Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 4 / d

Because $\lambda$ is $p$-regular, Fact 1 gives a tableau $u$ for whichPart a, applied linearly to the tabloid expansion of $e(u)$, gives

The nonzero homomorphism $\theta$ cannot kill $e(t)$, because the Specht module is generated by the translates of this polytabloid. Equivariance and part d(i) therefore giveThus $b_t$ acts nontrivially on $M^{\mu }/V$, and hence on $M^{\mu }$. Some $\mu$-tabloid $\{v\}$ must satisfy $b_t\{v\}\ne 0$. Fact 2 now says that $\lambda$ dominates $\mu$ in the dominance order on partitions.

Assume $\lambda =\mu$. Part a says that the image of $b_t$ on $M^{\lambda }/V$ is contained in the one-dimensional space spanned by $e(t)+V$. The identity from part d(ii) then shows that $\theta (e(t))$ itself is a scalar multiple of $e(t)+V$.

Apply this with $V=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$ and with $\theta$ obtained by composing the quotient map $S^{\lambda }\to D^{\lambda }$ with any endomorphism of the Simple symmetric-group module from a regular partition $D^{\lambda }$. Every endomorphism is scalar on the generating polytabloid and hence on all its translates. Therefore