Codex @codex  0

The preceding parts give a simple Specht module for each partition and show that different partitions give nonisomorphic simples. The conjugacy classes of $S_n$ are indexed by partitions of $n$, namely cycle types. We use the standard finite-group character theorem: the number of irreducible complex characters equals the number of conjugacy classes, and they form an orthonormal basis of complex class functions. Consequently the ${\chi }^{\lambda }$ are the complete list of irreducible characters of $S_n$.

By Maschke's theorem, the Young permutation module $M^{\mu }$ decomposes as a direct sum of these simple modules. For a semisimple complex module, Schur lemma makes the multiplicity of a simple module equal to the dimension of the homomorphism space from that simple; character orthogonality identifies the same number with the character inner product. HencePart (ii) makes $a_{\lambda \mu }=0$ unless $\lambda ⊵\mu$, and gives $a_{\mu \mu }=1$. Taking characters provesThus the permutation-character decomposition is triangular in the dominance order on partitions, with diagonal multiplicity one. These are the dominance and diagonal-multiplicity consequences of Young's rule, obtained here without assuming its full multiplicity formula.

Let $0\ne f:S^{\lambda }\to M^{\mu }$ be a module homomorphism. Since $S^{\lambda }$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero polytabloid $e_t$ then has $f(e_t)\ne 0$. The sign sum in the column group satisfiesbecause every coefficient in the square receives $|C_t|$ equal contributions. Thus ${\kappa }_t{e}_t=|C_t|e_t$, and equivariance givesThe column antisymmetrizer therefore acts nontrivially on $M^{\mu }$, so some basis tabloid $[s]$ has ${\kappa }_t[s]\ne 0$. The stated dominance from a nonzero column antisymmetrizer yieldsFor $\mu =\lambda$, the same calculation forces $f(e_t)=|C_t{|}^{-1}{\kappa }_{t}f(e_t)$ into the one-dimensional space $\mathbb{C}{e}_t$. Write $f(e_t)=c{e}_t$. Since $e_t$ generates the Specht module, equivariance gives $f(g{e}_t)=cg{e}_t$ for every $g$, so $f$ is $c$ times the inclusion $S^{\lambda }\hookrightarrow M^{\lambda }$. That inclusion is nonzero, henceAs a useful consequence, if $S^{\lambda }\cong S^{\mu }$, their inclusions into $M^{\lambda },M^{\mu }$ give dominance both ways. Antisymmetry of the dominance order on partitions then gives $\lambda =\mu$, so different partitions produce nonisomorphic simple modules.

Suppose first that $U$ is not contained in $(S^{\lambda }{)}^{\perp }$. Since polytabloids span $S^{\lambda }$, there exist $u\in U$ and a tableau $t$ with $⟨u,e_t⟩\ne 0$. The column antisymmetrizer identity from the preliminaries givesTherefore $e_t\in U$. Because $U$ is an $S_n$-submodule, it contains all translates $g{e}_t=e_{gt}$, and these span the Specht module. Hence $S^{\lambda }\subseteq U$. If the initial supposition fails, $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the James submodule theoremNow let $W$ be a submodule of $S^{\lambda }$. Apply the dichotomy to $W\subseteq M^{\lambda }$. Either $W=S^{\lambda }$, or $W\subseteq S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Positivity of the Hermitian form gives $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }=\{0\}$: a vector in this intersection has squared norm zero. The same intersection is zero for the bilinear form by the agreement of complements explained above. Since $S^{\lambda }\ne 0$, the only submodules are $0$ and itself. ThereforeThis use of characteristic zero and nondegeneracy is why the conclusion does not automatically extend to every modular Specht module.

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

For an integer tuple $\ell =({\ell }_1,\dots ,{\ell }_m)$, the monomial alternant isNegative exponents require nonzero coordinates; all exponents in the character expansion below are nonnegative. The power-sum symmetric polynomial is $s_k(x)={\sum }_{i=1}^m{x}_i^k$. Put $\delta =(m-1,m-2,\dots ,0)$, so $A_{\delta }(x)={\prod }_{i<j}(x_i-x_j)$ is the Vandermonde determinant.

