What is $S_n/S_{n-1}$? Before I begin $S_{n-1}$ is isomorphic to the subgroup of $S_n$ when $\pi(n)=n$ for all its elements, so I will just call that $S_{n-1}$ for convenience
So the first thing that I noticed was that $o(S_n/S_{n-1})=\frac{n!}{(n-1)!}=n$.  To understand the structure a bit more i tried to figure out $S_3/S_2$. I will denote the elements of $S_3$ like so $\{ 123, 231, 312, 321, 213,132\}$ where $132$ maps $1$ to $1$, $2$ to $3$ and $3$ to $2$. I don't know if this is a standard way to write them but its what I came up with. So $S_2 =\{123,213\}$. The equivalence relation $\pi  \sim \tau $ if $\pi^{-1}\tau \in S_2$ implies, $\pi^{-1}\tau(3)=3$ or $\tau(3)=\pi(3)$. WIth this in mind the cosets of $S_3$ mod $S_2$ should be $A=\{123,213\},B=\{132,312\},C=\{231,321\}$. The problem that i am having is that This doesn't appear to form a group for instance $BC$ seems to be different based on which element of c you pick for instance $BC=132 \circ 231 S_2=C$ and $BC=132 \circ 321 S_2=B$
Can someone explain what i am doing wrong and what the right way to do it is?
 A: You can take the quotient by a normal subgroup. Is $S_{n-1}$ normal in $S_n$?
A: I must differ with the previous answer.  While $S_n/S_{n-1}$ is not a group, it certainly exists....It is the set of left cosets of $S_{n-1}$ in $S_n$.  (The set of right cosets would be denoted $S_{n-1}\backslash S_n$ and the set of double cosets $S_{n-1}\backslash S_n/S_{n-1}$.)  It is a set with $n$ elements, corresponding to the set $\{1,2,...,n\}$ of objects on which $S_n$ acts in its natural action.  If $S_n$ acts on the left, then each left coset corresponds to the set of elements of $S_n$ mapping $n$ to $j$ for some $j$, and thus the cosets can be put in natural 1-1 correspondence with $\{1,2,...,n\}$.  In fact any transitive group action can be viewed as an action on $G/H$ where $H$ is the stabilizer of any point, and the $G/H$ will be in natural 1-1 correspondence with the set being acted upon.  
Some other information: The kernel of the action is the core of $H$ (the largest subgroup of $H$ that is normal in $G$).  Thus, the action is faithful if and only if there is no non-trivial subgroup of $H$ that is normal in $G$.  (Thus, $H$ normal corresponds to the "boring" case of an action of $G$ on $G/H$ that descends to the regular action of $G/H$ on $G/H$.)  The action is primitive if and only if $H$ is maximal, and the action is doubly transitive if and only if $|H\backslash G/H|=2$.
