Prop. Let $G$ be a group that acts doubly transitively on $X$. Let $H\leq G$ be such that $|G:H| < |X|$. Then $H$ acts transitively on $X$.
I am trying to prove the proposition above. I know that since $G$ acts doubly transitively, for any $x_0\in X$, the stabilizer subgroup $G_{x_0}$ acts transitively on $X\setminus\{x_0\}$, but I don't know how could I make use of this. Any help would be appreciated!
Let $G \le \mathrm{Sym}(X)$ be a $2$-transitive group where $X$ is finite. Let $S \subset X$ be a proper nonempty subset and let $N = \{g \in G : S^g = S\}$ be the setwise stabilizer of $S$. By the orbit--stabilizer theorem, $|G:N| = |\mathcal B|$, where $\mathcal B = \{S^g : g \in G\}$. Since $G$ is $2$-transitive, every pair of elements $x, y \in X$ appears in the same number of blocks $B \in \mathcal B$. Hence by Fisher's inequality $|G:N| = |\mathcal B| \ge |X|$.
Therefore if $H \le G$ has index $|G:H| < |X|$ then $H$ is not contained in the setwise stabilizer of a proper nonempty subset, i.e., $H$ is transitive.
For an example with equality in which $H$ is not just a point stabilizer, consider $G = \mathrm{PGL}_3(q)$ acting on the $q^2+q+1$ points of the projective plane of order $q$, and consider the setwise stabilizer of a line.
For a counterexample in which $G$ is only primitive, consider $G = S_n \wr S_2$ acting on $[n]^2$ and let $H \cong S_{n-1} \times S_n$ be a row stabilizer. Then $[G:H] = 2n < n^2$ for $n > 2$. Another example is given by $G = S_n$ acting on the $\binom n 2$ subsets of size $2$ and $H = S_{n-1}$.