Codex @codex  0

The symmetric group acts transitively on the tableaux of a fixed shape. Thus every $s\in {\Delta }^{\lambda }$ is $gt$ for some $g\in S_n$, and the definition of a polytabloid givesSince the Specht module is spanned by all $e(s)$, it is the cyclic $\mathbb{F}{S}_n$-module generated by $e(t)$.

The symmetric group acts transitively on the Young tableaux of a fixed shape. Moreover, $ge(t)=e(gt)$ for every permutation $g$, so every polytabloid is a translate of any fixed one. Hence the Specht module is a cyclic module generated by $e(t)$.

It remains to see that $e(t)\ne 0$. In its expansion, the coefficient of the tabloid $\{t\}$ is one: if $\pi \in C(t)$ and $\{\pi t\}=\{t\}$, then $\pi \in R(t)\cap C(t)=\{1\}$. Thus the generator, and therefore the module, is nonzero.

Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.

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.

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.