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Let $G$ be a finite group an $F(G)$ the algebra of functions on $G$. Let $N\lhd G$ be a normal subgroup and consider the ideal: $$J_N=\{f\in F(G)\,|\,\forall\,n\in N,\,f(n)=0\}.$$ Consider the quotient algebra $F(G)/J_N$ and the quotient map: $$\theta_N:F(G)\rightarrow F(G)/J_N\cong F(N),$$ $f\mapsto f+J_N=:[f]$.

Let $\rho$ be an irreducible representation of dimension $d$ of $G$. Where $$\chi=\sum_{i=1}^d\rho_{ii}\in F(G),$$ is the character of $\rho$, I believe that

$$\frac{1}{|N|}\sum_{t\in N}[\chi](t)$$

is equal to $d$ or zero.

Question 1: Is this true? It is a Frobenius-type result but I am interested, if possible, in a more direct proof.

Question 2: Does this fail for $N$ non-normal?

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  • $\begingroup$ I am sure I am failing to see something obvious, but how do $[\chi](t)$ and $\chi(t)$ compare for $t \in N$? $\endgroup$ – Andreas Caranti Jun 19 at 9:01
  • $\begingroup$ Can't you identify $F(G)/J_N$ with $F(N)$ and hence just use Frobenius and $\operatorname{Ind}^G_N 1=1^{\oplus [G:N]}$ (since $N$ normal)? $\endgroup$ – user10354138 Jun 19 at 9:06
  • $\begingroup$ @AndreasCaranti my instinct is that the thing to compare is $\sum_{t\in N}[\chi](t)$ and $\sum_{t\in N}\chi(t)$. If you are asking what is $[\chi](t)$ I guess I am invoking an isomorphism $F(G)/J\cong F(N)$. I suppose that $\theta_N(\delta_g)=\delta_g$ for $g\in N$, and zero otherwise. $\endgroup$ – JP McCarthy Jun 19 at 9:07
  • $\begingroup$ @user10354138 I had in a previous edit said that this is a Frobenius type result but I was interested in a more direct proof if possible. $\endgroup$ – JP McCarthy Jun 19 at 9:08
  • $\begingroup$ I have taken some of these comments back into the question. $\endgroup$ – JP McCarthy Jun 19 at 9:10
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Here is a proof in the language of representations of quantum groups. This is work of Piotr Podlés and later Shouzou Wang.

Let $\rho=(\rho_{ij})_{i,j=1}^{d_\rho}$ be an irreducible representation and $N$ a subgroup of $N$. Let $\theta_N:F(G)\rightarrow F(N)$ be the map: $$\delta_g\mapsto \begin{cases}\delta_g&\text{ if }g\in N\\ 0 & \text{ otherwise}\end{cases}.$$

Note that where $\Delta:F(G)\rightarrow F(G)\otimes F(G)$ is the comultiplication: $$\delta_g\mapsto \sum_{t\in G}\delta_{gt^{-1}}\otimes\delta_t,$$ and $\Delta_N:F(N)\rightarrow F(N)\otimes F(N)$ the comultiplication in $F(N)$:

$$\delta_g\mapsto \sum_{n\in N}\delta_{gn^{-1}}\otimes \delta_n,$$ the map $\pi$ has the property that: $$(\pi\otimes\pi)\circ \Delta=\Delta_N\circ \pi.$$

That $\rho$ is a representation of $G$ on $V$ implies that $(\rho_{ij})_{i,j=1}^{d_\rho}\in M_{d_\rho}(F(G))$ is a corepresentation matrix for $F(G)$.

That $N$ is a subgroup of $G$ implies that $(\pi(\rho_{ij}))_{i,j=1}^{d_\rho}$ is a corepresentation matrix for $F(N)$.

The trivial corepresentation of $F(N)$ is the map $\kappa_{\tau_N}$ $\lambda\mapsto \lambda\otimes \mathbf{1}_N$. Let $n_\rho$ be the multiplicity of $\kappa_{\tau_N}$ in $\pi(\kappa_\rho)$. Podlés/Wang claim that $n_\rho$ is either $d_\rho$ or zero. Assume $1<n_\rho<d_\rho$.

Let $F(N\backslash G)$ be the algebra of functions constant on the right cosets of $N$ in $G$ and $F(G/N)$ the algebra of functions on the right cosets.

If $N$ is normal, then these algebras coincide.

Let $E$ be the projection $F(G)\mapsto F(G/N)$ that maps a function $f$ to $E(f)$ where $$E(f)(Ng)=\frac{1}{|N|}\sum_{n\in N}f(ng),$$ the average of $f$ on the coset $Ng$.

Let $$n_N=\frac{1}{|N|}\sum_{n\in N}\delta^n$$ be the averaging state on $F(N)$. We claim that $$E(\rho_{ij})=E_{N\backslash G}(\rho_{ij})=\sum_{k=1}^{d_\rho}h_N(\pi(\rho_{ik}))\,\rho_{kj},$$ and $$E(\rho_{ij})=E_{G/N}(\rho_{ij})=\sum_{k=1}h_N(\pi(\rho_{kj}))\rho_{ik}.$$ This follows from the fact that $$E=\underbrace{(h_N\pi\otimes I_{F(G)})\circ \Delta}_{E_{N\backslash G}}=\underbrace{(I_{F(G)}\otimes h_N\pi)\circ \Delta}_{E_{G/N}},$$ as can be seen by showing that this maps $$\delta_g\mapsto \frac{1}{|N|}\mathbf{1}_{Ng}.$$

Now choose a basis such that the $n_\rho$ trivial representations of $N$ in $\pi(\rho)$ occur in the top left hand corner so that:

$$\pi(\rho)=\left(\begin{array}{cccc}\mathbf{1}_N & \dots & 0 & 0 \\ \vdots & \ddots & 0 & 0\\ 0 & 0 & \mathbf{1}_N & 0 \\ 0 & 0 & 0 & \varrho\end{array}\right),$$ where $\varrho$ is a sum of non-trivial representations of $N$.

Now, beyond this point we have matrix elements $\pi(\rho_{ij})$ of non-trivial representations of $N$. From corepresentation theory, we have that the average of a non-trivial representation is zero, and so, outside the $n_\rho\times n_\rho$ upper corner, $n_N(\pi(\rho_{ij}))=0$.

Beyond this square there must be a non-zero matrix element $\rho_{ij}$

Recall that the map $E$ is a projection from functions on $G$ onto functions on the quotient group.

Note that, from the above discussion $$E_{N\backslash G}(\rho_{ij})=\begin{cases}\rho_{ij}&1\leq i\leq n_\rho,\,1\leq j\leq d_\rho\\0 & \text{otherwise}\end{cases},$$ and $$E_{G/N}(\rho_{ij})=\begin{cases}\rho_{ij}&1\leq i\leq d_\rho,\,1\leq j\leq n_\rho\\0 & \text{otherwise}\end{cases}$$

Take an element $\rho_{ij}$ such that $j>n_\rho$.

Now $$0\neq \rho_{ij}=E_{N\backslash G}(\rho_{ij})=E_{G/N}(\rho_{ij})=0.$$

This implies that there is no such $j$ and so $n_\rho=d_\rho$. Alternatively $n_\rho=0$.

The result above should follow.

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