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Presumably the argument ilya was driving at is this: If one had a surjective map of group schemes $G \to H$, then consider the preimage of the nilradical of $H$. This is an algebraic group with a surjective map to a unipotent group. Since there are no group homomorphisms from reductive groups to unipotent ones, the nilradical of this preimage (which is contained in the nilradical of $G$) must surject onto the nilradical of $H$. So if the former is trivial, so is the latter.