Innocent question on tensor products of modular representations Let $K$ be a field (of course, of positive characteristic, unless you want a trivial question). Let $G$ be a finite group, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ over $K$. Is the (interior) tensor product $V\otimes_K W$ a completely reducible representation of $G$ ?
I know that this holds for exterior tensor products:
Let $K$ be a field. Let $G$ and $H$ be two finite groups, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ and $H$, respectively, over $K$. Then, the tensor product $V\otimes_K W$ is a completely reducible representation of $G\times H$.
(Proof: Combine Curtis/Reiner "Methods of Representation Theory I" Theorems 7.10 and 10.38 (i).)
Note that my above question is equivalent to the Jacobson radical of the group ring $KG$ being a coideal (the coalgebra structure on $KG$ is the canonical one, of course: $\Delta g=g\otimes g$). It may be total nonsense but unfortunately I don't have any nontrivial examples of modular representations to check with.
 A: Jim has already given the correct answer "no". 
Here is a hopefully instructive example. Let $k$ be an alg. closed field of pos. char $p$, and let $G = SL_2(k)$. Write $V = k^2$ for the "natural" 2-dimensional representation of $G$ say with basis $e_1,e_2$. Let $W = S^pV$ be the $p$-th symmetric power of $V$. Then $W$ contains a 2 dimensional submodule $A$ spanned by the $p$-th powers $e_1^p$ and $e_2^p$; the module $A$ is isom. to the "first Frobenius twist" of $V$.
It is an exercise to check that there is no $G$-stable complement to $A$ in $W$; i.e. the SES
$$0 \to A \to W \to W/A \to 0$$
is not split.
Thus
$W$ is not completely reducible. Evidently there is a surjective mapping
$V^{\otimes p} \to W$, thus also the $p$-th tensor power $V^{\otimes p}$ is not completely
reducible.
But $V$ is a simple (hence completely reducible) $G$-module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact,
the $(p-1)$-th tensor power $V^{\otimes p-1}$ is completely reducible;
arguing as before, one sees that $V \otimes (V^{\otimes p-1})$ is not completely
reducible; thus in general the tensor product of two completely reducible modules
is not completely reducible.
I gave some further remarks about semisimplicity of tensor products in an answer to 
this question.
A: The answer to your question is usually no (which is fortunate because the lack of complete reducibility gives modular representation theorists something to do), starting for example with the tensor product of two irreducible representations of $G$ over an algebraically closed field whose prime characteristic divides the group order.   Examples for finite groups of Lie type are legion and come up naturally when you tensor the Steinberg representation with an arbitrary one: then you get a projective module whose indecomposable direct summands are rarely irreducible.   Textbooks like those by Jon Alperin, Curtis-Reiner, Serre, or me on modular representations illustrate such outcomes of tensoring.
ADDED: Concerning failure of complete reducibility in general, see also the related MO question 18280.
For references to some older literature on tensoring with the Steinberg representation, see the third section of my 1987 AMS Bulletin survey 
here.
http://www.ams.org/journals/bull/1987-16-02/S0273-0979-1987-15512-1/S0273-0979-1987-15512-1.pdf">here.
A: I claim that semisimple KG-modules are closed under tensor product in the modular setting iff G has a unique p-Sylow subgroup where p is the characteristic. 
Pf. Let P be the p-radical of G. That is P is the largest normal p-subgroup of G. It is well known that P is the intersection of the kernels of all irreps of G over K. So we have $$KG\to K[G/P]\to KG/Rad(KG).$$ 
If P is a p-Sylow then $K[G/P]$ is semisimple by Maschke and so the last map is an isomorphism. Thus Rad(KG) is a Hopf ideal and so the completely reducible reps are closed under tensor product. 
On the other hand if the completely reducible reps are closed under tensor, the radical is a Hopf ideal. Since the Hopf algebra quotients of a group algebra are the algebras of quotient groups it follows the last map is an iso (since g-1 is in the radical iff g is in P). But then by Maschke p does not divide the order of G/P so P is a p-Sylow. 
