#We find all primitive subgroups of $GL(4,61)$ with parameters $e=2$, $a=2$
#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 $61^2-1$, $F=U\circ E$ and $E$ is extraspecial of order $2^{1+2}$, moreover
#in this case $|H:A|$ divides $2$. Since $b=1$ divides $a=2$, Lemma 5.3 implies that there is up to conjugation there is one $F$, and
#$N_{L}(F)/F\simeq Sp(2,2)\simeq S_3$, where $L=C_{GL(4,61)}(U)$.

gap> e:=2;; p:=61;; d:=4;; a:=2;;

#Now we generate the list of all primitive subgroups in $GL(4,61)$, 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 $\vert A/F\vert\in \{6\}$ and collect these subgroups in the list l2.

gap> l2:=Filtered(l1,x-> (Order(x)/((p^a-1)*e*e)) in  [6, 6*2]);; Size(l2);
8

#We define the vector $v$ and verify that $v^g$ is a regular orbit for all groups $g$ in l2, so Corollary 2.5 can be applied

gap> v:=[Z(61),0*Z(61),0*Z(61),Z(61)^5];;
gap> result:=[];; i:=0;;
gap> for H in l2 do
> n:=Size(H);; m:=Size(v^H);; i:=i+1;;
> result[i]:=[i,n,n/m];;
> Display(result[i]);
> od;
[ 1, 178560, 1 ]
[ 2, 89280, 1 ]
[ 3, 89280, 1 ]
[ 4, 89280, 1 ]
[ 5, 89280, 1 ]
[ 6, 89280, 1 ]
[ 7, 89280, 1 ]
[ 8, 89280, 1 ]




gap> so:=List(l2,x->[Size(v^x),Size(x)]);
[ [ 89280, 89280 ], [ 89280, 89280 ], [ 89280, 89280 ], [ 89280, 89280 ], [ 89280, 89280 ], [ 89280, 89280 ], [ 29760, 29760 ], [ 29760, 29760 ], [ 29760, 29760 ], [ 29760, 29760 ],
  [ 29760, 29760 ], [ 29760, 29760 ], [ 29760, 29760 ], [ 89280, 89280 ], [ 29760, 29760 ] ]
gap> Size(so);
15
gap> Number(so,x-> (x[2] <> x[1]) );
0
gap>