The sum of the signs of conjugacy classes in the symmetric group S_n [duplicate]

Question

Let $r$ be the number of conjugacy classes of the symmetric group $S_n$ whose sign is $1$, i.e. \begin{equation} r := \#\{c \in \text{Conj} (S_n): \text{sgn} (c) = 1 \}. \end{equation} Let $s$ be the number of conjugacy classes of $S_n$ whose sign is $-1$, i.e. \begin{equation} s := \#\{c \in \text{Conj} (S_n): \text{sgn} (c) = -1 \}. \end{equation} I want to prove that $r - s$ is the number of self-conjugate partitions of $n$.

In other words, I want to prove that \begin{equation} \sum_{c \in \text{Conj} (S_n)} \text{sgn} (c) = \#\{\lambda: \lambda \ \text{is a self-conjugate partition of } n\}. \end{equation}

Answer 1

We'll can prove this with Clifford theory and some counting, for the normal subgroup of $A_n$ in $S_n$. Let $\tau$ be a transposition in $S_n$.

Splitting into simple lemmas:

1. $r$ is the number of orbits of $\tau$ on conjugacy classes of $A_n$, which is the number of $\tau$ orbits on the set of $A_n$ irreps.

2. $s$ is the number of orbits of $A_n$ on the coset $A_n\tau$, which equals the number of $\tau$ fixed irreps of $A_n$.

3. $r-s$ is the number of $\tau$ orbits of size two on the conjugacy classes, which is number of size two orbits $\tau$ orbits on the irreps of $A_n$.

By Clifford theory, the free orbits of $\tau$ on the irreps of $A_n$ are in bijection (by induction) with irreps of $S_n$ with character vanishing on $A_n\tau$. This is equivalent to being fixed by the bijection $V\mapsto V\otimes \sigma$ given by tensoring with the sign representation. Since tensoring with the sign representation conjugates the associated partition, we get that $r-s$ counts self conjugate partitions.