can anybody please help me with this question? I have huge trouble even just start it.
Let H, K be subgroups of G and HK=G. If $\psi$ is a character of H, show that $(\psi^G)_K = (\psi_{H \cap K})^K$.
So basically $\psi^G = Ind^G_H(W)$, so W should be just H-module, and $(\psi^G)_K$ is just restricted to K. I have trouble to even just decompose this expression above. Can anybody please give me a hand? Thanks a lot.
I realized that it does take a bit more than just plugging into the formula. Here are the calculations.
Let $T\subseteq H$ be given such that $G = \bigcup_{t\in T}tK$ is a disjoint union. For $x\in K$ we have $$\varphi^G(x) = \frac{1}{|H|}\sum_{g\in G}\varphi^{\circ}(gxg^{-1}) = \frac{1}{|H|}\sum_{t\in T}\sum_{z\in K}\varphi^{\circ}(tzxz^{-1}t^{-1}) = \frac{1}{|H|}\sum_{t\in T}\sum_{z\in K}\varphi^{\circ}(zxz^{-1})$$ $$= \frac{1}{|H|}|T|\sum_{z\in K}\varphi^{\circ}(zxz^{-1}) = \frac{1}{|H|}|G:K|\sum_{z\in K}\varphi^{\circ}(zxz^{-1}) = \frac{|G|}{|H||K|}\sum_{z\in K}\varphi^{\circ}(zxz^{-1})$$ $$= \frac{1}{|H\cap K|}\sum_{z\in K}\varphi^{\circ}(zxz^{-1}) = (\varphi_{H\cap K})^K(x)$$
Where $\varphi^{\circ}(h) = \varphi(h)$ if $h\in H$ and $0$ otherwise.