Value of Irreducible Character in Quotient Algebra 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?

 A: 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.
