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...
 A: In this case, doubly transitive means that $|\{1,\dots,n\}|\ge 2$ and for all $1\le i_1,i_2,j_1,j_2\le n$ with $i_1\ne i_2$ and $j_1\ne j_2$, there is a permutation $\sigma\in S_n$ such that $\sigma(i_1)=j_1$ and $\sigma(i_2)=j_2$. The first condition is just $n\ge 2$. To prove the second condition, we will take five cases.
Case I: $i_1,i_2,j_1,j_2$ are all (pairwise) distinct. Take $\sigma=(i_1j_1i_2j_2)$.
Case II: $i_1=j_2$ and $i_2\ne j_1$. Take $\sigma=(i_1j_1i_2)$.
Case III: $i_1=j_2$ and $i_2=j_1$. Take $\sigma=(i_1j_1)$.
Case IV: $i_1\ne j_1$ and $i_2=j_2$. Take $\sigma=(i_1j_1)$.
Case V: $i_1=j_1$ and $i_2=j_2$. Take $\sigma=\text{id}$.
Any other case would violate the distinctness conditions on $i_1,i_2,j_1,j_2$, so we are done.
A: It suffices to prove that $\mbox{Stab}_{S_n}(i)$ is transitive on $\{1,\ldots,n\}\setminus\{i\}$.  Since $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$ for any $i$ and symmetric groups are clearly transitive, the assertion follows immediately.  Note that we can easily prove that $S_n$ is $n$-transitive in the same way.
A: 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 $k$.
