Automorphisms of genus 6 surfaces I have found in Wikipedia that the number of automorphisms of a Riemann surface of genus 6 does not exceed 150. The page offers neither a proof nor a reference. Can someone help me?
 A: See "The locus of curves with prescribed automorphism group" by Magaard-Shaska-Shpectorov-Völklein, and the references therein.
A: The question is purely combinatorial - the Hurwitz bound comes from the observation that the quotient of a surface by its automorphism group is a hyperbolic orbifold, and the hyperbolic orbifold of smallest area is the $(2, 3, 7)$ triangle orbifold, Dividing the area of your surface by the area of the orbifold gives you the Hurwitz bound. However, for some genera, there is no triangulation where the vertices have the right degrees, then you have to look for bigger triangle groups. For the surface of genus $6,$ you get the $(2, 3, 10)$ triangle group, but in any case, this is just a combinatorial exercise (for any fixed genus).
A: In case you don't know the general context: There is a curve of genus $g$ with endomorphism group contained in $G$ if and only if $G$ can be generated by elements $g_1$, $g_2$, ..., $g_k$ with orders $c_1$, $c_2$, ..., $c_k$, satisfying $g_1 g_2 \cdots g_k = 1$, obeying
$$|G| = (g-1) \left( \frac{1}{2} \sum (1-c_i^{-1}) - 1 \right)^{-1}. $$
The largest possible value of the second factor occurs for $(c_1, c_2, c_3) = (2,3,7)$, in which case we get $|G| = 84 (g-1)$. However, this bound is only achieved if there actually is a group of order $84 (g-1)$ generated by elements of orders $2$, $3$ and $7$ with product $1$. For whatever reason1, there is no such group of order $420$.
To get order $150$, we want $\sum (1-c_i^{-1}) = 2+\tfrac{1}{15}$; this occurs for $(2,3,10)$. I believe you can achieve this inside $S_3 \ltimes (\mathbb{Z}/5)^2$, taking elements in $S_3$ with product $1$ and orders $(2,3,2)$ and lifting them to the semidirect product so that one of the transpositions stays order $2$ and the other becomes order $10$. 
If we want order $>150$ then we need $2 < \sum (1-c_i^{-1}) < 2+\tfrac{1}{15}$. 
Note that we must have three summands, since $(1/2)+(1/2)+(1/2)+(1/2)=2$ is too small and $(1/2)+(1/2)+(1/2) + (2/3) = 2+\tfrac{1}{6}$ is too large. So we can alternatively say that we want $\tfrac{14}{15} < c_1^{-1}+c_2^{-1}+c_3^{-1} < 1$. Will Sawin and Roy Smith work out the options in the comments to another answer: You need to check $(2,3,7)$, $(2,3,8)$, $(2,3,9)$ and $(2,4,5)$, corresponding to groups of orders $420$, $240$, $180$ and $200$. 
Footnote 1: According to the groupprops wiki, the only nonsolvable group of order $420$ is $A_5 \times \mathbb{Z}/7$ and thus every group of order $420$ has a nontrivial abelian quotient $A$. If you believe this reference, I can give a quick proof. If $A$ is abelian, then the only solutions to $g_1 g_2 g_3=1$ in $A$ with $g_1^2 = g_2^3=g_3^7=1$ are $g_1 =g_2=g_3=1$; prove this by considering $(g_1 g_2 g_3)^6$, $(g_1 g_2 g_3)^{14}$ and $(g_1 g_2 g_3)^{21}$. So any solution in $G$ lies in the kernel of $G \to A$, and thus can't generate.
