Permutations on $\{1,2,\dots,n\}$ fixing $n$. Let $G = S_n$, the symmetric group of order $n$, acting as permutations on the set $\{1,2,\dots,n\}$. Let $H = \{\sigma \in G \mid n \cdot \sigma = n\}$.
(i) Prove that $H$ is isomorphic to $S_{n-1}$.
(ii) Find a set of elements $a_1,\dots,a_n \in G$ such that $Ha_1,\dots,Ha_n$ give all the  right cosets of $H$ in $G$.
(iii) Find the coset representation of $G$ by $H$.
I am not sure on this proof. In my self study class my TA said do not worry about it since we are not going to cover it. I was just wondering how would you prove it though. It seems like a nice proof that might help me with other topics I might learn later. 
The only things I know about permutations is that two even permutations is even. Odd permutations is odd. Even and odd gives you odd. 
Every permutation is a product of two cycles. This is transposition.
Not sure if this will help with proof but I am still looking forward to it.
 A: (i) Note that $H$ is the set of all the permutations that fixes $n$. Thus, every element of $H$ is completely determined by how it acts on $1, 2, ..., n-1$. But this is precisely the symmetric group on $n-1$ letters; hence, $H\cong S_{n-1}$.
(ii) Using part i), we have $\left|\dfrac{G}{H}\right|=\dfrac{|G|}{|H|}=\dfrac{|S_n|}{|S_{n-1}|}=\dfrac{n!}{(n-1)!}=n$. Thus, $H$ has exactly $n$ cosets. One can now show that the representative elements for these cosets are $\sigma_1$, $\sigma_2$, ..., $\sigma_n$ where $\sigma_i(n)=i$ for $1\le i\le n$. In other words, each coset of $H$ is determined by where $n$ is sent. 
(iii) I am not familiar with the term "coset representation". I would guess it means that $G$ is partitioned into $n$ cosets of $H$ (via proof of Lagrange's Theorem). Again this makes sense intuitively: each coset of $H$ represents those permutations that tells us where $n$ is sent.
A: Let's look at part (i) first. What do you know about the elements of $H$. In particular, how have you been writing elements of $S_{n}$. Have you been writing them in double row notation, say
$\sigma=\left(\begin{array}{ccccc}
1 & 2 & 3 & \cdots & n \\
1\cdot\sigma & 2\cdot\sigma & 3\cdot\sigma & \cdots & n\cdot\sigma
\end{array}\right),$
or have you been writing them as a product of disjoint cycles, say $\sigma=(1,1\cdot\sigma,1\cdot\sigma^{2}, \cdots)\cdots$. 
In either case if $\sigma\in H$ (so $n\cdot\sigma=n$), what form will $\sigma$ have?
