= Solution
Fix a <Young tableau> $t$ of shape $\lambda$, let $C(t)$ be its column stabilizer, and put
$$
b_t=\sum_{g\in C(t)}\operatorname{sgn}(g)g,\qquad e_t=b_t\{t\}.
$$
These are the <Column antisymmetrizer of a Young tableau> and its <polytabloid>. The <Specht module> is generated by $e_t$ as a <cyclic module>, so the nonzero <R-module homomorphism> $\phi$ has $\phi(e_t)\ne0$. Because its extension $\Phi$ commutes with the <group algebra> action,
$$
0\ne\phi(e_t)=\Phi(b_t\{t\})=b_t\Phi(\{t\}).
$$
Expand $\Phi(\{t\})$ in the <tabloid> basis of the <Young permutation module> $M^\mu$. At least one <tabloid> $\{u\}$ therefore satisfies $b_t\{u\}\ne0$.
If a row of $u$ contains two entries from the same column of $t$, their transposition $\tau\in C(t)$ fixes $\{u\}$. Pairing the terms indexed by $g$ and $g\tau$ in $b_t\{u\}$ makes them cancel. This remains valid in <characteristic> two, since the paired coefficients sum to zero there as well. Thus each row of $u$ meets each column of $t$ in at most one entry.
The first $r$ rows of $u$ consequently contain at most $\min(r,\lambda'_j)$ entries from column $j$ of $t$, where $\lambda'$ is the <conjugate partition>. Hence
$$
\sum_{i=1}^r\mu_i\leq\sum_j\min(r,\lambda'_j)=\sum_{i=1}^r\lambda_i
\qquad(r\geq1).
$$
By the <dominance order on partitions>, this proves
$$
\boxed{\lambda\unrhd\mu.}
$$
This is <dominance from a nonzero column antisymmetrizer>. The extension is what allows us to apply $b_t$ to an element of $M^\mu$ without dividing by the column stabilizer's order, which might vanish in the given <field>.
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