##################################################################
############THE COMPUTATIONS FOR THE CORES IN TYPE D_6############
##################################################################
###################################
############FOR [3,9,6]############
###################################
test:=ll[1][1];;
quat1:=getRoots(test[5]);
#[ 2, 8, 10, 11, 14, 15, 18, 19, 24 ]
z1:=cenNR(test[5]);
#[ 18, 19, 24 ]
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 2, 10, 11 ], [ 8, 14, 15 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 8, 14, 15 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, a_18, a_19 ], [ -a_18, 0, -a_24 ], [ -a_19, -a_24, 0 ] ]
det(last);
#2*a_18*a_19*a_24
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_8, a_18*t_10+a_19*t_11 ], [ s_14, -a_18*t_2-a_24*t_11 ],
# [ s_15, -a_19*t_2-a_24*t_10 ] ]
#If p>=3, then we are left with z1 after inflation and
#induction. The family is \chi^{a_{18}, a_{19}, a_{24}}
#of (q-1)^3 characters of degree q^3.
#If p=2, we compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_8, -a_24*t_15/a_18 ], [ t_14, -a_19*t_15/a_18 ], [ t_15, t_15 ] ]
el_old:=[t_15];; el_new:=[a_18*t];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_8, -a_24*t ], [ t_14, -a_19*t ], [ t_15, a_18*t ] ]
#Hence we have Y':=\{x_{8, 14, 15}(t) \mid t \in \F_q\}, where
#x_{8, 14, 15}(t):=x_{8}(a_{24}t)x_{14}(a_{19}t)x_{15}(a_{18}t)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_2, -a_24*t_11/a_18 ], [ t_10, -a_19*t_11/a_18 ], [ t_11, t_11 ] ]
ea_old:=[t_11];; ea_new:=[a_18*t];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_2, -a_24*t ], [ t_10, -a_19*t ], [ t_11, a_18*t ] ]
#Hence we have X':=\{x_{2, 10, 11}(t) \mid t \in \F_q\}, where
#x_{2, 10, 11}(t):=x_{2}(a_{24}t)x_{10}(a_{19}t)x_{11}(a_{18}t)
#If p=2, we then get a family
#\chi_{b_{2, 10, 11}, b_{8, 14, 15}}^{a_{18}, a_{19}, a_{24}}
#of q^2(q-1)^3 characters of degree q^3.
###################################
###########FOR [3,10,9]############
###################################
test:=ll[2][1];;
quat1:=getRoots(test[5]);
#[ 1, 2, 3, 7, 8, 12, 24, 26, 27, 28 ]
z1:=cenNR(test[5]);
#[ 12, 27, 28 ]
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 1, 3, 24 ], [ 7, 8, 26 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 7, 8, 26 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, a_12, a_27 ], [ -a_12, 0, -a_28 ], [ -a_27, -a_28, 0 ] ]
det(last);
#2*a_12*a_27*a_28
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_7, t_3*a_12+t_24*a_27 ], [ s_8, -t_1*a_12-t_24*a_28 ],
# [ s_26, -t_1*a_27-t_3*a_28 ] ]
#If p>=3, then we are left with z1 after inflation and
#induction. The family is \chi_{b_2}^{a_{12}, a_{27}, a_{28}}
#of q(q-1)^3 characters of degree q^3.
#If p=2, we compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_7, -t_26*a_28/a_12 ], [ t_8, -t_26*a_27/a_12 ], [ t_26, t_26 ] ]
el_old:=[t_26];; el_new:=[a_12*t];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_7, -t*a_28 ], [ t_8, -t*a_27 ], [ t_26, t*a_12 ] ]
#Thus Y':=\{x_{7, 8, 26}(t) \mid t \in \F_q\}, where
#x_{7, 8, 26}(t):=x_{7}(a_{28}t)x_{8}(a_{27}t)x_{26}(a_{12}t)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_1, -t_24*a_28/a_12 ], [ t_3, -t_24*a_27/a_12 ], [ t_24, t_24 ] ]
ea_old:=[t_24];; ea_new:=[a_12*t];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_1, -t*a_28 ], [ t_3, -t*a_27 ], [ t_24, t*a_12 ] ]
#Hence X':=\{x_{1, 3, 24}(t) \mid t \in \F_q\}, where
#x_{1, 3, 24}(t):=x_{1}(a_{28}t)x_{3}(a_{27}t)x_{24}(a_{12}t)
#We compute the action of X_2 on estA.
diagDirectCom([2], estA, signcom1);
#Reordered, gives:
#[ 7, s_2*t*a_28 ], [ 8, -s_2*t*a_27 ], [ [ 26, s_2*t*a_12 ],
# [ 12, -s_2*t^2*a_27*a_28 ], [ 27, s_2*t^2*a_12*a_28 ] ],
#[ 28, s_2*t^2*a_12*a_27 ],
#Hence evaluating by lambda gives
#\lambda([x_2(s), x_{1, 3, 24}(t)])=\phi(c_{7, 8, 26}st+a_{12}a_{27}a_{28}st^2)=0.
#If c_{7, 8, 26}=0, then get family
#F_1 consisting of the \chi^{a_{12}, a_{27}, a_{28}}
#of (q-1)^3 characters of degree q^3.
#If a_{7, 8, 26}:=c_{7, 8, 26} \ne 0, then get
#X_{1, 3, 24}'=\{1, x_{1, 3, 24}(c_{7, 8, 26}/(a_{12}a_{27}a_{28}))\}
#X_2'=\{1, x_{2}(a_{12}a_{27}a_{28}/c_{7, 8, 26}^2)\}.
#The family F_2 consisting of the
#\chi_{d_{2}, d_{1, 3, 24}}^{a_{7, 8, 26}, a_{12}, a_{27}, a_{28}}
#of 4(q-1)^4 characters of degree q^3/2.
###################################
###########FOR [4,18,18]###########
###################################
test:=ll[3][1];;
quat1:=getRoots(test[5]);
#[ 1, 2, 4, 5, 6, 7, 10, 11, 12, 14, 15, 16, 17, 18, 19, 21, 22, 28 ]
z1:=cenNR(test[5]);
#[ 16, 21, 22, 28 ]
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 1, 5, 6, 12, 14, 15 ],
# [ 7, 10, 11, 17, 18, 19 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 7, 10, 11, 17, 18, 19 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, 0, 0, 0, a_21, a_22 ], [ 0, 0, -a_16, -a_21, 0, 0 ],
# [ 0, -a_16, 0, -a_22, 0, 0 ], [ 0, a_21, a_22, 0, 0, 0 ],
# [ -a_21, 0, 0, 0, 0, -a_28 ], [ -a_22, 0, 0, 0, -a_28, 0 ] ]
det(last);
#4*a_16*a_21^2*a_22^2*a_28
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_7, t_14*a_21+t_15*a_22 ], [ s_10, -t_6*a_16-t_12*a_21 ],
# [ s_11, -t_5*a_16-t_12*a_22 ], [ s_17, t_5*a_21+t_6*a_22 ],
# [ s_18, -t_15*a_28-t_1*a_21 ], [ s_19, -t_14*a_28-t_1*a_22 ] ]
#If p>=3, then we can apply the Reduction Lemma with
#arm [ 1, 5, 6, 12, 14, 15 ] and leg [ 7, 10, 11, 17, 18, 19 ].
#Let us have a look at the rest of the commutator relations.
signcom0:=Filtered(signcom1, x-> not(AbsInt(x[1]) in Concatenation(arm, leg) or AbsInt(x[2]) in Concatenation(arm, leg) or AbsInt(x[3]) in Concatenation(arm, leg)));
#[ ]
#Then in this case we get a family
#\chi_{b_2, b_4}^{a_{16}, a_{21}, a_{22}, a_{28}}
#of q^2(q-1)^4 irreducible characters of degree q^6.
#Let us now suppose p=2. We compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_7, -t_19*a_28/a_21 ], [ t_10, -t_17*a_22/a_16 ],
# [ t_11, -t_17*a_21/a_16 ], [ t_17, t_17 ], [ t_18, -t_19*a_22/a_21],
# [ t_19, t_19 ] ]
el_old:=[t_17, t_19];; el_new:=[a_16*t_2, a_21*t_1];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_7, -t_1*a_28 ], [ t_10, -t_2*a_22 ], [ t_11, -t_2*a_21 ],
# [ t_17, t_2*a_16 ], [ t_18, -t_1*a_22 ], [ t_19, t_1*a_21 ] ]
#Hence we have Y'=Y_1'Y_2'=
#\{x_{7, 18, 19}(t_1)x_{10, 11, 17}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{7, 18, 19}(t_1):=x_{7}(a_{28}*t_1)x_{18}(a_{22}t_1)x_{19}(a_{21}t_1)
#x_{10, 11, 17}(t_2):=x_{10}(a_{22}*t_2)x_{11}(a_{21}t_2)x_{17}(a_{16}t_2)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_1, -t_15*a_28/a_21 ], [ t_5, -t_12*a_22/a_16 ],
# [ t_6, -t_12*a_21/a_16 ], [ t_12, t_12 ], [ t_14, -t_15*a_22/a_21 ],
# [ t_15, t_15 ] ]
ea_old:=[t_12, t_15];; ea_new:=[a_16*t_2, a_21*t_1];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_1, -t_1*a_28 ], [ t_5, -t_2*a_22 ], [ t_6, -t_2*a_21 ],
# [ t_12, t_2*a_16 ], [ t_14, -t_1*a_22 ], [ t_15, t_1*a_21 ] ]
#Thus we get X'=X_1'X_2'=
#\{x_{1, 14, 15}(t_1)x_{5, 6, 12}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{1, 14, 15}(t_1):=x_{1}(a_{28}*t_1)x_{14}(a_{22}t_1)x_{15}(a_{21}t_1)
#x_{5, 6, 12}(t_2):=x_{5}(a_{22}*t_2)x_{6}(a_{21}t_2)x_{12}(a_{16}t_2)
#Notice that Y is now central, since
diagDirectCom(estL, estL, signcom1);
#[ ]
LiNotArmLeg:=Filtered(quat1, x-> not (x in leg or x in arm));;
DiLiNotArmLeg:=ListI(LiNotArmLeg);;
diagDirectCom(estL, DiLiNotArmLeg, signcom1);
#[ ]
#Also, we notice that X' is a subgroup, since
diagDirectCom(estA, estA, signcom1);
#[ ]
#Let us first see the behavior of X_2
#with the other bits of the quotient.
