[Math] About transitive subgroups of symmetric group $S_n$

Question

When I am studying Galois theory I came across some problems:
Let $S_n $ be the symmetric group on $n$ letters($|S_n|=n!$).How to determine all the transitive group $G$ of $S_n $ ( A subgroup $G$ of $S_n $ is called transitive if for each $i,j\in \{1,2,\dots,n\}$,there is a $\tau \in G $ with $\tau(i)=j$).With some knowledge about group action, I can proof that if $G$ is a transitive group of $S_n $ then $n$ divide the order of $G$.For $n=3$,I can easily find out all the transitive subgroups of $S_3$ because we can check each of its subgroup.But when $n$ is bigger ,for instance $n=4,5$ do we have to first find out all the subgroups of $S_n$ and then check which one is transitive.I want to know if we have a better method.\
Another question is given a transitive group $G$ of $S_n $ ,how to find a field $F$,and an irreducible polynomial $f(x)$ over $F$,so that if $K$ is the splitting of $f(x)$,then we have :
$$ G\cong Gal(K/F)$$
I even don't know how to get down to this question.Hope someone can help me.Thank you very much!

Answer 1

Here’s what Ian Stewart has to say on p. 268 of his book Galois Theory*:

The transitive subgroups [of $S_n$], up to conjugacy, have been classified for low values of $n$ by Conway, Hulpke, and MacKay (1998). … There is only one such subgroup when $n=2$, two when $n=3$, and five when $n=4,5$. The magnitude of the task becomes apparent when $n=6$: in this case there are 16 transitive subgroups up to conjugacy. The number drops to seven when $n=7$; in general prime $n$ lead to fewer conjugacy classes of transitive subgroups than composite $n$ of similar size.

In view of this I doubt if there is a simple formula or method to compute transitive subgroups or their conjugacy classes for general $n$. For $n\le30$ you could check out this page of the GAP data library.

*Ian Stewart, Galois Theory (Fourth Edition), 2015, CRC Press.