Let $C$ be a Hurwitz surface, $G=\text{Aut}(C)$ and $N$ is a proper normal subgroup of $G$. Is there a simple argument (without using of classification theorems) for the fact that $N$ acts on $C$ freely?
I found this fact here see Section 3.
Let $C$ be a Hurwitz surface, $G=\text{Aut}(C)$ and $N$ is a proper normal subgroup of $G$. Is there a simple argument (without using of classification theorems) for the fact that $N$ acts on $C$ freely? I found this fact here see Section 3. 


I think that Klim wants to talk about proper normal subgroups $N$ of $G$. In that case, $N$ cannot contain an inertia generator: $G$ is generated by $a,b,c$ with relatively prime orders $2,3,7$ and $abc=1$. So if for instance $a\in N$, then modulo $N$ we have $bc=1$, and the order of $b$ divides $3$ and $7$. So $a,b,c\in N$, hence $G=N$. 