LiRem2:=ListRemI(quat1, Concatenation([2], arm, leg));
diagDirectCom([2], Concatenation(LiRem2, estA, estL), signcom1);
#[ [ 19, s_2*t_1*a_21 ], [ 18, -s_2*t_1*a_22 ], [ 7, s_2*t_1*a_28 ],
# [ 21, -s_2*t_1^2*a_22*a_28 ], [ 28, s_2*t_1^2*a_21*a_22 ],
# [ 22, s_2*t_1^2*a_21*a_28 ] ]
#We do the same with X_4
LiRem4:=ListRemI(quat1, Concatenation([4], arm, leg));;
diagDirectCom([4], Concatenation(LiRem4, estA, estL), signcom1);
#[ [ 17, -t_2*s_4*a_16 ], [ 11, -t_2*s_4*a_21 ], [ 10, -t_2*s_4*a_22 ],
# [ 16, -t_2^2*s_4*a_21*a_22 ], [ 22, t_2^2*s_4*a_16*a_21 ],
# [ 21, t_2^2*s_4*a_16*a_22 ] ]
#Since [2] (resp. [4]) just interacts with X_1' (resp. X_2'),
#we have [2, 4] as candidate for a leg, and X_1', X_2' as
#candidate for an arm. Let us compute that alltogether.
diagDirectCom([2, 4], estA, signcom1);
#Reordering, we get the following.
#[ 7, s_2*t_1*a_28 ], [ 18, -s_2*t_1*a_22 ], [ 19, s_2*t_1*a_21 ],
#[ 10, -t_2*s_4*a_22 ], [ 11, -t_2*s_4*a_21 ], [ 17, -t_2*s_4*a_16 ],
#[ 16, -t_2^2*s_4*a_21*a_22 ],
#[ 21, -s_2*t_1^2*a_22*a_28 ], [ 21, t_2^2*s_4*a_16*a_22 ],
#[ 22, s_2*t_1^2*a_21*a_28 ], [ 22, t_2^2*s_4*a_16*a_21 ],
#[ 28, s_2*t_1^2*a_21*a_22 ]
#This writes:
#\lambda([x_{2}(s_2)x_{4}(s_4), x_{1, 14, 15}(t_1)x_{5, 6, 12}(t_2)])=
#=\phi(
#s_2t_1(+c_{7, 18, 19}+a_{21}a_{22}a_{28}t_1)
#s_4t_2(+c_{10, 11, 17}+a_{16}a_{21}a_{22}t_2)
#)=1.
#If c_{7, 18, 19}=0 and c_{10, 11, 17}=0, then we can apply
#the Reduction Lemma with arm X' and leg X_2X_4.
#We then get a family F_1 that consists of the
#\chi^{a_{16}, a_{21}, a_{22}, a_{28}}
#which are (q-1)^4 characters of degree q^6.
#Let now a_{7, 18, 19}:=c_{7, 18, 19}\ne 0 and c_{10, 11, 17}=0. Define
#X_1'':=\{1, x_{1, 14, 15}(c_{7, 18, 19}/(a_{21}a_{22}a_{28}))\}, and
#Y_1'':=\{1, x_2(a_{21}a_{22}a_{28}/(c_{7, 18, 19}^2))\}.
#By the Reduction Lemma, we reduce to the subquotient X_1''Y_1''Z, which
#is abelian. Then we get a family F_2, whose elements are labelled by
#\chi_{d_{1, 14, 15}, d_2}^{a_{7, 18, 19}, a_{16}, a_{21}, a_{22}, a_{28}}.
#These are 4(q-1)^5 characters of degree q^6/2.
#If c_{7, 18, 19}=0 and a_{10, 11, 17}:=c_{10, 11, 17} \ne 0, the situation
#is similar. We define
#X_2'':=\{1, x_{5, 6, 12}(c_{10, 11, 17}/(a_{16}a_{21}a_{22}))\}, and
#Y_2'':=\{1, x_4(a_{16}a_{21}a_{22}/(c_{10, 11, 17}^2))\}.
#By the Reduction Lemma, we reduce to the abelian subquotient X_2''Y_2''Z.
#We get the family F_3, consisting of the
#\chi_{d_4, d_{5, 6, 12}}^{a_{10, 11, 17}, a_{16}, a_{21}, a_{22}, a_{28}},
#which again are 4(q-1)^5 irreducible characters of degree q^6/2.
#Finally, if a_{7, 18, 19}:=c_{7, 18, 19}\ne 0 and
#a_{10, 11, 17}:=c_{10, 11, 17} \ne 0, defining X_1'', X_2'', Y_1'' and Y_2''
#as above we reduce to the abelian subquotient X_1''Y_1''X_2''Y_2''Z.
#This yields the last family F_4, of the irreducible characters
#\chi_{d_{1, 14, 15}, d_2, d_4, d_{5, 6, 12}}^{a_{7, 18, 19}, a_{10, 11, 17}, a_{16}, a_{21}, a_{22}, a_{28}},
#which are 16(q-1)^6 irreducible characters of degree q^6/4.
###################################
###########FOR [4,21,28]###########
###################################
test:=ll[4][1];;
quat1:=getRoots(test[5]);
#[ 1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
# 26 ]
z1:=cenNR(test[5]);
#[ 20, 21, 22, 26 ]
suggestArmLeg(test);
#[ 4, "NOT-ABELIAN,BUT CLOSE,NORMAL:arm+leg", [ 1, 5, 6, 7, 10, 11 ],
# [ 12, 14, 15, 17, 18, 19 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 12, 14, 15, 17, 18, 19 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, 0, 0, 0, a_21, a_22 ], [ 0, 0, -a_20, -a_21, 0, 0 ],
# [ 0, -a_20, 0, -a_22, 0, 0 ], [ 0, a_21, a_22, 0, 0, 0 ],
# [ -a_21, 0, 0, 0, 0, -a_26 ], [ -a_22, 0, 0, 0, -a_26, 0 ] ]
det(last);
#4*a_20*a_21^2*a_22^2*a_26
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_12, t_10*a_21+t_11*a_22 ], [ s_14, -t_6*a_20-t_7*a_21 ],
# [ s_15, -t_5*a_20-t_7*a_22 ], [ s_17, t_5*a_21+t_6*a_22 ],
# [ s_18, -t_11*a_26-t_1*a_21 ], [ s_19, -t_10*a_26-t_1*a_22 ] ]
#If p>=3, then we can apply the Reduction Lemma with
#arm [ 1, 5, 6, 7, 10, 11 ] and leg [ 12, 14, 15, 17, 18, 19 ].
#Let us have a look at the rest of the commutator relations.
signcom0:=Filtered(signcom1, x-> not(AbsInt(x[1]) in Concatenation(arm, leg) or AbsInt(x[2]) in Concatenation(arm, leg) or AbsInt(x[3]) in Concatenation(arm, leg)));
#[ [ 3, 16, 20 ], [ 13, 16, -26 ] ]
Filtered(test[5], x-> not(
1 in x or 5 in x or 6 in x or 7 in x or 10 in x or 11 in x or
12 in x or 14 in x or 15 in x or 17 in x or 18 in x or 19 in x
));
#[ [ 3, 16, 20 ], [ 13, 16, 26 ] ]
#We then want to apply the Reduction Lemma again, with X_{16} as leg
#and X_3X_{13} as arm.
directCom([16], [3, 13], signcom0);
#[ 26*t_13*s_16, -20*t_3*s_16 ]
#Applying \lambda, we get:
#\lambda([x_{16}(s_{16}), x_3(t_3)x_{13}(t_{13})])=\phi(s_{16}(a_{20}t_3-a_{26}t_{13}))=1.
#We get that X'=1, and Y'=\{x_{3, 13}(t) \mid t \in \F_q\}, where
#x_{3, 13}(t):=x_3(a_{26}t)x_{13}(a_{20}t).
#We have already reduced to an abelian subquotient.
#If p \ge 3, then we have the family of the
#\chi_{b_{3, 13}, b_8, b_9}^{a_{20}, a_{21}, a_{22}, a_{26}}
#which are q^3(q-1)^4 irreducible characters of degree q^7.
#Let now p=2. We observe the following "layers", examined later in order.
# - The triangles [ 17, 5, 15, 7, 14, 6 ] and [ 19, 1, 18, 11, 12, 10 ]. These
#give 12 of the 28 relations.
# - The 2 relations 5--(16)--11 and 6--(16)--10 would hopefully disappear.
# - The butterfly 3--(20)-->16--(26)-->13. This gives 2 relations.
#It would be useful to make 3/13 diagonal.
# - The intertwining relations between triangles involving 3 and 13,
# namely 3--(12)--7, 3--(14)--10, 3--(15)--11
# and 13--(17)--1, 13--(18)--5, 13--(19)--16. These are 6 relations.
# - Finally, there are the relations that involve 9 and 8, the
#respective hearts, namely
#8--(12)--1, 8--(18)--10, 8--(19)--11, and
#9--(14)--5, 9--(15)--6, 9--(17)--7. These 6 relations
#also give rise to quadratic terms by summing again elements in the arm.
#We compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_12, -t_19*a_26/a_21 ], [ t_14, -t_17*a_22/a_20 ],
# [ t_15, -t_17*a_21/a_20 ], [ t_17, t_17 ], [ t_18, -t_19*a_22/a_21 ],
# [ t_19, t_19 ] ]
el_old:=[t_17, t_19];; el_new:=[-a_20*t_2, -a_21*t_1];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_12, t_1*a_26 ], [ t_14, t_2*a_22 ], [ t_15, t_2*a_21 ],
# [ t_17, -t_2*a_20 ], [ t_18, t_1*a_22 ], [ t_19, -t_1*a_21 ] ]
#Hence we have Y'=Y_1'Y_2'=
#\{x_{12, 18, 19}(t_1)x_{14, 15, 17}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{12, 18, 19}(t_1):=x_{12}(a_{26}*t_1)x_{18}(a_{22}t_1)x_{19}(a_{21}t_1)
#x_{14, 15, 17}(t_2):=x_{14}(a_{22}*t_2)x_{15}(a_{21}t_2)x_{17}(a_{20}t_2)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_1, -t_11*a_26/a_21 ], [ t_5, -t_7*a_22/a_20 ],
# [ t_6, -t_7*a_21/a_20 ], [ t_7, t_7 ], [ t_10, -t_11*a_22/a_21 ],
# [ t_11, t_11 ] ]
ea_old:=[t_7, t_11];; ea_new:=[-a_20*t_2, -a_21*t_1];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_1, t_1*a_26 ], [ t_5, t_2*a_22 ], [ t_6, t_2*a_21 ],
# [ t_7, -t_2*a_20 ], [ t_10, t_1*a_22 ], [ t_11, -t_1*a_21 ] ]
#Thus we get X'=X_1'X_2'=
#\{x_{1, 10, 11}(t_1)x_{5, 6, 7}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{1, 10, 11}(t_1):=x_{1}(a_{26}*t_1)x_{10}(a_{22}t_1)x_{11}(a_{21}t_1)
#x_{5, 6, 7}(t_2):=x_{5}(a_{22}*t_2)x_{6}(a_{21}t_2)x_{7}(a_{20}t_2)
#Notice that Y is now central, since
diagDirectCom(estL, estL, signcom1);
#[ ]
LiNotArmLeg:=Filtered(quat1, x-> not (x in leg or x in arm));;
DiLiNotArmLeg:=ListI(LiNotArmLeg);;
diagDirectCom(estL, DiLiNotArmLeg, signcom1);
#[ ]
#Also, we notice that X' is a subgroup, since
diagDirectCom(estA, estA, signcom1);
#[ [ 16, -t_2*t_1*a_21*a_22 ], [ 16, t_2*t_1*a_21*a_22 ],
# [ 16, -t_2*t_1*a_21*a_22 ], [ 16, t_2*t_1*a_21*a_22 ] ]
evalResCom(last);
#0
#We first examine the action of X_16 onto all the other roots.
