Transitive subgroup of symmetric group I'm working on the following question, and honestly have no idea how to begin.  Any hints would be greatly appreciated!
Let $H$ be a subgroup of $S_n$, the symmetry group of the set $\{1,2,\dots, n\}$. Show that if $H$ is transitive and if $H$ is generated by some set of transpositions, then $H=S_n$.  

Proof by induction on $n$. Case $n\le 2$ is trivial. 
Let $X$ be the set of transpositions in $H$, $X_1$ the set of transpositions in $X$ fixing 1 and $K$ the subgroup generated by $X_1$.
We have to prove that $K$ is transitive on $\{2,\dots,n\}$. If true, induction hypothesis implies $K$ is the full symmetry group on $\{2,\dots,n\}$, and since $H$ properly contains $K$ which is a maximal subgroup in $S_n$, it's immediate to conclude.
So let's prove this transitivity claim on $K$. Otherwise $K$ has at least two orbits $I,J$ in $\{2,\dots,n\}$. Since $H$ does not stabilize $I$ and is generated by transpositions, it contains a transposition $t=(u,v)$ with $u\notin I$ and $v\in I$. If $u\neq 1$, we deduce that $t\in X_1\subset K$ and deduce that $u$ is in the same $K$-orbit as $v$, a contradiction. So $u=1$ and $H$ contains the transposition $(1,v)$, $v\in I$. Similarly, $H$ contains a transposition $(1,w)$ with $w\in J$. Hence $H$ contains $(1,v)(1,w)(1,v)=(v,w)$. This is again a contradiction since this transposition would have to belong to $K$ and contradict that $v,w$ are in distinct $K$-orbits. So $K$ is transitive on $\{2,\dots,n\}$. This finishes the proof.
