Let $G$ be a connected reductive group split over a field $k$. Let $T$ be a maximal split torus of $G$. Consider $N_G(T)$, the normalizer of $T$ in $G$, we have $N_G(T)/T \cong W$, the Weyl group of $G$. Assume $|k|$ is large enough and consider $N_G(T)(k)$, the $k$-points of $N_G(T)$. Given $w \in W$, does there exist a regular semisimple element (w.r.t. $G$) $n \in N_G(T)(k)$ whose image in $W$ is $w$? It is easy to see this for classical groups by some matrix calculations but I wonder if there is a nice, uniform way to prove this in the positive.
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$\begingroup$ Not if $k$ is finite and its characteristic divides the order of $w$. In that case the order of $n$ would be of finite order divisible by the characteristic, which is impossible for a semisimple element. $\endgroup$– LSpiceCommented Oct 1, 2018 at 23:35
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$\begingroup$ If $w$ fixes a non-$0$ element of $\operatorname{Lie}(T)$, then we should be done by induction on the semisimple rank. Thus the difficult case is when $w$ fixes no such element, i.e., when it's elliptic. In this case it seems to me heuristically that either no lift should work or all should, but (a) I can't prove this and (b) I can't prove that it's not the case that no lift works. $\endgroup$– LSpiceCommented Oct 2, 2018 at 0:25
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$\begingroup$ @LSpice: I don't see how this induction process can be carried out in the first case, could you elaborate a little bit? The second case is true for GL_n and indeed all lifts work, by an elementary matrix calculation. $\endgroup$– ShawnCommented Oct 2, 2018 at 2:56
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$\begingroup$ If $M$ is a Levi subgroup of $G$, then (by looking at the action of simple roots outside $M$ on elements of $Z(M)^\circ$), for every $M$-regular semisimple element $t \in M$, there is a $G$-regular semisimple element of $t Z(M)^\circ$. If $w$ fixes a non-$0$ element of $\operatorname{Lie}(T)$, then we can take $M = \operatorname C_G(\operatorname{Lie}(T)^w)$ and work inductively. $\endgroup$– LSpiceCommented Oct 2, 2018 at 11:47
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$\begingroup$ (This assumes that centralisers of Lie-algebra elements are Levis, which is a large- or zero-characteristic thing.) $\endgroup$– LSpiceCommented Oct 2, 2018 at 11:47
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