I am struggling with the following problem: If $G$ is a transitive permutation group of even degree $>2$, then 4 divides $|G|$.
What I mean by $G$ being a transitive permutation group of even degree $>2$ is that $G$ is a subgroup of the symmetric group of degree $n$, $S_n$, with $n$ being an even number greater than 2. And that $G$ acts transitively on the set $\{1,\dots, n\}$ (note that the action comes from the action of $S_n$ on $\{1,\dots,n\}$).
My first approach has been to use that for ant $x\in \{1,\dots, n\}$, the index $|G:G_x|=|\text{Orb}_G(x)|=|\{1,\dots,n\}|=n$.
Here I distinguish two cases. If $4$ divides $n$ then $4$ divides $|G:G_x|$ and I am done.
So I can assume that $4$ does not divide $n$. We know that $2$ divides $n$, and so 2 divides $|G:G_x|$ and also $|G|$. And I have to show that two divides $|G_x|$.
I am struggling with this. I have tried to see why $G_x$ has order divisible by 2 but I have run out of ideas.
I know that I have to use the fact that $G$ is a transitive permutation group because this is not true in general. For example, any group $G$ acts transitively over itself so you can find groups with an order divisible by $2$ and not by $4$ that act transitively over a set of even order greater than 2.
What if $G$ is the transitive subgroup generated by the standard 6-cycle in $S_6$: $G = \langle(1,2,3,4,5,6)\rangle \le S_6$? This subgroup has order 6, which isn’t divisible by 4.