Relationship between Frobenius reciprocity, Frobenius algebras, Frobenius extensions and Frobenius functors

Question

Let me start by explaining what I mean by the terms in the question. Let $\Bbbk$ be a fixed field (for the sake of simplicity). As far as I know

1. Frobenius reciprocity says that for $H$ a subgroup of $G$, the induction functor $\mathsf{Ind}_H^G$ is left adjoint to the restriction functor $\mathsf{Res}^G_H$. This can be seen as an immediate consequence of the hom-tensor relation, since $$ \mathsf{Hom}_{\Bbbk G}(\Bbbk G\otimes_{\Bbbk H} V,U) \cong\mathsf{Hom}_{\Bbbk H}(V,\mathsf{Hom}_{\Bbbk G}(\Bbbk G,U)). $$
2. A Frobenius algebra is a $\Bbbk$-algebra $A$ such that $A\cong \mathsf{Hom}_\Bbbk(A,\Bbbk)$ as regular left (equiv. right) $A$-modules.
3. An extension of $\Bbbk$-algebras $R\to S$ is Frobenius if and only if there exists an $R$-bilinear morphism $\epsilon:S\to R$ and (a finite number of) elements $x_i,y_i\in S$ such that $$\sum_ix_i\epsilon(y_is) = s = \sum_i\epsilon(sx_i)y_i$$ for all $s\in S$.
4. A pair of functors $\mathcal{F}:\mathcal{C}\to \mathcal{D}$, $\mathcal{G}:\mathcal{D}\to \mathcal{C}$ is Frobenius if $\mathcal{G}$ is at the same time left and right adjoint to $\mathcal{F}$.

Now, one can show that an extension $R\to S$ is Frobenius if and only if the restriction of scalars functor $\mathsf{Hom}_S(S,-)$ is Frobenius. A $\Bbbk$-algebra is Frobenius if and only if the extension $\Bbbk \to A$ is Frobenius. What is not clear to me at the present moment is:

Question: How are the last three notions realted with the Frobenius reciprocity?

For example, I have been told that the fact that $\Bbbk G$ is Frobenius for a finite group $G$ is essentially the Frobenius reciprocity, but at the present moment I don't see the connection.