Number of transitive acting groups on four letters? Constructing a group (a permutation group) which acts on a set of 4 letters transitively is easy. for example $G_{1}=\{id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$ < $S_{4}$. Can it be verified how many groups we have like $G_{1}$ ?
 A: Although this is not an easy problem, finding all subgroups of $S_n$ first is not necessary, and is not the best way to proceed. The classification of all subgroups of $S_n$ has been completed for $n \le 18$, whereas the transitive subgroups are known for $n \le 32$. For $n=32$ there are 2801324 conjugacy classes of transitive subgroups, and their calculation is described in the paper: J. Cannon and D. Holt, The transitive groups of degree 32, Experimental Mathematics 17 (2008), 307-314.
The primitive subgroups of $S_n$ have been classified up to $n=4095$, so for a given $n$ the problem reduces to finding the imprimitive groups preserving a block system of $r$ blocks of size $b$, where $rb=n$. The possible induced actions on the block systems are just the transitive groups of degree $r$, which you can assume are already known. The idea is to first find all possible kernels of the actions on the block systems, which (in nearly all cases) are subdirect products of transitive groups of degree $b$, and then classify the groups with a given kernel and action group. This involves a large amount of computation, which becomes rapidly less feasible with increasing degree. Testing two subgroups for conjugacy in $S_n$ can be a particularly slow in difficult cases, so the principal aim is try to organise the computations so that you do as little conjugacy testing as possible.
A: It's pretty straightforward to count them using a table of subgroups of $S_4$. We have three cyclic groups of order $4$, one normal Klein four-subgroup, three copies of $D_8$, the alternating group $A_4$, and $S_4$ itself, for a total of nine.
