Let us call a **subgroup** an injective homomorphism between groups. 



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> **I warn the reader that a *subgroup* designates here an inclusion $(H \subset G)$, not $H$ alone.** 
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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](https://mathoverflow.net/questions/136021/is-there-a-natural-definition-of-equivalence-for-groups-inclusions-inner-to-the)** (defined [here](https://mathoverflow.net/questions/136021/is-there-a-natural-definition-of-equivalence-for-groups-inclusions-inner-to-the)).  

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 $6$](https://mathoverflow.net/questions/135806/are-subfactor-planar-algebras-hard-to-classify-at-index-6/135994#135994) ? 



An upper bound for the maximal subgroups at fixed index?
I'm absolutely ignorant about subfactors, but if I correctly understand what you said [there](https://mathoverflow.net/q/136021/6451) (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*](http://en.wikipedia.org/wiki/Primitive_permutation_group) (in particular transitive) finite permutation subgroups of $S\_n$. 


