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A standard Young tableau has its entries increasing from left to right in each row and from top to bottom in each column. For the convention $h_t=c_t{r}_t$, use column-reading order of standard Young tableaux: read columns from left to right, each from top to bottom, and compare the resulting words lexicographically. This makes the requested product direction explicit. The order and the symmetrizer multiplication convention must be chosen together.

We prove the vanishing claim. Suppose the rows of $t$ and the columns of $u$ have no collision. The argument in Question 1 shows that every column of $u$ contains one entry from each eligible row of $t$. Its first column thus selects one entry from every row. Each selected entry is at least the first entry of that row in $t$. Sorting the selected entries, as the standard tableau $u$ does, gives a column componentwise at least the first column of $t$: increasing individual entries cannot decrease any order statistic. If the columns are equal, equality of their sums forces every selected entry to be that row's first entry. Delete this common column and repeat. At the first differing column, its first differing entry in $u$ is therefore larger; otherwise $u=t$. Thus absence of a collision implies $u\ge t$ in column-reading order of standard Young tableaux.

If $t>u$, there must instead be a transposition in $R_t\cap C_u$. It fixes $r_t$ and negates $c_u$, giving $r_t{c}_u=0$. HenceThis is triangular vanishing of Young-symmetrizer products; it does not assert vanishing in the opposite order.

Normalize to $e_t=h_t/H_{\lambda }$. List all standard tableaux in increasing shape dictionary order on integer partitions, and within each shape in increasing column-reading order of standard Young tableaux. Then $e_i{e}_j=0$ for $i>j$, using Question 1 between different shapes and the result above within a shape. The left ideals $A{e}_i$ have an internal direct sum: if ${\sum }_i{z}_i=0$ with $z_i\in A{e}_i$, multiply on the right by $e_1$ to get $z_1=0$, since $z_1{e}_1=z_1$ and every later $z_i{e}_1=0$. Repeat with $e_2,e_3,\dots$. This proves directness without incorrectly treating all the idempotents as mutually orthogonal.

Let $f_{\lambda }$ count the standard tableaux of shape $\lambda$ and let $d_{\lambda }=\dim S^{\lambda }$. In the regular representation a simple module of dimension $d_{\lambda }$ occurs $d_{\lambda }$ times, by the Artin–Wedderburn theorem. The direct sum just constructed contains $f_{\lambda }$ copies of $S^{\lambda }$, so $f_{\lambda }\le d_{\lambda }$ for every shape. We supply the needed counting identity independently of the dimension conclusion.

The Robinson–Schensted correspondence bijects permutations with pairs of standard tableaux of the same shape. Here is its row insertion construction and inverse. Insert the successive permutation entries into an increasing row by replacing its first entry larger than the incoming entry, bumping that replaced entry into the next row; if no entry is larger, append at the row end. Continue until a new cell is created. Record the insertion time in that cell of a second tableau. For completeness, the successive bumped entries strictly increase, and their column indices weakly decrease: an entry below a bumped entry was originally larger, so the next replacement occurs no farther right. The entry newly placed in each row is smaller than the entry removed and larger than the entry above it. At a strictly earlier column, that last inequality follows from row increase in the preceding row; at the same column, it follows from the preceding bump. Hence the insertion tableau keeps increasing rows and columns. If a new cell is appended below the first row, the preceding bump guarantees that the row above reaches that column, so the shape stays a Young diagram. Recording times also increase in rows and columns, since every new cell is an outer corner of the current diagram. Thus both tableaux are standard at the end. Conversely, remove the cell with the largest recording label. Reverse its bumping path upward, replacing in each preceding row the rightmost entry smaller than the moving entry and moving the displaced entry upward. This recovers the last inserted letter; iterating recovers the entire permutation. The two rules undo one another at each row. ThusSemisimplicity also gives $n!={\sum }_{\lambda }{d}_{\lambda }^2$. Since $0\le f_{\lambda }\le d_{\lambda }$ termwise, equality of these sums forces $f_{\lambda }=d_{\lambda }$ for every shape. The internal direct sum has dimension $n!$ and hence fills $A$:There is also a useful numerical form. Right multiplication by $e_t$ is an idempotent with image $A{e}_t$. In the permutation basis of $A$, each diagonal coefficient is the coefficient of $1$ in $e_t$, namely $1/H_{\lambda }$. Its trace equals its rank, so $\dim S^{\lambda }=n!/H_{\lambda }$. Together with the count just obtained this recovers the hook-length formula.

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.

A near Young tableau is a filling of a Young diagram by distinct entries from a totally ordered alphabet, increasing along rows and down columns; the entries need not be $1,\dots ,n$. For row insertion of $x$, scan the first row for its leftmost entry larger than $x$. If one exists, replace it by $x$ and insert the displaced entry into the next row by the same rule. Otherwise append $x$ to the row and stop. Continue until a new outer-corner box is appended.

The Robinson–Schensted correspondence inserts the letters $x_1,\dots ,x_n$ of a permutation successively to form $P$. Whenever the $k$th insertion creates a new cell, put $k$ in that cell of a second tableau $Q$. Row insertion preserves increasing rows and columns, so $P$ is standard; the sequence of growing diagrams makes $Q$ standard, with the same shape. To reverse the construction, remove the box carrying the largest label in $Q$, and reverse the bumping in $P$: in each row above, exchange the carried entry with the rightmost smaller entry, continuing up to the first row. The final expelled entry is the last letter of the permutation. Repeating recovers the whole permutation, establishing the bijection with pairs of standard Young tableaux of a common shape.

For the bumping inequality, each displaced entry is the first entry strictly larger than the incoming one. Thus every step replaces a larger entry with a smaller entry and carries that larger value downwards. Consequently the bumped values strictly increase:

Insert $x$ into the first row of a Young tableau by replacing its leftmost entry strictly greater than $x$, and carry the displaced entry into the next row. If no entry is greater, append $x$ and stop. The procedure preserves a semistandard Young tableau; when all entries are distinct it preserves a near Young tableau. It underlies the Robinson–Schensted correspondence.

Right multiplication by $e_t=h_t/H_{\lambda }$ on the group algebra is an idempotent with image $A{e}_t$. Every diagonal entry in the permutation basis is the coefficient of the identity in $e_t$, namely $1/H_{\lambda }$. Thus its rank equals its trace, giving $\dim S^{\lambda }=n!/H_{\lambda }$. Triangular standard-tableau ideals and the Robinson–Schensted correspondence identify this dimension with the standard-tableau count.