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Lee Mosher
Lee Mosher
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You can also use Serre's theorem which says that kernel of the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite index subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.

You can also use Serre's theorem which says that the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite index subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.

You can also use Serre's theorem which says that kernel of the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.

Source Link
Lee Mosher
Lee Mosher
  • 15.4k
  • 2
  • 42
  • 81

You can also use Serre's theorem which says that the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite index subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.