Past exam of the representation Theory of Symmetric Groups course of the University of Oxford / 2011 / 2 / c / Solution

Identify $M=M_F^{(n-1,1)}$ with $F^n$, whose basis vectors record the singleton row. PutIf the characteristic does not divide $n$, then $M=T\oplus U$, and $U$ is simple. If it divides $n$, then $0<T<U<M$ is the submodule chain, with $T$ and $M/U$ trivial and $U/T$ simple of dimension $n-2$. In both cases there are no additional submodules: any vector with unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$, and the orbit of such a difference spans $U$. A submodule containing no nonconstant vector is contained in $T$.

Now suppose $p\nmid n$. When $p\ne 2$, part (b)'s valid vanishing ${\mathrm{Hom}}_{S_n}(M,W)=0$ implies ${\mathrm{Hom}}_{S_n}(U,W)=0$: extend a putative map using the projection $M\to U$.

When $p=2$, the assumption $p\nmid n$ makes $n$ odd. The corrected calculation in part (b) says every map $M\to W$ sends its singleton-row basis vectors to the same multiple of the globally invariant vector $v$. Hence it kills the augmentation submodule $U$. Extending a map $U\to W$ along the direct-sum projection again makes it zero. Thus, despite the characteristic-two error in the earlier claim, the requested conclusion remains valid:

The structural input is the submodule lattice of the standard symmetric-group permutation module.