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.
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$\begingroup$ 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. $\endgroup$– VenkataramanaCommented Feb 17, 2013 at 4:20
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$\begingroup$ 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/Hurwitz%2527s_theorem_on_automorphisms $\endgroup$– Klim PuhovCommented Feb 17, 2013 at 9:36
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$\begingroup$ In that case, $N$ could contain elements corresponding to the inertia group of these ramifications, and therefore cannot act freely either. $\endgroup$– VenkataramanaCommented Feb 17, 2013 at 9:45
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$\begingroup$ Could you provide an example? I found this fact here heldermann-verlag.de/gcc/gcc02/gcc028.pdf in Section 3. $\endgroup$– Klim PuhovCommented Feb 17, 2013 at 10:07
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1 Answer
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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, 2013 at 10:15
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$\begingroup$ 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. $\endgroup$ Commented Feb 17, 2013 at 10:26
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$\begingroup$ 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. $\endgroup$ Commented Feb 17, 2013 at 10:39
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$\begingroup$ @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. $\endgroup$ Commented Feb 17, 2013 at 11:18
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$\begingroup$ @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. $\endgroup$ Commented Feb 17, 2013 at 14:02
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$\begingroup$ @Peter: Ok, maybe "classification" is not a right word. So, you use the description of finite branched Galois covers of $P^1$. Where is the proof can be found? $\endgroup$ Commented Feb 17, 2013 at 14:34