Young Diagrams Tableaux and Permutation Modules

Partitions. Let $n\ge 0$ be an integer. A partition of $n$ is a finite weakly decreasing sequence $\lambda =({\lambda }_1,\dots ,{\lambda }_k)$ of positive integers with ${\lambda }_1+\cdots +{\lambda }_k=n$. Its entries are the parts of $\lambda$ and $k$ is the number of parts. For $n=0$ the only partition is the empty partition $\mathrm{\varnothing }$, the empty sequence with $k=0$; for $n\ge 1$ one has ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_k\ge 1$ and $k\le n$. We write $\lambda ⊢n$. Trailing zeros are not parts: a finite sequence of nonnegative integers ending in $0$ is not a partition, so the number of parts of $\lambda$ is determined by $\lambda$ and $({\lambda }_1,\dots ,{\lambda }_k)$ is a different data type from $({\lambda }_1,\dots ,{\lambda }_k,0)$.

English Young diagrams. The (English) Young diagram of a nonempty $\lambda$ is $[\lambda ]:=\{(i,j):1\le i\le k,1\le j\le {\lambda }_i\}\subseteq \mathbb{N}\times \mathbb{N},$ where $i$ numbers the rows downward and $j$ numbers the columns rightward. An element of $[\lambda ]$ is a node, or box, of the diagram. Row $i$ of $[\lambda ]$ carries ${\lambda }_i$ nodes, and column $j$ carries ${\lambda }_j^{\mathrm{\prime }}:=\mathrm{\#}\{i:{\lambda }_i\ge j\}$ nodes, a quantity that is $0$ beyond the last column (and for every $j\ge 1$ when $\lambda =\mathrm{\varnothing }$). Since a partition is weakly decreasing, $[\lambda ]$ determines $\lambda$: if $[\mu ]=[\lambda ]$, then row $i$ of $[\lambda ]$ is row $i$ of $[\mu ]$, so ${\mu }_i={\lambda }_i$ for every $i$ and $\mu =\lambda$. In particular $[\mathrm{\varnothing }]=\mathrm{\varnothing }$, the empty diagram.

Conjugation. The conjugate partition of a nonempty $\lambda$ is ${\lambda }^{\mathrm{\prime }}:=({\lambda }_1^{\mathrm{\prime }},{\lambda }_2^{\mathrm{\prime }},\dots ,{\lambda }_{{\lambda }_1}^{\mathrm{\prime }}),$ the sequence of column heights of $[\lambda ]$, with ${\mathrm{\varnothing }}^{\mathrm{\prime }}:=\mathrm{\varnothing }$ for the empty partition. This is again a partition of $n$, and $[{\lambda }^{\mathrm{\prime }}]$ is the transpose of $[\lambda ]$: for $1\le i\le {\lambda }_1$ and $1\le j\le k$, $(i,j)\in [{\lambda }^{\mathrm{\prime }}]⟺j\le {\lambda }_i^{\mathrm{\prime }}⟺{\lambda }_j\ge i⟺(j,i)\in [\lambda ]\mathrm{.}$ Outside these bounds neither diagram contains the corresponding node; for the empty partition both diagrams are empty. Conjugation is an involution, ${\lambda }^{\mathrm{\prime }\mathrm{\prime }}=\lambda$, because ${\lambda }_i^{\mathrm{\prime }\mathrm{\prime }}=\mathrm{\#}\{j:{\lambda }_j^{\mathrm{\prime }}\ge i\}$ counts the columns of $[\lambda ]$ of height at least $i$, that is, the columns $j$ with at least $i$ rows of length $\ge j$, which by weak decrease is exactly the set of $j\le {\lambda }_i$. Thus $\lambda \mapsto {\lambda }^{\mathrm{\prime }}$ is a bijection on the partitions of $n$, exchanging the number of parts with the largest part for nonempty partitions.

