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.



Innocent question on tensor products of modular representations
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.](https://mathoverflow.net/questions/22707/if-a-tensor-product-of-modules-is-semi-simple-are-the-tensor-factors-semi-simple/31477#31477)


