Show that $ (φ^G )_K = (φ_{H∩K})^K $ with Mackey's theorem

Question

Suppose H,K ≤ G e θ $ ϵ $ Char(H). Show that Z(θ)≤H. Suppose H,K ≤ G and HK = G. Se $ φ $ ϵ Char(H) show that $ (φ^G )_K = (φ_{H∩K})^K $. For the proof I have to use the Mackey's theorem. How do I proceed on this? Thanks

Answer 1

I suppose that your superscript means the induced character and subscript means the restricted character?

Then it follows immediately from the statement of Mackey's theorem: $(\varphi^G)_K=\sum_{H\backslash G/K}({}^g\varphi_{^gH\cap K})^K$ (so here $g$ runs over a set of coset representatives for $H\backslash G/K$, and $^g\varphi$ denotes the character obtained by pre-conjugating). But now $G=HK$ and so the trivial coset is the only coset; hence the sum simplifies to give $(\varphi^G)_K=(\varphi_{H\cap K})^K$.