Exercise 6.3.6

Proof. A subgroup $H$ of $S_n$ defines an action on $\text{[[}1,n\text{]]}$ by $h\cdot x=h(x),h\in H,x\in \text{[[}1,n\text{]]}$. By definition $H$ is a transitive subgroup of $S_n$ if this action is transitive, i.e. if the only orbit is ${\mathcal{}O}_i=\text{[[}1,n\text{]]},i=1,\cdots ,n$. If we write $H_i={\mathrm{Stab}}_H(i)$ the stabilizer in $H$ of a fixed element $i$, then $(H:H_i)=|{\mathcal{}O}_i|=n$, thus $|H|=|H_i|\times n$:

□