i have to show
if n is odd then there are exactly two conjugacy classes of n cycles in An each of which contains (n-1)!/2 elements.
also there is a hint says let An act on itself
i know the fact that since An acts on itself, x doesnt commute with any odd permutation. So it splits into two Ccl An (x) and CCl An (12)x(12).
First of all i have no idea about how to find the elements of CCl An (12)x(12). For sure there are n!/2 elements of An and it will have the half of the number of elements that Sn does.
the number of the elements in CCl Sn (x) is n (n-1)...(n-k+1)/ k
so we better get k=n!/2
but somehow i cant reach the answer
thank you for your help