Column collision cancels antisymmetrization

Put $\tau =(ab)$. Since $a,b$ are in the same column of $t$, $\tau$ preserves every column set and lies in $C_t$ by [F1]. Since they are in the same row of $s$, it preserves every row set of $s$; therefore $\tau \cdot \{s\}=\{s\}$ by [F1,F2].

The labels $a,b$ are distinct, so $n\ge 2$. The canonical relabelling in [F6] preserves order and inversion number. If $a<b$, the one-line permutation for $\tau =(ab)$ has one inversion from $(a,b)$ and two inversions for each intermediate label, so its inversion number is $1+2(b-a-1)=2(b-a)-1$, which is odd. Thus $\mathrm{sgn}(\tau )=-1$ by [F5]. Order permutations by one-line notation and take the least element in each right coset $g⟨\tau ⟩$ of $C_t$. These canonical representatives partition $C_t$ into pairs $g,g\tau$, whence ${\kappa }_t={\sum }_g(\mathrm{sgn}(g)g+\mathrm{sgn}(g\tau )g\tau )={\sum }_g\mathrm{sgn}(g)g(1-\tau )$ by [F3,F4]. The least representative is uniquely defined in each finite coset, so no choice principle is used.

Applying the last expression to $\{s\}$, every term is zero because $(1-\tau )\cdot \{s\}=\{s\}-\{s\}=0$ by step 1.1. Thus ${\kappa }_t\cdot \{s\}=0$, as claimed.