# Action of $SL_2(Z)$ on Markoff quadratic forms

My setting is as follows: Fix a Markoff form $f_m(x,y)$ (see definition in the link below). If $f_m$ has the form ${\alpha}x^2+{\beta}xy+{\gamma}y^2$ then each element $A\in SL_2(Z)$ acts on $f_m$ in the following way: $Af_m=f_m(ax+by, cx+dy)$ ($a,b,c,d$ are the elements of the matrix $A$). Denote $G:=\{A\in SL_2(Z) : Af_m=f_m\}/\{I,-I\}$.

Question: Why is $G$ an infinite cyclic group?

For the definition of a Markoff form, see the first page here: http://arxiv.org/pdf/1106.1844v1.pdf

Edit: In order to avoid conufsion, my original question was about general quadratic forms. As the general question seems to have a negative answer (see rschwieb's answer below), I've changed it to the case of Markoff forms.

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I'm having trouble believing $G$ has to be infinite.
Let's take the simplest case of the quadratic form $f(x,y)=x^2+y^2$. You can check that a matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in M_2(\mathbb{Z})$ satisfies $Af=f$ iff it is an orthogonal matrix.
But there are only so many possibilities for this when you are working over $\mathbb{Z}$! One relation is $a^2+c^2=1$. How many pairs of positive integers can you add together to get 1? Clearly $a$ and $c$ have to be in $\{-1,0,1\}$, and exactly one has to be 0. The same can be said when you consider $b^2+d^2=1$. That said, a very rough estimate of the number of matrices here is "less than 81", and the quotient by $\{I,-I\}$ will be even smaller. So, I don't think this group has to be infinite.
(At the moment I'm counting that the actual number of matrices is 8, and $|G|=4$. Moreover, $G$ is the Klein four group rather than a cyclic group, so I'm having trouble believing the cyclic part now too.)
Thanks. I'm actually interested in the case where $f$ is a Markoff form. I guess asking about general quadratic forms was too optimistic. Anyway, I've now edited the question accordingly. – diophantus May 30 '12 at 13:17