#We find all primitive subgroups of $GL(12,3)$ with parameters $e=4$, $a=3$. 
#In the notations of Theorem 5.1, if $H\leq GL(12,3)$ is a primitive solvable subgroup, then it possesses a series

#$$1<U<F<A<H,$$

#where $U$ is cyclic of order $26$, $F=U\circ E$ and $E$ is extraspecial of order $2^{1+4}$, moreover
#in this case $|H:A|$ divides $3$. Since $b=2$ does not divide $a=3$, Lemma 5.3 implies that there are two nonconjugate subgroups $F_1$ and $F_2$, and
#$N_{GL(12,3)}(F_1)/F_1\simeq Q^+(4,2)$, while  $N_{GL(12,3)}(F_2)/F_2\simeq Q^-(4,2)$

gap> e:=4;; p:=3;; d:=12;; a:=3;;

#Now we generate the list of all primitive subgroups in $GL(12,3)$, using standart function of the package IRREDSOL

gap> l0:=AllIrreducibleSolvableMatrixGroups(Degree,d,Field,GF(p),IsPrimitiveMatrixGroup,true);;

#In this list we choose subgroups, whose order is divisible by the order of $F$ in the notations of Theorem 5.1

gap> l1:=Filtered(l0,x-> (Order(x) mod ((p^a-1)*e*e)) = 0);;

#In view of Lemma 5.3, we choose subgroups satisfying $\vert A/F\vert\in \{12,20,72\}$ and collect these subgroups in the list l2.

gap> l2:=Filtered(l1,x-> (Order(x)/((p^a-1)*e*e)) in  [12,20,72,3*12,3*20,3*72]);;

#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> v:=[Z(3),0*Z(3),0*Z(3),Z(3),0*Z(3),0*Z(3),0*Z(3),Z(3),0*Z(3),0*Z(3),0*Z(3),2*Z(3)];;
gap> result:=[];;
gap> i:=0;;
gap> for j in l2 do
> n:=Size(j);;
> m:=Size(v^j);
> i:=i+1;
> Append(result,[[i,n,n/m]]);
> Display([i,n,n/m]);
> od;
[ 1, 14976, 1 ]
[ 2, 89856, 1 ]
[ 3, 24960, 1 ]
[ 4, 4992, 1 ]
[ 5, 14976, 1 ]
[ 6, 29952, 1 ]
[ 7, 14976, 1 ]
[ 8, 14976, 1 ]
[ 9, 8320, 1 ]