Understanding $1$-line proof that if $V$ is irreducible $\mathbb{C}S_n$-module, then $\operatorname{res}_{S_{n-1}}V$ is multiplicity free?

Question

I'm having trouble understanding a terse proof of a theorem.

Theorem. If $V$ is an irreducible $\mathbb{C}S_n$-module, then the restriction $\operatorname{res}^{S_n}_{S_{n-1}}V$ is multiplicity free.

Proof: The centralizer of $\mathbb{C}S_{n-1}$ in $\mathbb{C}S_n$ is commutative.

I know how to show that the centralizer above is commutative, but I don't see the connection with the restricted module being multiplicity free. I know that a criterion for a $\mathbb{C}G$-module to be commutative is that its endomorphism algebra be commutative. Is $\operatorname{End}_{S_{n-1}}(\operatorname{res}^{S_n}_{S_{n-1}}V)$ seen to be commutative somehow using the fact that the centralizer is commutative?

Answer 1

I'll do this more generally to show that there is nothing funny going on with group algebras, but for reference my $A$ will be $\mathrm{C}S_n$ and my $B$ will be $\mathrm{C}S_{n-1}$.

Let $B \subseteq A$ be finite-dimensional algebras over $\mathbb{C}$, and $C$ the centraliser of $B$ in $A$, $C = \{c \in A \mid cb = bc \text{ for all } b \in B\}$. Now let $V$ be an $A$-module and $W$ a $B$-module. The hom-space $\mathrm{Hom}_B(W, \mathrm{res}^A_B V)$ is a priori just a vector space, but it actually has a left action by $C$, by $(c \cdot f)(w) = c f(w)$. We can check that the map $c \cdot f$ is again $B$-equivariant: $$ (c \cdot f)(bw) = cf(bw) = cb f(w) = bc f(w) = b (c \cdot f)(w) $$

So $\mathrm{Hom}_B(W, \mathrm{res}^A_B V)$ is indeed a $C$ module. Now suppose further that $V$ and $W$ are irreducible: we would like to show that $\mathrm{Hom}_B(W, \mathrm{res}^A_B V)$ must be irreducible as a $C$ module.

By the Wedderburn-Artin theorem, we may assume that $A = \mathrm{End}_\mathbb{C}(V)$, a matrix ring. Note that this makes $C = \mathrm{End}_B(V)$. Now we split $\mathrm{res}_B V$ as a $B$-module: $\mathrm{res}_B V = W^{\oplus k} \oplus X$, where $X$ has no composition factor of $W$. The algebra $\mathrm{End}_B(W^{\oplus k})$ is naturally contained in $C$, since $W^{\oplus k} \subseteq V$. Then $\mathrm{Hom}_B(W, \mathrm{res}^A_B V)$, and acts on the space $\mathrm{Hom}_B(W, \mathrm{res}_B V)$ (by post-composition) as the full endomorphism algebra. Hence $\mathrm{Hom}_B(W, \mathrm{res}_B V)$ is irreducible, or zero.

Now, if we also assume that $C$ is commutative, then $\mathrm{Hom}_B(W, \mathrm{res}^A_B V)$ is irreducible or zero.