Jim has already given the correct answer "no".
Here is a hopefully instructive example. Let kk be an alg. closed field of pos. char pp, and let G=SL2(k)G=SL2(k). Write V=k2V=k2 for the "natural" 2-dimensional representation of GG say with basis e1,e2e1,e2. Let W=SpVW=SpV be the pp-th symmetric power of VV. Then WW contains a 2 dimensional submodule AA spanned by the pp-th powers ep1ep1 and ep2ep2; the module AA is isom. to the "first Frobenius twist" of VV.
It is an exercise to check that there is no GG-stable complement to AA in WW; i.e. the SES
0toAtoWtoW/Ato00toAtoWtoW/Ato0
is not split.
Thus
WW is not completely reducible. Evidently there is a surjective mapping
VotimesptoWVotimesptoW, thus also the pp-th tensor power VotimespVotimesp is not completely
reducible.
But VV is a simple (hence completely reducible) GG-module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact,
the (p−1)(p−1)-th tensor power Votimesp−1Votimesp−1 is completely reducible;
arguing as before, one sees that Votimes(Votimesp−1)Votimes(Votimesp−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.
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