Centralizer of $\mathbb{C}[G]$ in $\mathbb{C}[H]$

Solution 1:

Let's do it slightly more generally in the following

Context. Let $A\subseteq B$ be finite-dimensional, semisimple ${\mathbb C}$-algebras.

Also, put $\text{C}_A(B) := \{b\in B\ |\ ab=ba\text{ for all }a\in A\}$, the centralizer of $A$ in $B$.

Proposition. If $N$ resp. $M$ are finite-dimensional irreducible modules over $A$ resp. $B$, then $\text{Hom}_A(N,\text{res}^B_A M)$ is an irreducible $\text{C}_A(B)$-module for the canonical action $(b.\varphi)(n) := b.\varphi(n)$.

This is the essential part in establishing your claim.

Corollary. $\text{C}_A(B)$ is commutative iff $\text{res}_A^BM$ is multiplicity free for all irreducible $B$-modules $M$.

Proof of corollary:

$\Rightarrow$: Suppose $\text{C}_A(B)$ is commutative. Then all its irreducible modules are $1$-dimensional, and in particular $\text{mult}_N(\text{res}^B_AM)=\text{dim}_{\mathbb C}\text{Hom}_A(N,\text{res}^B_AM)\leq 1$ in the situation of the proposition.

$\Leftarrow$: Conversely, suppose that all $\text{Hom}_A(N,\text{res}^B_AM)$ are of dimension $\leq 1$. Then, decomposing $_AA=N_1\oplus\ldots\oplus N_n$ and $_BB = M_1\oplus\ldots\oplus M_m$, we have $\text{C}_A(B)\subseteq B$ acting faithfully on $B = \text{Hom}_A(A,\text{res}^B_A B) = \bigoplus_{i,j} \text{Hom}_A(N_i,\text{res}^B_AM_j)$, so that the algebra homomorphism $$\text{C}_A(B)\longrightarrow\prod_{i,j}\text{End}_{\mathbb C}\left(\text{Hom}_A(N_i,\text{res}^B_AM_j)\right)\cong {\mathbb C}\times\dots\times{\mathbb C}$$ is injective. Since the right hand side is commutative, we deduce that so is $\text{C}_A(B)$.

Proof of proposition. This is an application of averaging.

1. It suffices to show that for any two $\alpha,\beta\in\text{Hom}_A(N,\text{res}^B_AM)\setminus\{0\}$ there ex. $x\in \text{C}_A(B)$ with $x\alpha=\beta$.

2. To begin, $\text{Hom}_{\mathbb C}(N,M)\supset\text{Hom}_A(N,\text{res}^B_AM)$ is an irreducible $A^{\text{op}}\otimes_{\mathbb C} B$-module, so there exists $\tilde{x}\in A^{\text{op}}\otimes_{\mathbb C} B$ with $\tilde{x}\alpha=\beta$. Somehow we need to get the desired element $x\in\text{C}_A(B)$ from $\widetilde{x}$.

   Addendum concerning $\text{Hom}_{\mathbb C}(N,M)\cong N^{\ast}\otimes_{\mathbb C} M$ being irreducible over $A^{\text{op}}\otimes_{\mathbb C} B$: By the Artin-Wedderburn Structure Theorem, $A^{\text{op}}\to\text{End}_{\mathbb C}(N)$ and $B\to\text{End}_{\mathbb C}(M)$ are surjective, and hence so is $A^{\text{op}}\otimes_{\mathbb C} B\to\text{End}_{\mathbb C}(N^{\text{op}}\otimes_{\mathbb C} M)$; then, note that any finite-dimensional vector space is simple over its endomorphism ring.

3. Now, since $A$ is semisimple and we are working over ${\mathbb C}$, it is separable, meaning that there exists some separability idempotent $e = \sum_i e_i\otimes f_i\in A\otimes_{\mathbb C} A$ satisfying $$\sum_i e_i f_i = 1\qquad\text{ and }\qquad\sum_i a e_i\otimes f_i = \sum_i e_i\otimes f_i a\text{ for all }a\in A.$$ Such a separability idempotent allows to project any $A-A$-bimodule $M$ onto its $A$-symmetric elements $\{m\in M\ |\ a.m=m.a\text{ for all} a\in A\}$ by $m\mapsto \sum_i e_i.m.f_i$.

   Note that in case $A={\mathbb C}[H]$ is the group algebra of a finite group, the separability idempotent is given by $e := \frac{1}{|H|}\sum\limits_{h\in H} h\otimes h^{-1}$, and we recover the classical averaging map $m\mapsto \frac{1}{|H|}\sum\limits_{h\in H} g.m.g^{-1}$ from the proof of Maschke's Theorem, for example.

4. We apply this to the $A-A$-bimodule $\text{Hom}_{\mathbb C}(N,\text{res}^B_AM)$, the $A$-symmetric elements of which are precisely $\text{Hom}_A(N,\text{res}^B_AM)$. Now $\alpha$ and $\beta$ are already $A$-linear, so $$\beta\ =\ e\beta\ =\ e\widetilde{x}\alpha\ =\ (e\widetilde{x}e)\alpha.$$ Consider the element $e\widetilde{x}e\in A^\text{op}\otimes_{\mathbb C} B$: The right multiplication by $e$ causes it to be $A$-symmetric with respect to the $A-A$-bimodule structure on $A^{\text{op}}\otimes_{\mathbb C} B$ given by $a.(s\otimes t).a^{\prime} := (as)\otimes (ta^{\prime})$, while the left multiplication by $e$ causes it to also be $A$-symmetric for the $A-A$-bimodule structure given by $a.(s\otimes t).a^{\prime} := (sa^{\prime})\otimes (at)$. The set of such elements is a subalgebra of $A^{\text{op}}\otimes_{\mathbb C} B$ mapping to $\text{C}_A(B)$ under multiplication $\mu: A^{\text{op}}\otimes_{\mathbb C} B\to B$. As $(e\tilde{x}e)\alpha = \mu(e\tilde{x}e)\alpha$ by the $A$-linearity of $\alpha$, putting $x := \mu(e\tilde{x}e)\in \text{C}_A(B)$ does the job. $\Box$