Row and column stabilizers

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 $S_{\lambda }$.