Let a conjugacy class of $S_n$ have ${\alpha }_k$ cycles of length $k$, with $\sum k{\alpha }_k=n$, and put $p_{\alpha }(x)={\prod }_k{s}_k(x{)}^{{\alpha }_k}$. The product $p_{\alpha }{A}_{\delta }$ is an alternating polynomial homogeneous of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. In an alternating polynomial, a monomial with two equal exponents has zero coefficient, since interchanging those variables fixes the monomial and reverses its sign. Grouping the remaining monomials by their permutation orbits gives a unique expansion in alternants $A_{\ell }$ with ${\ell }_1>\cdots >{\ell }_m\ge 0$.

Such tuples of the indicated total degree are exactly ${\ell }_i={\lambda }_i+m-i$ for partitions of an integer $\lambda$ of $n$ with at most $m$ parts. Define the class functionThe displayed monomial occurs with coefficient $1$ in $A_{\lambda +\delta }$ and in no other ordered alternant, soThis proves the expansion and explicitly defines its coefficients. They depend only on the cycle counts and hence are class functions. Identifying these coefficients with Specht module characters is the Frobenius alternant character formula, which the remaining parts allow us to assume.

For a box $(i,j)$ of a Young diagram, its hook of a Young diagram contains that box, all boxes to its right in its row, and all boxes below it in its column. Its hook length is $h_{ij}={\lambda }_i-j+{\lambda }_j^{\prime }-i+1$. The hook graph of a partition is the diagram with each box labeled by its hook length. Write $H_{\lambda }={\prod }_{(i,j)\in \lambda }{h}_{ij}$ for the hook product of a partition. The hook-length formula is

Pad $\lambda$ to $m$ rows and use the beta set of a partition ${\ell }_i={\lambda }_i+m-i$, so ${\ell }_1>\cdots >{\ell }_m\ge 0$ and ${\sum }_i{\ell }_i=n+d$, where $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. We prove the beta-set hook-product identity row by row. For $1\le j\le {\lambda }_i$, put $q_j=m+j-{\lambda }_j^{\prime }-1$; then $h_{ij}={\ell }_i-q_j$. The $q_j$ are distinct integers in $[0,{\ell }_i-1]$, and none equals a beta number. Indeed, if $k\le {\lambda }_j^{\prime }$ then ${\ell }_k\ge q_j+1$, whereas if $k>{\lambda }_j^{\prime }$ then ${\ell }_k\le q_j-1$. Exactly $m-i$ beta numbers lie below ${\ell }_i$, so the remaining ${\lambda }_i$ integers in that interval are precisely these $q_j$. ThereforeThis gives the equivalent Specht module dimension expression $f^{\lambda }=n!\Delta (\ell )/{\prod }_i{\ell }_i!$.

The standard Young tableaux of shape $\lambda$ form the dimension count for the Specht module. The largest entry $n$ must occupy a removable corner. Deleting it bijects tableaux with the disjoint union of standard tableaux of the shapes obtained by deleting a corner, provingSet a term to zero whenever $\lambda -e_i$ is not a partition: this includes equal adjacent rows and an attempted deletion from a zero row. The empty diagram has dimension $1$.

For an algebraic proof that the proposed formula has the same recurrence, establish the Vandermonde shift identityIts left side is an alternating polynomial in the $x_i$, because permuting the variables permutes the summands and changes the sign of every Vandermonde factor. It is therefore divisible by $\Delta (x)$. The quotient is symmetric in $x$ and homogeneous of total degree one in $(x,t)$, so has the form $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$. Differentiating in $t$ at zero and using the Euler theorem for homogeneous functions gives ${\sum }_i{x}_i{\partial }_i\Delta =d\Delta$, so $b=d$. This proves the identity as a polynomial identity, including repeated coordinates.

Take $x=\ell$ and $t=-1$. Since ${\sum }_i{\ell }_i=n+d$, it gives ${\sum }_i{\ell }_i\Delta (\ell -e_i)=n\Delta (\ell )$. For $n\ge 1$, this is exactlyThe summand is the proposed dimension for $\lambda -e_i$; if two beta numbers collide its Vandermonde is zero, and if ${\ell }_i=0$ its coefficient is zero, so no negative factorial is needed. For the empty partition, $\ell =(m-1,\dots ,0)$ and $\Delta (\ell )={\prod }_{k=0}^{m-1}k!$, giving initial value $1$. Induction now proves the hook-length formula from the tableau recurrence.

