Can we prove $ Aut(S_g) , g \geq 2 $  is finite in the following way ? I was trying to prove that $ Aut( S_g $), g$ \geq 2 $ [ orientation preserving isometries ] is finite in the following way :
For fixed $M $ ( positive ) there are finitely many , say $ k $ number of simple closed geodesics ( with repeated multiplicities ), say $c_1...c_k $ with length $\leq M$. Consider the group $Aut(S_g$) acting on this finite set $S$ of geodesics ( it acts, since an isometry preserve the length of geodesics ). Now the number of permutations of $S$ is $k!$ ( k-factorial ). So, if we can prove that :
Lemma : An automorphism of surface of genus $\geq 2 $ is fully and uniquely determined by its action on finitely many ( say $k$ ,depending on genus g ) simple closed geodesics ( forming the set $S$ ),
then we can have only $k! $ permutations of $S$ and hence we would have only at most $ k! $ distinct automorphism of $ S_g$, if the Lemma is true . Is the lemma true and easily provable ? Or is there any other way to prove the main question ( $ Aut(S_g) $  is finite ) ?
 A: The answer is yes.  It's a little easier if we consider oriented geodesics (which is fine for your argument).  Let $\alpha$ and $\beta$ be two oriented simple closed curves on $S_g$ that intersect once.  Let $a$ and $b$ be geodesics that are homotopic to $\alpha$ and $\beta$, respectively.  Then $a$ and $b$ only intersect once, say at a point $x \in S$.  Now let two isometries $f_1,f_2 : S \rightarrow S$ be given.  Assume that $f_1(a) = f_2(a)$ and $f_1(b) = f_2(b)$ (as oriented geodesics).  I claim that $f_1=f_2$.  Indeed, it is clear that $f_1(x)=f_2(x)$.  Moreover, if $v$ is the unit tangent vector at $x$ pointing in the direction of $a$ (remember, $a$ has an orientation), then we also know that $f_1(v)=f_2(v)$.  Since an element of $SO(2)$ is determined by its value at a single non-zero vector, this implies that both the values and the derivatives of $f_1$ and $f_2$ agree at $x$, and a standard result of Riemannian geometry (using the exponential map) says that then $f_1 = f_2$.
This implies that you can take $S$ to be the set of all geodesics whose lengths are at most the lengths of $a$ or $b$.
