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 $H=S_n$.  

So I am convinced that I am miss understanding the question and its probably due to the fact that I am miss understanding the definition of $H$ being a transitive subgroup of $S_n$
Since $H$ is a transitive subgroup of $S_n$ I believe that means for any $\sigma, \tau \in S_n$ there exists an $h \in H$ such that $\sigma h = \tau$.
If this definition is correct then we can consider an arbitrary $\sigma \in S_n$. Let $h_1 \in H$ since $H$ is a subgroup of $S_n$ we have $h_1 \in S_n$ and so since $H$ is a transitive subgroup of $S_n$ we get that there is an $h_2 \in H$ such that $h_2h_1 = \sigma$ but since $H$ is a group and hence closed under multiplication we have $\sigma = h_2h_1 \in H$ and since $\sigma \in S_n$ was arbitrary $S_n = H$
I am definitly not understanding something about this question because otherwise this problem was to easy and the fact that  $H$ is generated by some set of transpositions was completely irrelevant information.
