Conjugacy class size formula in symmetric group

Statement

Suppose $n$ is a natural number and $\lambda$ is an unordered integer partition of $n$ such that $\lambda$ has $a_j$ parts of size $j$ for each $j$. In other words, there are $a_1$ $1$s, $a_2$ $2$s, $a_3$ $3$s, and so on. Let $c$ be the conjugacy class in the symmetric group of degree $n$ comprising the elements whose Cycle type (?) is $\lambda$, i.e., those elements whose Cycle decomposition (?) has $a_j$ cycles of length $j$ for each $j$. Then:

$\left|c\right|=\frac{n!}{{\prod }_j(j{)}^{a_j}(a_j!)}$

Note that those $j$ where $a_j=0$ contribute a $1$ in the denominator and can be ignored from the product, while for those $j$ where $a_j=1$, the ,math>a_j!</math> term can be omitted.

Equivalently, if $C$ is the centralizer of any element of $c$, then:

$\left|C\right|={\prod }_j(j{)}^{a_j}(a_j!)$

These are equivalent because size of conjugacy class equals index of centralizer, which follows from the identification of the conjugacy class with the left coset space of the centralizer via the action of the group on itself as automorphisms by conjugation.