I have encountered a situation in a proof of a result of which I am struggling to figure out.
Suppose we have subgroups $H,K$ and $N$ of a group $G$ such that $N\unlhd K$ and $K \leq NH$. If $\psi \in \text{Irr}(N)$ such that no element of $K \cap H$ stabilizes $\psi$, then $\psi^K \in \text{Irr}(K)$
The result mentions that is due to Clifford's Correspondence but I can't seem to reason why.
The only information I know off: Let $T = \text{Stab}_K(\psi) = \{x \in K \;|\; \psi^x = \psi \}$. It is clear that $N \leq T$. If $\chi \in \text{Irr}(T)$ is an irreducible constituent of $\psi^T$, then $\chi^K \in \text{Irr}(K)$
It suffices to show that $T = N$, as then $\psi$ is the only constituent of $\psi^T = \psi$ and the information you have gives the desired result.
So let $x \in T \leq K \leq N H$ and write $x = n h$. You have $ h = n^{-1} x \in K \cap H$ and so $\psi = \psi^{x} = \psi^{n h} = \psi^{h}$ which implies $h = 1$ (no element of $K \cap H$ except the identity stabilizes $\psi$) and so $x = n \in N$ which shows $T = N$ and we are done.