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I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One an occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One can occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behavior perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One such is $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One an occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One such is when $q = p^{a}$ and $p \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$.

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One such is $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behaviour perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).