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 $C_{G}(y)$?

  • $\begingroup$ No. It is false in ${\rm PSL}(4,3)$ for example because you can choose $y$ such that $|C_G(y)|$ does not divide $|C_G(x)|$ . (In fact it is false in ${\rm PSL}(4,2) \cong A_8$ but that's less obvious.) $\endgroup$ – Derek Holt Feb 14 '16 at 12:10
  • $\begingroup$ What is the meaning of "involves" in the question? $\endgroup$ – YCor Feb 14 '16 at 13:33
  • $\begingroup$ Usually $G$ involves $H$ means that some quotient of some subgroup of $G$ is isomorphic to $H$. Equivalently you can say that $H$ is a section of $G$. $\endgroup$ – Derek Holt Feb 14 '16 at 13:38

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.

  • $\begingroup$ Thank you very much Professor Holt. Indeed, also I have realized that $A_{8}$ has two classes of involutions, the centralizer of the central involution is isomorphic to $C_{2}^{4} \rtimes A_{4}$ while the other involution has centralizer isomorphic to $D_{8}\times A_{4}$. Indeed, it is easy to see that my conjecture is false by this example too. $\endgroup$ – Kıvanç Ersoy Feb 17 '16 at 21:39

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