Let $G$ be an connected reductive algebraic group over $k=\bar{\mathbb{F}\_p}$. Suppose $G$ is defined over $\mathbb{F}\_q$. Let $G^{F}$ be the corresponding finite group associated to $G$. Suppose $s\in G^{F}$ is a regular semisimple element. Now, $s$ is contained in a unique maximal Torus $T$, and $T$ is necessarily $F$-stable. Let $T^F$ denote the set of $F$-rational points of $T$. 


It is clear that $T^F\subseteq C\_{G^F}(s)$. My question is whether $T^F=C\_{G^F}(s)$? In the case that this isn’t true is there a simple description of the quantity $C\_{G^F}(s)$?


Thank you.



Centraliser of regular semisimple element in $G^F$, for a connected reductive algebraic group $G$
It is a good question! The answer is NO, see the counter-example below.


Take $p=3$; then $\mathbb F\_3=\{0,1,-1\}$.
Write $L=\mathbb F\_3(i)$, where $i^2=-1$; then $L\simeq \mathbb F\_9$.


Take $$G={\rm GL}\_{2,L}\,,\quad G'=G/\{\pm 1\}.$$
Let $T\subset G$ denote the subgroup of diagonal matrices.
Take 
$$ s={\rm diag}(i,-i)\in T(L)\subset G(L).$$
Then the centralizer of $s$ in $G$ is $T$, hence $s$ is a regular semisimple element of $G(L)$.


Write 
$$
n=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\in G(L).
$$
Then
$$ n s n^{-1} ={\rm diag}(-i,i)= -s.$$
This means that if we denote by $s'$ and $n'$ 
the images in $G'(L)$ of $s$ and $n$, respectively, then
$$ n' s' (n')^{-1} = s'.$$
Thus 
$$ n'\in C\_{G'}(s')(L),$$
but $n'\notin T'(L)$, where $T'$ denotes the image of $T$ in $G'$.
We see that $$C\_{G'(L)}(s')\supsetneqq T'(L).$$
In the notation of the question, we obtain that 
$$C\_{G^{\prime F}}(s')\supsetneqq T^{\prime F}.$$ 


