Transitive subgroups of the symmetric group

Let $S_n$ denote the symmetric group on $A=\{1,2,\ldots n\}$. A subgroup $G\subseteq S_n$ is said to be a transitive subgroup if the natural action on the set $A$ is transitive.

Q : Find all isomorphism classes of transitive subgroups of size n of $S_n$.

Let us try to find one subgroup .., well, $\mathbb{Z}_n$ generated by the $n$-cycle $(1 2 \ldots n)$. Clearly it is transitive, and in fact, more is true, namely, the stabilizer of any element is the trivial permutation.

Can we think of more ?

Let us look at $S_{2n}$ and look at $p=(1, 2, \ldots n)(n+1, \ldots, 2n)$ [this is the cycle notation of permutations]. $p$ has order $n$. Now if we can find a $q$ which somehow can make 1 go to n+1, then the subgroup generated by $p,q$ would be transitive.

In fact, we can find $q$ such that $qp=pq$ or rather $qpq^{-1}=p$. Just look at $q=(1, n+1)(2 , n+2)(3 , n+3)\ldots (n , 2n)$. Then $qpq^{-1}$ = $(1, 2 ,\ldots, n)(n+1, \ldots, 2n)=p$

The subgroup generated by $p,q$ is isomorphic to $\mathbb{Z}_n\times \mathbb{Z}_2$, so we see that $\mathbb{Z}_{n}\times \mathbb{Z}_{2}$ is also a transitive subgroup of size 2n in $S_{2n}$

In fact, more is true !

Claim : Let $G$ be ANY subgroup of size $n$. Then it can be embedded as a transitive subgroup of $S_n$.

Proof : Let $G=\{g_1,\ldots,g_n\}$ and let $S_n$ be the symmetric group on a set of size n, which we can assume to be G ! We have the following (usual) embedding of $G$ into $S_n$ got by the action of $G$ on itself by a modified left translation, namely

$f:G\to S_n$ which sends $g\to [g_i\to g^{-1}g_i]$, where $S_n$ is thought of as set bijections of G and itself.

1. The map $[g_i\to g^{-1}g_i]$ is a set bijection because $g^{-1}g_i=g^{-1}g_j\iff g_i=g_j$
2. $f$ is a group homomorphism – easy check. [This is why we need $g_i\to g^{-1}g_i$ instead of $g_i\to gg_i$
3. $f$ is a 1-1 function (or an embedding) because the map $[g_i\to g^{-1}g_i]$ is identity if and only if $g^{-1}g_i=g_i\forall g_i$ which happens if and only if $g=e$

So G can be identified as a subgroup of $S_n$ via $f$. Now we check that it is a transitive subgroup !

Any $g_i$ can be moved to $g_j$ by the element $a=g_ig_j^{-1}$ because $a\star g_i = a^{-1}g_i = g_jg_i^{-1}g_i=g_j$.

Infact, what is the stabilizer of any $g_i$ under this action of G ? It is the identity ! So this action of G  is in fact regular (or transtive+free = simply transitive, namely given $g_i$ and $g_j$, there is a unique element in G which takes one to the other)