Let $ G$ be a finite group, and let $ X$ be a family of subgroups of $ G$ closed under conjugation and under passage to subgroups. Suppose further that $ G$ is the union of the elements of $ X,$ and denote by $ R(G)$ the representation ring. By Artin’s theorem, we have a surjective map $ $ Ind: \oplus_{H \in X} \mathbb{Q} \otimes R(H) \rightarrow \mathbb{Q} \otimes R(G),$ $ given by taking the induced character and extending by linearity. I will now define two special classes of elements in the source of this map.

**Elements of Class 1:** Let $ H,H’ \in X, H’ \subset H,$ and let $ \psi’ \in R(H’).$ Let $ \psi = Ind_{H’}^H \psi’.$ Then elements of class $ 1$ are of the form $ \psi-\psi’.$

**Elements of class 2:** Let $ H \in X, s \in G$ and set $ H^s = sHs^{-1}.$ Let $ \psi \in R(H),$ and define $ \psi^s \in R(H^s)$ by $ \psi^s(shs^{-1})= \psi(h).$ Then elements of class $ 2$ are of the form $ \psi-\psi’.$

It is easy to see that all elements of class $ 1$ and $ 2$ lie in the kernel of the map $ Ind.$ I want to prove that in fact, the kernel is generated by elements of class $ 1$ and $ 2.$ The hint I have seem is that one should base change go $ \mathbb{C},$ and then use duality. It is then claimed that one is reduced to proving that if for every $ H \in X$ one has a class function $ f_H$ satisfying restriction snd conjugation conditions as above, then there exists a class function $ f$ on $ G$ restricting to $ f_H$ for all $ X.$ .

I supppse we should use Frobenius reciprocity in some clever way, but I do not see how the statement reduces to what I just claimed, so I would appreciate help, or full solutions.