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Expand the Column antisymmetrizer of a Young tableau:If two entries in one column of $v$ lie in the same row of $w$, their transposition belongs to both $C(v)$ and the row stabilizer of $w$, so the terms cancel in pairs. The assumption $b_v\{w\}\ne 0$ therefore says that every row of $w$ meets every column of $v$ in at most one entry.

The first row of $w$ has ${\lambda }_1$ entries, while $v$ has exactly ${\lambda }_1$ nonempty columns. It must consequently contain exactly one entry from each column of $v$. Permuting within each column puts these entries in the first row positions of $v$. Delete the matched first rows and repeat the argument on the remaining Young diagram. The product of the resulting column permutations is an element $h\in C(v)$ for which the row sets of $hv$ are those of $w$. ThusThis is the nonzero column antisymmetrizer criterion.

The assumed one-dimensional-image property givesfor some $c\in \mathbb{F}$. The coefficient of $\{t\}$ in $e(t)$ is one, soThe tabloid bilinear form is invariant, and the involution on the group algebra fixes the Column antisymmetrizer of a Young tableau because inversion preserves sign. Thereforewhich proves the formula.

Choose a copy of $S^{\alpha }$ inside $M^{\lambda }$ and a nonzero polytabloid in it. Its image under a suitable Column antisymmetrizer of a Young tableau is nonzero. By the fact allowed in the question, this can happen only if $\alpha$ dominates $\lambda$. Hence $m_{\alpha }>0$ implies $\alpha ⊵\lambda$.

Let $b_t$ be the Column antisymmetrizer of a Young tableau. For a tabloid $\{u\}$, if two entries from one column of $t$ lie in one row of $u$, their column transposition fixes $\{u\}$ and pairs every term of $b_t\{u\}$ with its negative. Thus $b_t\{u\}=0$.

Otherwise each row of $u$ meets each column of $t$ at most once. Matching entries within columns then gives a column permutation $c\in C(t)$ for which $\{cu\}=\{t\}$. Reindexing the antisymmetrizer gives $b_t\{u\}=\mathrm{sgn}(c)e(t)$. Hence every basis tabloid maps into $\mathbb{F}e(t)$, while $b_t\{t\}=e(t)$, and therefore