LiRem16:=ListRemI(quat1, Concatenation([16], arm, leg));;
diagDirectCom([16], Concatenation(LiRem16, estA, estL), signcom1);
#[ [ 26, s_13*s_16 ], [ 20, -s_3*s_16 ] ]
#Hence X_{16} would be a candidate for the leg, if X_3X_{13} would
#be a candidate for an arm, that is, if we would obtain a closed
#set by removing the roots 3 and 13. Now notice the following.
Filtered(signcom1, x-> 3 in x or 13 in x);
#[ [ 1, 13, 17 ], [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ],
# [ 3, 16, 20 ], [ 5, 13, -18 ], [ 6, 13, -19 ], [ 13, 16, -26 ] ]
#Hence the source is indeed closed. We apply the lemma.
diagDirectCom([3, 13], [16], signcom1);
#[ [ 20, t_16*s_3 ], [ 26, -t_16*s_13 ] ]
#We can then apply the reduction lemma. We reduce to the
#subquotient such that X_{16} is in the kernel, and X_3X_{13} is
#replaced by X_{3, 13}=\{x_{3, 13}(t) \mid t \in \F_q\}, where
#x_{3, 13}(t):=x_3(a_{26}t)x_{13}(a_{20}t).
st313:=[[s_3, a_26*s_1], [s_13, a_20*s_1]];;
#We are left with the following subgroups.
#(X_1'X_2')(X_{3, 13})(X_8X_9)(Y_1'Y_2')Z.
#We need to extend to the Y_1' and Y_2' in the usual way.
#In case of "nice" extensions, we notice a path like this.
#8--(12/18/19)--1/10/11--(12/18/19)--3/13--(14/15/17)--5/6/7--(14/15/17)--9
#We examine the action of 8, 9 and 3/13 on the other roots.
LiRem8:=ListRemI(quat1, Concatenation([8], arm, leg));;
diagDirectCom([8], Concatenation(LiRem8, estA, estL), signcom1);
#[ [ 19, -s_8*t_1*a_21 ], [ 18, s_8*t_1*a_22 ], [ 12, -s_8*t_1*a_26 ],
# [ 21, -s_8*t_1^2*a_22*a_26 ], [ 26, s_8*t_1^2*a_21*a_22 ],
# [ 22, s_8*t_1^2*a_21*a_26 ] ]
#So 8 interacts just with X_1'.
LiRem9:=ListRemI(quat1, Concatenation([9], arm, leg));;
diagDirectCom([9], Concatenation(LiRem9, estA, estL), signcom1);
#[ [ 17, t_2*s_9*a_20 ], [ 15, t_2*s_9*a_21 ], [ 14, t_2*s_9*a_22 ],
# [ 20, -t_2^2*s_9*a_21*a_22 ], [ 22, t_2^2*s_9*a_20*a_21 ],
# [ 21, t_2^2*s_9*a_20*a_22 ] ]
#So 9 interacts just with X_2'.
LiRem313:=ListRemI(quat1, Concatenation([3, 13], arm, leg));;
diagDirectCom([3, 13], Concatenation(LiRem313, estA, estL), signcom1);
#[ [ 15, -t_1*s_3*a_21 ], [ 14, t_1*s_3*a_22 ], [ 12, t_2*s_3*a_20 ],
# [ 20, s_3*s_16 ], [ 21, t_2*t_1*s_3*a_20*a_22 ],
# [ 22, -t_2*t_1*s_3*a_20*a_21 ], [ 19, t_2*s_13*a_21 ],
# [ 18, t_2*s_13*a_22 ], [ 17, -t_1*s_13*a_26 ], [ 26, -s_13*s_16 ],
# [ 21, -t_2*t_1*s_13*a_22*a_26 ], [ 22, -t_2*t_1*s_13*a_21*a_26 ],
# [ 26, -t_2*t_1*s_13*a_21*a_22 ], [ 26, t_2*t_1*s_13*a_21*a_22 ] ]
#Then we notice that apart from s_16, which is now in the kernel, the
#action of 3 and 13 is nontrivial just when hitting X_1' and X_2'. It
#could also combine the two of them.
#By the above, we can take X' as a candidate for the arm, and
#X_{3/13}X_8X_9 as a candidate for a leg. We get the following computation.
diagDirectCom(Concatenation(st313, [[s_8, s_8], [s_9, s_9]]), estA, signcom1);
#Reordered, this gives...
#[ 12, -s_8*t_1*a_26 ],[ 12, t_2*s_1*a_20*a_26 ],
#[ 18, s_8*t_1*a_22 ], [ 18, t_2*s_1*a_20*a_22 ],
#[ 19, -s_8*t_1*a_21 ], [ 19, t_2*s_1*a_20*a_21 ],
#[ 14, t_2*s_9*a_22 ], [ 14, t_1*s_1*a_22*a_26 ],
#[ 15, t_2*s_9*a_21 ], [ 15, -t_1*s_1*a_21*a_26 ],
#[ 17, t_2*s_9*a_20 ], [ 17, -t_1*s_1*a_20*a_26 ],
#[ 20, -t_2^2*s_9*a_21*a_22 ], [ 21, t_2^2*s_9*a_20*a_22 ],
#[ 22, t_2^2*s_9*a_20*a_21 ],
#[ 21, -s_8*t_1^2*a_22*a_26 ], [ 22, s_8*t_1^2*a_21*a_26 ],
#[ 26, s_8*t_1^2*a_21*a_22 ],
#[ 21, t_2*t_1*s_1*a_20*a_22*a_26 ], [ 21, -t_2*t_1*s_1*a_20*a_22*a_26 ],
#[ 22, -t_2*t_1*s_1*a_20*a_21*a_26 ], [ 22, -t_2*t_1*s_1*a_20*a_21*a_26 ],
# [ 26, -t_2*t_1*s_1*a_20*a_21*a_22 ], [ 26, t_2*t_1*s_1*a_20*a_21*a_22 ]
#(these last ones cancel!)
#Then the usual equation writes:
#\lambda([x_{3, 13}(s_1)x_8(s_8)x_9(s_9), x_{1, 5, 6}(t_1)x_{7, 10, 11}(t_2)])=
#=\phi(
#+s_1(c_{12, 18, 19}a_{20}t_2+c_{14, 15, 17}a_{26}t_1)
#+s_8(c_{12, 18, 19}t_1+a_{21}a_{22}a_{26}t_1^2)
#+s_9(c_{14, 15, 17}t_2+a_{20}a_{21}a_{22}t_2^2)
#)=1. (*)
#In the sequel, we call by \omega_{21, 22} the square root of a_{21}a_{22}.
#Let c_{12, 18, 19}=c_{14, 15, 17}=0. It is then easy to
#check, as usual, that (*) holds for every s iff t_1, t_2=0.
#It is also easy to check that s_1=0, s_8=a_{2}s, s_9=a_{26}s satisfy
#the above for every t. By the Reduction Lemma, we conclude X''=1 and
#Y''=X_{3, 13}. We finally reduce to an abelian subquotient. We get a family F_1 of the
#\chi_{b_{3, 13}}^{a_{20}, a_{21}, a_{22}, a_{26}}
#which consists of q(q-1)^4 characters of degree q^7.
#(CAREFUL WITH LABEL b SOMETIMES QUADRATIC!)
#Let a_{12, 18, 19}:=c_{12, 18, 19} \ne 0, and c_{14, 15, 17}=0. Equation (*) writes:
#=\phi(
#+s_1(c_{12, 18, 19}a_{20}t_2)
#+s_8(c_{12, 18, 19}t_1+a_{21}a_{22}a_{26}t_1^2)
#+s_9(a_{20}a_{21}a_{22}t_2^2)
#)=1.
#This holds for every s iff t_2=0, and t_1 is 0 or c_{12, 18, 19}/a_{21}a_{22}a_{26}.
#On the other hand, we write the above in this form,
#=\phi(
#+(c_{12, 18, 19}s_8t_1+a_{21}a_{22}a_{26}s_8t_1^2)
#+(c_{12, 18, 19}a_{20}s_1t_2+a_{20}a_{21}a_{22}s_9t_2^2)
#)=1.
#As usual, this gives s_1=\omega_{21, 22}s, s_9=c_{12, 18, 19}^2a_{20}s^2, and
#s_8=0 or s_8=a_{21}a_{22}a_{26}/c_{12, 18, 19}^2. Hence we have
#X''=\{x_{1, 10, 11}(\epsilon c_{12, 18, 19}/a_{21}a_{22}a_{26}) \mid \epsilon \in \F_2\},
#Y''=\{x_{3, 13}(\omega_{21, 22}s)x_8(\epsilon a_{21}a_{22}a_{26}/c_{12, 18, 19}^2)x_9(c_{12, 18, 19}^2a_{20}s^2) \mid s \in \F_q, \epsilon \in \F_2\}
#In this case, we get a family F_2 of characters, labelled by
#\chi_{b_{3, 9, 13}, d_{1, 10, 11}, d_8}^{a_{12, 18, 19}, a_{20}, a_{21}, a_{22}, a_{26}}
#which are 4q(q-1)^5 characters of degree q^7/2.
#The case c_{12, 18, 19} = 0, and a_{14, 15, 17}:=c_{14, 15, 17} \ne 0 is symmetric. Get
#X''=\{x_{5, 6, 7}(\epsilon c_{14, 15, 17}/a_{20}a_{21}a_{22}) \mid \epsilon \in \F_2\},
#Y''=\{x_{3, 13}(\omega_{21, 22}s)x_8(c_{14, 15, 17}^2a_{26}s^2)x_9(\epsilon a_{20}a_{21}a_{22}/c_{14, 15, 17}^2) \mid s \in \F_q, \epsilon \in \F_2\}
#We get a family F_3 of characters whose behavior is similar as for F_2, namely
#\chi_{b_{3, 8, 13}, d_{5, 6, 7}, d_9}^{a_{14, 15, 17}, a_{20}, a_{21}, a_{22}, a_{26}}
#again 4q(q-1)^5 characters of degree q^7/2.
#We finally assume:
#a_{12, 18, 19}:=c_{12, 18, 19} \ne 0, and a_{14, 15, 17}:=c_{14, 15, 17} \ne 0.
