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I have a basic question about doubly transitive subgroups of $S_n$.

Let $G$ be a doubly transitive subgroup of $S_n$ such that $G$ contains a simple transposition. Why does it follow that $G=S_n$?

This fact is freely used in the article "Galois groups of enumerative problems" by Joe Harris, but without any reference. So I assume that it is a relatively straightforward fact, but a proof is evading me.

Let me recall few definitions for the sake of context. If a group $G$ acts on a set $X$, the action of $G$ on $X$ is called transitive if for every $x, y\in X$, there exists $g\in G$ such that $g\cdot x=y$. The action is called doubly transitive if for any two pairs $(x, y)$ and $(x', y')$ with $x\neq y$ and $x'\neq y'$, there is a $g\in G$ such that $g\cdot x = x'$ and $g\cdot y=y'$. In other words, we can use an element of $G$ to move any distinct pair to any distinct pair. Some authors use the terminology 2-transitive for this concept.

So, when we say a doubly transitive subgroup of $S_n$, we are thinking of the natural action of the subgroup on $n$ letters (or symbols, etc.), say $X=\{1, 2, ..., n\}$. This is an action "inherited" from $S_n$. Finally, a simple transposition just means a permutation on $X$ that switches $i$ and $j$ ($i\neq j$) but fixes everything else.

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    If $G$ contains a transposition, say $(i,j)$ and $G$ is doubly transitive, then for any distinct $k,l \in X$, there is a $g \in G$ mapping $i \to k$ and $j \to l$, and then the conjugate of $(i,j)$ under $g$ is equal to $(k,l)$. So $G$ contains all transpositions, and hence $G=S_n$. – 2017-02-14
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    @DerekHolt Thank you so much. So slick! Feel free to post it in the answer box so that I can accept it. – 2017-02-14

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Suppose $G$ contains a transposition $\sigma=(i\ j)$. Then for any distinct $i',j'\in \{1,\dots,n\}$, there exists some $\tau\in G$ such that $\tau(i)=i'$ and $\tau(j)=j'$, since $G$ is doubly transitive. Now just notice that $\tau\sigma\tau^{-1}$ is the transposition $(i'\ j')$. Since $i'$ and $j'$ were arbitrary, this means $G$ contains all transpositions. Since transpositions generate $S_n$, this means $G=S_n$.

More generally, this shows that if $G$ contains a transposition and $N(G)$ is doubly transitive, then $G=S_n$.