Size zero. The symmetric group of $\{1,\dots ,n\}$ is $S_n:=\mathrm{Sym}(\{1,\dots ,n\})$ (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition) for every $n\ge 0$, and we fix $S_0:=\mathrm{Sym}(\mathrm{\varnothing })=\{1\},$ the trivial group; for $n=0$ the set $\{1,\dots ,n\}$ is empty and the group acts trivially on every size-zero object below. This is the convention used whenever the constructions of this page are read at $n=0$.

• Indexing conventions. Rows are numbered from top to bottom and columns from left to right, and the row lengths are weakly decreasing, so the diagram is left-aligned and top-aligned inside its bounding rectangle of ${\lambda }_1$ columns and $k$ rows. This is the English convention; the French convention (rows weakly increasing downward) is not used here.

• Conjugation transposes the diagram. The identity $[{\lambda }^{\mathrm{\prime }}]=[\lambda {]}^T$ says that summing over the parts of ${\lambda }^{\mathrm{\prime }}$ is summing over the columns of $[\lambda ]$: double counting the nodes of $[\lambda ]$ by rows gives $n$ and by columns gives ${\lambda }_1^{\mathrm{\prime }}+\cdots +{\lambda }_{{\lambda }_1}^{\mathrm{\prime }}$, so ${\lambda }^{\mathrm{\prime }}$ is a partition of $n$. A partition equal to its conjugate is self-conjugate; the diagonal nodes of $[\lambda ]$ are the fixed points of the transpose.

Let $\lambda ⊢n$ and let $[\lambda ]$ be its Young diagram (Partitions, English diagrams, and conjugation). A tableau of shape $\lambda$, or $\lambda$-tableau, is a bijection $t:[\lambda ]⟶\{1,2,\dots ,n\}\mathrm{.}$ We write $t(i,j)$ for the entry of $t$ in the node $(i,j)$, and we display $t$ as the diagram $[\lambda ]$ with each node carrying its entry. Thus a tableau places each of the numbers $1,\dots ,n$ in exactly one node of $[\lambda ]$. For $\lambda =\mathrm{\varnothing }$ the diagram is empty and the empty map is the unique tableau of shape $\mathrm{\varnothing }$, the empty tableau. The shape of a tableau $t$ is the partition $\lambda$ with $t:[\lambda ]\to \{1,\dots ,n\}$, which is determined by $t$ because $[\lambda ]$ determines $\lambda$.

Standard tableaux. A tableau $t$ of shape $\lambda$ is standard if its entries strictly increase along rows and down columns, that is, if $t(i,j)<t(i,j+1)\text{whenever }(i,j),(i,j+1)\in [\lambda ],$ $t(i,j)<t(i+1,j)\text{whenever }(i,j),(i+1,j)\in [\lambda ]\mathrm{.}$ The empty tableau is standard, because both conditions are vacuous, and it is the unique standard tableau of shape $\mathrm{\varnothing }$. Following the classical notation we write $f^{\lambda }$ for the number of standard $\lambda$-tableaux.

The left action on tableaux. For $\sigma \in S_n$ and a $\lambda$-tableau $t$, define $(\sigma \cdot t)(i,j):=\sigma (t(i,j))((i,j)\in [\lambda ])\mathrm{.}$ Since $t$ is a bijection onto $\{1,\dots ,n\}$ and $\sigma$ is a bijection of $\{1,\dots ,n\}$, the map $\sigma \cdot t$ is again a bijection $[\lambda ]\to \{1,\dots ,n\}$, hence again a $\lambda$-tableau, and $\mathrm{i}\mathrm{d}\cdot t=t$, $\sigma \cdot (\tau \cdot t)=(\sigma \tau )\cdot t$ for all $\sigma ,\tau \in S_n$: the rule is a left action of $S_n$ on the set of $\lambda$-tableaux. For every $\sigma \in S_n$ the row sets of $\sigma \cdot t$ are the images under $\sigma$ of the row sets of $t$, and the column sets of $\sigma \cdot t$ are the images under $\sigma$ of the column sets of $t$.

• A tableau is not a tabloid. A tableau records a position for every entry; the row-equivalence classes of tableaux, called tabloids, are defined later on this page and forget the order of the entries inside each row. The action above is the one that descends to tabloids.

