$ G \leq S_n$ transitive and n prime $\Rightarrow$ $n$ divides $ord(G)$

Question

I am doing a task and therefore I need to show the following:

$ G \leq S_n$ transitive and n prime $\Rightarrow$ $n$ divides $ord(G)$

I think it's quite obvious, but I don't know, how to proof it formally right.

Transitivity means that for all $x,y \in G$ exists $g \in G$ such that $x^g=y$, where $x^g$ denotes the conjugation $g^{-1}xg$.

Answer 1

If $G$ is transitive on $n$ elements, then $G$ has a subgroup of index $n$ by the orbit stabilizer theorem, so $n$ divides the order of $G$.

Answer 2

When you talk about a group being "transitive", what you really mean is that the group is acting on a set, and the action is transitive. So you need to be really clear about what set the group is acting on, and how it is acting.

It is impossible for a group $G$ to act transitively on itself by conjugation. To see this, notice that if you take the identity element $e$, you have $e^{g} = e$ for all $g \in G$. So most likely, if you found this problem in a book, what is meant is that since $G$ is a subgroup of $S_{n}$, $G$ is a collection of permutations of $\{1,2,\ldots,n\}$ (so the action is simply by permuting these numbers).

To say $G$ is transitive means that for any $i,j \in \{1,\ldots,n\}$, there is a $g$ in $G$ with $i^{g} = j$, where $i^{g}$ means you apply the permutation $G$ to $i$ in the natural way. Then you can apply the orbit-stabilizer theorem to show that $n$ divides $|G|$. You don't need the fact that $n$ is prime.