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 (even-dimensional) cohomology ring of $G$. To a finite $kG$-module $M$ we can associate its support in $V$, namely the support of $Ext^(M,M)$ Ext(M,M) as a graded $H^{2}(G,k)$-module. 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$is just a single point, it is possible to have non-projective modules whose tensor product is projective. For$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$for some$n\leq 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$and is hence not projective. 1 If$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)$, the projective variety associated to the (even-dimensional) cohomology ring of$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$is just a single point, it is possible to have non-projective modules whose tensor product is projective. For$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$for some$n\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.