 Solution

Let $\theta \in {\mathrm{End}}_{\mathbb{F}{S}_n}(S^{\lambda })$ and putSince $\lambda$ is $p$-regular, every $a_j<p$, so $q\ne 0$ in $\mathbb{F}$. Parts b and c give $b_{t}e(t^{\ast })=qe(t)$. Since $\theta$ commutes with the group-algebra action,The left side belongs to $b_t{M}^{\lambda }=\mathbb{F}e(t)$. Division by $q$ shows that $\theta (e(t))$ is a scalar multiple of $e(t)$. Part a(i) says that $e(t)$ generates $S^{\lambda }$, so $\theta$ is that scalar multiple of the identity. This proves the endomorphism theorem for a regular Specht module.