Let $N_5$ denote the non-orientable surface of genus 5.

In Conder's database https://www.math.auckland.ac.nz/~conder/BigSurfaceActions-Genus2to101-ByGenus.txt we can see that the biggest finite group $F$ acting on $N_5$ has order 120. Moreover, the quotient has signature $(0; +; [-]; \{(2,4,5)\})$.

Is there a very concrete description of this group $F$?

To be even more concrete. I would like to the length $n$ of the largest chain of subgroups $1=F_0<F_1<\cdots< F_n=F$ of $F$. Since $120=(2^3)(3)(5)$ then $n\leq 5$. Is it $n=5$? Is $n$ strictly less than 5?

Thank you!

  • 1
    $\begingroup$ I don't understand the notation well enough to answer your first question. As to the second, it seems to me that, no matter which group on the genus 5 list one considers, the only possible composition factors are groups of prime order and the alternating group $A_5$. As $A_5$ has a chain $1<Z_2<Z_2 \times Z_2<A_4<A_5$, it follows that every one of the groups in question has a chain of subgroups that is as long as Lagrange's Theorem will allow. $\endgroup$ Jul 1, 2020 at 19:41
  • $\begingroup$ Eventually it's just a basic question about a certain group of order 120, which you should make more explicit, but actually from John Shareshian's answer, it follows that every group of order 120 has a chain of the largest possible size (as it's always true for solvable groups, and for $A_5$). $\endgroup$
    – YCor
    Jul 1, 2020 at 20:04

1 Answer 1


The group $F$ is isomorphic to the symmetric group $S_5$.

In fact, since $N_5$ is non-orientable of genus $5$, both $F$ and the extended group $F^*$ (of order twice the order of $F$) act on its orientable double cover, that has genus $4$. In Conder's database, this is expressed by saying that the action of $F$ in genus $4$ is reflexible and that there is a non-orientable quotient.

Looking at the actions of groups of order $120$ on orientable surfaces of genus $4$ that have these properties, we can find the presentation of $F$; in fact, there is a unique such a group:

enter image description here

Now a simple computation with GAP4 does the job.

enter image description here

  • $\begingroup$ Thank you Francesco. Your answer is exactly the kind of answer I was expecting. $\endgroup$
    – Luis Jorge
    Jul 1, 2020 at 23:15

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