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We need to prove that the symmetric group $S_n$ acts transitively in it's usual action as permutations on $ A = \{ 1,2,3,....,n \}$.

We know that the action of a group $G$ is called transitive if there is only one orbit(Number of equivalence classes of an element).

Also , index of a stabilizer of an element gives the number of elements in the equivalence class of that element.

So , my approach is , if I somehow show that $G_i$ which is the stabilizer of any point $i$ in $A$ has index $n$ then our job is done.

$G_i = s \in S_n | s \cdot i = i $ ,

But how to show that it has index $n$ ?

Or how to show it has $n$ number of cosets ? Could anyone help ?

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    $\begingroup$ Take $a,b\in A=\{1,2,3,\ldots,n\}$. Can you find $\sigma\in S_n$ such that $\sigma(a)=b$? $\endgroup$ – Christoph Nov 11 '15 at 13:43
  • $\begingroup$ Yeah, you don't need any bells and whistles or theorems about stabilizers, just definitions. $\endgroup$ – pjs36 Nov 11 '15 at 13:50
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Hint: Notice that the cycle $(k,l)\in S_n$ for any $k,l\in \{1,2,..,n\}$.

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