TITLE: Conjugacy of Borel subgroups over arbitrary fields
QUESTION [8 upvotes]: Let $k$ be a field and $G$ a connected semisimple algebraic group over $k$. 
If $k$ is algebraically closed, then it is well known that all Borel subgroups of $G$ are conjugate by the action of $G(k)$. I would like to know whether this is also true over non-closed fields.


Are all Borel subgroups of $G$ over $k$ conjugate by the action of $G(k)$?


Recall that a Borel subgroup $B$ of $G$ over $k$ is a closed subgroup of $G$ over $k$ such that the base change of $B$ to the algebraic closure $\bar{k}$ is a Borel subgroup of $G \times_k \bar{k}$. 
Of course Borel subgroups need not exist in general; when they do one says that $G$ is quasi-split. If $G$ is not quasi-split then the statement is vacuously true.

REPLY [11 votes]: Yes. This follows directly from Theorem 20.9 (i) in "Armand Borel, Linear Algebraic Groups, Second enlarged edition, 1991" which goes like this:
Theorem: Let $ G $ be a connected reductive group over a field $ k $. The minimal parabolic $k$-subgroups of $G$ are conjugate under $G(k)$.