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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)?

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    $\begingroup$ Quick side note: We say that $d$ is 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
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Yes. Consider the map sending $b \in B$ to $bdb^{-1}d^{-1}$. The fibers are cosets of $T$, the centralized of $d$, so the image is isomorphic to $U$. The image is also clearly contained in $U$, so by Ax-Grothendieck, or, in the finite field case, counting, the image is all of $U$.

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  • $\begingroup$ Thanks Will. That's an elegant and fun argument. And in the context of a finite $G$, that shows that this uses nothing more than $U$ containing the derived subgroup of $B$. $\endgroup$ – Joël Apr 19 '13 at 19:01
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The answer to your question is yes. Your element $d$ is regular semisimple since its centralizer is assumed to have minimal dimension equal to the rank (being the maximal torus $T$). In turn, it's a standard fact that $du$ is also regular semisimple and in fact conjugate in $B$ to $d$. This requires only some basic structure theory of semisimple groups; see for example Prop. 2.4 in my 1995 AMS book Conjugacy Classes in Semisimple Groups. It's only stated at first for a connected semisimple group but generalizes immediately to the reductive case (no reference needed to fields of definition or being quasi-split).

When placed in the axiomatic setting of split BN-pairs, I think the story is much the same provided you build in some simplicity assumption on $G$. But here I'd have to check a bit, since Steinberg's theory of regular elements isn't explicitly placed in that setting. [ADDED: Probably the axioms for a split BN-pair are too weak to yield a natural version of the elementary argument used in the Borel-Chevalley structure theory of algebraic groups. But for finite groups of Lie type, the algebraic group methods involving Jordan decomposition should apply directly.]

P.S. As Jeff Adler notes, a regular element (including Steinberg's wider sense) is just defined to be one whose centralizer has the smallest possible dimension: the rank of the given semisimple (or reductive) group. Connectedness is more delicate even for semisimple elements and may depend on the isogeny type.

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