Lecture 18

Solution: Let $g\in C(t)$, then if $g$ stabilises the columns of $t$ so does $g^{-1}$. Additionally suppose $h\in G$, then $(gh)\cdot t=g\cdot(h\cdot t)=g\cdot t=t$. Thus $C(t)$ is a subgroup of $S_{n}$.

Given $g\in C(t)$, we have that $\sigma g\sigma^{-1}\cdot(\sigma\cdot t)=\sigma\cdot(g\cdot t)$. Since $g\in C(t)$, the entries of the columns of $g\cdot t$ are the same as those of $t$, hence the entries of the columns of $\sigma\cdot(g\cdot t)$ are the same as those of $\sigma\cdot t$, i.e. $\sigma g\sigma^{-1}\in C(\sigma\cdot t)$. This shows that $\sigma C(t)\sigma^{-1}\subset C(\sigma\cdot t)$. By symmetry (i.e. replacing $t$ with $\sigma\cdot t$, and $\sigma$ with $\sigma^{-1}$ in the previous argument), we then obtain $\sigma^{-1}C(\sigma\cdot t)\sigma\subset C(t)$, hence $\sigma C(t)\sigma^{-1}=C(\sigma\cdot t)$.

Solution: If ${\mathcal{S}}^{\lambda}\subseteq W$ then we are done so suppose ${\mathcal{S}}^{\lambda}\nsubseteq W$. As ${\mathcal{S}}^{\lambda}$ is irreducible then $W\nsubseteq{\mathcal{S}}^{\lambda}$ and thus ${\mathcal{S}}^{\lambda}\cap W=\{0\}$. By Problem 92, $(\pi,{\mathcal{M}}^{\lambda})$ is unitary so ${\mathcal{M}}^{\lambda}={\mathcal{S}}^{\lambda}\oplus\left({\mathcal{S}}^{\lambda}\right)^{\perp}$. Thus $W\subseteq\left({\mathcal{S}}^{\lambda}\right)^{\perp}$.