= 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 $e_1,\ldots,e_n$, let $N=M^{(n-2,2)}$ have basis $w_{\{i,j\}}$, and define $S_n$-homomorphisms
$$
p_1(v_{ij})=e_i,\qquad
p_2(v_{ij})=e_j,\qquad
q(v_{ij})=w_{\{i,j\}}.
$$
Put
$$
U=\ker p_1,\qquad V=U\cap\ker p_2,
$$
and let $\varepsilon:M\to\mathbb F$ send every basis vector to one.
The augmentation submodule
$$
A=\left\{\sum_i a_ie_i:\sum_i a_i=0\right\}
$$
is $S^{(n-1,1)}$. The map $p_1$ identifies $(\ker\varepsilon)/U$ with $A$, and $p_2$ identifies $U/V$ with another copy of $A$. Also $M/\ker\varepsilon\cong S^{(n)}$.
The subspace $V$ is generated by rectangle differences
$$
v_{ij}-v_{i\ell}-v_{kj}+v_{k\ell}.
$$
Their 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 $(n-2,1,1)$ gives
$$
\ker(q|_V)\cong S^{(n-2,1,1)}.
$$
These descriptions use coefficients $0$, $1$, and $-1$ only, so they remain valid over every field.
For $n\geq4$, a required strict filtration is
$$
0<\ker(q|_V)<V<U<\ker\varepsilon<M,
$$
with successive quotients
$$
S^{(n-2,1,1)},\quad S^{(n-2,2)},\quad
S^{(n-1,1)},\quad S^{(n-1,1)},\quad S^{(n)}.
$$
For $n=3$, $q(V)=0$, 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>.
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