Sylow $p$-group of a transitive subgroup of $S_p$

Question

How do I show that any transitive subgroup of $S_p$ contains a non-trivial Sylow $p$ subgroup, of cardinality $p$?

I am trying to prove a result of Galois and the only hint I have is that if $p$ is a prime number, then any non-trivial normal subgroup of a transitive subgroup of $S_p$ also acts transitively on $\{1,\ldots, p\}$.

Answer 1

If $G\subseteq S_p$ acts transitively on $\{1,\dots,p\}$, the subgroup $G_1=\{g\in G:g1=1\}$ that fixes $1$ has index $p$, since the function $gG_1\in G/G_1\mapsto g1\in\{1,\dots,p\}$ is a bijection.

It follows that $p$ divides the order of $G$ and, therefore, that the $p$-Sylow subgroup of $G$ is non-trivial.

On the other hand, since $p^2$ does not divide the order of $G$, because it already does not divide the order of $S_p$, which is $p!$ the $p$-Sylow subgroup of $G$ cannot have order larger than $p$.

Answer 2

Sketch: Use the Orbit-Stabilizer theorem, Lagrange's theorem, Sylow's theorem, a little arithmetic.