We find all primitive subgroups of $GL(8,11)$ with parameters $e=8$, $a=1$. Since in this case $b=2$ does not divide $a=1$, Lemma 5.3 implies that there exists two 
#nonconjugate subgroups $F_1$ and $F_2$ such that $N_{GL(8,11)}(F_1)/F_1\simeq O^+(6,2)$, while $N_{GL(8,11)}(F_2)/F_2\simeq O^-(6,2)$. This log-file contains
#calculations for $F_2$.
#In the notations of Theorem 5.1, if $H$ is a primitive solvable subgroup, then it possesses a series


$$1<U<F<A<G.$$

#$U$ is cyclic of order $10$, $F$ is of type $2^{1+6}$ is such that $N_{GL(8,11)}(F)/F\simeq O^-(6,2)$, and $G=A$. First we take the list of generators of solvable 
#subgroups of $GL(8,11)$, containing all maximal primitive subgroups of $GL(8,11)$ from the file   generators-GL(8,11)-.txt.


#Now we run the following loop. Each entry of the list result has three items: the first is the
#number of the group in the list genS, the second is the order of the group, and the third is the order of the stabilizer of v in H. 
#If the order of the stabilizer is 1, then |v^H=|H| and Corollary 2.5 can be applied.

gap> z:=Z(11);;
gap> v:=[z,z^3,z^2,0*z,z^4,z^5,z^7,z^9];;
gap> result:=[];;
gap> for j in [1..Size(gens)] do
> H:=Subgroup(GL(8,11),gens[j]);;
> n:=Size(H);;
> m:=Size(v^H);
> result[j]:=[j,n,n/m];
> Display(result[j]);
> od;
[ 1, 5760, 1 ]
[ 2, 5760, 1 ]
[ 3, 5760, 1 ]
[ 4, 6400, 1 ]
[ 5, 6400, 1 ]
[ 6, 11520, 1 ]
[ 7, 11520, 1 ]
[ 8, 11520, 1 ]
[ 9, 11520, 1 ]
[ 10, 11520, 1 ]
[ 11, 11520, 2 ]
[ 12, 11520, 1 ]
[ 13, 11520, 1 ]
[ 14, 11520, 1 ]
[ 15, 11520, 1 ]
[ 16, 11520, 1 ]
[ 17, 12800, 1 ]
[ 18, 12800, 1 ]
[ 19, 17280, 1 ]
[ 20, 17280, 1 ]
[ 21, 17280, 1 ]
[ 22, 23040, 1 ]
[ 23, 23040, 1 ]
[ 24, 23040, 1 ]
[ 25, 23040, 1 ]
[ 26, 23040, 1 ]
[ 27, 23040, 1 ]
[ 28, 23040, 1 ]
[ 29, 23040, 1 ]
[ 30, 23040, 1 ]
[ 31, 23040, 1 ]
[ 32, 23040, 1 ]
[ 33, 23040, 1 ]
[ 34, 25600, 1 ]
[ 35, 34560, 2 ]
[ 36, 34560, 1 ]
[ 37, 34560, 1 ]
[ 38, 34560, 1 ]
[ 39, 34560, 1 ]
[ 40, 34560, 2 ]
[ 41, 34560, 1 ]
[ 42, 34560, 1 ]
[ 43, 51840, 1 ]
[ 44, 46080, 1 ]
[ 45, 46080, 1 ]
[ 46, 46080, 1 ]
[ 47, 46080, 1 ]
[ 48, 46080, 1 ]
[ 49, 46080, 1 ]
[ 50, 46080, 1 ]
[ 51, 69120, 1 ]
[ 52, 69120, 2 ]
[ 53, 69120, 2 ]
[ 54, 69120, 1 ]
[ 55, 69120, 1 ]
[ 56, 69120, 1 ]
[ 57, 69120, 1 ]
[ 58, 69120, 1 ]
[ 59, 69120, 1 ]
[ 60, 69120, 1 ]
[ 61, 69120, 2 ]
[ 62, 103680, 1 ]
[ 63, 103680, 1 ]
[ 64, 103680, 2 ]
[ 65, 92160, 1 ]
[ 66, 138240, 1 ]
[ 67, 138240, 2 ]
[ 68, 138240, 1 ]
[ 69, 138240, 2 ]
[ 70, 138240, 2 ]
[ 71, 138240, 1 ]
[ 72, 138240, 2 ]
[ 73, 138240, 1 ]
[ 74, 138240, 1 ]
[ 75, 138240, 1 ]
[ 76, 207360, 1 ]
[ 77, 207360, 2 ]
[ 78, 276480, 1 ]
[ 79, 276480, 2 ]
[ 80, 414720, 2 ]
[ 81, 414720, 2 ]
[ 82, 414720, 1 ]
[ 83, 414720, 1 ]
[ 84, 829440, 2 ]
[ 85, 829440, 2 ]


#Now we choose groups such that |v^H|<>|H| in the list res1

gap> res1:=Filtered(result,x->x[3]<>1);;              
gap> Size(res1);
17

#We choose another vector v and repeat the cycle for the remaining groups 


gap> v:=[z,z^3,z^2,0*z,z^4,z^5,0*z^7,0*z^9];;
gap> result1:=[];;        
gap> for j in res1 do
> n:=j[2];; 
> H:=Subgroup(GL(8,11),gens[j[1]]);;
> m:=Size(v^H);;
> Append(result1,[[j[1],n,n/m]]);
> Display([j[1],n,n/m]);
> od;
[ 11, 11520, 1 ]
[ 35, 34560, 1 ]
[ 40, 34560, 1 ]
[ 52, 69120, 1 ]
[ 53, 69120, 2 ]
[ 61, 69120, 1 ]
[ 64, 103680, 1 ]
[ 67, 138240, 1 ]
[ 69, 138240, 2 ]
[ 70, 138240, 1 ]
[ 72, 138240, 1 ]
[ 77, 207360, 1 ]
[ 79, 276480, 1 ]
[ 80, 414720, 2 ]
[ 81, 414720, 1 ]
[ 84, 829440, 1 ]
[ 85, 829440, 1 ]

#Now we choose groups such that |v^H|<>|H| in the list res1

gap> res2:=Filtered(result1,x->x[3]<>1);;              
gap> Size(res1);


#We choose another vector v and repeat the cycle for the remaining groups 

gap> v:=[z,z^3,z^2,0*z,z^4,z^5,0*z^7,0*z^9];;
gap> result1:=[];;        
gap> for j in res1 do
> n:=j[2];; 
> H:=Subgroup(GL(8,11),gens[j[1]]);;
> m:=Size(v^H);;
> Append(result1,[[j[1],n,n/m]]);
> Display([j[1],n,n/m]);
> od;
[ 53, 69120, 1 ]
[ 69, 138240, 1 ]
[ 80, 414720, 1 ]