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$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$In the finite group $G = \operatorname{PGL}_8(5)$, there is an elementary abelian $2$-subgroup $E$ of rank 5. $E = A_{1} \times A_{2} $ where $A_{1}$ has rank $2$ and $A_{2}$ rank $3$. Direct computation shows that $C_{G}(E) \cong A_{1} \times (5-1)^3$ and $N_{G}(E)/C_{G}(E) \cong \begin{pmatrix} \Sp_{2}(2) & 0 \\ *_{3\times2} & S_{4} \end{pmatrix}.$ It's obvious that $\Sp_{2}(2)$ acts on $A_{1}$ and $S_{4}$ on $(5-1)^3$.

I believe this $A_{2}$ corresponds to a maximal elementary abelian $2$-subgroup $EK$ contained in a maximal torus of $K = \operatorname{PGL}_{4}(\mathbb{C})$ with centraliser $T_{3}$ and $N_{K}(EK)/C_{K}(EK)$ isomorphic to $S_{4}$. Hence we have $S_{4}$ acting on $(5-1)^3$ in the finite group.

What's interesting is that computation shows that $E$ has a $G$-conjugate $E'$ contained in a Class $2$ maximal subgroup $M = \SL_{2}(5)^4.(5-1)^3.S_{4}$ of $G$. Let's identify $E$ with $E'$. And direct computation shows that $N_{G}(E)$ can be realized in this $M$. I suppose $A_{1}.\Sp_{2}(2)$ is embedded in the $\SL_{2}(5)^4$ part since $\SL_{2}(5)$ has a quaternion subgroup. I can't see how $*_{3\times2}$ is realized in this $M$. And this has the potential to be generalized to $\operatorname{PGL}_{n}$ where n is even.

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    $\begingroup$ I know the notation, e.g., $4^3$, but what does $(5 - 1)^3$ mean? $\endgroup$
    – LSpice
    Commented Aug 23, 2022 at 11:36
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    $\begingroup$ Sorry. It does mean $4^{3}$. I put it this way because it comes from $(q-1)^3$. Thank you! $\endgroup$
    – user488802
    Commented Aug 23, 2022 at 20:23
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    $\begingroup$ I think you need to define the subgroup $E$ more precisely, perhaps by giving generating matrices in ${\rm GL}(8,5)$. $\endgroup$
    – Derek Holt
    Commented Aug 24, 2022 at 16:47
  • $\begingroup$ I think you have understated $M$. The normalizer of one of your $SL_2(5)$'s in $M$ induces all of $PGL_2(5)$ on that $SL_2(5)$. $\endgroup$
    – Richard Lyons
    Commented Aug 26, 2022 at 14:25
  • $\begingroup$ Sorry for the late response. I thought not many interested. Do you mean like: $N_{G}(SL_{2}(5))$ induces all of $PGL_{2}(5)$? Would you please elaborate more? Thank you. $\endgroup$
    – user488802
    Commented Aug 28, 2022 at 23:42

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