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Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_{q^3}$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Denote the Fricke polynomial by $$\kappa(x, y, z) = x^2 + y^2 + z^2 - xyz - 2.$$ It is surjective (since any element in a finite field can be written as a sum of two squares). So the solutions $\kappa^{-1}(a)$ partition the affine space. The permutations $h, u, v$ fix each level set $\kappa^{-1}(a)$, so the action is not transitive, orbits stay within the same level sets.

The structure of the generated permutation group will depend on $q$. Based on computational experiments, it appears that the permutation $(x, y, z) \mapsto (z, y, yz -x)$ will have order $p(p^{2n}-1)/2$ over $\mathbb{F}_{p^n}$.

Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_{q^3}$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Denote the Fricke polynomial by $$\kappa(x, y, z) = x^2 + y^2 + z^2 - xyz - 2.$$ It is surjective (since any element in a finite field can be written as a sum of two squares). So the solutions $\kappa^{-1}(a)$ partition the affine space. The permutations $h, u, v$ fix each level set $\kappa^{-1}(a)$, so the action is not transitive, orbits stay within the same level sets.

It interesting to note that $Aut(\kappa)$ is generated by $h, u, v$ and maps that change the signs of pairs of coordinates.

The structure of the generated permutation group will depend on $q$. Based on computational experiments, it appears that the permutation $(x, y, z) \mapsto (z, y, yz -x)$ will have order $p(p^{2n}-1)/2$ over $\mathbb{F}_{p^n}$ for $p$ odd, and $2(2^{2n}-1)$ for $p = 2$.

Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_{q^3}$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Edit: The structure of the generated permutation group will depend on $q$. When I go home, I can give an example of an element whose order is $\mathbb{F}_p^3$ is $p(p^2-1)/2$.

Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_{q^3}$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Denote the Fricke polynomial by $$\kappa(x, y, z) = x^2 + y^2 + z^2 - xyz - 2.$$ It is surjective (since any element in a finite field can be written as a sum of two squares). So the solutions $\kappa^{-1}(a)$ partition the affine space. The permutations $h, u, v$ fix each level set $\kappa^{-1}(a)$, so the action is not transitive, orbits stay within the same level sets.

The structure of the generated permutation group will depend on $q$. Based on computational experiments, it appears that the permutation $(x, y, z) \mapsto (z, y, yz -x)$ will have order $p(p^{2n}-1)/2$ over $\mathbb{F}_{p^n}$.

Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_q^3$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Edit: The structure of the generated permutation group will depend on $q$. When I go home, I can give an example of an element whose order is $\mathbb{F}_p^3$ is $p(p^2-1)/2$.

Let $\mathbb{A}^3 = \mathbb{F}_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$: \begin{eqnarray*} h: (x, y, z) &\mapsto& (x, y, xy - z) \\ u: (x, y, z) &\mapsto& (y, x, z) \\ v: (x, y, z) &\mapsto& (x, z, y) \end{eqnarray*} It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S_{q^3}$.

If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Edit: The structure of the generated permutation group will depend on $q$. When I go home, I can give an example of an element whose order is $\mathbb{F}_p^3$ is $p(p^2-1)/2$.