$G \le S_n$ G operates on the set $X = \{(i,j)| i \ne j , 1 \le i , j \le n \}$ by $\mu (i,j) = (\mu (i), \mu (j))$.
If $(12) \in G$ and G operates transitively on X (there exists only one orbit), show that $G = S_n$.
What I tried:
I know that (12)(23)..(n-1 n) are all in X.
If I can prove they are also in G I'm done. But I had trouble showing that.
I tried to use the orbit stabiliser theorem but also without success.
Any help will be appreciated.
Since $S_n$ is generated by its transpositions, it is enough to show that $G$ contains all transpositions. Now, if $\tau\in G$, then $$\tau(12)\tau^{-1}=(\tau(1),\tau(2)).$$ Since $G$ acts transitively on $X$, for any $(i,j)\in X$, we can find $\tau\in G$ such that $\tau(1,2)=(i,j)$. Then, $\tau(1)=i$ and $\tau(2)=j$. This means $$\tau(12)\tau^{-1}=(ij)\in G.$$ Since this can be done for any $i\neq j$, every transposition $(ij)\in G$. Hence, $G=S_n$.