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 $p = 2$.

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    $\begingroup$ You might mention that these generate automorphisms of the affine variety defined by the Markoff equation $x^2+y^2+z^2=xyz$. Your group $\mathbb Z_2*\mathbb Z_2*\mathbb Z_2$ is (up to finite index) $\mathbb Z*\mathbb Z$, the free group on two generators. Do you have any reason to suspect that its action on $\mathbb F_q^3$ won't depend on $q$ in a reasonably random fashion? (Have you done some experiments?) If you do a Google search on "markoff equation finite field", you'll find lots of articles studying (generalized) Markoff equations over finite fields. $\endgroup$ – Joe Silverman Aug 5 '15 at 19:54
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    $\begingroup$ Markoff equation... interesting. The action will depend on q (I have done some experiments). I found elements whose order in $\mathbb{F}_p^3$ is $p(p^2-1)/2$. $\endgroup$ – user4324 Aug 5 '15 at 19:58
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    $\begingroup$ In fact the permutations generate a subgroup of $S_{q^3-1}$, since $(0,0,0)$ is fixed. Do you know if the group they generate is transitive / primitive on non-zero vectors? It would also help to know the answer for $q=2,3,4,5$ for instance... In general if you pick permutations at random, you'll either get $A_n$ or $S_n$, so it would be worth seeing if this is the case for small $q$... $\endgroup$ – Nick Gill Aug 6 '15 at 13:45
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    $\begingroup$ Note that the action on non-zero vectors is not transitive. The triples with exactly two of $\{x,y,z \}$ equal to $0$ are invariant under the three given involutions. $\endgroup$ – Geoff Robinson Aug 6 '15 at 13:55
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    $\begingroup$ @NickGill The Markoff equation has been much studied as an algebraic variety with an interesting automorphism group, especially in term of integer solutions. For example, every solution in positive integers to $x^2+y^2+z^2=xyz$ is obtained by starting with $(3,3,3)$ and applying automorphisms. Zagier has a beautiful paper in which he estimates the number of solutions of size at most $T$. The "Unicity Conjecture" says that if solutions are listed as $(a,b,c)$ with $a\ge b\ge c$, then no $a$ value appears more than once. (BTW, the classical equation is $x^2+y^2+z^2=3xyz$.) $\endgroup$ – Joe Silverman Aug 6 '15 at 14:54

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.

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    $\begingroup$ Thank you for the answer! I should note that this question arose from a presently ongoing summer project involving undergraduates at the Mason Experimental Geometry Lab (meglab.wikidot.com). $\endgroup$ – Sean Lawton Aug 6 '15 at 21:09
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    $\begingroup$ Exactly the idea I was looking for. Thank you. : ) $\endgroup$ – user4324 Aug 7 '15 at 18:59

EDIT: THIS IS WRONG. I based it on the statment in the question that $\langle v,h \rangle \cong D_4$, which I thought meant a dihedral group of order 8. But some quick experiments suggests that the dihedral group one gets has larger order, depending on $q$.

I am basing this answer on your remarks assuming "$q$ large enough", with the assumption that for finitely many "small" $q$ you can solve this by some other means.

Since your generators are involutions, and since any two of them generate Coxeter groups if $q$ is large enough (of types $A_1\times A_1$, $A_2$ and $B_2$, respectively), your group must be a quotient of the Coxeter group of type $B_3$, i.e. of $W(B_3)=\langle r,s,t\mid r^2=s^2=t^2=(rt)^2=(rs)^3=(st)^4 \rangle$. That's the group of automorphisms of a cube, and it has size 48.

Moreover, you already have subgroups of order 6 and 8, so the order of the whole group myst be a multiple of $lcm(6,8)=24$.

Now $W(B_3)$ has exactly one normal subgroup of order 2, namely its center. The only non-trivial element in it is $r(srt)^2st$.

So, it suffices to look at that element, and see if it acts trivial (then you get the quotient of order 24), or non-trivial (then you get all of $W(B_3)$.

  • $\begingroup$ Maybe I am misunderstanding what you mean by $\langle h,v \rangle \cong D_4$? I thought you meant a dihedral group of order 8 -- but upon doing some quick computations, I don't see how you could conclude this. So, what does $D_4$ mean? $\endgroup$ – Max Horn Aug 6 '15 at 16:24

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