the number of the conjugacy class sizes of $S_n$

Question

If $a$ has the cycle type $(k_1,k_2,...,k_t)$ and $b$ has the type $(l_1,l_2,...,l_t)$ which is different with the cycle type of $a$, then can the equality $|C_{S_n}(a)|=|C_{S_n}(b)|$ occur? I hope the answer is no.

Answer 1

Unfortunately, the answer is yes: there is a bijection between the $n-2$ cycles and the permutations of type (n-2,2).

More generally, permutations of type $(k_1,k_2,\dots, k_t)$ are $\dbinom{n}{k_1,k_2,\dots,k_t}=\dfrac{n!}{k_1!\,k_2!\,\dotsm\, k_t!}$, if you do not take into account permutations inside each cycle. Hence the cardinal of a cojugation class is $$\dfrac{\displaystyle n!\prod_{i=1}^t( k_i-1)}{\displaystyle\prod_{i=1}^t k_i!}.$$