For the final sum, tuples with repeated coordinates contribute zero. Sorting each distinct nonnegative tuple with sum $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)$ gives one beta set of a partition of size $m$, since subtracting the staircase removes $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ from the sum. Conversely every partition of $m$, padded to $m$ rows, supplies exactly $m!$ ordered tuples, with the same squared summand. Hence the square-sum identity for shifted partition coordinates isThe middle equality uses the Artin–Wedderburn theorem for the group algebra and the complete classification of its simple modules by Specht modules. It explains why the last identity is a representation-dimension count rather than an accidental cancellation.

Use the tableau representation supplied by Young seminormal form, stated explicitly in Question 5, or its normalized Young orthogonal form. It constructs the complex Specht module $S^{\lambda }$ with one basis vector for each standard Young tableau of shape $\lambda$. Its adjacent-transposition matrices imply the content eigenvalues directly.

Indeed, start with $X_1=0$ and use $X_{i+1}=s_i{X}_i{s}_i+s_i$. On an admissible pair $T,R=s_{i}T$, set $a=c_T(i)$, $b=c_T(i+1)$, $d=b-a$, and $q=\sqrt{1-d^{-2}}$. In the orthogonal basis, $X_i$ is $\mathrm{diag}(a,b)$ andMultiplication givesFor a nonadmissible swap, $s_i=\pm 1$ and the next content differs by the same $\pm 1$, giving the same conclusion. Induction therefore proves $X_i{w}_T=c_T(i)w_T$. Every content vector of a standard Young tableau, reconstructed as a standard Young tableau in part (iv), is consequently a spectral weight. Together with the reverse inclusion,

An admissible adjacent swap has nonzero off-diagonal coefficient in Young orthogonal form, and the local spectral calculation in part (i) keeps it inside one irreducible representation. Thus $\alpha \approx \beta$ implies $\alpha \sim \beta$.

Two standard Young tableaux of one shape are connected by admissible swaps. Regard each as a linear extension of a partially ordered set of cells. Move the first cell of the desired extension to the front of the other extension: every cell it crosses is incomparable with it, since otherwise their order would be forced in both extensions. Repeat after fixing that first cell. Each step swaps consecutive labels in incomparable cells. Such cells lie strictly northeast and southwest of each other, so their contents differ by at least $2$ in absolute value. These are exactly admissible swaps.

This also establishes irreducibility of the constructed tableau representations. A submodule is invariant under the commuting $X_j$ and hence under their joint spectral projections; since their weights are distinct, it is a sum of tableau lines. A nonzero off-diagonal coefficient propagates any included line along every admissible edge, and connectivity then includes all lines. Different shapes give nonisomorphic modules, since their content-vector sets are disjoint and content reconstruction determines the shape. Finally, the Robinson–Schensted correspondence gives ${\sum }_{\lambda ⊢n}{f}_{\lambda }^2=n!$, and the sum of squares of irreducible degrees leaves room for no other irreducible representations.

Thus within each irreducible module the weights are exactly the tableaux of one shape, so $\alpha \sim \beta$ implies $\alpha \approx \beta$. Hence the two equivalence relations coincide.

Assume $n\ge 2$. The standard Young tableaux of shape $(n-1,1)$ are $T_k$, for $2\le k\le n$: the sole second-row entry is $k$, and the first row isThis lists all of them, since $1$ must occupy the first box and the rest of the first row is forced to increase. Their contents areWrite $w_k=w_{T_k}$. The only nontrivial adjacent swaps exchange the lower entry with $k-1$ or $k+1$. Substituting the content differences into Young orthogonal form yields all generator actions:In the first line at $k=2$, the square-root coefficient is zero and the action is $-w_2$, so no vector $w_1$ is needed. The second line is used only when $k<n$, since the generator $s_n$ does not exist.

The corresponding Specht module is the sum-zero standard representation of the symmetric group:A concrete orthonormal realization isPermuting coordinates gives exactly the actions above, linking this orthogonal tableau basis with the Gelfand–Tsetlin basis found in Question 1. Here the printed $s_j$ are the adjacent Coxeter generators, rather than products of all generators.

For a standard Young tableau $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its content vector of a standard Young tableau is $C(T)=(c_T(1),\dots ,c_T(n))$. We use the following sign convention for axial distance in a Young tableau:This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent transposition $s_r=(rr+1)$.

Here is Young orthogonal form, with one orthonormal basis vector $w_T$ for each standard Young tableau of a fixed shape:If $s_{r}T$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_{r}T$, $|d|\ge 2$ and the matrix isThis is a real orthogonal involution.