• Counting tableaux. Fixing any listing of the $n$ nodes of $[\lambda ]$, a tableau is the same thing as an ordering of the entries $1,\dots ,n$ along that listing, so there are $n!$ tableaux of shape $\lambda$; for $n=0$ the empty tableau is the single one and $0!=1$. For $n\ge 1$ the extreme shapes have exactly one standard tableau each: the single row $12\cdots n$ for $(n)$, and the single column with $1,2,\dots ,n$ from top to bottom for $(1^n)$, so $f^{(n)}=f^{(1^n)}=1$.

Let $\lambda ⊢n$ with Young diagram $[\lambda ]$, and let $\mu ⊢n-1$ and $\nu ⊢n+1$ denote partitions of the neighbouring sizes (Partitions, English diagrams, and conjugation).

A node $x\in [\lambda ]$ is removable if deleting it leaves a Young diagram, that is, if there is a partition $\mu ⊢n-1$ with $[\mu ]=[\lambda ]\setminus \{x\};$ such a $\mu$ is unique because a partition is determined by its diagram. A point $y\mathrm{\notin }[\lambda ]$ is addable for $\lambda$ if inserting it leaves a Young diagram, that is, if there is a partition $\nu ⊢n+1$ with $[\nu ]=[\lambda ]\cup \{y\};$ again $\nu$ is unique. We write $\mathrm{Rem}(\lambda )$ and $\mathrm{Add}(\lambda )$ for the sets of removable and of addable nodes of $\lambda$.

Since $[\lambda ]=\{(i,j):1\le i\le k,1\le j\le {\lambda }_i\}$ is determined by the inequalities $1\le j\le {\lambda }_i$, the two conditions have the following row form. Write ${\lambda }_{k+1}:=0$ for a partition $\lambda =({\lambda }_1,\dots ,{\lambda }_k)$. A node $(i,{\lambda }_i)$ is removable if and only if ${\lambda }_i>{\lambda }_{i+1}$: deleting the last node of row $i$ keeps the row lengths weakly decreasing exactly when row $i+1$ is strictly shorter, and no node $(i,j)$ with $j<{\lambda }_i$ can be deleted, since the node $(i,j+1)$ would then have no node to its left. Similarly a node $(i,{\lambda }_i+1)$ with $1\le i\le k$ is addable if and only if $i=1$ or ${\lambda }_{i-1}>{\lambda }_i$, the node $(k+1,1)$ opening a new row is always addable, and these are all the addable nodes. In particular a row endpoint of $[\lambda ]$ is removable only if no node of $[\lambda ]$ lies immediately below it.

For the empty partition the diagram is empty, so $\mathrm{Rem}(\mathrm{\varnothing })=\mathrm{\varnothing }$, while $\mathrm{Add}(\mathrm{\varnothing })=\{(1,1)\}$ is a single node, whose insertion produces the partition $(1)$.

• Removable nodes are exactly the corners. The removable nodes of $\lambda$ are the row endpoints $(i,{\lambda }_i)$ with ${\lambda }_i>{\lambda }_{i+1}$; the lowest row always qualifies, because ${\lambda }_{k+1}=0<{\lambda }_k$. Every removable node has hook length $1$ in the usual terminology, and $(k+1,1)$ is always addable, so $\lambda$ arises from exactly $\mathrm{\#}\mathrm{Rem}(\lambda )$ partitions of $n-1$ by inserting one node, and is contained in exactly $\mathrm{\#}\mathrm{Add}(\lambda )$ partitions of $n+1$.

• Nodes versus row endpoints. Ending a row is necessary but not sufficient for removability: in $\lambda =(3,3,1)$ the node $(1,3)$ ends the first row but the node $(2,3)$ lies directly below it, so deleting $(1,3)$ leaves the row lengths $(2,3,1)$, which are not weakly decreasing and are not the row lengths of a partition.

For every $n\ge 1$, the box occupied by $n$ in a standard tableau of size $n$ is removable, and deleting it leaves a standard tableau of size $n-1$.

