I have encountered a problem while going through a proof of a result. Here is the gist of the problem:
Suppose that $G = PN$ is a group with $P\leq G$ and $N\unlhd G$. Let $\psi \in \text{Irr}(N)$.
Then using Mackey's Theorem, one can get the relation that $((1_P)^G)_N = \rho_N$, which is the regular character of $N$. Hence, we must have that $\langle ((1_P)^G)_N, \psi \rangle \neq 0$. The proof then concludes that there exists $\chi \in \text{Irr}(G)$ such that $\langle \chi_N, \psi \rangle \neq 0$ and $\langle \chi_P, 1_P \rangle \neq 0$. I am unable to figure out why this is the case.
Here is my attempt so far:
I know that since $N\unlhd G$, we have $$\psi^G = \sum\limits_{\chi \in \text{Irr}(\psi^G)} \langle\psi^G,\chi\rangle \chi $$ and $$\text{Irr}(\psi^G) = \{ \chi \in \text{Irr}(G) \; | \;\langle \chi_N, \psi \rangle \neq 0 \} $$
Hence, we choose some $\chi \in \text{Irr}(G)$ such that $\langle \chi_N, \psi \rangle \neq 0 $.
Now, since $\langle ((1_P)^G)_N, \psi \rangle \neq 0$, the characters $((1_P)^G)_N$ and $\chi_N$ share a common irreducible constituent, viz, $\psi$. Therefore, we must have that $\langle \chi_N, ((1_P)^G)_N \rangle \neq 0$. If I can show that $\langle \chi, (1_P)^G \rangle \neq 0$, I will be done but I'm having a difficulty seeing it at this moment
Instead of looking at the decomposition of $\psi^G$, take the decomposition of $(1_P)^G$ into irreducibles, $$ (1_P)^G = \sum_{\chi \text{Irr}(G)} \langle \chi, (1_P)^G \rangle \chi .$$ You have $$ 0 \neq \langle ((1_P)^G)_N, \psi \rangle = \sum_{\chi \text{Irr}(G)} \langle \chi, (1_P)^G \rangle \langle \chi_N, \psi \rangle $$ and thus at least one of the summands is non-zero, so you find $\chi \in \text{Irr}(G)$ such that $\langle \chi, (1_P)^G \rangle \neq 0 \neq \langle \chi_N, \psi \rangle$. Finally use that $\langle \chi, (1_P)^G \rangle = \langle \chi_P, 1_P \rangle$ by Frobenius reciprocity.