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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.

A filling $T$ of shape $\lambda$ and content $\alpha$ defines a module homomorphism ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: given a numbered tableau, sum over distinct assignments obtained by permuting $T$ within its rows, and put each numbered entry into output row $j$ when its assigned label is $j$. Restricting to $S^{\lambda }$ gives ${\hat{\Theta }}_T$. For Semistandard Young tableaux $T$, these restrictions are linearly independent over every field. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ if the characteristic of a field is not two, or if $\lambda$ is a two-regular partition of an integer. For two-singular shapes in characteristic two, they need not span. Column antisymmetrization and triangular leading tabloids establish independence; straightening the images of a cyclic polytabloid establishes spanning under the stated hypotheses.

The polytabloids $e_t$ indexed by standard Young tableaux of shape $\lambda$ form a basis of $S^{\lambda }$ over every field, and a basis of its integral lattice over $\mathbb{Z}$. Column antisymmetry and Garnir relations straighten any polytabloid into this set. Ordering leading tabloids by the rows containing successive entries makes the corresponding coefficient matrix triangular with diagonal entries one, proving independence. In particular the dimension is the standard-tableau count in every characteristic.

The character of a Young permutation module counts fixed tabloids and is integer-valued. Dominance order on partitions makes its decomposition into Specht modules triangular with diagonal multiplicity one. Descending induction in any order refining dominance therefore expresses each irreducible Specht character as an integer combination of permutation characters, proving integrality.

Let $M(\mu ,\lambda )$ be the multiplicity of $V^{\mu }$ in the complex Young permutation module $M^{\lambda }$. For $\lambda ⊢n$ and $\rho ⊢n-1$,Here $\nearrow$ means adding one cell, and zero parts are omitted. Equal row lengths contribute separately on the right. Restrict $M^{\lambda }$ to $S_{n-1}$ and separate the tabloid orbits by the row containing $n$ to obtain the right side. Decompose into irreducibles and use the restriction branching rule for a symmetric group to obtain the left side. Equating multiplicities proves the formula.

A Young tableau is a filling of a Young diagram with entries, typically positive integers. The shape is the partition of an integer giving its row lengths. We draw Young tableaux in the English convention of Young diagram, with left-aligned rows arranged from top to bottom. A filling need not obey increasing conditions unless these are specified: standard Young tableaux use each of $1,\dots ,n$ exactly once and strictly increase along rows and down columns; Semistandard Young tableaux permit repetitions and require weakly increasing rows and strictly increasing columns.

In the constructions for a symmetric group involving Young symmetrizers, tabloids, and polytabloids, an unqualified numbered Young tableau means a bijective filling by $1,\dots ,n$, with no increasing condition imposed. This narrower convention is compatible with the broader use needed for Semistandard Young tableaux. Historically, Young's work connected these fillings to representation theory of the symmetric group; in particular, the equality between standard Young tableau counts and irreducible representation degrees over the complex numbers predates the modern hook-length formula. See Stanley's Algebraic Combinatorics, historical notes to Chapter 8.