Transitive subgroup of $S_n$, with $n$ prime, containing a transposition is always the entire $S_n$

Question

This is a question from a worksheet I'm working at for several days, I'm having huge problems figuring out a nice solution, because I'm always stuck at the solution and don't know where to use that $n$ is prime. The question is as follows:

Let $G \leq S_n$, with $n$ prime, be a transitive subgroup ($\forall k, l \in \{ 1,...,n\}\ \exists \ \sigma \in G:\sigma (k) = l$). If $G$ contains a transposition $(k,l)$, then $G = S_n$.

I have tried constructing several generators for $S_n$, but I couldn't construct an n-cycle (which would be the easiest way of proving it). The second attempt was to construct all possible transpositions, but I also fail here and I have no intuition on where to use that n is prime. The last thing we learned in the lecture were group operations, and the operation of $G$ on $\{ 1,...,n\} $ should also be transitive, but I got nothing out of this approach.

Any help appreciated!

Answer 1

The fact that the action is transitive implies that $n$ divides the order of $G$. Cauchy's theorem then implies that $G$ contains an element of order $n$ because $n$ is prime. What are the elements of order $n$ in $S_n$?