Let $\lambda ,\mu ⊢n$ be partitions of the same integer $n$ (Partitions, English diagrams, and conjugation). We say that $\lambda$ dominates $\mu$, and write $\lambda ⊵\mu$, exactly when ${\sum }_{i=1}^r{\lambda }_i\ge {\sum }_{i=1}^r{\mu }_i\text{for every }r\ge 1,$ where each sequence is padded with zeros beyond its number of parts; since both partitions have total $n$, both sides equal $n$ for all $r$ at least the number of parts of either, so the condition is a finite family of inequalities between integers. We write $\lambda ⊳\mu$ when $\lambda ⊵\mu$ and $\lambda \mathrm{\ne }\mu$. The relation $⊵$ is the dominance order on the partitions of $n$; we call $\lambda$ and $\mu$ incomparable when neither $\lambda ⊵\mu$ nor $\mu ⊵\lambda$ holds.

Because it is defined by a family of non-strict inequalities between integers, $⊵$ is reflexive and transitive. It is also antisymmetric: if $\lambda ⊵\mu$ and $\mu ⊵\lambda$, then the prefix sums of $\lambda$ and of $\mu$ are equal for every $r\ge 1$, and subtracting consecutive prefix sums gives ${\lambda }_i={\mu }_i$ for every $i$ (both sequences are eventually zero). Hence $⊵$ is a partial order on the set of partitions of $n$. For $n\ge 1$, the partition $(n)$ is its unique maximum and the partition $(1^n)$ its unique minimum: for every $\mu ⊢n$ and every $r\ge 1$ one has $\min (r,n)={\sum }_{i\le r}(1^n{)}_i\le {\sum }_{i\le r}{\mu }_i\le n={\sum }_{i\le r}(n{)}_i\mathrm{.}$

For $n=0$ the order is the trivial order on the one-element set $\{\mathrm{\varnothing }\}$.

• Partial, not total. Dominance is in general a proper partial order, not a total order: the partitions $(4,1,1)$ and $(3,3)$ of $6$ are incomparable, because their prefix sums $4,5,6$ and $3,6,6$ cross, and so are their conjugates $(3,1,1,1)$ and $(2,2,2)$. For each $n\le 5$, by contrast, all partitions of $n$ are comparable. The companion examples page lists the chains through size five and this first incomparable pair.

• Not the lexicographic order. Dominance must not be identified with the lexicographic order on partitions, which orders $\lambda$ and $\mu$ by their first differing part and is total. The two relations agree on all partitions of $n$ for $n\le 5$, but lexicographic order is total by definition while dominance is not, so the relations are distinct; a dominance step never follows from a comparison of single parts alone, only from all the prefix sums.

• Conjugation reverses the order. Transposing diagrams turns prefix sums of row lengths into prefix sums of column heights, and this reverses dominance: $\lambda ⊵\mu$ holds if and only if ${\mu }^{\mathrm{\prime }}⊵{\lambda }^{\mathrm{\prime }}$. This is proved on this page as Conjugation reverses dominance and is used to keep row and column versions of every later statement consistent.

For all partitions $\lambda$ and $\mu$ of the same integer $n$, $\lambda ⊵\mu ⟺{\mu }^{\mathrm{\prime }}⊵{\lambda }^{\mathrm{\prime }}\mathrm{.}$

Let $\lambda ⊢n$ and let $t$ be a $\lambda$-tableau, with entries $t(i,j)$ in the nodes of $[\lambda ]$ (Tableaux and standard tableaux). If $n=0$, there are no rows or columns; define $R_t=C_t=S_0=\{1\}$ for the unique empty tableau. For the row and column formulas below assume $n\ge 1$, so ${\lambda }_1$ exists. For a row $i$ write $A_i:=\{t(i,j):1\le j\le {\lambda }_i\}$ for the set of entries in that row, and for a column $j$ write $B_j:=\{t(i,j):1\le i\le {\lambda }_j^{\mathrm{\prime }}\}$ for the set of entries in that column. The sets $A_1,\dots ,A_k$ are pairwise disjoint and partition $\{1,\dots ,n\}$, and likewise the sets $B_1,\dots ,B_{{\lambda }_1}$ are pairwise disjoint and partition $\{1,\dots ,n\}$, because $t$ is a bijection $[\lambda ]\to \{1,\dots ,n\}$.

