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 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 ) ?

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 ) ?