Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / B / 5 / Solution

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.