$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$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.

$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$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.

$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$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 $M$.

$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$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.

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 $M$.

$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$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 $M$.