= Solution
For <integer partitions> $\lambda,\mu$ of $n$, padded by zero parts, the <dominance order on partitions> is
$$
\boxed{\lambda\unrhd\mu\iff\sum_{i=1}^r\lambda_i\geq\sum_{i=1}^r\mu_i\quad\text{for every }r\geq1.}
$$
A $\lambda$-tableau fills the <Young diagram> of $\lambda$ with $1,\ldots,n$, each once. A <tabloid> $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The <Young permutation module> $M^\lambda=\mathbb CX_\lambda$ has these <tabloids> as basis, with $S_n$ acting by permuting the entries.
The <row and column stabilizers> $R_t,C_t$ independently permute entries within the rows and columns of $t$. The <column antisymmetrizer> is $\kappa_t=\sum_{c\in C_t}\operatorname{sgn}(c)c$, and $e_t=\kappa_t[t]$ is the associated <polytabloid>. Since $C_{gt}=gC_tg^{-1}$ and conjugation preserves permutation sign,
$$
\kappa_{gt}=g\kappa_tg^{-1},\qquad e_{gt}=\kappa_{gt}[gt]=g\kappa_t[t],\qquad \boxed{ge_t=e_{gt}.}
$$
Thus the <Specht module> is the submodule
$$
\boxed{S^\lambda=\operatorname{span}_{\mathbb C}\{e_t:t\text{ is a }\lambda\text{-tableau}\}\subseteq M^\lambda.}
$$
Every tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The <tabloids> $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne0$.
We use the following precise <column antisymmetrizer> facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal <tabloids> with opposite signs in $\kappa_t[s]$, so $\kappa_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min(r,\lambda'_j)$ entries from column $j$ of $t$, whence
$$
\sum_{i=1}^r\mu_i\leq\sum_j\min(r,\lambda'_j)=\sum_{i=1}^r\lambda_i.
$$
This states and explains <dominance from a nonzero column antisymmetrizer>: $\kappa_t[s]\ne0$ implies $\lambda\unrhd\mu$.
When $\mu=\lambda$, the <nonzero column antisymmetrizer criterion> says more: either $\kappa_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and $\kappa_t[s]=\pm e_t$. Thus $\kappa_tM^\lambda=\mathbb Ce_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.
For the <inner product>, use the positive Hermitian extension of the orthonormal <tabloid> basis, linear in its first argument. It has the supplied self-adjointness of $\kappa_t$. Its orthogonal complement of $S^\lambda$ agrees with that obtained from the complex <tabloid bilinear form>, since the spanning <polytabloids> have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identity
$$
\boxed{\kappa_tu=\langle u,e_t\rangle e_t\qquad(u\in M^\lambda).}
$$
Indeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of $\kappa_tu$, which is $\langle\kappa_tu,[t]\rangle=\langle u,\kappa_t[t]\rangle=\langle u,e_t\rangle$.
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