Let $G$ be a finite group and $H \le G$ such that $gHg^{-1} \cap H = 1$ for any $g \in G \setminus H$. Let $\pi$ be the permutation representation of $G$ associated to the left regular action of $G$ on $G/H$, let $\rho$ be a simple $H$-representation and \begin{align*} \widetilde{\chi}_\rho(g) = \begin{cases} \chi_\rho(xgx^{-1})& xgx^{-1} \in H \\ \chi_\rho(1),& \text{otherwise}. \end{cases} \end{align*}
Show that
(a) $\widetilde{\chi}_\rho \in F(G)_{class}$ and $||\widetilde{\chi_\rho}||_G^2 =||\chi_\rho||_H^2= 1$
(b) $\widetilde{\chi}_\rho = \chi_{\operatorname{ind}(\rho)} - \chi_\rho(1)(\chi_\pi - \chi_{\operatorname{triv}_G})$.
I can show that the character is in $\mathcal{F}_{class}(G)$, but I do not know how to show that the inner product is $1$. If I could show that the associated irreducible representation was simple, then that would show it. I also tried working with the inner product directly, and I get $\frac{|H|}{|G|} + \frac{1}{|G|} \sum\limits_{g \in G \setminus H} \widetilde{\chi}_\rho(g) \overline{\widetilde{\chi}}_\rho(g)$, but I get stuck here. For part b), I just don't even know where to start.