# The largest group acting on a non-orientable surface of genus 5

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 $$1=F_0 of $$F$$. Since $$120=(2^3)(3)(5)$$ then $$n\leq 5$$. Is it $$n=5$$? Is $$n$$ strictly less than 5?

Thank you!

• 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. Jul 1, 2020 at 19:41
• 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 $A_5$).
– YCor
Jul 1, 2020 at 20:04

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: