Question: Let $H,K\subseteq G$ with $HK=G$ and suppose $\phi$ is a class function of $H$. Show that $(\phi^G)_K=(\phi_{H\cap K})^K$
Thoughts: Define $\dot{\phi}(g)=\phi(g)$ if $g\in H$ and $0$ otherwise. If I want to play with the LHS first, then my idea was to take $\displaystyle (\phi^G)(g)=\frac{1}{|H|}\sum_{x\in G}\dot{\phi}(x^{-1}gx)$, and then restrict it to $K$, but I am not getting anywhere with that. So, I think it may be best to play with the RHS first. So, $\displaystyle (\phi_{H\cap K})^K(k)=\frac{1}{|H\cap K|}\sum_{y\in K}\dot{\phi}(y^{-1}ky)$. Now, $|H\cap K|=\frac{|H||K|}{|G|}$, so if I can split this sum up a bit and write it as an induced character where I can pull out $\frac{1}{|K|}$ and $\frac{1}{|H|}$, then that could clear up the constant stuff, but I can't quite "make a map" that would help clear some of this up. I've fully convinced myself this is true (woop-dee-do) but I can't make anymore progress on it. Any help?