Q:

Modules "projective in a subcategory"

In my research I have come up with the following notion which I would like to learn more about. It may be very naive.
Let $R$ be a ring, $M$ an $R$-module and $S$ a class or $R$-modules closed under coproducts. We say that $M$ is projective relative to $S$ if every surjection from an element of $S$ to a coproduct of copies of $M$ admits a section. We say that $M$ is strongly projective relative to $S$ if every such surjection splits with summands in $S$. (The generality in this definition is just to make the exposition clearer, I have a fairly specific example in mind, see below.)
The easiest example is if we let $I$ be an ideal in $R$ and $M=R/I$. Then we may take $S$ to be the class of modules with annihilator containing $I$ (I.e. the $R/I$-modules) and $M$ is strongly projective relative to $S$. If $R/I$ is local then we can take $S$ to be coproducts of $M$ and $M$ is strongly projective relative to $S$ (by kaplansky's theorem).
The example I have in mind is where $G$ is a finite $p$-group, $H$ is a normal subgroup, $k$ a field of characteristic $p$, $R= kG$ is the group algebra (which is local) and $M=kG/H$. Then $M$ is strongly projective to the same $S$ as before (I.e. the coproducts of $M$).
But what if $H$ is not normal? Is $M$ still strongly projective relative to $S$? If not what goes wrong?
Note that I have no particular reason to believe that this should be true, other than that I could not find a counterexample (likely because of my lack of group theory fu) and that it would be helpful for something I want to prove.

A:

More generally, the following is true:
Let $A$ be a finite-dimensional $k$-algebra ($k$ a field), $M$ a finite-dimensional indecomposable (right) $A$-module, and $S$ the category of coproducts of copies of $M$. Then $M$ is "strongly projective" relative to $S$.
In your example, $M=k[G/H]$ is indecomposable, since, as it is a quotient of $kG$, it has a simple head.
Let $E=\operatorname{End}_A(M)$, which is local since $M$ is indecomposable, and $\operatorname{Proj}(E)$ the category of projective (and hence free) $E$-modules. Then
$$R=\operatorname{Hom}_A(M,-): S\to \operatorname{Proj}(E)$$
is an equivalence of categories, with inverse
$$L=M\otimes_E-:\operatorname{Proj}(E)\to S,$$
since the functors are an adjoint pair, both preserving coproducts (since $M$ is finitely generated), and clearly the natural maps $LR(M)\to M$ and $E\to RL(E)$ are isomorphisms.
Also, $R$ preserves surjections, since if $\alpha:X\to Y$ is a map in $S$ for which $R\alpha:RX\to RY$ is not surjective, then there is a nonzero map from the cokernel to $E$ (since $E$ is a local finite-dimensional $k$-algebra, and so every non-zero $E$-module has a non-zero map to the socle of $E$), and hence a non-zero map $\beta:RY\to E$ such that $\beta\circ R\alpha=0$. Applying $L$ gives a non-zero map $L\beta:Y\to M$ such that $L\beta\circ\alpha=0$, so $\alpha$ can't be surjective.
So this translates the problem into the question about projective modules for local rings that you know how to deal with. 

