Prove that $S_n$ is doubly transitive on $\{1, 2,..., n\}$ for all $n \geqslant 2$. 
Prove that $S_n$ is doubly transitive on $\{1, 2,\ldots, n\}$ for all $n \geqslant 2$.  

I understand that transitive implies only one orbit, but...

In fact, $S_n$ acts $n$-fold transitive on $\{1,\ldots,n\}$ (hence the claim follows from $n\ge 2$), i.e. for $n$ different elements (which could that be?) $i_1,\ldots ,i_n$ you can prescribe any $n$ different elements $j_1,\ldots,j_n$ and dan find (exactly) one element $\sigma\in S_n$ such that $\sigma(i_k)=j_k$ for all $k$.
