= Solution
Let $\theta\in\operatorname{End}_{\mathbb F S_n}(S^\lambda)$ and put
$$
q=\prod_{j=1}^n(a_j!)^j.
$$
Since $\lambda$ is <regular partition>[$p$-regular], every $a_j<p$, so $q\ne0$ in $\mathbb F$. Parts b and c give $b_t e(t^*)=q e(t)$. Since $\theta$ commutes with the group-algebra action,
$$
b_t\theta(e(t^*))
=\theta(b_t e(t^*))
=q\,\theta(e(t)).
$$
The left side belongs to $b_tM^\lambda=\mathbb F e(t)$. 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>.
Back to article page
