Right adjoint to forgetful functor from $G$-Sets to $H$-Sets? Let $H$ be a subgroup of group $G$.
Let $F_1$ be a forgetful functor from $G$-Sets (sets with action of group $G)$ to $H$-Sets.
Is it true that right-adjoint functor to $F_1$ exists if and only if $H$ is a normal subgroup of $G$ ?
 A: No, $F_1$ has adjoints on both sides, whether $H$ is normal or not.
Indeed, the category of $G$-sets is isomorphic to the category of functors $[G,\mathbf{Set}]$ (where $G$ is seen as a one-object category), and the functor $F_1$ coincides with the precomposition with the inclusion functor/homomorphism $i:H\to G$. Thus $F_1$ has left/right adjoints, which are constructed as left/right Kan extensions along $i$; these Kan extensions exist because $\mathbf{Set}$ is complete and cocomplete.
In fact this is true even for an arbitrary morphism $\varphi:H\to G$, or if $G$ and $H$ are monoids, or if one replaces sets with vector spaces, modules, topological spaces, or any (co)complete category (although it could make the constructions more complex).
A: This is a complement to Arnaud D's excellent answer (+1). In this situation, it is actually easy to compute explicitly the right adjoint $R$.
For any $G$-set $X$ and any $H$-set $Y$, you want a natural isomorphism :
$$\operatorname{Hom}_G(X,R(Y))=\operatorname{Hom}_H(F_1(X),Y)$$
If this holds for any $X$, in particular, it should hold for $X=G$ where the $G$-action is given by left multiplication. Thus :
$$\operatorname{Hom}_G(G,R(Y))=\operatorname{Hom}_H(F_1(G),Y)$$
But $\operatorname{Hom}_G(G,R(Y))=R(Y)$ as sets by the bijection $\varphi\mapsto \varphi(e)$ whose inverse is $y\mapsto (g\mapsto gy)$. It follows that, as sets $R(Y)=\operatorname{Hom}_H(F_1(G),Y)=\operatorname{Hom}_H(G,Y)$ (where $G$ is a $H$-set via left multiplication). It remains to find the correct $G$-action tu put on $R(Y)$ for the adjunction to hold. Well, there is nothing fancy here, the only reasonable action is the following : for any $\psi\in\operatorname{Hom}_H(G,Y)$, the action of $g$ is $g\psi=(g'\mapsto \psi(g'g))$ (check that this is indeed a left action).
So you have it : $R(Y)=\operatorname{Hom}_H(G,Y)$ with the action $g\psi=(g'\mapsto\psi(g'g))$. It remains to check the adjunction property. I leave it to you.
