Distinct complex Specht modules are inequivalent

Let $t_{\lambda }$ be the canonical tableau from [F1]. By [F2]-[F3], $e_{t_{\lambda }}\mathrm{\ne }0$ in $S^{\lambda }$. Let ${\iota }_{\mu }:S^{\mu }\hookrightarrow M^{\mu }$ be inclusion and set $\psi :={\iota }_{\mu }\circ f$. Since $f$ and inclusion are injective, $\psi (e_{t_{\lambda }})\mathrm{\ne }0$. By [F4]-[F5], both maps are $S_n$-equivariant, so [F6] makes $\psi$ a $\mathbb{C}[S_n]$-module homomorphism.

The inverse $f^{-1}$ is equivariant: for $y=f(x)$, surjectivity and equivariance of $f$ give $f^{-1}(\sigma y)=f^{-1}(\sigma f(x))=f^{-1}(f(\sigma x))=\sigma x=\sigma f^{-1}(y)$ for every $\sigma \in S_n$. Let $t_{\mu }$ be the canonical tableau from [F1] and let ${\iota }_{\lambda }:S^{\lambda }\hookrightarrow M^{\lambda }$ be inclusion. By [F2]-[F3], $e_{t_{\mu }}\mathrm{\ne }0$ in $S^{\mu }$; since $f^{-1}$ and inclusion are injective, ${\psi }^{\mathrm{\prime }}:={\iota }_{\lambda }\circ f^{-1}:S^{\mu }\to M^{\lambda }$ is nonzero. By [F4]-[F6], it is a $\mathbb{C}[S_n]$-module homomorphism.

Apply [F7] to the nonzero map $\psi :S^{\lambda }\to M^{\mu }$ from step 1.1; it gives $\lambda ⊵\mu$.

Apply [F7] with $\alpha =\mu$ and $\beta =\lambda$ to ${\psi }^{\mathrm{\prime }}$ from step 1.2. It gives $\mu ⊵\lambda$.

Steps 2.1 and 2.2 give both dominance relations, so [F8] yields $\lambda =\mu$.