(ii) and (iii), $k(b) = k(b')$.

(Separation condition) Let $a'$ be a $p$-element of $G'$ and let $\Pi'_{a'}$ be the set of elements $\pi$ of $\Pi$ which is $G'$-conjugate to $a'$. For $\pi \in \Pi'_{a'}$, let $c'_{\pi a'}\in G'$ be such that $a'^{c'_{\pi a'}} = (a')^{c'_{\pi a'}} = \pi$. Similarly for a $p$-element $a$ of $G$, we define $\Pi_a$ and $c_{\pi a}$ for $\pi \in \Pi_{a}$. Now for $\rho \in G_{p'}$ and ${\rho'} \in C_{G'}(a')_{p'}$, from (1) and by the assumption and [@NT], Theorem 5.4.11 for basic sets, $$\sum_{i= 1}^{k(b)} \varepsilon_{i} \chi_i (\rho)\overline{{\chi}'_i (a' {\rho}'}) = \sum_{i = 1}^{k(b)} \chi_i(\rho) \left( \sum_{\pi \in \Pi'_{a'}} \sum_{j =1}^{l(b_{\pi})} \ \overline{d^{x}_{\chi_i \varphi_{j}^{(\pi)}}}\overline{{\varphi'}_{j}^{(\pi)} ({\rho}'^{c'_{\pi a'}})}\ \right)$$ $$=  \sum_{\pi \in \Pi'_{a'}} \sum_{j = 1}^{l(b_{\pi})} \left( \sum_{i= 1}^{k(b)} \chi_i(\rho) \overline{d^{\pi}_{\chi_i \varphi_{j}^{(\pi)}}}\ \right) \overline{\varphi'^{(\pi)}_{j} ({\rho'}^{{c'}_{\pi a'}})} = 0.$$ (see [@Navarro], Lemma 7.5) Similarly for a $p$-singular element $x \in G$ and $y' \in G'_{p'}$,\
$\displaystyle\sum_{i=1}^{k(b)} \varepsilon_{i}\chi_i(x)\overline{\chi'_{i}(y')} = 0.$ (Integrality condition) Let $\rho \in G_{p'}$ and ${\rho'}\in G'_{p'}$. By the above, $\displaystyle\sum_{i= 1}^{k(b)} \varepsilon_i \overline{\chi'_i({\rho'})}\chi_i$ vanishes on the $p$-singular elements of $G$. Hence $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i(\rho)\overline{\chi'_i ({\rho'})}$ is divisible by $|C_{G}(\rho)|$ in ${\cal O}$ by [@NT], Theorems 3.6.10 and 3.6.13. Similarly $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i(\rho)\overline{\chi'_i ({\rho'})}$ is divisible by 