Let $t$ be a (not necessarily standard) tableau of shape $\lambda \vdash n$ consisting of the numbers $1, ..., n$, each used exactly once.
Notation:
(i) $\mathfrak{S}_{n}$ is the symmetric group on $n$ letters, and $R_t, C_t$ are the row and column subgroups of $\mathfrak{S}_n$ corresponding to $t$, respectively. i.e. consists of all row (resp. column) stabilizers under the action of $\mathfrak{S}_n$ on $t$.
(ii) $a_t = \displaystyle\sum_{\sigma \in R_t} \sigma$, $b_t = \displaystyle\sum_{\pi \in C_t} \text{sgn}(\pi)\pi$, $c_t = b_ta_t$.
(iii) $f^\lambda$ denotes the number of standard Young tableau with shape $\lambda$.
(from Fulton's Young Tableaux, p. 96-97 (Lemma 5))
I desire to show that $c_t^2=\displaystyle\frac{n!}{f^\lambda} c_t$.
Writing $c_tc_t = \displaystyle\sum_{\pi_1 \in C_t}\displaystyle\sum_{\sigma_1 \in R_t}\displaystyle\sum_{\pi_2 \in C_t}\displaystyle\sum_{\sigma_2 \in R_t} \text{sgn}(\pi_1\pi_2) \pi_1\sigma_1\pi_2\sigma_2 = \displaystyle\sum_{\tau \in C_t}\displaystyle\sum_{\omega \in R_t} n_{\tau\omega}\text{sgn}(\tau)\tau\omega$ and matching coefficients, we have $n_{\tau\omega} = \displaystyle\sum \text{sgn}(\tau^{-1}\pi_1\pi_2)$ summed over all $(\pi_1, \sigma_1, \pi_2, \sigma_2)$ satisfying $\pi_1\sigma_1\pi_2\sigma_2 = \tau\omega$.
Note that $\tau_1\omega_1 = \tau_2\omega_2$ implies $\tau_1 = \tau_2$, $\omega_1 = \omega_2$ since $R_t \cap C_t = 1$. Thus we may simply fix a pair $(\tau, \omega)$ and replace $\tau^{-1}\pi_1$ with $\pi_1$ and $\sigma_2\omega^{-1}$ with $\sigma_2$ to rewrite the condition to $\pi_1\sigma_1\pi_2\sigma_2 = 1$. In particular, the number of such quadruples do not depend on the choice of $(\tau, \omega)$, and is also independent of choice of $t$ since $R_{\eta t} = \eta R_t \eta^{-1}$ and $C_{\eta t} = \eta C_t \eta^{-1}$ for any $\eta \in \mathfrak{S}_n$. Rewrite $n_{\tau\omega} = n_\lambda, c_t^2 = n_\lambda c_t$ to indicate dependence on $\lambda$ only.
Fulton says that $n_\lambda$ is the number of $(\pi_1, \sigma_1, \pi_2, \sigma_2)$ satisfying $\pi_1\sigma_1\pi_2\sigma_2 = 1$ and $\text{sgn}(\pi_1\pi_2) = 1$, which seems to me that the possibility of $\text{sgn}(\pi_1\pi_2) = -1$ is being ignored entirely. What am I missing here? In particular, my question here is:
Does $\pi_1\sigma_1\pi_2\sigma_2 = 1$ imply $\text{sgn}(\pi_1\pi_2) = 1$ automatically?
Edit: The answer to this question is no, as shown by a counterexample here.
Exercises 18, 19 in Fulton p. 103 outline a representation-theoretic approach to determine the coefficient $n_\lambda$ by evaluating the trace of the map $\mathbb{C}\mathfrak{S}_n \rightarrow \mathbb{C}\mathfrak{S}_n$ defined as left multiplication by $c_t$. This raises another question:
Can we directly evaluate the sum $\displaystyle\sum \text{sgn}(\pi_1\pi_2)$ over $\pi_1\sigma_1\pi_2\sigma_2 = 1$ to obtain $\displaystyle\frac{n!}{f^\lambda}$?