#We refer to (*). We solve for every s. Putting s_8=s_9=0, we first get that
#t_1=c_{12, 18, 19}a_{20}t and t_2=c_{14, 15, 17}a_{26}t for t \in \F_q. By
#putting (s_1, s_8, s_9)=(0, 1, 0) or (0, 0, 1), we obtain the same equation, namely
#t(1+a_{20}a_{21}a_{22}a_{26}t)=0. Hence we obtain
#X''=\{x_{1, 10, 11}(\epsilon c_{12, 18, 19}/(a_{21}a_{22}a_{26}))
#x_{5, 6, 7}(\epsilon c_{14, 15, 17}/(a_{20}a_{21}a_{22})) \mid \epsilon \in \F_2\}
#We then write (*) in this form, in order to solve for every t_1, t_2 \in \F_q,
#=\phi(
#+t_1(c_{14, 15, 17}a_{26}s_1+c_{12, 18, 19}s_8+a_{21}a_{22}a_{26}s_8t_1)
#+t_2(c_{12, 18, 19}a_{20}s_1+c_{14, 15, 17}s_9+a_{20}a_{21}a_{22}s_9t_2)
#)=1.
#We observe that this is equivalent to solve simultaneously the following two equations,
#c_{14, 15, 17}^2a_{26}^2s_1^2+c_{12, 18, 19}^2s_8^2=a_{21}a_{22}a_{26}s_8, and
#c_{12, 18, 19}^2a_{20}^2s_1^2+c_{14, 15, 17}^2s_9^2=a_{20}a_{21}a_{22}s_9.
#Multiplying by c_{14, 15, 17}^2a_{26}^2 the second equation and by
#c_{12, 18, 19}^2a_{20}^2 the first equation, and taking their sum, we get that
#s_1=\omega_{21, 22}s+c_{12, 18, 19}c_{14, 15, 17}s^2,
#s_8=c_{14, 15, 17}^2a_{26}s^2,
#s_9=\epsilon a_{20}a_{21}a_{22}/(c_{14, 15, 17}^2) + c_{12, 18, 19}^2a_{20}s^2.
#Hence we have that
#Y'':=\{x_{3, 13}(\omega_{21, 22}s+c_{12, 18, 19}c_{14, 15, 17}s^2)
#x_8(c_{14, 15, 17}^2a_{26}s^2)
#x_9(\epsilon a_{20}a_{21}a_{22}/(c_{14, 15, 17}^2) + c_{12, 18, 19}^2a_{20}s^2)
#\mid s \in \F_q, \epsilon \in \F_2\}.
#Notice that we could have put the parameter \epsilon into X_8 instead. We will then
#refer to the parameter d for that as d_{8, 9} in this case.
#Finally, in this case we also reduce to an abelian subquotient, and we get
#the last family F_4 of the irreducible characters
#\chi_{d_{1, 5, 6, 7, 10, 11}, b_{3, 8, 9, 13}, d_{8, 9}}^{a_{12, 18, 19}, a_{14, 15, 17}, a_{20}, a_{21}, a_{22}, a_{26}},
#which are 4q(q-1)^6 irreducible characters of degree q^7/2.
###################################
###########FOR [4,24,43]###########
###################################
test:=ll[5][1];;
quat1:=getRoots(test[5]);
#[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
# 21, 22, 23, 24 ]
#Now: a problem.
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 1, 5, 6, 7, 10, 11 #], [ 12, 14, 15, 17, 18, 19 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 17, 18, 19 ]
#WARNING!! Not the whole of the leg is central in the source!
#Just the one of the "triangle on top".
#In order to apply the Reduction Lemma as planned, we should go
#just triangle by triangle. But first, we notice that we can get
#rid of the following.
Filtered(signcom1, x-> 20 in x or -20 in x);
#[ [ 2, 20, 23 ], [ 3, 16, 20 ], [ 4, 20, 24 ], [ 5, 15, 20 ], [ 6, 14, 20 ]
# ]
Filtered(signcom1, x-> 2 in x or 4 in x);
#[ [ 1, 2, 7 ], [ 2, 3, 8 ], [ 2, 9, 13 ], [ 2, 14, 18 ], [ 2, 15, 19 ],
# [ 2, 20, 23 ], [ 3, 4, 9 ], [ 4, 5, 10 ], [ 4, 6, 11 ], [ 4, 8, -13 ],
# [ 4, 12, -17 ], [ 4, 20, 24 ] ]
#Hence:
# 1) If we remove 2 and 4, the root 20 would be central;
# 2) The only central hooks involving 20 and something else just play
#with 2 and 4, that in turn are not centrally hooked.
directCom([2, 4], [20], signcom1);
#[ 23*s_2*t_20, 24*t_20*s_4 ]
#We apply the Reduction Lemma with leg X_{20} and arm X_{2, 4}.
#We get to a subquotient that
#has X_{20} in the quotient, and where we replace X_2X_4 by
#X_{2, 4}=\{x_{2, 4}(t) \mid t \in \F_q\}, where
#x_{2, 4}:=x_2(a_{24}t)x_4(-a_{23}t)
st24:=[[t_2, a_24*s], [t_4, -a_23*s]];;
signcom2:=Filtered(signcom1, x-> not 20 in x);;
Length(last);
#38
#We first look at the triangle, such that the leg is central in the source.
#Namely, this is...
arm:=[1, 5, 6];;leg:=[17, 18, 19];;
matrixAction(leg, arm, signcom1);
#[ [ 0, a_21, a_22 ], [ -a_21, 0, -a_23 ], [ -a_22, -a_23, 0 ] ]
det(last);
#2*a_21*a_22*a_23
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_17, t_5*a_21+t_6*a_22 ], [ s_18, -t_1*a_21-t_6*a_23 ],
# [ s_19, -t_1*a_22-t_5*a_23 ] ]
#Let us first assume p>=3. Then we can apply the Reduction
#Lemma with arm [1, 5, 6] and leg [17, 18, 19].
#We produce new comrels.
signcom3:=Filtered(signcom2, x-> not(
(x[1] in arm and x[2] in leg) or (x[1] in leg and x[2] in arm)));;
Length(last);
#32
#Since...
Filtered(signcom1, x->12 in x{[1, 2]} or 14 in x{[1, 2]} or 15 in x{[1, 2]});
#[ [ 2, 14, 18 ], [ 2, 15, 19 ], [ 4, 12, -17 ], [ 5, 15, 20 ],
# [ 6, 14, 20 ], [ 7, 14, 21 ], [ 7, 15, 22 ], [ 10, 12, -21 ],
# [ 10, 15, 24 ], [ 11, 12, -22 ], [ 11, 14, 24 ] ]
#Then we can apply the Reduction Lemma with arm [7, 10, 11] and leg [12, 14, 15]
arm2:=[7, 10, 11];;leg2:=[12, 14, 15];;
matrixAction(leg2, arm2, signcom3);
#[ [ 0, a_21, a_22 ], [ -a_21, 0, -a_24 ], [ -a_22, -a_24, 0 ] ]
det(last);
#2*a_24*a_21*a_22
stabL:=stabAction(leg2, arm2, signcom3);
#[ [ s_12, t_10*a_21+t_11*a_22 ], [ s_14, -a_24*t_11-t_7*a_21 ],
# [ s_15, -a_24*t_10-t_7*a_22 ] ]
#Hence we can pply again the Reduction Lemma, this time
#with arm [7, 10, 11] and leg [12, 14, 15]. We get:
signcom4:=Filtered(signcom3, x-> not(
x[1] in arm or x[1] in arm2 or x[1] in leg or x[1] in leg2
or x[2] in arm or x[2] in arm2 or x[2] in leg or x[2] in leg2));;
Length(last);
#6
signcom4;
#[ [ 2, 3, 8 ], [ 2, 9, 13 ], [ 3, 4, 9 ], [ 4, 8, -13 ], [ 8, 16, 23 ],
# [ 9, 16, -24 ] ]
#Let us extend our character to a_{13} nontrivially. Then we have a circle
# 2/4 --(13)-- 8 --(23)-- 16 --(24)-- 9 --(13)-- 2/4.
#We take X_8X_9 as a candidate for a leg, and X_{2,4}X_{16} as a candidate
#for an arm.
diagDirectCom(Concatenation(st24, [[t_16, t_16]]), [[s_8, s_8], [s_9, s_9]], signcom1);
#[ [ 13, a_24*t*s_9 ], [ 13, s_8*t*a_23 ], [ 24, t_16*s_9 ], [ 23, -s_8*t_16 ] ]
#The equation writes:
#\lambda([x_{2, 4}(t)x_{16}(t_{16}), x_8(s_8)x_9(s_9)])=
#=\phi(
#a_{13}t(a_{24}s_9+a_{23}s_8)+t_{16}(a_{24}s_9-a_{23}s_8)
#)=1.
#Since p \ne 2, we have that the above is satisfied for
#every t, t_{16} if and only if s_8=s_9=0. Similarly, by writing
#=\phi(
#s_8(a_{13}a_{23}t-a_{23}t_{16})+s_9(a_{13}a_{24}t+a_{24}t_{16})
#)=1,
#the above holds for every s_8, s_9 if and only if t=t_{16}=0.
#In this case, we apply the Reduction Lemma with arm X''=X_{2,4}X_{16}
#and leg X_8X_9. We reduce of course to an abelian quotient, with the
#root 3 in direct product. We get the first family F_1 of the
#\chi_{b_3}^{a_{13}, a_{21}, a_{22}, a_{23}, a_{24}}
#which are q(q-1)^5 irreducible characters of degree q^9.
#Let us now assume that X_{13} is in the kernel of our character. Then...
signcom5:=Filtered(signcom4, x -> not (13 in x or -13 in x));
#[ [ 2, 3, 8 ], [ 3, 4, 9 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
#We first apply a Reduction Lemma with leg X_8X_9 and arm X_{16}.
directCom([8, 9], [16], signcom1);
#[ 23*s_8*t_16, -24*t_16*s_9 ]
#Then we reduce to a subquotient with X_{16} in the kernel, and X_8X_9 is replaced by
#X_{8, 9}=\{x_{8, 9}(t) \mid t \in \F_q\}, where
#x_{8, 9}(t)=x_8(a_{24}t)x_9(a_{23}t).
#We extend \lambda nontrivially onto a_{8, 9}. over X_{8, 9}. We get
diagDirectCom(st24, [[s_3, s_3]], signcom1);
#[ [ 8, a_24*t*s_3 ], [ 13, -a_24*t^2*s_3*a_23 ], [ 9, t*s_3*a_23 ] ]
Filtered(last, x-> not x[1]=13);
#[ [ 8, a_24*t*s_3 ], [ 9, t*s_3*a_23 ] ]
#Hence [x_{2, 4}(s), x_3(t)]=x_{8, 9}(st). We can apply the
#Reduction Lemma with arm X_{2, 4} and leg X_{3}.
#This gives a second family of irreducible characters F_2, which consists of the
#\chi^{a_{8, 9}, a_{21}, a_{22}, a_{23}, a_{24}},
#that is, (q-1)^5 irreducible characters of degree q^9.
#Finally, if X_{8, 9} is also in the kernel of our character, we
#have already an abelian subquotient modulo the kernel of such character.
