#We find all primitive subgroups of $GL(6,11)$ with parameters $e=3$, $a=2$
#In the notations of Theorem 5.1, if $H$ is a primitive solvable subgroup, then it possesses a series

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

#where $U$ is cyclic of order $120$, $F=U\circ E$ and $E$ is extraspecial of order $3^{1+2}$, moreover
#in this case $|H:A|$ divides $2$.


gap> e:=3;; p:=11;; d:=6;; a:=2;;

#Now we generate the list of all primitive subgroups in $GL(6,11)$, 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 $\vert A/F\vert\in \{24\}$ and collect these subgroups in the list l2.

gap> l2:=Filtered(l1,x-> (Order(x)/((p^a-1)*e*e)) in  [24, 2*24]);; Size(l2);
4

#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> z1:=Z(11);; z0:=0*Z(11);; v:=[z1,z1,z0,z1,z1,z0];;
gap> result:=[];; i:=0;;
gap> for H in l2 do
> i:=i+1;; n:=Size(H);; m:=Size(v^H);;
> result[i]:=[i,n,n/m];
> Display(result[i]);
> od;
[ 1, 51840, 1 ]
[ 2, 25920, 1 ]
[ 3, 25920, 1 ]
[ 4, 25920, 1 ]