2
$\begingroup$

Let us call a subgroup an injective homomorphism between groups.

I warn the reader that a subgroup designates here an inclusion $(H \subset G)$, not $H$ alone.

A subgroup $H \subset G$ is maximal if for all intermediate subgroups $H \subset K \subset G$, then $K=H$ or $G$.

Let $\sim$ be the equivalence of subgroups (defined here).
Note that the maximality is invariant under $\sim$.

Let $n$ be a fixed integer. For each equivalence class of index $n$ maximal subgroups, we choose a representative $(H \subset G)$ with $G$ of minimal order. Let $R_{n}$ be the set of all these representatives.

Interrelated questions :

  • Does $R_{n}$ is a finite set ?
  • $\forall (H \subset G) \in R_{n}$, is $ord(G)$ bounded ?
  • Is $G$ always a finite group ? A counter-example ?

Original motivation: What's the list of all the maximal subgroups at index $6$ ?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

I'm absolutely ignorant about subfactors, but if I correctly understand what you said there (the theorem in the question), the answer to all three questions is yes, even with non maximal subgroups.

Namely, let $H\subset G$ be of finite index $n$. Since the intersection $K$ of all conjugates of $H$ in $G$ has index dividing $n!$ in $G$ (it is the kernel of the obvious morphism $G\to S_n$) and is of course normal in $G$, you have $(H\subset G)\sim (H/K\subset G/K)$.

The case of maximal subgroups is then about primitive (in particular transitive) finite permutation subgroups of $S_n$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thank you @BS. for this nice answer. It's wonderful that the problem reduced to primitive permutation groups. I found this paper in the wiki-page you link, where all is explained in details and classified for n<2500 (also 4096 in a more recent paper). $\endgroup$
    – Sebastien Palcoux
    Commented Jul 9, 2013 at 14:14
  • $\begingroup$ The last question to answer is whether or not some of these objects give equivalent maximal subgroups ? $\endgroup$
    – Sebastien Palcoux
    Commented Jul 9, 2013 at 15:22
  • $\begingroup$ @Sebastien : there seems to be a criterion for von Neumann equivalence in the IMRN article imrn.oxfordjournals.org/content/2002/34/1791.short but I don't have access to it. $\endgroup$
    – BS.
    Commented Jul 9, 2013 at 20:18
  • $\begingroup$ Yes @BS. You can access to this paper here via a ps2pdf converter. The group theoretic characterization of the equivalence is quite technical, so that it seems not obvious to answer this last question for a non group theorist. $\endgroup$
    – Sebastien Palcoux
    Commented Jul 9, 2013 at 21:42
  • $\begingroup$ BS: Your answer works, but only for finitely-generated groups. In the infinitely-generated case $Hom(G,S_n)$ could be infinite and an answer would require opening up the definition of "equivalence" in the OP. Another caveat is that, on the face of it, the OP uses very unusual definition of a "subgroup." With this definition, $Z^2$ contains infinitely many subgroups of index 1. Maybe this does not matter though (again, it depends on the definition of equivalence). $\endgroup$
    – Misha
    Commented Jul 9, 2013 at 23:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .