 Solution

Identify $M=M^{(n-2,1^2)}$ with the vector space having basis $v_{ij}$ for ordered pairs $i\ne j$. Let $P$ be the permutation module with basis $e_1,\dots ,e_n$, let $N=M^{(n-2,2)}$ have basis $w_{\{i,j\}}$, and define $S_n$-homomorphismsPutand let $\epsilon :M\to \mathbb{F}$ send every basis vector to one.

The augmentation submoduleis $S^{(n-1,1)}$. The map $p_1$ identifies $(\ker \epsilon )/U$ with $A$, and $p_2$ identifies $U/V$ with another copy of $A$. Also $M/\ker \epsilon \cong S^{(n)}$.

The subspace $V$ is generated by rectangle differencesTheir images under $q$ are the standard polytabloid generators of $S^{(n-2,2)}$, so $q(V)=S^{(n-2,2)}$. The kernel of $q{|}_V$ is generated by the alternating oriented-triangle relations; identifying these with the column antisymmetrizations of shape $(n-2,1,1)$ givesThese descriptions use coefficients $0$, $1$, and $-1$ only, so they remain valid over every field.

For $n\ge 4$, a required strict filtration iswith successive quotientsFor $n=3$, $q(V)=0$, so omit the repeated term $V=\ker (q{|}_V)$; the factors are $S^{(1,1,1)}$, two copies of $S^{(2,1)}$, and $S^{(3)}$. This is the Specht filtration of the ordered-pair permutation module.