This is not a homework.
I am a beginner in studying the representation theory and character theory and I am doing some exercises from "A Course in the Theory of Groups by Robinson" and "character theory of finite groups by Isaacs". I have a problem with solving exercise 8.4.1 of Robinson's book that is similar to the exercise 5.6 of Isaacs's book:
Let $G$ be a finite group and $H,K \le G$ and . Let $\chi_1(H)$ and $\chi_1(K)$ denote the trivial characters of $H$ and $K$ over an algebraically closed field of characteristic 0. I want to prove that $<\chi_1(H)^G,\chi_1(K)^G>$ equals the number of $(H, K)$-double cosets.
I know that I should use Mackey's formula but I don't know what is that. After some search I found this proof in Wikipedia but I can not understand all parts:
I will be so appresiate for any helpful answer.
Here's how I would view the problem; I don't actually use Mackey's formula, but, in this special case, Frobenius reciprocity leads to the same thing.
Inducing the trivial character of $H$ up to $G$ produces the permutation character of $G$ acting on a transitive $G$-set with $H$ as one of the point-stabilizers, i.e., the $G$-set $G/H$. The inner product you want is, by Frobenius reciprocity, the inner product of the trivial character of $K$ times the restriction to $K$ of this permutation character of $G$. In other words, it is the dimension of the fixed-point space in the permutation representation of $K$ given by this restriction, $(G/H)\upharpoonright K$. But the dimension of the fixed-point space in a permutation representation is the number of orbits, so what you need is the number of $K$-orbits in $G/H$. Fortunately, these orbits are essentially the double-cosets. (More precisely, if you take any $K$-orbit in $G/H$, that's a set of $H$-cosets, and the union of those $H$-cosets is an $(H,K)$-double-coset.)