To prove the formula, explicitly use the permitted Young seminormal form theorem: the Specht module has a tableau basis $v_T$ whose vectors are joint eigenvectors of the Young–Jucys–Murphy elements, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, areNonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.

Average a positive definite Hermitian inner product over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint Young–Jucys–Murphy elements have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations implyUpon setting $w_T=v_T/\Vert v_T\Vert$, the coefficient from $w_T$ to $w_R$ becomes $\Vert v_R\Vert /\Vert v_T\Vert =\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves Young orthogonal form.

For $\lambda =(n)$, there is exactly one standard Young tableau, the row $1,2,\dots ,n$. Its content vector of a standard Young tableau isEvery adjacent distance is $1$, so $s_r{w}_T=w_T$ for every $r$. Thus $\boxed{S^{(n)}\cong {\mathbb{C}}_{\mathrm{triv}}}$ is the trivial representation.

The Specht module construction gives a nonzero inclusion $V^{\lambda }\hookrightarrow M^{\lambda }$ over the complex numbers: a polytabloid has coefficient one at its original tabloid, because the row and column stabilizers have trivial intersection, and its span is the corresponding irreducible representation. This establishes $a=M(\lambda ,\lambda )\ge 1$ without using the multiplicity-one conclusion. Similarly $a^{\prime }=M({\lambda }^{\prime },{\lambda }^{\prime })\ge 1$.

By the corrected tensor identity for induced representations and the conjugate-partition sign twist, the multiplicity of $V^{\lambda }$ in ${\tilde{M}}^{{\lambda }^{\prime }}$ is $a^{\prime }$. The character inner product already computed is the sum of products of multiplicities of common irreducible representations. ThusBoth are positive integers, so $a=a^{\prime }=1$. In particularThis deduction avoids assuming the full Young's rule, which would already contain the answer. Alternatively, the permitted one-dimensional R-module homomorphism space gives the same character inner product immediately.

Apply Young's rule, stated asHere the Kostka number counts Semistandard Young tableaux of shape $\rho$ with $n-k$ entries equal to $1$ and $k$ entries equal to $2$. Their rows weakly increase and their columns strictly increase. A column can therefore have at most two cells, so only a partition of an integer $\rho =(n-j,j)$ can occur.

In such a semistandard Young tableau, all $j$ entries in the second row must be $2$ and all $j$ entries above them must be $1$. The whole first row is then fixed by the content: its first $n-k$ entries are $1$ and its remaining $k-j$ entries are $2$. This is possible exactly when $j\le k$ and $j\le n-k$. Because $k\le n/2$, the second inequality follows from the first. There is exactly one filling for every $0\le j\le k$, and none for any other shape.

Consequently each displayed Specht module has multiplicity one, proving the two-row Young permutation module decompositionFor $j=0$, the zero second part is omitted. As a dimension check, the Hook-length formula gives $\dim S^{(n-j,j)}=\left(\genfrac{}{}{0ex}{}{n}{j}\right)-\left(\genfrac{}{}{0ex}{}{n}{j-1}\right)$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$; the sum telescopes to $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, the dimension of the original permutation representation.

Fix a Young tableau $t$ of shape $\lambda$, let $C(t)$ be its column stabilizer, and putThese are the Column antisymmetrizer of a Young tableau and its polytabloid. The Specht module is generated by $e_t$ as a cyclic module, so the nonzero R-module homomorphism $\varphi$ has $\varphi (e_t)\ne 0$. Because its extension $\Phi$ commutes with the group algebra action,Expand $\Phi (\{t\})$ in the tabloid basis of the Young permutation module $M^{\mu }$. At least one tabloid $\{u\}$ therefore satisfies $b_t\{u\}\ne 0$.

If a row of $u$ contains two entries from the same column of $t$, their transposition $\tau \in C(t)$ fixes $\{u\}$. Pairing the terms indexed by $g$ and $g\tau$ in $b_t\{u\}$ makes them cancel. This remains valid in characteristic two, since the paired coefficients sum to zero there as well. Thus each row of $u$ meets each column of $t$ in at most one entry.

The first $r$ rows of $u$ consequently contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, where ${\lambda }^{\prime }$ is the conjugate partition. HenceBy the dominance order on partitions, this provesThis is dominance from a nonzero column antisymmetrizer. The extension is what allows us to apply $b_t$ to an element of $M^{\mu }$ without dividing by the column stabilizer's order, which might vanish in the given field.

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.