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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 1 / d / Solution

If $\lambda=(n^{a_n},\ldots,1^{a_1})$ is $p$-regular, then
$$
\operatorname{End}_{\mathbb F S_n}(S^\lambda)=\mathbb F.
$$
For a tableau $t$ and its row reversal $t^*$,
$$
\langle e(t),e(t^*)\rangle=\prod_j(a_j!)^j,
$$
which is nonzero in characteristic $p$. Applying a column antisymmetrizer to an endomorphism at $e(t^*)$ therefore forces its value on the cyclic generator $e(t)$ to be scalar.


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 Incoming links: Endomorphism theorem for a regular Specht module