Find the number of cosets of $S_{n-1}$ in $S_n$

Question

Find the number of cosets of $S_{n-1}$ in $S_n$.

My attempt:

The number of elements in $S_{n-1}$ is $(n-1)!$ and in $S_n$ is $n!$ , so the number of cosets should be $n$ [by Lagrange's theorem]. How do I proceed from here ? How do I prove that the $n$ cosets will be distinct?

Answer 1

To represent the cosets explicitly:

To start: $S_n$ acts on $\{1,2,\ldots, n\}$ and let $H$ be the subgroup isomorphic to $S_{n-1}$ that fixes $n$ i.e., $H = \{\pi \in S_n: \pi(n)=n\}$.

Then the $k$-th coset of $H$; $k=1,\ldots, n-1$ are the set of elements $\pi$ that satisfy $\pi(n) = k$.

What GSofer said about Lagrange's Thm.

Answer 2

You are basically done, since by Lagrange's theorem - they are distinct.