Homomorphisms from Specht to Young permutation modules obey dominance

Given: $n\ge 0$, partitions $\lambda ,\mu ⊢n$, and a complex $\mathbb{C}[S_n]$-module homomorphism $\varphi :S^{\lambda }\to M^{\mu }$.

$S_n$ is the group of bijections of $\{1,\dots ,n\}$, with $S_0$ trivial (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition, $\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Young subgroups, tabloids, and permutation modules).

Each $M^{\nu }$ is a finite-dimensional complex representation of $S_n$ (Young subgroups, tabloids, and permutation modules).

The $\nu$-tabloids form a basis of $M^{\nu }$, with the linear extension of the left $S_n$-action (Young subgroups, tabloids, and permutation modules).

For a $\lambda$-tableau $t$, ${\kappa }_t={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\gamma$, $e_t={\kappa }_t\cdot \{t\}$, and $S^{\lambda }$ is the complex span of the $e_s$ (Column antisymmetrizers, polytabloids, and Specht modules).

Every $e_t$ is nonzero, since its coefficient at $\{t\}$ is $1$ (Column antisymmetrizers, polytabloids, and Specht modules).

$S^{\lambda }$ is an $S_n$-subrepresentation generated by any one $e_t$ (Polytabloid covariance and the column sign rule).

A subrepresentation is a linear subspace stable under each group element (Subrepresentations, direct sums of representations, and irreducibility).

For a finite group over a field whose characteristic does not divide the group order, every subrepresentation has a complementary subrepresentation; this applies over $\mathbb{C}$ (Maschke's theorem for finite groups over fields whose characteristic does not divide $\mathrm{\mid }G\mathrm{\mid }$).

The space ${\mathrm{Hom}}_{S_n}(V,W)$ consists of complex-linear $S_n$-equivariant maps and is a subspace of the complex vector space of linear maps (Intertwiners, the spaces ${\mathrm{Hom}}_G(V,W)$ and ${\mathrm{End}}_G(V)$, equivalent representations, and faithful representations).

A complex-linear map is $S_n$-equivariant exactly when it is a $\mathbb{C}[S_n]$-module homomorphism (For a commutative ring $R$, $R$-linear $G$-actions are exactly the compatible left $R[G]$-module structures).

If ${\kappa }_t{M}^{\mu }\mathrm{\ne }0$, then $\lambda ⊵\mu$ (Nonzero antisymmetrizer image detects dominance).

For every $\lambda$-tableau $t$, ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$ (The antisymmetrizer image in its own tabloid module is one-dimensional).

$\lambda ⊵\mu$ means every prefix sum of $\lambda$ is at least the corresponding prefix sum of $\mu$ (Dominance order on partitions).

Every shape has a canonical standard row-filled tableau $t_0$ (Young subgroups, tabloids, and permutation modules).

The label set $\{1,\dots ,n\}$ is finite (empty when $n=0$), and the set of bijections of a finite set is finite; hence $S_n$ is finite (The natural numbers $\mathbb{N}$ (von Neumann), Young subgroups, tabloids, and permutation modules, A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

No form of the Axiom of Choice is used. The proof uses one complement supplied by Maschke's theorem and existential witnesses from spanning families; it does not choose from an arbitrary indexed family.