If $n$ is odd, then the set of $n$-cycles consists of two conjugacy classes of equal size in $A_n$.

Question

If $n$ is odd, then the set of $n$-cycles consists of two conjugacy classes of equal size in $A_n$.

I have read split conjugacy class in $S_n$ into two equal size conjugacy classes in $A_n$, but I still cannot clearly figure out why. I want to know how to begin the proof.

When I saw this problem, I first thought of the split criterion, but I don't know how to do next.

Could you tell me how you think through this problem and the proof?

Answer 1

HINT:

The centralizer of an $n$-cycle has cardinality $n$ and equals the subgroup generated by it. Therefore, the conjugacy class of an $n$-cycle in $A_n$ has cardinality $\frac{(n-1)!}{2}$. Now, we have $(n-1)!$ $n$-cycles.