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Feb 6, 2014 at 14:48 answer added Jim Humphreys timeline score: 3
Feb 6, 2014 at 6:40 comment added Jay Taylor If $x$ is semisimple and $F$ is algebraically closed or a finite field then one can appeal to the theory of reductive algebraic groups to obtain the centralisers. In particular, $C_G(x)$ will be a Levi subgroup. So one simply needs to compute maximal Levi subgroups in $G$ (which can be determined by the root system). You may find Bonnaf\'e's paper "Quasi-Isolated Elements in Reductive Groups" helpful.
Feb 5, 2014 at 21:55 comment added Mark Wildon If $x$ is unipotent and $x-1$ is nilpotent of degree $d$, then the centralizer of $x$ is contained in the centralizer of $1+(x-1)^{d-1}$. So the unipotent case reduces to centralizers of non-identity elements $x$ such that $(x-1)^2 = 0$. See also mathoverflow.net/questions/148704/….
Feb 5, 2014 at 18:57 comment added j.p. For centralizers of elements $x$, whose order is coprime to the characteristic of the field, you should be able to find the maximal centralizers (you may assume $x$ to have prime order). The unipotent case is probably hard.
Feb 5, 2014 at 18:30 history asked maryam CC BY-SA 3.0