Let $G$ be a Hurwitz group, i.e. the automorphism group of some Hurwitz surface $C$. Then Hurwitz's automorphisms theorem shows that the quotient map of $C$ by $G$ has ramification points of indices $2$, $3$ and $7$. My question is how to deduce that $G$ is generated by elements $x$ and $y$ satisfying $x^2=y^3=(xy)^7=1$?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
By the definition, the quotient $C/G$ is the orbifold $O=S^2(2,3,7)$: Sphere with 3 cone points with point-stabilizers $Z_2, Z_3, Z_7$. Thus, the fundamental group $\pi$ of $O$ has the presentation $$ \langle a, b, c| a^2, b^3, c^7, abc\rangle. $$ (You can prove it by appealing to van Kampen theorem for instance, or by looking at the fundamental domain for the action of $\pi$ on the universal cover.) The usual covering theory applies to orbifolds, so every covering of $O$ is given by a subgroup in $\pi$. Regular coverings correspond to normal subgroups $N$ in $\pi$, so the quotient group is $\pi/N$ (same as the deck-group $G$ in your question) has the properties you need.
