Show that if $p$ is a prime number, and $G$ is a transitive subgroup of $S_p$, then $G$ must contain a cycle of length $p$.

Question

Note that a subgroup $G$ of $S_n$ is called a transitive subgroup of $S_n$ if it acts transitively on the set $\{1,2,...,n\}$.

I understand that for any $x$, $y$ $\in \{1,2,...,p\}$, there exists an element $g \in G$ such that $g x=y$, but I do not know how to use this to demonstrate the existence of a $p$-cycle.

This comes in a section of text about group actions, which is before discussion of Sylow Theorems, so I presume the proof will not require them. Any help is very appreciated.

Answer 1

By the orbit stabilizer theorem and transitivity, the number of cosets of the stabilizer of a point is $p$. In particular, as in a hint you received in a comment, the order of the group is divisible by $p$. By Cauchy's theorem there must be an element of order $p$. In general an element of the symmetric group of order $p$ can only be shown to be a product of disjoint $p$-cycles. However, no two $p$-cycles are disjoint in $S_p$, so the subgroup must contain a $p$-cycle.