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For $G$ a finite group, it is easy to construct a (connected, orientable) surface with a faithful action of $G$.
E.g.: take a disjoint union of $G$ many spheres, and add a 1-handle for every edge in the Cayley graph of $G$.

What is known about the minimal genus of a surface faithfully acted upon by the symmetric group $S_n$?

(Same question may be asked for other families of finite groups.)

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  • $\begingroup$ For example, I believe that $S_3$ acts faithfully on $T^2$. Think of $T^2$ as a regular hexagon where opposing edges are identified in an orientation preserving way. Then picking every second vertex gives a regular triangle. The canonical $S_3$ action can (also rather canonicallly) be extended to the entire $T^2$. Of course we have a global fixed point in the middle, but faithful only means that no group element acts as the identity. $\endgroup$
    – HenrikRüping
    Commented Jul 23 at 10:37
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    $\begingroup$ The case of $g=0$ and $g=1$ you can do by hand (and don't include $S_n$ for $n \ge 5$), but for higher genus you can't do better asymptotically than $O(|G|) = O(n!)$ (with an obvious explicit constant) by the Hurwitz bound. (compare mathoverflow.net/questions/179785/…). So the only question is whether you care about the constant. $\endgroup$
    – user491858
    Commented Jul 23 at 11:03
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    $\begingroup$ I just realized -- do you mean a holomorphic action, or do you also allow anti-holomorphic maps? Because I think that Condor's paper, which I cited below, should show that $S_n$ acts on a surface with $n! = 168 (g-1)$, if you allow anti-holomorphic maps. $\endgroup$
    – David E Speyer
    Commented Jul 23 at 12:02
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    $\begingroup$ @DavidESpeyer My intention was to allow anti-holomorphic maps, but I realise that I might have been ambiguous. $\endgroup$
    – André Henriques
    Commented Jul 23 at 12:06
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    $\begingroup$ About existence: $\pi_1(\Sigma_2)$ ($\Sigma_g$ closed surface of genus $g$) has a $S_n$ quotient, which yields a Galois covering of genus $1+n!$ with free action of $S_n$. $\endgroup$
    – YCor
    Commented Jul 23 at 14:12

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I don't have a precise answer, but the genus of $S$ has to grow like $n!$.

To see this, note that when $n$ is large enough, $S$ cannot be the sphere or torus. So $S$ admits hyperbolic metrics. By Nielsen realisation (due to Kerckhoff) the surface $S$ admits a hyperbolic metric invariant under the $S_n$ action. The quotient is an orbifold. So it has area greater than or equal to the area of the $(2, 3, 7)$-triangle in the hyperbolic plane. That is, it has area at least $\pi/42$. So $S$ has area at least $n! \cdot \pi/42$. Since the area of $S$ equals $2\pi(2g - 2)$ we deduce that the genus is bounded below by a linear function of $n!$.

Added by AndrΓ© Henriques:
David Speyer and Nick Gill have shown, in the comments, that for $n$ large enough (specifically $𝑛\ge 167$), the symmetric group $𝑆_𝑛$ acts faithfully on a surface of genus $g$, with $π‘”βˆ’1=\frac{𝑛!}{168}$. As argued above, this is optimal.

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    $\begingroup$ We can't get equality in the Hurwitz bound. Equality happens for a group $G$ iff $G$ is generated by elements $x$, $y$, $z$ obeying $x^2=y^3=z^7=xyz=1$. But $y^3=z^7=1$ forces $y$ and $z$ in $A_n$, and then $x = z^{-1} y^{-1}$ forces $x$ in $A_n$. $\endgroup$
    – David E Speyer
    Commented Jul 23 at 11:54
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    $\begingroup$ @AndréHenriques Higman's paper is unpublished, but Condor has a published paper where he makes Higman's bound explicit: Condor, "Generators for alternating and symmetric groups" (1980) shows that $A_n$ is a Hurwitz group for $n \geq 168$. $\endgroup$
    – David E Speyer
    Commented Jul 23 at 11:55
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    $\begingroup$ @DavidESpeyer, I think you mean Conder rather than Condor. $\endgroup$
    – Nick Gill
    Commented Jul 23 at 15:45
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    $\begingroup$ On p.745 of Regular maps with an alternating or symmetric group as automorphism group, Conder writes "It was shown by the author in his 1980 doctoral thesis [3] that both the alternating group $A_n$ and the symmetric group $S_n$ are smooth quotients of the extended $(2, 3, 7)$ triangle group for all $n\geq 167$". Note that the extended $(2,3,7)$ triangle group is the same as the $(2,3,7)$ Coxeter group. @AndréHenriques, does that deal with your query above? $\endgroup$
    – Nick Gill
    Commented Jul 23 at 15:51
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    $\begingroup$ @NickGill. Thank you to both Nick and David. Your comments completely answer my question. $\endgroup$
    – André Henriques
    Commented Jul 23 at 20:30

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