Young subgroups, tabloids, and permutation modules

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.