The row stabilizer of $t$ is $R_t:=\{\sigma \in S_n:\sigma (A_i)=A_i\text{ for every row }i\},$ and the column stabilizer of $t$ is $C_t:=\{\sigma \in S_n:\sigma (B_j)=B_j\text{ for every column }j\}\mathrm{.}$ Thus $R_t$ is the subgroup of $S_n$ consisting of the permutations that map each row set of $t$ onto itself, and $C_t$ is the subgroup of those that map each column set of $t$ onto itself. In terms of the left action $(\sigma \cdot t)(i,j)=\sigma (t(i,j))$, a permutation lies in $R_t$ exactly when $\sigma \cdot t$ can be obtained from $t$ by permuting the entries within each row, and in $C_t$ exactly when $\sigma \cdot t$ is obtained from $t$ by permuting the entries within each column.

Both sets are subgroups of $S_n$: the identity preserves every $A_i$ and every $B_j$, and if $\sigma$ and $\tau$ preserve each of these sets then so do $\sigma \tau$ and ${\sigma }^{-1}$. Moreover the sets $A_i$ are permuted onto themselves one by one, not merely as a family, and each $A_i$ determines a subgroup $S(A_i)\le S_n$ of permutations fixing the complement of $A_i$ pointwise; because the $A_i$ are pairwise disjoint and cover $\{1,\dots ,n\}$, every $\sigma \in R_t$ factors uniquely as $\sigma ={\sigma }_1\cdots {\sigma }_k$ with ${\sigma }_i\in S(A_i)$, so $R_t=S(A_1)\times \cdots \times S(A_k)\cong S_{{\lambda }_1}\times \cdots \times S_{{\lambda }_k},\mathrm{\mid }{R}_t\mathrm{\mid }={\lambda }_1!\cdots {\lambda }_k!\mathrm{.}$ The same argument with columns gives $C_t=S(B_1)\times \cdots \times S(B_{{\lambda }_1})\cong S_{{\lambda }_1^{\mathrm{\prime }}}\times \cdots \times S_{{\lambda }_{{\lambda }_1}^{\mathrm{\prime }}},\mathrm{\mid }{C}_t\mathrm{\mid }={\lambda }_1^{\mathrm{\prime }}!\cdots {\lambda }_{{\lambda }_1}^{\mathrm{\prime }}!\mathrm{.}$ For $n=0$ there is one tableau, the empty one, and $R_t=C_t=S_0=\{1\}$.

• Equal rows are still distinguished. The row sets are individual sets: each $A_i$ must be preserved individually, even when two rows have equal length. The factors $S(A_i)$ on distinct row sets are distinct when their common size is at least two; singleton row sets both give the trivial subgroup. This labelled-rows convention is what makes $R_t\cap C_t$ trivial: a permutation preserving every row set and every column set sends the entry $t(i,j)$ into $A_i\cap B_j=\{t(i,j)\}$, since row $i$ and column $j$ meet in the single box $(i,j)$; so it fixes every entry and is the identity, and $R_t\cap C_t=\{1\}$.

• Relation to Young subgroups. If $t_0$ is the standard row-filled $\lambda$-tableau, whose row $i$ carries the consecutive block of entries ${\lambda }_1+\cdots +{\lambda }_{i-1}+1,\dots ,{\lambda }_1+\cdots +{\lambda }_i$, then $R_{t_0}$ is the standard Young subgroup $S_{\lambda }$ of the next definition on this page; for an arbitrary tableau, $R_t$ is a conjugate of $S_{\lambda }$.

