$S_n$ acting transitively on $\{1, 2, \dots, n\}$ I am reading Dummit and Foote, and in Section 4.1: Group Actions and Permutation Representations they give the following example of a group action: 


  The symmetric group $G = S_n$ acts transitively in its usual action as permutations on $A = \{1, 2, \dots, n\}$. Note that the stabilizer in $G$ of any point $i$ has index $n = | A |$ in $S_n$ (My italics)


I am having trouble seeing why the stabilizer has index $n$ in $S_n$. Is this due to the fact that we have $n$ elements of $A$ and thus $n$ distinct (left) cosets of the stabilizer $G_i$?

And how would I see this if I were to work with the permutation representation of this action? The action is not very clear to me.
We have that $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$ for every $i\in \{1,\ldots,n\}$.  (Why?) 


   $S_n$ contains all possible permutations on $n$ vertices.  The set of those permutations which fix $i$ are simply the ones that permute the set $\{1,\ldots,n\}\setminus i$ in every possible way.  Thus  $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$.


Now, $|S_{n-1}|=(n-1)!$, so $[S_n:\mbox{Stab}_{S_n}(i)]=\ldots?$


   $$[S_n:\mbox{Stab}_{S_n}(i)]=\frac{n!}{(n-1)!}=n.$$


And how big is $A$?
$\sigma \in \textrm{ Stab } (i)$ for $1\le i \le n$ iff $\sigma $ fixes $i.$ You just have to count the number of permutations that fix $i$ - working in the usual order, there are $n-1$ choices for the image of the first element in the domain, $n-2$ for the second, and so on, so that $|\textrm{ Stab } (i) | = (n-1)!.$ Apply Lagrange's Theorem.