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 $H=S_n$.  
 A: This might not be at the appropriate level for you, but I think this is a cute argument.
Suppose we wanted to transpose $1$ and $n$ in the list $1,\cdots,n$ but we were only allowed to swap adjacent numbers each step, how would we do it? Well, we could use adjacent transpositions to move $1$ all the way to the right, which would shift the numbers $2,\cdots,n$ each one left. Then we could move $n$ to the left with adjacent transpositions until we get the list $n,2,\cdots,n-1,1$.
In other words, we have
$$ (12)(23)\cdots(n-2\,n-1)\cdot(n\,n-1)\cdots(32)(21)=(n1). \tag{$\ast$} $$
(Remember, the rightmost permutations are applied to an input first. Interpret $(u\,v)$ as "swap the numbers that are $u$th and $v$th in the list.") We can use this...
Form a graph as follows: given a generating set $S$ of $H$ consisting of transpositions, let $\{1,\cdots,n\}$ be the vertex set and form an edge $(i,j)$ for every transposition $(ij)\in S$. The hypotheses that $H$ acts transitively and is generated by $S$ is equivalent to saying this graph is connected, so between any two values $a,b\in\{1,\cdots,n\}$ there is some sequence of transpositions
$$ (a_1\,a_2),(a_2\,a_3),\cdots,(a_{k-1}\,a_k)\in S $$
where $a_1=a$ and $a_k=b$. Then we can reuse the $(\ast)$ trick:
$$ (a_1\,a_2)(a_2\,a_3)\cdots(a_{k-2}\,a_{k-1})\cdot(a_k\,a_{k-1})\cdots(a_3\,a_2)(a_2\,a_1)=(a_k\,a_1)=(ab)\in H.$$
That is, $H$ contains every transposition $(ab)$, so it must be $S_n$.
A: 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.
A: 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.