Let $n\ge 0$, let $\lambda ,\mu ⊢n$, let $t$ be a $\lambda$-tableau and let $s$ be a $\mu$-tableau such that every row of $s$ meets each column of $t$ in at most one entry. Then $\lambda ⊵\mu$. Moreover, if $\lambda =\mu$, then there are $\rho \in R_s$ and $\gamma \in C_t$ with $\rho \cdot s=\gamma \cdot t\mathrm{.}$

Let $n\ge 0$ and let $\lambda =({\lambda }_1,\dots ,{\lambda }_k)⊢n$ be a partition with Young diagram $[\lambda ]$ (Partitions, English diagrams, and conjugation). Throughout, $S_n$ denotes the symmetric group of $\{1,\dots ,n\}$, with $S_0=\{1\}$.

Standard Young subgroups. For $1\le i\le k$ let $B_i:=\{{\lambda }_1+\cdots +{\lambda }_{i-1}+1,{\lambda }_1+\cdots +{\lambda }_{i-1}+2,\dots ,{\lambda }_1+\cdots +{\lambda }_i\}$ be the $i$-th block of $\lambda$. The blocks are consecutive intervals of integers, they are pairwise disjoint, each has $\mathrm{\mid }{B}_i\mathrm{\mid }={\lambda }_i$, and together they partition $\{1,\dots ,n\}$. The standard Young subgroup of type $\lambda$ is $S_{\lambda }:=\{\sigma \in S_n:\sigma (B_i)=B_i\text{ for every }1\le i\le k\}\mathrm{.}$ This is a subgroup of $S_n$, namely the direct product $S(B_1)\times \cdots \times S(B_k)$ of the symmetric groups of the individual blocks, each acting on its block and fixing the remaining entries pointwise; as in Row and column stabilizers, the product decomposition is unique because the blocks are pairwise disjoint and cover $\{1,\dots ,n\}$, so $\mathrm{\mid }{S}_{\lambda }\mathrm{\mid }={\lambda }_1!\cdots {\lambda }_k!$. A Young subgroup of type $\lambda$ is a subgroup of $S_n$ conjugate to $S_{\lambda }$. For $k=0$, that is for $\lambda =\mathrm{\varnothing }$ and $n=0$, there are no blocks and $S_{\mathrm{\varnothing }}=S_0=\{1\}$.

Row equivalence and tabloids. Let $t$ and $u$ be $\lambda$-tableaux (Tableaux and standard tableaux). We say that $t$ and $u$ are row equivalent, and write $t\sim u$, when they have the same row sets: for every row $i$, $\{t(i,j):1\le j\le {\lambda }_i\}=\{u(i,j):1\le j\le {\lambda }_i\}\mathrm{.}$ Equivalently, $t\sim u$ if and only if $u=\rho \cdot t$ for some $\rho \in R_t$: if $u=\rho \cdot t$ then $\rho$ merely permutes the entries inside each row of $t$, and conversely, if the row sets agree, then $\rho (t(i,j)):=u(i,j)$ defines a permutation $\rho$ of $\{1,\dots ,n\}$ that preserves each row set of $t$, so $\rho \in R_t$ and $\rho \cdot t=u$. Hence $\sim$ is an equivalence relation on the $\lambda$-tableaux, and the equivalence class $\{t\}:=\{u:u\sim t\}=\{\rho \cdot t:\rho \in R_t\}$ is the tabloid of $t$. We draw $\{t\}$ as the diagram $[\lambda ]$ filled with the entries of $t$ and bars between the rows, recording that the order of the entries inside a row is forgotten. A tabloid is standard when it contains a standard tableau.

