Proof that conjugate partitions of n give irreducible characters of the symmetric group that differ by sgn. Consider two conjugate partitions $\lambda$ and $\lambda'$, which can just be viewed as having flipped rows and columns in the associated Young diagrams.
Why is it the case that $\chi^{\lambda}(\pi) = sgn(\pi)\chi^{\lambda'}(\pi)$ ?
Most places I have read simply claim this comes from the construction of $S^{\lambda}$, the irreducible representations of $S_n$. Looking at the construction, it seems like for conjugate partitions we have modules:
picking a tableau t of shape $\lambda$, we have $S^{\lambda}$ as the cyclic module generated by $\sum\limits_{\pi \in C_t} sgn(\pi)\pi (R_tt)$, and $S^{\lambda'}$ generated by $\sum\limits_{\pi \in R_t} sgn(\pi)\pi (C_tt)$.
Where $R_t$ is $S_{R_1} \times ... \times S_{R_k}$ and $C_t$ is $S_{C_1} \times ... \times S_{C_l}$ where these are the sets of numbers in the rows and columns of t.
I don't know where to go from here--is there a simple way to write out the action by some permutation and show it differs by a sign?
 A: Let $T$ be a standard tableau of shape $\lambda$, $R_T$ and $C_T$ the row and column subgroups, and $$c_\lambda = \sum_{r \in R_T, c \in C_T} (-1)^c r c$$ the symmetriser. Then $V_\lambda = \mathbb{C}[S_n]c_\lambda$ is the irreducible Specht module of shape $\lambda$.
Consider the linear map $\tau: \mathbb{C}[S_n] \to \mathbb{C}[S_n]$ acting by $\tau(\pi) = (-1)^\pi \pi$ for all $\pi \in S_n$. For any representation $\rho: \mathbb{C}[S_n] \to V$, the composition $\rho \circ \tau$ is a representation, and it is not hard to see that it is the tensor product of $\rho$ with the sign representation.
Now check what $\tau$ does to the symmetriser $c_\lambda$:
$$ \tau(c_\lambda) = \sum_{r \in R_T, c \in C_T} (-1)^c \tau(r) \tau(c) = \sum_{r \in R_T, c \in C_T} (-1)^r r c = \sum_{r \in R_{T'}, c \in C_{T'}} (-1)^c rc = c_{\lambda'}$$
where here I have used the fact that if $T'$ is the transpose of $T$, then $R_T = C_{T'}$ and $C_T = R_{T'}$. Hence the composite of $\tau$ with the representation $V_\lambda$ is the representation $V_{\lambda'}$.
