Past exam of the representation Theory of Symmetric Groups course of the University of Oxford / 2011 / 2

The permutation module $M_F^{\mu }$ is the $F$-vector space with the $\mu$-tabloids as basis, with $S_n$ acting by permuting their entries. For a tableau $t$ with column stabilizer $C_t$, its polytabloid isThe Specht module $S_F^{\mu }$ is the submodule of $M_F^{\mu }$ spanned by these polytabloids, equivalently generated by any one $e_t$ under the group action.

The standard basis theorem states thatis an $F$-basis of $S_F^{\mu }$ in every characteristic. Its dimension is the number of standard tableaux of shape $\mu$; in the hook shape used below the basis is exactly the displayed $e_{1ij}$ with $2\le i<j\le n$.

Put $W=S_F^{(n-2,1,1)}$ and let $H\cong S_{n-1}$ be the subgroup fixing one. For $\tau =(r,s)\in H$, let $u=\min \{\tau (i),\tau (j)\}$ and $v=\max \{\tau (i),\tau (j)\}$. Permuting a tableau's entries and then ordering its first column givesThe underlying basis element is $e_{1ij}$ exactly when $\tau$ preserves the unordered pair $\{i,j\}$: either $\{r,s\}=\{i,j\}$, giving the minus sign, or the two pairs are disjoint, giving the plus sign.

If $p\ne 2$, invariance of $x$ under $(i,j)$ forces its $e_{1ij}$ coefficient to satisfy $a_{ij}=-a_{ij}$, because no other basis vector is sent to that basis vector. Thus every coefficient vanishes. Consequently $W^H=0$, and in particular $W^{S_n}=0$, so there is no nonzero homomorphism from the trivial module. Any homomorphism $M_F^{(n-1,1)}\to W$ is determined by the image of its tabloid with singleton row $\{1\}$. That image is fixed by $H$ and is zero, making the whole homomorphism zero.

The printed vanishing assertion is false in characteristic two. Signs then coincide, so $H$ permutes the basis indexed by unordered pairs from $\{2,\dots ,n\}$. This action is transitive, andThe vector is nonzero by standard-basis independence. It also gives a nonzero map from the permutation module: send the singleton-row basis vector indexed by $r$ to $g_{r}v$, where $g_r(1)=r$. This is well defined because two choices of $g_r$ differ by an element of $H$, and it is $S_n$-equivariant by construction. Thus the claimed vanishing of permutation-module homomorphisms fails for every $n\ge 4$ in characteristic two.

The full-group fixed space can be determined explicitly. Denote a hook tabloid by $[a,b]$, recording its second- and third-row entries. In characteristic two, $e_{1ij}$ is the sum of the six tabloids using ordered pairs from $\{1,i,j\}$. In $v$, a tabloid whose two recorded entries both differ from one appears once; a tabloid using one appears $n-2$ times. Hence for odd $n$, $v$ is the sum of all hook tabloids and is fixed by $S_n$. For even $n$, it is the sum of precisely the tabloids avoiding one and is moved by $(1,2)$. Since every full-group fixed vector already lies in $Fv$, the corrected dimensions areThus even the trivial-module vanishing fails for odd $n$ in characteristic two; for example $n=5$ gives the explicit nonzero invariant $v$. These corrected fixed spaces allow the valid later conclusions to be proved in all characteristics.

The corrected parity dependence is described by fixed vectors in a hook Specht module in characteristic two.

Identify $M=M_F^{(n-1,1)}$ with $F^n$, whose basis vectors record the singleton row. PutIf the characteristic does not divide $n$, then $M=T\oplus U$, and $U$ is simple. If it divides $n$, then $0<T<U<M$ is the submodule chain, with $T$ and $M/U$ trivial and $U/T$ simple of dimension $n-2$. In both cases there are no additional submodules: any vector with unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$, and the orbit of such a difference spans $U$. A submodule containing no nonconstant vector is contained in $T$.

Now suppose $p\nmid n$. When $p\ne 2$, part (b)'s valid vanishing ${\mathrm{Hom}}_{S_n}(M,W)=0$ implies ${\mathrm{Hom}}_{S_n}(U,W)=0$: extend a putative map using the projection $M\to U$.

When $p=2$, the assumption $p\nmid n$ makes $n$ odd. The corrected calculation in part (b) says every map $M\to W$ sends its singleton-row basis vectors to the same multiple of the globally invariant vector $v$. Hence it kills the augmentation submodule $U$. Extending a map $U\to W$ along the direct-sum projection again makes it zero. Thus, despite the characteristic-two error in the earlier claim, the requested conclusion remains valid:

The structural input is the submodule lattice of the standard symmetric-group permutation module.

If $p\nmid n$, the preceding part says $U=S_F^{(n-1,1)}$ is simple and has no nonzero homomorphism into $W$. Its only nonzero simple quotient is itself, so $W$ cannot contain a submodule isomorphic to such a quotient.

If $p\mid n$, use the given nonzero homomorphism $\varphi :U\to W$. The submodule chain shows that $T$ is the unique maximal proper submodule of $U$ and $U/T$ is simple. Moreover ${\mathrm{Hom}}_{S_n}(T,W)=0$: for odd characteristic this follows from part (b), and in characteristic two the divisibility condition makes $n$ even, where the corrected fixed-space calculation also gives zero. Thus $\varphi$ kills $T$ and descends to a nonzero map $U/T\to W$. Simplicity makes this map injective.

Its image is the required simple submodule, provingIn the divisible case the quotient is explicitly $U/T$, of dimension $n-2$. The argument uses the supplied existence of $\varphi$ and only the fixed-space vanishing that remains valid in that case.