rt.representation theory - 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.