatrix}$ and $\begin{bmatrix}2 & 1 \ 2& 0\end{bmatrix}$ do generate. – Andreas Caranti Sep 29 '22 at 06:52
  • As to OP's last question, please check this answer for two solutions in GAP. – Andreas Caranti Sep 29 '22 at 06:57

1 Answers1

0

Multiplying a matrix with $A$ and $B$ gives you all possible row and column operations, so by row/column reduction you have an algorithm for writing any matrix of $SL_2(\mathbb{F}_3)$ as a product of $A$'s and $B$'s.

This works for $SL_2(\mathbb{F}_p)$ for any prime $p$ actually.

Anyway, if you can write $A$ and $B$ using $\begin{pmatrix}1 & 0\\ 0 & 2 \end{pmatrix}$ and $\begin{pmatrix}2 & 2\\ 1 & 0 \end{pmatrix}$, you also have an algorithm for writing matrices of $GL_2(\mathbb{F}_3)$ as products of $\begin{pmatrix}1 & 0\\ 0 & 2 \end{pmatrix}$ and $\begin{pmatrix}2 & 2\\ 1 & 0 \end{pmatrix}$.

spin
  • 12,267