Let $G$ be a finite group and $k$ a field, let us assume that char($k$) divides the group order. Let $kG$-mod denote the category of fintely generated $kG$-modules.
This category has as a tensor product $\otimes\_{k}$ with diagonal $G$-action.
Given now $M,N\in kG$-mod such that $M\otimes\_{k}N$ is projective, can we then conclude that either $M$ or $N$ had to be projective?
If not, can we ask for certain conditions on the field $k$ or the group $G$ such that the statement holds?



If the tensor product of two $kG$-modules is projeсtive, does either of them have to be projective?
The answer to the first question is no. Let $G$ be the group $C\_2 \times C\_2$ and let $k$ be a field of characteristic 2. Then $kG \cong k[x,y]/(x^2, y^2)$. Let $M = k[x]/(x^2)$ with $y$ acting trivially, and let $N = k[y]/(y^2)$ with $x$ acting trivially. Then neither $M$ nor $N$ is projective (for modules over a $p$-group in characteristic $p$, projective is the same as free), but $M \otimes\_k N \cong kG$.


Edit: I think that if the Sylow $p$-subgroup of $G$ is cyclic, then you can use Avrunin and Scott's tensor product theorem to prove that whenever $M \otimes\_k N$ is projective, then so is either $M$ or $N$, but that seems like a big tool to use for this.


