Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / B / 7

The permutation module $M^{\lambda }$ has a basis consisting of row tabloids of shape $\lambda$: tableaux are identified when each row has the same set of entries. A permutation acts by applying it to every entry, and the action is extended linearly over $F$. Its dimension is $n!/{\prod }_i{\lambda }_i!$.

Here the two singleton rows form an ordered pair of distinct entries, soA standard tableau of the same shape must put one in its top-left corner. Choose two entries from $2,\dots ,n$ for the two remaining boxes of the first column, in increasing order; all other entries then fill the top row in increasing order. Thus the standard-polytabloid basis gives

Write $v_{ij}$ for the tabloid with second-row entry $i$ and third-row entry $j$. Equivariance gives $\varphi (v_{ij})=w_i+w_j$ for every distinct pair. For a tableau $t$, the only nontrivial column antisymmetrizer is on the three entries in its first column. In its alternating sum, pair each column permutation with the permutation obtained by interchanging the two singleton-row positions. Their images under $\varphi$ are the same unordered sum of two $w$'s, while their signs are opposite. Hence $\varphi (e_t)=0$ for every polytabloid, proving $\boxed{S^{(n-2,1,1)}\subseteq \ker \varphi }$.

For distinct $i,j,k$,Each summand is an image and two is invertible in characteristic zero. Thus every basis vector of $M^{(n-1,1)}$ is in the image, proving surjectivity. rank-nullity theorem then gives

The stabilizer of an ordered pair is the subgroup ${\Sigma }_{n-2}$ fixing both entries, so the permutation module is ${\mathrm{Ind}}_{{\Sigma }_{n-2}}^{{\Sigma }_n}1$. Induce in two stages through ${\Sigma }_{n-1}$. The first stage is $S^{(n-1)}\oplus S^{(n-2,1)}$. The characteristic-zero induction branching rule for a symmetric group adds one box in every possible partition position, with multiplicity one. Thus the two second-stage decompositions areBy transitivity of induction this yieldsThe dimensions are $1$, $2(n-1)$, $n(n-3)/2$ and $(n-1)(n-2)/2$, whose sum is $n(n-1)$, checking the decomposition also at $n=4$.