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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.

edited Feb 17 at 10:28

asked Feb 16 at 19:59

Klim Puhov
207●7

	I am a bit puzzled by the question; is it not the case that $G$ acts freely on $C$? Then of course, $N$ acts freely as well. – Aakumadula Feb 17 at 4:20
	No. The quotient map by the $G$-action have ramification points of indexes $2$, $3$ and $7$ (see [wiki][1]). They have non-trivial stabilizers. [1]: en.wikipedia.org/wiki/… – Klim Puhov Feb 17 at 9:36
	In that case, $N$ could contain elements corresponding to the inertia group of these ramifications, and therefore cannot act freely either. – Aakumadula Feb 17 at 9:45
	Could you provide an example? I found this fact here heldermann-verlag.de/gcc/gcc02/gcc028.pdf in Section 3. – Klim Puhov Feb 17 at 10:07

Peter Mueller · Answer 1 · 2013-02-17 10:15:05Z UTC

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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 $G=N$.

answered Feb 17 at 10:15

Peter Mueller
3,780●1●9●18

	Could you explain, please, why $G$ is generated by a,b,c with relatively prime orders 2,3,7 and abc=1? I can't find the proof in literature. – Klim Puhov Feb 17 at 10:26
	Well, many people even take that as the definition, as in the paper you quoted in your comment. I believe that the Riemann-Hurwitz genus formula, together with the proof of the Hurwitz bound, gives this connection. – Peter Mueller Feb 17 at 10:39
	thanks. This clarifies the question a lot – Aakumadula Feb 17 at 10:43
	@Peter: Your answer is exactly what I called "using of classification theorems". So, unfortunately, it is not helpfull for me. Using Riemann-Hurwitz genus formula, together with the proof of the Hurwitz bound, I can show that the quotient map by the $G$-action have ramification points of indexes $2$, $3$ and $7$, but I can't figure out that $G$ have the description that you give in your answer. – Klim Puhov Feb 17 at 11:18
	@Klim: I do not agree that I use any classification theorems. If $G$ is the automorphism group of a Hurwitz surface $C$, then $C/G$ is a projective line $P^1$ (should be a by-product in the proof of the Hurwitz bound). Finite branched Galois covers with group $G$ of $P^1$ are described in terms of generating systems $g_1,\dots,g_r$ of $G$ with $g_1g_2\dots g_r=1$, where the $g_i$ are the generators of the local monodromies. Knowing that in the Hurwitz case there are three generators of orders $2$, $3$, and $7$ respectively is more or less rephrasing that the Hurwitz bound is sharp. – Peter Mueller Feb 17 at 14:02

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On the group actions on Hurwitz surfaces

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