1
$\begingroup$

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 $C_{G}(y)$?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.