I am currently working through this question. I have completed part (a) and (c), however I am unable to make any progress with (b). I know $S_n$ is the symmetric group on n symbols, and that it has order n!, so it is clear that there will be $n!/2$ disjoint pairs, however I'm not sure why they are of the form given, or even what the form actually represents. I believe it is something to do with the fact $A^{(i)}=A^{(j)}$ but cannot say much more than that. Can anyone show me how best to approach this?
The alternating group $A_n$ is the subgroup of $S_n$ which contains all the even permutations of $S_n$ i.e. with signature $1$ and let
$$\phi: S_n\to S_n,\quad \sigma\mapsto \sigma(i,j)$$ then it's easy to see that $\phi$ is a bijection hence $A_n$ and $\phi(A_n)$ have the same cardinal and since they are disjoint then the cardinal of $A_n$ is $\frac{n!}{2}$ and if $\sigma\in A_n$ then $\phi(\sigma)=\sigma(i,j)\in\phi(A_n)$ and the result follows.