Let B=TU be a Borus of G defined over F<sub>q</sub>. Let H be the image of T(F<sub>q</sub>) in G(F<sub>q</sub>)/G(F<sub>q</sub>)<sup>+</sup>. If H has index k, then the preimage of H in G(F<sub>q</sub>) has index k. The preimage of H contains the large cell U<sup>-</sup>BU, which is dense in G. At least for a fixed type of G and q>>0, this immediately implies that k=1 (which of course implies the required abelianility). In general, a finer analysis of the formula for the number of points in a reductive group over a finite field should get finer control over for which q and G this argument works.

Actually (Alternatively, using Bruhat and idea of looking at image of T(F<sub>q</sub>) it suffices to show that every Weyl group element has a representative in the subgroup generated by unipotent elements. The Weyl group is generated by simple reflections, so this question is reduced to a computation in rank one, where it suffices to consider the simply connected groups SL<sub>2</sub> and SU<sub>3</sub>.