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

If $p\nmid n$, the preceding part says $U=S_F^{(n-1,1)}$ is simple and has no nonzero homomorphism into $W$. Its only nonzero simple quotient is itself, so $W$ cannot contain a submodule isomorphic to such a quotient.

If $p\mid n$, use the given nonzero homomorphism $\varphi :U\to W$. The submodule chain shows that $T$ is the unique maximal proper submodule of $U$ and $U/T$ is simple. Moreover ${\mathrm{Hom}}_{S_n}(T,W)=0$: for odd characteristic this follows from part (b), and in characteristic two the divisibility condition makes $n$ even, where the corrected fixed-space calculation also gives zero. Thus $\varphi$ kills $T$ and descends to a nonzero map $U/T\to W$. Simplicity makes this map injective.

Its image is the required simple submodule, provingIn the divisible case the quotient is explicitly $U/T$, of dimension $n-2$. The argument uses the supplied existence of $\varphi$ and only the fixed-space vanishing that remains valid in that case.