#We get a last family of irreducible characters, namely F_3, of the
#\chi_{b_{2, 4}, b_3}^{a_{21}, a_{22}, a_{23}, a_{24}}
#which are q^2(q-1)^4 characters of degree q^8.
#Let us now assume that p=2. We compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_17, -t_19*a_23/a_21 ], [ t_18, -t_19*a_22/a_21 ], [ t_19, t_19 ] ]
el_old:=[t_19];;
el_new:=[-a_21*t];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_17, t*a_23 ], [ t_18, t*a_22 ], [ t_19, -t*a_21 ] ]
#Hence we have Y_1'=\{x_{17, 18, 19}(t) \mid t \in \F_q\}, where
#x_{17, 18, 19}(t):=x_{17}(a_{23}t)x_{18}(a_{22}t)x_{19}(a_{21}t)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_1, -t_6*a_23/a_21 ], [ t_5, -t_6*a_22/a_21 ], [ t_6, t_6 ] ]
ea_old:=[t_6];;
ea_new:=[-a_21*t];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_1, t*a_23 ], [ t_5, t*a_22 ], [ t_6, -t*a_21 ] ]
#Hence we have X_1'=\{x_{1, 5, 6}(t) \mid t \in \F_q\}, where
#x_{1, 5, 6}(t):=x_{1}(a_{23}t)x_{5}(a_{22}t)x_{6}(a_{21}t)
#Let us first extend to X_{17, 18, 19} trivially.
#First notice the following.
Filtered(signcom1, x-> 13 in x);
#[ [ 1, 13, 17 ], [ 2, 9, 13 ], [ 5, 13, -18 ], [ 6, 13, -19 ] ]
#Hence the root 13 is central once we remove 1, 5, 6. Moreover, notice that
Filtered(signcom1, x-> (1 in x or 5 in x or 6 in x) and (AbsInt(x[3]) in [17, 18, 19]));
#[ [ 1, 13, 17 ], [ 5, 13, -18 ], [ 6, 13, -19 ] ]
#We can then apply the Reduction Lemma with arm [1, 5, 6] and leg [13]. We get...
diagDirectCom([13], estA, signcom1);
#Reordered, this gives...
#[[ 17, -t*s_13*a_23 ], [ 18, t*s_13*a_22 ], [ 19, -t*s_13*a_21 ],
# [ 21, -t^2*s_13*a_22*a_23 ], [ 22, t^2*s_13*a_21*a_23 ], [ 23, t^2*s_13*a_21*a_22 ]]
#By applying \lambda, this gives...
#\lambda([x_{13}(s_{13}), x_{1, 5, 6}(t)])=
#=\phi(
a_{21}a_{22}a_{23}s_{13}t^2
#)
#=1.
#We then reduce to a subquotient without X_{1, 5, 6} and X_{13} .
#So far, the degree is q^4.
#We can then work with the following set.
signcoma1:=Filtered(signcom1, x -> not (1 in x or 5 in x or 6 in x));;
#We carry on with the following.
Filtered(signcoma1, x -> AbsInt(x[3]) in [17, 18, 19, 21, 22, 23, 24]);
#Reordered, this gives...
#[ 8, 16, 23 ], [ 9, 16, -24 ],
#[ 2, 14, 18 ], [ 2, 15, 19 ], [ 4, 12, -17 ],
# [ 7, 14, 21 ], [ 7, 15, 22 ], [ 10, 12, -21 ], [ 11, 12, -22 ], [ 11, 14, 24 ], [ 10, 15, 24 ],
# [ 7, 9, 17 ],[ 8, 10, 18 ],[ 8, 11, 19 ], ]
#[ 2, 20, 23 ], [ 4, 20, 24 ]
#Let us look at the relations involving 16, and 8, 9.
Filtered(signcoma1, x -> 16 in x or -16 in x);
#[ [ 3, 16, 20 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
#(We notice that the 20 is not here anymore). We then have:
Filtered(signcoma1, x -> 8 in x or 9 in x or -8 in x or -9 in x);
#[ [ 2, 3, 8 ], [ 2, 9, 13 ], [ 3, 4, 9 ], [ 4, 8, -13 ], [ 7, 9, 17 ],
# [ 8, 10, 18 ], [ 8, 11, 19 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
#Notice that the 13 has disappeared, as well as the 17, 18, 19, which is
#now in the kernel. This was the only thing stopping 8, 9 to be central, once
#we remove 16. By the above, also notice that the source is closed if we
#remove 16. We then apply the Reduction Lemma with arm X_{16} and leg X_8X_9.
diagDirectCom([8, 9], [16], signcom1);
#[ [ 23, s_8*t_16 ], [ 24, -t_16*s_9 ] ]
#Hence we apply the Reduction Lemma, getting to degree q^5, and to a
#subquotient such that it does not contain X_{16} and replaces X_8X_9
#with X_{8, 9}:=\{x_{8, 9}(t) \mid t \in \F_q\}, where
#x_{8, 9}(t)=x_8(a_{24}t)x_9(a_{23}t).
st89:=[[t_8, a_24*t], [t_9, a_23*t]];;
#We also notice that the roots [12, 14, 15] are now central
#once we remove 7, 10, 11, since the other hurdle would have been 2/4,
#commutating them to 17, 18, 19, which is not there anymore. We then
#apply the Reduction Lemma with arm 7, 10, 11 and leg 12, 14, 15.
arm2:=[7, 10, 11];;leg2:=[12, 14, 15];;
matrixAction(leg2, arm2, signcom1);
#[ [ 0, a_21, a_22 ], [ -a_21, 0, -a_24 ], [ -a_22, -a_24, 0 ] ]
#Hence we apply the Reduction Lemma getting arm
#X_{7, 10, 11}=\{x_{7, 10, 11}(t) \mid t \in \F_q\}, where
#x_{7, 10, 11}(t)=x_7(a_{24}t)x_{10}(a_{22}t)x_{11}(a_{21}t), and leg
#X_{12, 14, 15}=\{x_{12, 14, 15}(t) \mid t \in \F_q\}, where
#x_{12, 14, 15}(t)=x_{12}(a_{24}t)x_{14}(a_{22}t)x_{15}(a_{21}t).
stA2:=[[t_7, a_24*t], [t_10, a_22*t], [t_11, a_21*t]];;
stL2:=[[t_12, a_24*t], [t_14, a_22*t], [t_15, a_21*t]];;
#The degree is now q^7.
#We are now left with fewer relations, namely the following.
signcoma2:=Filtered(signcoma1, x -> not 20 in x);
#[ [ 2, 3, 8 ], [ 2, 9, 13 ], [ 2, 14, 18 ], [ 2, 15, 19 ], [ 3, 4, 9 ],
# [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ], [ 4, 8, -13 ], [ 4, 12, -17 ],
# [ 7, 9, 17 ], [ 7, 14, 21 ], [ 7, 15, 22 ], [ 8, 10, 18 ], [ 8, 11, 19 ],
# [ 8, 16, 23 ], [ 9, 16, -24 ], [ 10, 12, -21 ], [ 10, 15, 24 ],
# [ 11, 12, -22 ], [ 11, 14, 24 ] ]
signcoma3:=Filtered(signcoma2, x -> not(x[1] in arm2 and x[2] in leg2));
#[ [ 2, 3, 8 ], [ 2, 9, 13 ], [ 2, 14, 18 ], [ 2, 15, 19 ], [ 3, 4, 9 ],
# [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ], [ 4, 8, -13 ], [ 4, 12, -17 ],
# [ 7, 9, 17 ], [ 8, 10, 18 ], [ 8, 11, 19 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
signcoma4:=Filtered(signcoma3, x -> not(13 in x or -13 in x));
#[ [ 2, 3, 8 ], [ 2, 14, 18 ], [ 2, 15, 19 ], [ 3, 4, 9 ], [ 3, 7, -12 ],
# [ 3, 10, 14 ], [ 3, 11, 15 ], [ 4, 12, -17 ], [ 7, 9, 17 ], [ 8, 10, 18 ],
# [ 8, 11, 19 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
signcoma5:=Filtered(signcoma4, x -> not(17 in x or 18 in x or 19 in x
or -17 in x or -18 in x or -19 in x));
#[ [ 2, 3, 8 ], [ 3, 4, 9 ], [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ],
# [ 8, 16, 23 ], [ 9, 16, -24 ] ]
signcoma6:=Filtered(signcoma5, x -> not(16 in x));
#[[ [ 2, 3, 8 ], [ 3, 4, 9 ], [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ] ]
#We notice that now 8/9 is central. Moreover, of course 3 interacts
#just with 2/4 and 7/10/11, which do not interact with each other.
#We have X_3 as a candidate for a leg, and X_{2, 4} and X_{7, 10, 11}
#as a candidate for an arm. We get...
st24:=[[t_2, a_24*t_1], [t_4, -a_23*t_1]];;
stA2:=[[t_7, a_24*t_2], [t_10, a_22*t_2], [t_11, a_21*t_2]];;
diagDirectCom([3], Concatenation(st24, stA2), signcom1);;
Filtered(last, x -> not x[1] in [13, 17, 18, 19]);
#Reordered, this gives...
#[ [ 8, -a_24*t_1*s_3 ], [ 9, -t_1*s_3*a_23 ],
#[ 12, -a_24*t_2*s_3 ], [ 14, t_2*s_3*a_22 ], [ 15, t_2*s_3*a_21 ],
#[ 21, -a_24*t_2^2*s_3*a_22 ], [ 22, -a_24*t_2^2*s_3*a_21 ], [ 24, -t_2^2*s_3*a_21*a_22 ]]
#Let us extend to c_{8, 9} and to c_{12, 14, 15} in the usual way. We can then write:
#\lambda([x_3(s_3), x_{2, 4}(t_1)x_{7, 10, 11}(t_2), x_2(s_2)x_3(s_3)x_5(s_5)])=
#=\phi(
s_3(
c_{8, 9}t_1+c_{12, 14, 15}t_2+a_{21}a_{22}a_{24}t_2^2
)
#)=1.
#Let us first assume that c_{8, 9}=c_{12, 14, 15}=0. Hence the Reduction Lemma
#applies with leg X_3 and arm X_{7, 10, 11}. We reduce finally to an abelian
#subquotient, the degree now being q^8. We get a first family F_1, of the
#\chi_{b_{2, 4}}^{a_{21}, a_{22}, a_{23}, a_{24}}
#which consists of q(q-1)^4 characters of degree q^8.
#Then we assume that c_{8, 9}=0 and a_{12, 14, 15}:=c_{12, 14, 15}\ne 0.
#In this case, we can apply the Reduction Lemma with leg a complement of
#X_3':=\{1, x_3(a_{21}a_{22}a_{24}/(c_{12, 14, 15}^2)), and arm a complement of
#X_{7, 10, 11}':=\{1, x_{7, 10, 11}(c_{12, 14, 15}/(a_{21}a_{22}a_{24}))\}.
