Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 / 3 / ii / Solution

Let $0\ne f:S^{\lambda }\to M^{\mu }$ be a module homomorphism. Since $S^{\lambda }$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero polytabloid $e_t$ then has $f(e_t)\ne 0$. The sign sum in the column group satisfiesbecause every coefficient in the square receives $|C_t|$ equal contributions. Thus ${\kappa }_t{e}_t=|C_t|e_t$, and equivariance givesThe column antisymmetrizer therefore acts nontrivially on $M^{\mu }$, so some basis tabloid $[s]$ has ${\kappa }_t[s]\ne 0$. The stated dominance from a nonzero column antisymmetrizer yieldsFor $\mu =\lambda$, the same calculation forces $f(e_t)=|C_t{|}^{-1}{\kappa }_{t}f(e_t)$ into the one-dimensional space $\mathbb{C}{e}_t$. Write $f(e_t)=c{e}_t$. Since $e_t$ generates the Specht module, equivariance gives $f(g{e}_t)=cg{e}_t$ for every $g$, so $f$ is $c$ times the inclusion $S^{\lambda }\hookrightarrow M^{\lambda }$. That inclusion is nonzero, henceAs a useful consequence, if $S^{\lambda }\cong S^{\mu }$, their inclusions into $M^{\lambda },M^{\mu }$ give dominance both ways. Antisymmetry of the dominance order on partitions then gives $\lambda =\mu$, so different partitions produce nonisomorphic simple modules.