What are the transitive groups of degree $4$? How can I find all of the transitive groups of degree $4$ (i.e. the subgroups $H$ of $S_4$, such that for every $1 \leq i, j \leq 4$ there is $\sigma \in H$, such that $\sigma(i) = j$)? I know that one way of doing this is by brute force, but is there a more clever approach? Thanks in advance!
 A: Let $G$ be a transitive subgroup of $S_4$. Since the orbit of $1$ under the action of $G$ is $\{1, 2, 3, 4\}$, the order of $G$ must be divisible by $4$, and so must be equal to one of $4, 8, 12, 24$. 
An order $4$ $G$ would be either cyclic (generated by a $4$-cycle, giving 3 subgroups) or Klein-Four. There are two $V_4$ subgroups of $D_8$, but only one of them is transitive in each $D_8$, and they're all equal to $\{1, (12)(34), (13)(24), (14)(23)\}$
The order $8$ subgroups are Sylow-$2$'s, so they're all conjugate to each other and isomorphic to $D_8$ (it's easy to find a subgroup isomorphic to $D_8$ by just looking at the symmetries of the vertices of a square). The number of them is either $1$ or $3$ by a Sylow Theorem, and it's $3$ because $D_8$ is not normal in $S_4$. 
The order $12$ subgroup must be $A_4$, and the order $24$ subgroup is then $S_4$ itself.  
A: We know subgroups of $S_4$ come in only a few different orders: $1,2,3,4,6,8,12,$ and $24$.
If we use the orbit stabilizer theorem, we have that $|H| = |\operatorname{Orb}_H(x)| \cdot |\operatorname{Stab}_H(x)|$; this limits us to only four possible subgroup orders, as $|\operatorname{Orb}_H(x)|$ can only be one thing.
Among those four orders, we can find $5$ non-isomorphic groups fulfilling the criteria. I don't know off-hand how many copies of such groups $S_4$ has offhand, but it shouldn't be too hard to pin that down.
Note that just because two subgroups of $S_4$ may be isomorphic, it doesn't mean that they all (do or don't) act transitively on $\{1,2,3,4\}$. For example, I can only think of one particular Klein four-group in $S_4$ that does, although there are several isomorphic subgroups in $S_4$.