#Hence we have a family F_2 of characters, namely
#\chi_{b_{2, 4}, d_3, d_{7, 10, 11}}^{a_{12, 14, 15}, a_{21}, a_{22}, a_{23}, a_{24}}
#which are 4q(q-1)^5 characters of degree q^8/2.
#We then assume a_{8, 9}:=c_{8, 9}\ne 0, and b_{12, 14, 15}:=c_{12, 14, 15}
# is arbitrary. Then we can apply
#the Reduction Lemma with leg X_3 and arm a complement of
#X_{2, 4, 7, 10, 11}=\{x_{2, 4}((c_{12, 14, 15}t_2+t^2*a_{21}a_{22}a_{24})/(c_{8, 9}^2))
#x_{7, 10, 11}(t) \mid t \in \F_q\}. Hence we get a family F_3 of the
#\chi_{b_{2, 4, 7, 10, 11}}^{a_{8, 9}, b_{12, 14, 15}, a_{21}, a_{22}, a_{23}, a_{24}}
#which are q^2(q-1)^5 characters of degree q^8.
#Let us extend to a_{17, 18, 19}:=c_{17, 18, 19} nontrivially. Notice the following.
Filtered(signcom1, x -> AbsInt(x[3]) in [17, 18, 19, 21, 22, 23, 24]);
#Reordered, and removing the ones of [1, 5, 6] interacting with [17, 18, 19],
#this gives...
#[ [ 1, 13, 17 ], [ 5, 13, -18 ], [ 6, 13, -19 ],
# [ 4, 12, -17 ], [ 2, 14, 18 ], [ 2, 15, 19 ],
# [ 7, 14, 21 ], [ 7, 15, 22 ], [ 10, 12, -21 ], [ 10, 15, 24 ], [ 11, 12, -22 ],
# [ 11, 14, 24 ] ]
#[ 8, 10, 18 ], [ 8, 16, 23 ], [ 8, 11, 19 ], [ 9, 16, -24 ], [ 7, 9, 17 ],
#[ 2, 20, 23 ], [ 4, 20, 24 ],
#We want to proceed in the above order indeed. (REALLY?)
#First notice the following.
Filtered(signcom1, x-> 13 in x or -13 in x);
#[ [ 1, 13, 17 ], [ 2, 9, 13 ], [ 5, 13, -18 ], [ 6, 13, -19 ] ]
#Hence the root 13 is central once we remove 1, 5, 6. Moreover, notice that
Filtered(signcom1, x-> (1 in x or 5 in x or 6 in x) and (AbsInt(x[3]) in [17, 18, 19]));
#[ [ 1, 13, 17 ], [ 2, 9, 13 ], [ 4, 8, -13 ], [ 5, 13, -18 ], [ 6, 13, -19 ] ]
#We can then apply the Reduction Lemma with arm [1, 5, 6] and leg [13]. We get...
diagDirectCom([13], estA, signcom1);
#Reordered, this gives...
#[[ 17, -t*s_13*a_23 ], [ 18, t*s_13*a_22 ], [ 19, -t*s_13*a_21 ],
# [ 21, -t^2*s_13*a_22*a_23 ], [ 22, t^2*s_13*a_21*a_23 ], [ 23, t^2*s_13*a_21*a_22 ]]
#By applying \lambda, this gives...
#\lambda([x_{13}(s_{13}), x_{1, 5, 6}(t)])=
#=\phi(
c_{17, 18, 19}s_{13}t+a_{21}a_{22}a_{23}s_{13}t^2
#)
#=1.
#We then reduce to a subquotient with X_{1, 5, 6}' (resp. X_{13}') in place of
#X_{1, 5, 6} (resp. X_{13}), where
#X_{1, 5, 6}'=\{1, x_{1, 5, 6}(c_{17, 18, 19}/(a_{21}a_{22}a_{23}))\}, and
#X_{13}'=\{1, x_{13}(a_{21}a_{22}a_{23}/(c_{17, 18, 19}^2))\}.
#So far, the degree is q^4/2.
st156:=[[t_1, s_50/(a_21*a_22)], [t_5, s_50/(a_21*a_23)],
[t_6, s_50/(a_22*a_23)]];;
#Now, notice the following.
Filtered(signcom1, x-> (12 in x or 14 in x or 15 in x));
#Reordered, this gives:
#[ [ 4, 12, -17 ], [ 2, 14, 18 ], [ 2, 15, 19 ],
#[ 7, 14, 21 ], [ 7, 15, 22 ], [ 10, 12, -21 ], [ 10, 15, 24 ], [ 11, 12, -22 ], [ 11, 14, 24 ]
#[ 1, 8, 12 ], [ 3, 10, 14 ], [ 3, 11, 15 ],
#[ 5, 15, 20 ], [ 6, 14, 20 ],
#CHECK THE BIT IN BETWEEN THE (*****) CAREFULLY!!!
#(*****)
#IMPORTANT: NOTICE THAT
#[ [ 2, 20, 23 ], [ 3, 16, 20 ], [ 4, 20, 24 ], [ 5, 15, 20 ], [ 6, 14, 20 ] ]
#HENCE 20 ACTS JUST WITH 2 AND 4, WHICH WE ARE GOING TO REMOVE.
#Hence 12, 14, 15 is central once we remove 2/4, 7, 10, 11.
#Moreover, notice the following.
Filtered(signcom1, x-> AbsInt(x[3]) in [2, 4, 7, 10, 11]);
#[ [ 1, 2, 7 ], [ 4, 5, 10 ], [ 4, 6, 11 ] ]
#Thus since 2/4 are removed as well, the source is a closed set.
#All of the above imply that the conditions (1)--(4) of the Reduction Lemma
#are satisfied with X_{12}X_{14}X_{15} as candidate for a leg, and
#X_{2/4}X_7X_{10}X_{11} as candidate for an arm. We get:
st24:=[[t_2, a_24*t], [t_4, -a_23*t]];;
diagDirectCom([12, 14, 15], Concatenation(st24, [[t_7, t_7], [t_10, t_10], [t_11, t_11] ]),
signcom1);
#[ [ 22, t_11*s_12 ], [ 21, t_10*s_12 ], [ 17, -t_1*s_12*a_23 ],
# [ 24, -t_11*s_14 ], [ 21, -s_14*t_7 ], [ 18, -a_24*s_14*t_1 ],
# [ 24, -t_10*s_15 ], [ 22, -s_15*t_7 ], [ 19, -a_24*s_15*t_1 ] ]
#This writes:
#\lambda([x_{12}(s_{12})x_{14}(s_{14})x_{15}(s_{15}),
#x_{2, 4}(t)x_7(t_7)x_{10}(t_{10})x_{11}(t_{11})])=
#=\lambda(x_{17}(a_{23}s_{12}t_1)x_{18}(a_{24}s_{14}t_1)x_{19}(a_{24}s_{15}t_1)
#\phi(
#+s_{12}(a_{21}t_{10}+a_{22}t_{11})
#+s_{14}(a_{21}t_{7}+a_{24}t_{11})
#+s_{15}(a_{22}t_{7}+a_{24}t_{10})
#)=1.
#It is easy that in this case the Reduction Lemma applies with
#arm X_{7, 10, 11}:=\{x_{7, 10, 11}(t) \mid t \in \F_q\}, where
#x_{7, 10, 11}(t)=x_7(a_{24})x_{10}(a_{22})x_{11}(a_{21}), and
#the leg is the whole of X_{12}X_{14}X_{15}.
#The degree now goes up straight to q^7/2.
stA2:=[[t_7, a_24*t], [t_10, a_22*t], [t_11, a_21*t]];;
#(*****)
#Observe now that since...
Filtered(signcom1, x -> 3 in x);
#[ [ 2, 3, 8 ], [ 3, 4, 9 ], [ 3, 7, -12 ], [ 3, 10, 14 ], [ 3, 11, 15 ],
# [ 3, 16, 20 ] ]
#And all of 2, 4, 12, 14, 15, 20 are now gone, then X_3 is a central root.
#We finally notice that
Filtered(signcom1, x -> (8 in x or 9 in x or -8 in x or -9 in x or 7 in x or 10 in x or 11 in x or -7 in x or -10 in x or -11 in x) and (not(2 in x or 4 in x or 12 in x or 14 in x or 15 in x or 20 in x)));
#[ [ 3, 7, -12 ], [ 5, 9, -14 ], [ 5, 11, 16 ], [ 6, 9, -15 ], [ 6, 10, 16 ],
# [ 7, 9, 17 ], [ 8, 10, 18 ], [ 8, 11, 19 ], [ 8, 16, 23 ], [ 9, 16, -24 ] ]
#(notice that [ 5, 11, 16 ] and [ 6, 10, 16 ] collapse)
#Hence we have a circle of the form
#8 --(23)-- 16 --(24)-- 9 --(17/18/19)-- 7/10/11 --(17/18/19)-- 8.
#Notice the following.
diagDirectCom([8, 9], [1, 3, 5, 6, 13, 17, 18, 19, 21, 22, 23, 24], signcom1);
#[ [ 12, -s_8*t_1 ], [ 15, t_6*s_9 ], [ 14, t_5*s_9 ], [ 20, -t_5*t_6*s_9 ] ]
#...and since 12, 15, 16, 20 are in the kernel, that means that X_8X_9 is central
#once we remove 7/10/11 and 16. Moreover, we get
Filtered(signcom1, x -> AbsInt(x[3]) in [7, 10, 11, 16]);
#[ [ 1, 2, 7 ], [ 4, 5, 10 ], [ 4, 6, 11 ], [ 5, 11, 16 ], [ 6, 10, 16 ] ]
#Hence as 2, 4 are removed and the other relations collapse, we have that
#the source is closed.
#We have X_8X_9 as candidate for a leg, and X_{7, 10, 11}X_{16}
#as candidate for an arm. We get...
diagDirectCom([8, 9], Concatenation(stA2, [[t_16, t_16]]), signcom1);
#[ [ 23, s_8*t_16 ], [ 19, s_8*t*a_21 ], [ 18, s_8*t*a_22 ], [ 24, -t_16*s_9 ],
# [ 17, -a_24*t*s_9 ] ]
#This writes...
#\lambda([x_8(s_8)x_9(s_9), x_{7, 10, 11}(t)x_{16}(t_{16})])=
#\lambda(x_{17}(a_{24}s_9t)x_{18}(a_{22}s_8t)x_{19}(a_{21}s_8t))
#\phi(
a_{23}s_8t_{16}+a_{24}s_9t_{16}
#)=1.
#As c_{17, 18, 19}\ne 0, we have that the whole arm is X_{7, 10, 11}X_{16}, and
#the whole leg is X_8X_9. We finally get down to an abelian subquotient,
#and the degree now reached q^9/2.
#This means that we have the final, following family F_4, of the
#\chi_{d_{1, 5, 6}, b_3, d_{13}}^{a_{17, 18, 19}, a_{21}, a_{22}, a_{23}, a_{24}},
#which are 4q(q-1)^5 irreducible characters of degree q^9/2.
