Adjacent-column Garnir relation over C

Set $Z=X\cup Y$ and $A_Z={\sum }_{z\in S_Z}\mathrm{sgn}(z)z$, where $S_Z$ fixes labels outside $Z$. The columns are disjoint, so $\mathrm{\mid }Z\mathrm{\mid }=\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }>{\lambda }_j^{\mathrm{\prime }}$. If ${\lambda }_j^{\mathrm{\prime }}=0$, both $X$ and $Y$ are empty, contradicting this inequality; hence $\mathrm{\mid }Z\mathrm{\mid }\ge 2$. For each $\gamma \in C_t$, every label in $Z$ lies in one of the first ${\lambda }_j^{\mathrm{\prime }}$ rows of $\gamma \cdot \{t\}$: its preimage under $\gamma$ is in column $j$ or $j+1$, and ${\lambda }_{j+1}^{\mathrm{\prime }}\le {\lambda }_j^{\mathrm{\prime }}$ by [F4]. Thus two labels of $Z$ lie in one row of that tabloid. Form $\nu =(n-\mathrm{\mid }Z\mathrm{\mid }+1,1^{\mathrm{\mid }Z\mathrm{\mid }-1})$ and the $\nu$-tableau $u$ that places the labels of $Z$ in increasing order down its first column and all remaining labels in increasing order along its first row; this is a tableau by [F6]. Its other columns are singletons, so $C_u=S_Z$ by [F5] and ${\kappa }_u=A_Z$. Applying [F3] to $u$ and each tabloid $\gamma \cdot \{t\}$ gives $A_Z(\gamma \cdot \{t\})=0$. Expanding $e_t$ by [F1] now gives $A_Z{e}_t=0$.

Since $H$ permutes labels within the two respective columns, $H\subseteq C_t$ by [F5]. Define $A_H={\sum }_{h\in H}\mathrm{sgn}(h)h$; [F2] gives $A_H{e}_t=\mathrm{\mid }H\mathrm{\mid }{e}_t$. The left-coset decomposition $S_Z={⨆}_{g\in T}gH$ and multiplicativity [F7] give $A_Z=G_{X,Y}{A}_H$. Therefore step 1.1 yields $0=A_Z{e}_t=G_{X,Y}{A}_H{e}_t=\mathrm{\mid }H\mathrm{\mid }{G}_{X,Y}{e}_t$. The positive integer $\mathrm{\mid }H\mathrm{\mid }$ is nonzero in $\mathbb{C}$, so division gives $G_{X,Y}{e}_t=0$. The proof works for every supplied transversal $T$ and makes no further choice.