I have a somewhat technical question about conjugacy in quasi-reductive groups.

Let $k$ be a field (in my main case interest, $k$ is finite), $G$ be a connected quasi-split reductive group over $k$. As is customary, let $S$ be a maximal $k$-split torus in $G$, $T$ its centralizer in $G$ (which is a maximal $k$-torus), $B$ a Borel containing $T$, $U$ the radical unipotent of $B$ (so that $B=UT$).

Let $d$ be an element of $T$ whose centralizer in $G$ is $T$. (in passing, first question:
is this condition the definition of $d$ being *regular*?). Let $u \in U$.

Is it true that $du$ is conjugate to $d$ in $G$? in $B$?

The reason for which I think it is true is this special case: Let $G=Gl_n(k)$, $T$ the diagonal torus, $B$ the upper triangular Borel, and $U$ the unipotent matrices in $B$. Then the condition on $d$ says that $d$ is diagonal with distinct eigenvalues. Then $du$ is triangular with the same diagonal terms as $d$ (in the same order), so it is easy to see that is conjugate to $d$ in $B$.

If not false, this should be a rather elementary lemma, and a reference or a proof would be equally welcome. I have a second question, but the solution of the first might answer it as well: If $G$ is more generally a split $BN$-pair, that is a $BN$-pair (or Tits system) together with a normal unipotent subgroup $U$ of $B$ such that $B=UT$, then is the answer to the question still yes (assuming it was yes to begin with for quasi-split reductive group)?

`regular'' if the connected part of $C_G(d)$ is $T$. If $C_G(d) = T$, then $d$ is`

strongly regular''. From Steinberg (Endomorphisms...), these notions are the same if $G$ is simply connected, but not in general. $\endgroup$ – Jeff Adler Apr 20 '13 at 5:50