Let $\mathrm{G}$ be any infinite discrete group, and $\mathrm{H}$ be any finite index subgroup of $\mathrm{G}.$ I do not know if the Frobenius reciprocity theorem is true for the infinite groups. I want to say that, given any irreducible finite-dimensional complex representation $\rho$ of $\mathrm{G}$ there exists an irreducible finite-dimensional complex representation $\psi$ of $\mathrm{H}$ such that $\text{Ind}\_{\mathrm{H}}^{\mathrm{G}}(\psi)$ contains $\rho.$ How to prove this then?



Frobenius reciprocity theorem for infinite groups
A semi-answer, too long for a comment.


"the Frobenius reciprocity theorem" for finite groups is just a special case of the Hom-Tensor adjunction if you phrase it as
$$\operatorname{Hom}\_{\mathbb{C}[H]}(X,\operatorname{Res}\_H^G(Y)) = \operatorname{Hom}\_{\mathbb{G}[G]}(\operatorname{Ind}\_H^G(X),Y)$$
because $\operatorname{Ind}\_H^G(X) = \mathbb{C}[G] \otimes\_{\mathbb{C}[H]} X$. So in this sense it is also true for infinite groups (and any number of other algebraic objects) because it is "just abstract nonsense".


But the answer to your question depends on what you mean by "contains". If you mean "is a subquotient" then the usual proof using that adjunction goes through unchanged; in fact you get an epimorphism $\operatorname{Ind}\_H^G(\psi)\twoheadrightarrow \rho$ if you choose $\psi$ to be any (irreducible) $H$-submodule of the restriction of $\rho$.
If "contains" means "is a submodule" however, then things get tricky, because representations of infinite group need not be semisimple. Even uncomplicated groups like $G=\mathbb{Z}$ have indecomposable, but not irreducible modules: Jordan blocks.


And because of that, in some areas of mathematics, people restrict the meaning of "representation" to mean only "unitary representation" or something like that so that more representations are semisimple. But again, not all infinite groups have this nice property. Non-amenable groups have already been mentioned as a source of problems here.


