State and prove Frobenius reciprocity for group representations.
Solution
Let \(H\le G\), let \(R\) be a commutative coefficient ring, let \(V\) be an \(R[H]\)-module, and let \(W\) be an \(R[G]\)-module. Frobenius reciprocity is the natural isomorphism \[ \operatorname{Hom}_{R[G]}\!\left(R[G]\otimes_{R[H]}V,W\right) \cong \operatorname{Hom}_{R[H]}\!\left(V,\operatorname{Res}_H^G W\right). \] Equivalently, \[ \operatorname{Hom}_G(\operatorname{Ind}_H^G V,W) \cong \operatorname{Hom}_H(V,\operatorname{Res}_H^G W). \]
Given a \(G\)-map \[ F:R[G]\otimes_{R[H]}V\to W, \] define \[ \Phi(F)(v)=F(1\otimes v). \] For \(h\in H\), \[ \Phi(F)(hv)=F(1\otimes hv)=F(h\otimes v)=hF(1\otimes v), \] so \(\Phi(F)\) is \(H\)-linear.
Conversely, given an \(H\)-map \(f:V\to W\), define \[ \Psi(f)(g\otimes v)=g f(v). \] This is well defined because \[ \Psi(f)(gh\otimes v)=ghf(v)=g f(hv)=\Psi(f)(g\otimes hv), \] and it is visibly \(G\)-linear.
Finally, \[ \Phi(\Psi(f))(v)=f(v) \] and \[ \Psi(\Phi(F))(g\otimes v)=gF(1\otimes v)=F(g\otimes v). \] Thus \(\Phi\) and \(\Psi\) are inverse natural isomorphisms.