Cardinals of transitive permutation groups acting on $\{1,\dots,n\}$

Is there a nice description of all divisors of $n!$ which can be realized as cardinals of permutation groups acting transitively on $\{1,\dots,n\}$?

A necessary condition is of course that such a divisor is a multiple of $n$.

$n$ (cyclic) and $2n$ (dihedral) are always possible but, without mistake on my behalf, $3n$ is not feasible unless $n\equiv 1\pmod 3$ (it can then be realized as a semi-direct product analogous to the dihedral group). can only be realized if $3$ divides the number of invertible integers modulo $n$.

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This is almost certainly too difficult to answer in general. What you write about $3n$ is not true. There are transitive groups of order $3n$ for $n=9$ and $n=14$ for example. On the other hand, there are none for $n=10$. The transitive groups have been enumerated for all $n \le 32$ by the way. Could you try asking a more specific question? – Derek Holt Apr 5 '12 at 12:23
Thank you for the correction: A semi-direct product needs of course divisibility by 3 of the number of invertible elements modulo $n$. I think you suggest that the answer is messy! – Roland Bacher Apr 5 '12 at 13:25

As Derek suggests in his comment, this question is too difficult to answer in general. However one could limit the question as follows: clearly if $K$ is a transitive permutation group then $|K|$ divides $|M|$ where $M$ is a maximal transitive subgroup of ${\mathrm Sym}(n)$; thus we can ask about the cardinality of a maximal transitive subgroup $M$ of ${\mathrm Sym}(n)$.
The O'Nan-Scott theorem is the main tool here. Roughly speaking it asserts that such a subgroup $M$ is either imprimitive (and hence a wreath product, with order formula easy), or else it is in a bunch of primitive families. Most of these families have a geometric description and, as such, it is easy to calculate their order.