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} However I know the proof of this problem using the generating function of $\sum_{c \in \text{Conj} (S_n)} \text{sgn} (c)$, I want to know another proof if there is.

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}

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}

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} However I know the proof of this problem using the generating function of $\sum_{c \in \text{Conj} (S_n)} \text{sgn} (c)$, I want to know another proof if there is.