Studying Gabriel Navarro's book "Character Theory and the McKay Conjecture", I've come across the following problem. First, let's fix some notation: $G$ will be a finite group, $R$ will denote the algebraic integers of $\mathbb{C}$ over $\mathbb{Q}$, $p$ will be a prime, $M$ will be a maximal ideal of $R$ containing $pR$ and $\mathbb{F}$ will denote the field $R/M$, where $*:R \to \mathbb{F}$ is the canonical projection.
Let $\chi\in \operatorname{Irr}(G)$ and write $\chi^*$ for the function $\chi^*(g) = \chi(g)^*$. Then:
- If $\chi_1, ..., \chi_t \in \operatorname{Irr}(G)$ are all distinct and of $p$-defect zero, then $\chi_1^*, ..., \chi_t^*$ are $\mathbb{F}$-linearly independent;
- If $\chi \in \operatorname{Irr}(G)$ has $p$-defect zero and $x \in G$ is such that $\chi(x)^* \neq 0$, then $|G|_p = |x^G|_p$ (i.e., the greatest power of $p$ that divides the order of $G$ also divides the size of the conjugacy class of $x$);
- The number of irreducible $p$-defect zero characters of $G$ is bounded above by the number of conjugacy classes $K$ of $G$ such that $|K|_p = |G|_p$;
My attempt
1. Suppose $\sum_i \lambda_i \chi_i^* = 0$. Then we may take $\mu_i \in R$ such that $\mu_i^* = \lambda_i$. Our goal is $\mu_i \in M$. Take $\theta = \sum_i \mu_i \chi_i$. Our hypothesis guarantees that $\theta(G) \subset M$. We will use 2. and its notation. We have: $$\sum_{i=1}^t \lambda_i \chi_i(1)_{p'}^* \lambda_{\chi_i}( \hat{K}) = \sum_{i=1}^t \lambda_i \chi_i(1)_{p'}^* \left( \frac{|K|\chi_i(x_K)}{\chi_i(1)} \right)^* = \sum_{i=1}^t \left( \frac{|K|\mu_i \chi_i(x_K)}{p^a} \right)^* = \left( \frac{|K|\theta(x_K)}{p^a} \right)^*$$ where $x_K$ is a representative of the conjugacy class $K$ and $p^a$ is the $p$-part of $|G|$.
Now, either $\chi_i(x_K)^* = 0$ for all $i$, or there exists some $i$ such that $\chi_i(x_K)^* \neq 0$. In this last case, by 2., the expression above will yield $0$. All that's left is to show the same for the first case from which it will follow, from the linear independence of the $\lambda_{\chi_i}$ (which can be shown for characters of $p$-defect zero), that $\lambda_i = 0$ for all $i$.
2. Let $\lambda_\chi : Z(\mathbb{F}G) \to \mathbb{F}$ be the algebra homomorphism defined by $\lambda_\chi(\hat{K}) = \left( \frac{|K|\chi(x)}{\chi(1)}\right)^*$, where $K$ is a conjugacy class of $G$ with representative $x$ (that this is well-defined and is a homomorphism was done previously in the book). We can write: $$\left( \frac{|K|\chi(x)}{\chi(1)}\right)^* = \left( \frac{|K|_p |K|_{p'}\chi(x)}{\chi(1)_p \chi(1)_{p'}}\right)^* \in \mathbb{F}$$
i.e., separating out the power of $p$ of the two integers from the rest. Now, by hypothesis, $\left( \frac{\chi(1)_{p'}}{|K|_{p'}\chi(x)}\right)^*$ is a well-defined element of $\mathbb{F}$. Thus: $$\left( \frac{|K|_p |K|_{p'}\chi(x)}{\chi(1)_p \chi(1)_{p'}}\right)^* \cdot \left( \frac{\chi(1)_{p'}}{|K|_{p'}\chi(x)}\right)^* = \left( \frac{|K|_p}{ \chi(1)_p}\right)^* \in \mathbb{F}$$
This then proves $|K|_p \geq \chi(1)_p$. As the latter is equal to $|G|_p$, we get $|K|_p = |G|_p$, as desired.
3. Here, my guess would be, using 2., to find an element $x$ of $G$ such that $\chi(x)^* \neq 0$ for one defect zero caracter $\chi$, but not any other. This, I have no idea how to do. It will probably involve 1. somehow, but I can't see it...
Could anyone please provide some hint on how to finish item 1. and how to approach 3.? Is my intuition for this last item correct and, if so, how to go about applying it?
Thanks in advance!
PS: I would hazard a guess that this is a particular version of some much more general result for modular characters, but I don't know much of the latter...
