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?
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. 


There are interesting but isolated examples for finite groups of Lie type to consider. Here the simple Lie type and the characteristic of the defining field combine to give factorizations of some projective modules for finite Chevalley groups as tensor products of two nonprojective simple modules. This combines work of Chevalley on special (or exceptional) isogenies between simple algebraic groups with work of Steinberg on related simple modules and their twisted tensor products. It's all a bit complicated to explain, but relates to the simple groups discovered by Suzuki and by Ree. For instance, in type $G_2$ with $p=3$ the usual Chevalley group over a field of order $q=p^r$ has a smallest possible projective module of dimension $q^6$ (the Steinberg module) which factors when $r$ is odd as the tensor product of two nonprojective simple modules whose highest weights are supported on long (resp. short) simple roots; take $r>1$ here to get an associated simple Ree group. Besides the original literature, there is a short account in my LMS Lecture Note volume Modular representations of finite groups of Lie type (see 20.3). Obviously what is going on in such examples requires specific primes for the specific groups involved as well as specific simple or projective modules. But these examples indicate some of the complicated behavior found in modular representation theory and caution against oversimplified expectations based on older characteristic 0 theory. 


If $G$ is a $p$group, this is almost never true (and I believe but may be wrong that for general $G$, it's completely governed by the $p$Sylow). In that case, we can understand modules in terms of the support variety $V={\rm Proj} H^{2*}(G,k)$, the projective variety associated to the (evendimensional) cohomology ring of $G$. To a finite $kG$module $M$ we can associate its support in $V$, namely the support of Ext(M,M) as a graded $H^{2*}(G,k)$module. This support is a closed subset of $V$, and conversely every closed subset is the support of some finite $kG$module. Finally, the support of $M\otimes N$ is the intersection of the support of $M$ and the support of $N$, and a module is projective iff its support is empty. Thus $M\otimes N$ is projective iff $M$ and $N$ have disjoint support. Thus unless $V$ is just a single point, it is possible to have nonprojective modules whose tensor product is projective. For $V$ to be a point, the cohomology of $G$ must be a polynomial ring in one variable, up to nilpotent elements. By Quillen's theorem, this is the case iff all elementary abelian subgroups of $G$ are conjugate and rank 1. In particular, it is true for cyclic groups, but otherwise it is almost always false. There's a simple argument to see directly that it holds for cyclic groups of order $p$: in that case, $kG$ can be identified with $k[x]/x^p$, and every indecomposable module is of the form $M_i=k[x]/x^i$ for some $i\leq p$. Such a module $M_i$ is projective iff $i=p$. If $M_i$ and $M_j$ are not projective, then $M_i\otimes M_j$ has dimension $ij$, which is not divisible by $p$. Thus $M_i \otimes M_j$ cannot be a sum of copies of $M_p$ and is hence not projective. 

