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There is a maximal nontoral (not contained in a conjugate of a fixed maximal torus) elementary abelian $2$-subgroup of rank 6 in $G = PGL(8,7)$. I denote this group by $A$. Its normaliser $N_{G}(A)$ is $A.N$ where $N$ is a maximal subgroup of $Sp(6,2)$ of order $40320$. And this $A.N$ is itself a subgroup of a Class 8 maximal subgroup of $G$ and is isomorphic to $SO_{8}^{+}(7).2$, I believe. I've been trying to construct this $A$ in $G$ for two days. The following is what I did but didn't work.

     > d:=ClassicalMaximals("L",8,7:general:=true,classes:={8});
     > X:=GL(8,7);
     > G:=PGL(8,7);
     > ro:=hom<X->G|G.1,G.2>;
     > m3:=ro(d[3]);m1:=ro(d[1]);
     > SUBOFM3:=Subgroups(m3:OrderEqual:=2580480);

     >> SUBOFM3:=Subgroups(m3:OrderEqual:=2580480);
                 ^
     Runtime error in 'Subgroups': Cannot compute subgroups of all composition
     factors of this group.

I intended to get this $A.N$ and then use the command of pCore to find A. I suppose I need to narrow down the searching range a bit... If I get the Sylow $2$-sub of $m3$ and then try to find elementary abelian subgroups of order $2^6$, it'd be killed by Magma. Any suggestions would be greatly appreciated!

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1 Answer 1

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Sorry, my previous answer was not right. The subgroup $A$ that I constructed was toral. The one you are looking for is Aschbacher class C6, not C2. There are two different maximal subgroups of ${\rm \Omega}^+(8,7)$ with the same structure. Here is the corrected version:

> c:=ClassicalMaximals("O+",8,7:classes:={6},all:=false);
> AN:=ro(c[1]);
> AN subset m3;
true
> ChiefFactors(AN);

G
|  Cyclic(2)
*
|  Alternating(8)
*
|  Cyclic(2) (6 copies)
1

> A := pCore(AN,2);
> #A;
64
> Normaliser(G,A) eq AN;
true
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  • $\begingroup$ cannot thank you enough! two days trying to find a way. Seems like I still have a long long way to go... $\endgroup$
    – scsnm
    Commented Aug 21, 2022 at 9:01
  • $\begingroup$ Note that I have now corrected an error in my previous attempt! $\endgroup$
    – Derek Holt
    Commented Aug 21, 2022 at 13:02
  • $\begingroup$ Thank you for putting in more time. I've noticed this when computing the centraliser orders. But still, very enlightening! $\endgroup$
    – scsnm
    Commented Aug 21, 2022 at 21:56

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