Proof any Finite Doubly Transitive Permutation Group containing a transposition is a Full Symmetric Group?

Question

This is the special case I am working on (from this wikipedia page on symmetric groups). How does one show that if $\operatorname{Gal}(f(x, s)/Q(s))$ acts doubly transitive on the roots of $f$ and if it contains a transposition, then $\operatorname{Gal}(f(x, s)/Q(s))$ must be the full symmetric group $S_n$. The more general case of this theorem is in the title above. I know how to prove that a transitive subgroup with a transposition of $S_p$ ($p$ prime) must be equal to $S_p$, but don't know how to prove this for general n integer.

Answer 1

Let $G \subseteq S_n$ be a group acting doubly transitive on $\{1,\dots,n\}$ and suppose $(1,2) \in G$. Now consider $i,j \in \{1,\dots, n\}$, $i \neq j$. Since $G$ acts doubly transitive, there exists $\sigma \in G$ such that $\sigma(1) = i$ and $\sigma(2) = j$. We have $$(i,j) = \sigma (1,2) \sigma^{-1} \in G.$$ It follows that $G$ contains all transpositions and since $S_n$ is generated by transpositions, we have $G = S_n$.