###################################
###########FOR [5,18,18]###########
###################################
test:=ll[6][1];;
quat1:=getRoots(test[5]);
#[ 2, 3, 4, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 24, 25 ]
z1:=cenNR(test[5]);
#[ 17, 18, 19, 24, 25 ]
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 2, 4, 7, 10, 11, 16 ],
# [ 8, 9, 12, 14, 15, 20 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 8, 9, 12, 14, 15, 20 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, 0, 0, a_18, a_19, 0 ], [ 0, 0, -a_17, 0, 0, -a_24 ],
# [ 0, a_17, 0, 0, 0, a_25 ], [ -a_18, 0, 0, 0, -a_24, 0 ],
# [ -a_19, 0, 0, -a_24, 0, 0 ], [ 0, -a_24, -a_25, 0, 0, 0 ] ]
det(last);
#4*a_18*a_19*a_24^2*a_17*a_25
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_8, a_18*t_10+a_19*t_11 ], [ s_9, -a_24*t_16-t_7*a_17 ],
# [ s_12, t_4*a_17+t_16*a_25 ], [ s_14, -a_18*t_2-a_24*t_11 ],
# [ s_15, -a_19*t_2-a_24*t_10 ], [ s_20, -a_24*t_4-t_7*a_25 ] ]
#If p>=3, then we can apply the Reduction Lemma with
#arm [ 2, 4, 7, 10, 11, 16 ] and leg [ 8, 9, 12, 14, 15, 20 ].
#Let us have a look at the rest of the commutator relations.
signcom0:=Filtered(signcom1, x-> not(AbsInt(x[1]) in Concatenation(arm, leg) or AbsInt(x[2]) in Concatenation(arm, leg) or AbsInt(x[3]) in Concatenation(arm, leg)));
#[ ]
#In the case of p odd prime, we get just one family
#\chi_{b_3}^{a_{17}, a_{18}, a_{19}, a_{24}, a_{25}}
#of q(q-1)^5 irreducible characters of degree q^6.
#Let us now suppose p=2. We compute the stabilizer in arm and leg.
#Stab in leg:
stL:=solveStab(arm, leg, signcom1, 2);
#[ [ t_8, -a_24*t_15/a_18 ], [ t_9, -t_20*a_25/a_17 ],
# [ t_12, -a_24*t_20/a_17 ], [ t_14, -a_19*t_15/a_18 ], [ t_15, t_15 ],
# [ t_20, t_20 ] ]
el_old:=[t_15, t_20];; el_new:=[-a_18*t_1, -a_17*t_2];;
estL:=evalStabAct(stL, el_old, el_new);
#[ [ t_8, a_24*t_1 ], [ t_9, t_2*a_25 ], [ t_12, a_24*t_2 ],
# [ t_14, a_19*t_1 ], [ t_15, -a_18*t_1 ], [ t_20, -t_2*a_17 ] ]
#Hence we have Y'=Y_1'Y_2'=
#\{x_{8, 14, 15}(t_1)x_{9, 12, 20}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{8, 14, 15}(t_1):=x_{8}(a_{24}*t_1)x_{14}(a_{19}t_1)x_{15}(a_{18}t_1)
#x_{9, 12, 20}(t_2):=x_{9}(a_{25}*t_2)x_{12}(a_{24}t_2)x_{20}(a_{17}t_2)
#Stab in arm:
stA:=solveStab(leg, arm, signcom1, 2);
#[ [ t_2, -a_24*t_11/a_18 ], [ t_4, -t_16*a_25/a_17 ],
# [ t_7, -a_24*t_16/a_17 ], [ t_10, -a_19*t_11/a_18 ], [ t_11, t_11 ],
# [ t_16, t_16 ] ]
ea_old:=[t_11, t_16];; ea_new:=[-a_18*t_1, -a_17*t_2];;
estA:=evalStabAct(stA, ea_old, ea_new);
#[ [ t_2, a_24*t_1 ], [ t_4, t_2*a_25 ], [ t_7, a_24*t_2 ],
# [ t_10, a_19*t_1 ], [ t_11, -a_18*t_1 ], [ t_16, -t_2*a_17 ] ]
#Thus we get X'=X_1'X_2'=
#\{x_{2, 10, 11}(t_1)x_{4, 7, 16}(t_2) \mid t_1, t_2 \in \F_q\}, where
#x_{2, 10, 11}(t_1):=x_{2}(a_{24}*t_1)x_{10}(a_{19}t_1)x_{11}(a_{18}t_1)
#x_{4, 7, 16}(t_2):=x_{4}(a_{25}*t_2)x_{7}(a_{24}t_2)x_{16}(a_{17}t_2)
#Notice that Y is now central, since
diagDirectCom(estL, estL, signcom1);
#[ ]
LiNotArmLeg:=Filtered(quat1, x-> not (x in leg or x in arm));;
DiLiNotArmLeg:=ListI(LiNotArmLeg);;
diagDirectCom(estL, DiLiNotArmLeg, signcom1);
#[ ]
#Also, we notice that X' is a subgroup, since
diagDirectCom(estA, estA, signcom1);
#[ ]
#We want to see the behavior of X_3, the heart, with the other bits of the quotient.
LiRem3:=ListRemI(quat1, Concatenation([3], arm, leg));;
diagDirectCom([3], Concatenation(LiRem3, estA, estL), signcom1);
#Reordered, gives...
#[ 8, -a_24*t_1*s_3 ],[ 14, a_19*t_1*s_3 ], [ 15, -a_18*t_1*s_3 ],
#[ 18, -a_19*a_24*t_1^2*s_3 ],[ 19, a_18*a_24*t_1^2*s_3 ],[ 24, a_18*a_19*t_1^2*s_3 ],
#[ 9, t_2*s_3*a_25 ], [ 12, -a_24*t_2*s_3 ], [ 20, -t_2*s_3*a_17 ],
#[ 17, -a_24*t_2^2*s_3*a_25 ], [ 24, t_2^2*s_3*a_17*a_25 ], [ 25, a_24*t_2^2*s_3*a_17 ]
#We have that [3] interacts with both X_1' and X_2', separately.
#Thus we can take [3] as candidate for a leg, and X_1'X_2' as
#candidate for an arm. Let us compute that alltogether.
diagDirectCom([3], estA, signcom1);
#Reordered, gives... (basically the expression obtained before)
#[ 8, -a_24*t_1*s_3 ], [ 14, a_19*t_1*s_3 ], [ 15, -a_18*t_1*s_3 ],
#[ 18, -a_19*a_24*t_1^2*s_3 ], [ 19, a_18*a_24*t_1^2*s_3 ], [ 24, a_18*a_19*t_1^2*s_3 ],
#[ 9, t_2*s_3*a_25 ], [ 12, -a_24*t_2*s_3 ], [ 20, -t_2*s_3*a_17 ],
#[ 17, -a_24*t_2^2*s_3*a_25 ], [ 24, t_2^2*s_3*a_17*a_25 ], [ 25, a_24*t_2^2*s_3*a_17 ]
#This writes:
#\lambda([x_{3}(s_3), x_{2, 10, 11}(t_1)x_{4, 7, 16}(t_2)])=
#=\phi(
#s_3t_1(+c_{8, 14, 15}+a_{18}a_{19}a_{24}t_1)
#s_3t_2(+c_{9, 12, 20}+a_{17}a_{24}a_{25}t_2)
#)=1. (*)
#Now we recall a result, to label as LEMMA. Suppose we have
#the following equation on the field with char=2.
#(at_1+bt_2)^2+ct_1+dt_2=0.
# - Let us suppose that ad \ne bc. Then all the solutions are
#t_1=g_1(t)=g_1^{a, b, c, d}(t), t_2=g_2(t)=g_2^{a, b, c, d}(t) for t \in \F_q, where
#g_1(t)=\frac{t(d+bt)}{ad+bc}, and g_2(t)=\frac{t(c+at)}{ad+bc}.
# - Let us now suppose that ad=bc. Then all the solutions are
#t_1=h_1(t)=h_1^{a, b, c, d}(t) and (t_2=h_2(t, \eps)=h_2^{a, b, c, d}(t, \eps)
#for t \in \F_q and \eps \in \F_2, where
#h_1(t)=bt and h_2(t, \eps)=at+\eps d/(b^2).
#Notice that there are 2q solutions if c \ne 0 and d \ne 0, and
#q solutions if c=d=0.
#Let us first find the arm, that is, we want to solve (*)
#for every s \in \F_q. Let \omega_{18, 19, 24} be such that
#\omega_{18, 19, 24}^2=a_{18}a_{19}a_{24}, similarly for \omega_{17, 24, 25}.
#Equation (*) writes
#(\omega_{18, 19, 24}t_1+\omega_{17, 24, 25}t_2)^2+c_{8, 14, 15}t_1+
#+c_{9, 12, 20}t_2=0.
#We apply the LEMMA above.
#Let us first suppose that
#c_{8, 14, 15}/\omega_{18, 19, 24} \ne c_{9, 12, 20}/\omega_{17, 24, 25}, equivalent to
#c_{8, 14, 15}^2/(a_{18}a_{19}a_{24}) \ne c_{9, 12, 20}^2/(a_{17}a_{24}a_{25}). Then
#X''=\{x_{2, 4, 7, 10, 11, 16}(t) \mid t \in \F_q\}, where
#x_{2, 4, 7, 10, 11, 16}(t):=x_{2, 10, 11}(g_1(t))x_{4, 7, 16}(g_2(t))
#and g_1(t), g_2(t) are as above. By the Reduction Lemma, we necessarily
#have that Y''=1.
#In this case, we get a family F_1 of the
#\chi_{b_{2, 4, 7, 10, 11, 16}}^{c_{8, 14, 15}^a, c_{9, 12, 20}^a, a_{17}, a_{18}, a_{19}, a_{24}, a_{25}}
#which are q^2(q-1)^6 characters of degree q^5.
#(CAREFUL WITH LABEL b SOMETIMES QUADRATIC!)
#Let us then suppose that
#c_{8, 14, 15}^2/(a_{18}a_{19}a_{24}) = c_{9, 12, 20}^2/(a_{17}a_{24}a_{25}). (**)
#Then X''=\{x_{2, 4, 7, 10, 11, 16}(t, \eps) \mid t \in \F_q\}, where
#x_{2, 4, 7, 10, 11, 16}(t, \eps):=x_{2, 10, 11}(h_1(t))x_{4, 7, 16}(h_2(t, \eps))
#and h_1(t), h_2(t, \eps) are as above. We have |X''|=2q if c_{8, 14, 15} \ne 0
#and c_{9, 12, 20}\ne 0, and |X''|=q if c_{8, 14, 15}=c_{9, 12, 20}=0. We
#solve (*) for every t_1, t_2 \in \F_q in order to find Y''.
#Let us assume (**) and c_{8, 14, 15} \ne 0, hence c_{9, 12, 20}\ne 0. Then
#(*) yields s_3=0 or s_3 is the simultaneous solution of these two following equations,
#s_3c_{8, 14, 15}^2=a_{18}a_{19}a_{24} and s_3c_{9, 12, 20}^2=a_{17}a_{24}a_{25}.
