Let $G$ be a finite simple group of Lie type and $x$ be a central involution (that is, an involution which is contained in the center of a Sylow $2$-subgroup).


Is it true that, if $y$ is another involution in $G$, then $C\_{G}(x)$ involves a group isomorphic to $C\_{G}(y)$?



Centralizer of a central involution in a simple group of Lie type
Just to add a bit more detail to my comment, let $G = {\rm PSL}(4,3)$, which has order $6065280 = 2^7.3^6.5.13$.


The image $x$ of the diagonal matrix $t$ with entries $(-1, -1, 1, 1)$ is a central involution whose centralizer is the image of a subgroup of index $2$ in ${\rm GL}(2,3) \wr C\_2$. (Note that elements that conjugate $t$ to $-t$ lie in this centralizer.) 
So $C\_G(x)$ solvable, and has order $1152 = 2^7.3^2$.


There is another involution $y \in G$ which is the image of an element of order $4$ in ${\rm SL}(4,3)$,and whose centralizer is the image of a group containing ${\rm SL}(2,9)$ in its non-absolutely irreducible representation of degree $4$ of ${\mathbb F}\_3$. So $C\_G(y)$ is not solvable and cannot be involved in $C\_G(x)$. In fact $|C\_G(y)| = 2880 = 2^6.3^2.5$.


As I said in my comment, ${\rm PSL}(4,2) \cong A\_8$ is a smaller counterexample, but it is harder to do this by hand.


