Proving that $H$ subgroup is equal to $S_n$

Question

Let $H$ a subgroup of $S_n$ such that for all $a,b\in \{1,2,...,n\}$ there exist $\tau\in H$ such that $\tau(a)=b$. Show that if $H$ contains a transposition and a $n-1$ cycle then $H=S_n$.

Answer 1

Okay since you have a transposition (lets call it $\tau=(i,j)$ ) in H and for any $a,b\in\{1,...,n\}$ their exist a permutation in $\beta_{a,b}\in H$ such that $\beta_{a,b}(a)=b$, it follows that for any $a\in \{1,...,n\}$ we have that $\beta_{i,a}\tau \beta_{i,a}^{-1}=(a,j)$. In this way we generate all transpositions and thus all permutations.