The permutation module. Let ${\mathrm{\Omega }}_{\lambda }$ be the finite set of $\lambda$-tabloids. For $\sigma \in S_n$ and a tabloid $\{t\}$ define $\sigma \cdot \{t\}:=\{\sigma \cdot t\}\mathrm{.}$ This is well defined: if $t,u$ are $\lambda$-tableaux with $t\sim u$, then the row sets of $\sigma \cdot u$ are the $\sigma$-images of the row sets of $u$, which are the $\sigma$-images of the row sets of $t$, and these are exactly the row sets of $\sigma \cdot t$ (Tableaux and standard tableaux); so $\sigma \cdot u\sim \sigma \cdot t$ and the tabloids $\{\sigma \cdot u\}$ and $\{\sigma \cdot t\}$ coincide. Because the action on tableaux is a left action, the induced rule on tabloids satisfies $\mathrm{i}\mathrm{d}\cdot \{t\}=\{t\}$ and $\sigma \cdot (\tau \cdot \{t\})=(\sigma \tau )\cdot \{t\}$, so $S_n$ acts on ${\mathrm{\Omega }}_{\lambda }$ from the left. The Young permutation module attached to $\lambda$ is the complex vector space $M^{\lambda }:={\mathbb{C}}^{({\mathrm{\Omega }}_{\lambda })}$ with the tabloids as basis, again written $\{t\}$ for the basis vector of the tabloid $\{t\}$, equipped with the linear extension of the action above: $\sigma \cdot {\sum }_{\{t\}\in {\mathrm{\Omega }}_{\lambda }}{a}_{\{t\}}\{t\}:={\sum }_{\{t\}\in {\mathrm{\Omega }}_{\lambda }}{a}_{\{t\}}\{\sigma \cdot t\}\mathrm{.}$ This is the permutation representation of $S_n$ on the finite set ${\mathrm{\Omega }}_{\lambda }$ (The trivial representation, the regular representation, and permutation representations from finite $G$-sets), so $M^{\lambda }$ is a finite-dimensional complex representation of $S_n$.

The stabilizer of a tabloid. For every $\lambda$-tableau $t$ one has $\sigma \cdot \{t\}=\{t\}⟺\sigma \cdot t\sim t⟺\sigma \in R_t,$ so the stabilizer in $S_n$ of the tabloid $\{t\}$ is exactly the row stabilizer $R_t$ (Row and column stabilizers). The action on ${\mathrm{\Omega }}_{\lambda }$ is transitive: given tabloids $\{t\}$ and $\{s\}$, the permutation $\sigma$ determined by $\sigma (t(i,j)):=s(i,j)$ satisfies $\sigma \cdot t=s$, hence $\sigma \cdot \{t\}=\{s\}$. In particular, for the standard row-filled $\lambda$-tableau $t_0$, whose row $i$ carries the entries of $B_i$ in increasing order, the row sets of $t_0$ are precisely the blocks $B_1,\dots ,B_k$, so $R_{t_0}=S_{\lambda }$ and the stabilizer of the tabloid $\{t_0\}$ is the standard Young subgroup $S_{\lambda }$.

• Two descriptions of a tabloid. A tabloid of shape $\lambda$ is the same data as an unordered partition of $\{1,\dots ,n\}$ into $k$ labelled classes of sizes ${\lambda }_1,\dots ,{\lambda }_k$: the class number $i$ is the set of entries in row $i$. The tabloids of shape $\lambda$ are exactly the images $\{\sigma \cdot t_0\}$ of the single tabloid $\{t_0\}$ under $S_n$, so ${\mathrm{\Omega }}_{\lambda }$ is a transitive $S_n$-set with point stabilizer $S_{\lambda }$ in the sense just described.

• Notation. Some sources write $M^{\lambda }$ for the induced module ${\mathrm{Ind}}_{S_{\lambda }}^{S_n}\mathbf{1}$; the next lemma on this page proves that this is the same representation as the tabloid module defined above, and the identification also shows that $M^{\lambda }$ is generated by the single tabloid $\{t_0\}$.

• Extreme shapes. For $\lambda =(n)$ all $\lambda$-tableaux have the same single row set $\{1,\dots ,n\}$, so there is exactly one tabloid and $M^{(n)}$ is one-dimensional with trivial action. For $\lambda =(1^n)$ each row is a single box, so the row-equivalence classes are singletons and ${\mathrm{\Omega }}_{(1^n)}$ is the set of all $\lambda$-tableaux; the companion examples page uses this to identify $M^{(1^n)}$ with the regular representation.