6
$\begingroup$

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!

$\endgroup$
  • 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. $\endgroup$ – John Shareshian Jul 1 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 at 20:04
9
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.