Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 8

The simple modules are indexed by the p-regular partitions $\lambda$ of $n$, meaning no part is repeated $p$ or more times. Equip the tabloid permutation module with the bilinear form invariant under the symmetric group making its tabloid basis orthonormal. The construction isThe standard modular Specht theorem says this quotient is nonzero precisely for p-regular $\lambda$, is then simple, and these quotients give all simple modules.

To verify distinctness of labels, use the standard dominance triangularity theorem: if $D^{\nu }$ occurs as a composition factor of $S^{\lambda }$, then $\nu \ge \lambda$; for a p-regular label the diagonal multiplicity is one. If $D^{\lambda }\cong D^{\mu }$, then $D^{\lambda }$ occurs in $S^{\mu }$, so $\lambda \ge \mu$, and the symmetric argument gives $\mu \ge \lambda$. Antisymmetry of the dominance order on partitions yields $\boxed{\lambda =\mu }$. These are the general representation-theoretic results allowed in this part.

Realize $S^{(n-1,1)}$ as the augmentation submoduleof the natural permutation module: its polytabloids are the differences $e_i-e_j$, which span exactly $W$. Let $U$ be a submodule. If it contains a vector $v$ with two unequal coordinates $v_i,v_j$, thenThe coefficient is nonzero, so $e_i-e_j\in U$; its symmetric-group orbit spans $W$, forcing $U=W$. Therefore every proper nonzero submodule consists only of constant-coordinate vectors. Those form at most a one-dimensional trivial submodule. If such a trivial submodule is absent, no proper nonzero submodule exists, so the Specht module is simple.

On a one-dimensional representation, $s_i$ acts by a nonzero scalar ${\epsilon }_i$. Its square is one. The braid relation becomes ${\epsilon }_i^2{\epsilon }_{i+1}={\epsilon }_{i+1}^2{\epsilon }_i$, and hence ${\epsilon }_{i+1}={\epsilon }_i$. Thus all adjacent transpositions act by one common scalar $\epsilon$ with ${\epsilon }^2=1$.

In characteristic different from two, $\epsilon =1$ or $-1$. Since adjacent transpositions generate the group, these choices give exactly the trivial representation and sign representation. For $n\le 1$ they coincide, leaving only the trivial representation.

In characteristic three with $n=5$, a constant vector $(c,c,c,c,c)$ belongs to the augmentation submodule only if $5c=0$, which forces $c=0$. Thus $S^{(4,1)}$ has no trivial submodule. Part (b) proves that it is simple, of dimension four, and therefore equals $D^{(4,1)}$.

The standard polytabloid basis is defined over the integers and remains a basis after reduction modulo three. Thus $\dim S^{(3,2)}=f^{(3,2)}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)-\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$. The given submodule has dimension four by part (d), so the quotient is one-dimensional and part (c) leaves only the trivial or sign action.

A transposition has ordinary trace one on $S^{(3,2)}$: the Murnaghan–Nakayama rule has only the horizontal bottom-row two-strip, leaving shape $(3)$ with dimension one. Reduction of its integral action matrix preserves this trace in $F$. Its trace on $D^{(4,1)}=S^{(4,1)}$ is $3-1=2$, since the natural five-point permutation representation has three fixed coordinates and contains the trivial line. Therefore its trace on the one-dimensional quotient is $1-2=-1$, distinct from one in characteristic three. This gives the characteristic-three sign quotient of the Specht module of shape three two. The quotient is the sign representation: