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Peter Mueller
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The experimental observation about the order of $w:(x, y, z) \mapsto (z, y, yz-x)$ isn't hard to verify: By induction, one showshows that $w^m(x,y,z)=(a_mx+b_mz,y,c_mx+d_mz)$ with $\begin{pmatrix}a_m&b_m\\c_m&d_m\end{pmatrix}=\begin{pmatrix}0&1\\-1&y\end{pmatrix}^m$. Looking at the eigenvalues of $\begin{pmatrix}0&1\\-1&y\end{pmatrix}$ (which are in $\mathbb F_{q^2}$), and not forgetting the non-diagonalizable cases $y=1$$y=2$ and $y=-1$$y=-2$, yields the assertion after a short calculation.

The experimental observation about the order of $w:(x, y, z) \mapsto (z, y, yz-x)$ isn't hard to verify: By induction, one show that $w^m(x,y,z)=(a_mx+b_mz,y,c_mx+d_mz)$ with $\begin{pmatrix}a_m&b_m\\c_m&d_m\end{pmatrix}=\begin{pmatrix}0&1\\-1&y\end{pmatrix}^m$. Looking at the eigenvalues of $\begin{pmatrix}0&1\\-1&y\end{pmatrix}$ (which are in $\mathbb F_{q^2}$), and not forgetting the non-diagonalizable cases $y=1$ and $y=-1$, yields the assertion after a short calculation.

The experimental observation about the order of $w:(x, y, z) \mapsto (z, y, yz-x)$ isn't hard to verify: By induction, one shows that $w^m(x,y,z)=(a_mx+b_mz,y,c_mx+d_mz)$ with $\begin{pmatrix}a_m&b_m\\c_m&d_m\end{pmatrix}=\begin{pmatrix}0&1\\-1&y\end{pmatrix}^m$. Looking at the eigenvalues of $\begin{pmatrix}0&1\\-1&y\end{pmatrix}$ (which are in $\mathbb F_{q^2}$), and not forgetting the non-diagonalizable cases $y=2$ and $y=-2$, yields the assertion after a short calculation.

Source Link
Peter Mueller
  • 22.5k
  • 1
  • 75
  • 107

The experimental observation about the order of $w:(x, y, z) \mapsto (z, y, yz-x)$ isn't hard to verify: By induction, one show that $w^m(x,y,z)=(a_mx+b_mz,y,c_mx+d_mz)$ with $\begin{pmatrix}a_m&b_m\\c_m&d_m\end{pmatrix}=\begin{pmatrix}0&1\\-1&y\end{pmatrix}^m$. Looking at the eigenvalues of $\begin{pmatrix}0&1\\-1&y\end{pmatrix}$ (which are in $\mathbb F_{q^2}$), and not forgetting the non-diagonalizable cases $y=1$ and $y=-1$, yields the assertion after a short calculation.