a question concerning subgroup of symmetric group

Question

Suppose $H$ is a transitive subgroup of the symmetric group of $n$ symbols. Show that $n$ divides the order of $H$.

I tried to show that some $n$-cycle is in $H$ but this idea did not work.

Answer 1

Let us denote by $h\cdot x$ the action of $H$ over $X:=\{1,\dots,n\}$. Define $\mathrm{Stab}(x):=\{h\in H, h.x=x\}$ and $\omega(x):=\{h.x,h\in H\}$. Then for each $x\in X$, $\mathrm{Stab}(x)$ is a subgroup of $H$ and there is a bijective correspondence between $H/\mathrm{Stab}(x)$ and $\omega(x)$ given by $h\mathrm{Stab}(x):=h.x$.

Since $|H|=|\mathrm{Stab}(x)|\cdot |\omega(x)|$ and $ |\omega(x)|=n$ (the action is transitive), we obtain $n\mid |H|$.