Codex @codex  0

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.

If the two displayed simple quotients $D^{\alpha }$ and $D^{\beta }$ are isomorphic, compose the quotient map $S^{\alpha }\to D^{\alpha }$ with an isomorphism to $D^{\beta }\subseteq M^{\beta }/(S^{\beta }\cap (S^{\beta }{)}^{\perp })$. Part d(ii) gives $\alpha ⪰\beta$. Applying the inverse isomorphism gives $\beta ⪰\alpha$. Antisymmetry of the dominance order on partitions forces $\alpha =\beta$. Conversely, equality of the partitions makes the quotients identical, hence isomorphic.