#By (**), those two equations are dependent, hence we get another vaue of s_3. This yields
#Y''=\{1, x_3(a_{18}a_{19}a_{24}/(c_{8, 14, 15}^2)).
#This gives the family F_3, parameterized by the
#\chi_{b_{2, 4, 7, 10, 11, 16}, d_{2, 4, 7, 10, 11, 16}, d_3}^{c_{8, 14, 15}^b, c_{9, 12, 20}^b, a_{17}, a_{18}, a_{19}, a_{24}, a_{25}}
#which are 4q(q-1)^6 characters of degree q^5/2.
#Finally, if c_{8, 14, 15}=c_{9, 12, 20}=0, the Reduction Lemma implies
#|Y''|=1 since |X''|=q. Hence we get the last family F_3, consisting of the
#\chi_{b_{2, 4, 7, 10, 11, 16}}^{a_{17}, a_{18}, a_{19}, a_{24}, a_{25}}
#that is, q(q-1)^5 characters of degree q^5.
###################################
###########FOR [6,19,20]###########
###################################
test:=ll[7][1];;
quat1:=getRoots(test[5]);
#[ 2, 3, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, #24, 25 ]
z1:=cenNR(test[5]);
#[ 13, 17, 18, 19, 24, 25 ]
suggestArmLeg(test);
#[ 1, "ABELIAN+CLOSE+NORMAL:arm+leg", [ 2, 4, 7, 10, 11, 16 ],
# [ 8, 9, 12, 14, 15, 20 ] ]
arm:=last[3];;leg:=last2[4];;
signcom1:=getLie(test, signcomrel);;
#Check leg central in the source.
cenNRleg(signcom1, leg, arm);
#[ 8, 9, 12, 14, 15, 20 ]
#Then it is.
matrixAction(leg, arm, signcom1);
#[ [ 0, a_13, 0, a_18, a_19, 0 ], [ -a_13, 0, -a_17, 0, 0, -#a_24 ],
# [ 0, a_17, 0, 0, 0, a_25 ], [ -a_18, 0, 0, 0, -a_24, 0 ],
# [ -a_19, 0, 0, -a_24, 0, 0 ], [ 0, -a_24, -a_25, 0, 0, 0 ] ]
det(last);
#-a_13^2*a_24^2*a_25^2+4*a_17*a_18*a_19*a_24^2*a_25
stabL:=stabAction(leg, arm, signcom1);
#[ [ s_8, t_4*a_13+t_10*a_18+t_11*a_19 ], [ s_9, -t_2*a_13-#t_7*a_17-t_16*a_24 ]
# , [ s_12, t_4*a_17+t_16*a_25 ], [ s_14, -t_2*a_18-#t_11*a_24 ],
# [ s_15, -t_2*a_19-t_10*a_24 ], [ s_20, -t_4*a_24-t_7*a_25 ] ]
#Let p=2. Then the determinant is zero. We observe that
#just one root is left now, namely 3. We then get the family
#\chi_{b_3}^{a_{13}, a_{17}, a_{18}, a_{19}, a_{24}, a_{25}}
#of q(q-1)^6 irreducible characters of degree q^6.
#Let us now assume that p \ge 3.
#Simplified determinant: -a_13^2*a_25+4*a_17*a_18*a_19.
#Let us assume a_{25}^*:=a_{25} \ne 4*a_17*a_18*a_19/(a_{13}^2).
#Since the determinant is nonzero, we apply the Reduction
#Lemma, and we have the family F1, that consists of the
#\chi_{b_3}^{a_{13}, a_{17}, a_{18}, a_{19}, a_{24}, a_{25}^*}
#which are q(q-1)^5(q-2) irreducible characters of degree q^6.
#From now on, we assume that a_{25} = 4a_17a_18a_19/(a_{13}^2).
#We need to be more explicit on stabActions for arms and legs,
#and we define the old expression a_{23} to be
#substituted by the new expression -a_{15}a_{17}a_{20}/(a_{13}#a_{14}).
e_old:=[a_25];; e_new:=[4*a_17*a_18*a_19/((a_13)^2)];;
#Leg:
stabL:=stabAction(arm, leg, signcom1);
#[ [ s_2, t_9*a_13+t_14*a_18+t_15*a_19 ],
# [ s_4, -t_8*a_13-t_12*a_17+t_20*a_24 ], [ s_7, t_9*a_17+t_20*a_25 ],
# [ s_10, -t_8*a_18+t_15*a_24 ], [ s_11, -t_8*a_19+t_14*a_24 ],
# [ s_16, t_9*a_24-t_12*a_25 ] ]
estabL:=evalStabAct(stabL, e_old, e_new);
#[ [ s_2, t_9*a_13+t_14*a_18+t_15*a_19 ],
# [ s_4, -t_8*a_13-t_12*a_17+t_20*a_24 ],
# [ s_7, (t_9*a_13^2*a_17+4*t_20*a_17*a_18*a_19)/a_13^2 ],
# [ s_10, -t_8*a_18+t_15*a_24 ], [ s_11, -t_8*a_19+t_14*a_24 ],
# [ s_16, (t_9*a_13^2*a_24-4*t_12*a_17*a_18*a_19)/a_13^2 ] ]
#Check this.
e1_old:=[t_8, t_9, t_12, t_14, t_15, t_20];;
e1_new:=[2*a_17*a_24*t,
Value(-a_13*a_25*t, e_old, e_new),
-a_13*a_24*t,
2*a_17*a_19*t, 2*a_17*a_18*t, a_13*a_17*t];;
evalStabAct(estabL, e1_old, e1_new);
#[ [ s_2, 0 ], [ s_4, 0 ], [ s_7, 0 ], [ s_10, 0 ], [ s_11, 0 ], #[ s_16, 0 ] ]
#Hence we have that Y’:=\{x_{8, 9, 12, 14, 15, 20}(t) \mid t \in \F_q\}, where
#x_{8, 9, 12, 14, 15, 20}(t):=x_{2}(2a_{17}a_{24}t)x_{4}(-a_{13}a_{25}t)
#x_{7}(-a_{13}a_{24}t)x_{10}(2a_17a_19t)x_{11}(2a_17a_18t)x_{16}(a_13a_17t).
e1:=[[t_8, 2*a_17*a_24*s_1], [t_9, -a_13*a_25*s_1], [t_12, -a_13*a_24*s_1], [t_14, 2*a_17*a_19*s_1], [t_15, 2*a_17*a_18*s_1], [t_20, a_13*a_17*s_1]];;
#Arm:
stabA:=stabAction(leg, arm, signcom1);
#[ [ s_8, t_4*a_13+t_10*a_18+t_11*a_19 ], [ s_9, -t_2*a_13-t_7*a_17-t_16*a_24 ]
# , [ s_12, t_4*a_17+t_16*a_25 ], [ s_14, -t_2*a_18-t_11*a_24 ],
# [ s_15, -t_2*a_19-t_10*a_24 ], [ s_20, -t_4*a_24-t_7*a_25 ] ]
estabA:=evalStabAct(stabA, e_old, e_new);
#[ [ s_8, t_4*a_13+t_10*a_18+t_11*a_19 ], [ s_9, -t_2*a_13-t_7*a_17-t_16*a_24 ]
# , [ s_12, (t_4*a_13^2*a_17+4*t_16*a_17*a_18*a_19)/a_13^2 ],
# [ s_14, -t_2*a_18-t_11*a_24 ], [ s_15, -t_2*a_19-t_10*a_24 ],
# [ s_20, (-t_4*a_13^2*a_24-4*t_7*a_17*a_18*a_19)/a_13^2 ] ]
#Check this.
e2_old:=[t_2, t_4, t_7, t_10, t_11, t_16];;
e2_new:=[2*a_17*a_24*t,
Value(a_13*a_25*t, e_old, e_new),
-a_13*a_24*t, -2*a_17*a_19*t, -2*a_17*a_18*t, -a_13*a_17*t];;
evalStabAct(estabA, e2_old, e2_new);
#[ [ s_8, 0 ], [ s_9, 0 ], [ s_12, 0 ], [ s_14, 0 ], [ s_15, 0 ], [ s_20, 0 ] ]
#Hence we have X’:=\{x_{2, 4, 7, 10, 11, 16}(t) \mid t \in \F_q\}, where
#x_{2, 4, 7, 10, 11, 16}(t):=x_{2}(2a_{17}a_{24}t)x_{4}(a_{13}a_{25}t)
#x_{7}(-a_{13}a_{24}t)x_{10}(-2a_17a_19t)x_{11}(-2a_17a_18t)x_{16}(-a_13a_17t).
e2:=[[t_2, 2*a_17*a_24*s_2], [t_4, a_13*a_25*s_2], [t_7, -a_13*a_24*s_2], [t_10, -2*a_17*a_19*s_2], [t_11, -2*a_17*a_18*s_2], [t_16, -a_13*a_17*s_2]];;
#We check immediately that Y’ is central. Namely
directCom([3], [8, 9, 12, 14, 15, 20], signcom1);
#[ ]
#We extend our character in the usual way. We are just left to compute
#the commutator between X_3 and X’. We use the function e1, that we will
#evaluate substituting a_25 with its value.
diagDirectCom([[t_3, t_3]], e2, signcom1);
#Reordered, gives:
#[ 8, -2*t_3*s_2*a_17*a_24 ],[ 9, t_3*s_2*a_13*a_25 ],[ 12, t_3*s_2*a_13*a_24 ],
#[ 14, -2*t_3*s_2*a_17*a_19 ], [ 15, -2*t_3*s_2*a_17*a_18 ],
#[ 20, -t_3*s_2*a_13*a_17 ],
#—> this term is exactly the opposite of x_{8, 9, 12, 14, 15, 20}(t_3).
#[ 13, -2*t_3*s_2^2*a_13*a_17*a_24*a_25 ], [ 17, t_3*s_2^2*a_13^2*a_24*a_25 ],
#[ 18, 4*t_3*s_2^2*a_17^2*a_19*a_24 ],[ 19, 4*t_3*s_2^2*a_17^2*a_18*a_24 ],
#[ 24, -4*t_3*s_2^2*a_17^2*a_18*a_19 ], [ 24, t_3*s_2^2*a_13^2*a_17*a_25 ],
#[ 25, -t_3*s_2^2*a_13^2*a_17*a_24 ]
#We evaluate that quickly using the programs.
cenToEval:=[[ 13, -2*t_3*s_2^2*a_13*a_17*a_24*a_25 ], [ 17, t_3*s_2^2*a_13^2*a_24*a_25 ],
[ 18, 4*t_3*s_2^2*a_17^2*a_19*a_24 ],[ 19, 4*t_3*s_2^2*a_17^2*a_18*a_24 ],
[ 24, -4*t_3*s_2^2*a_17^2*a_18*a_19 ], [ 24, t_3*s_2^2*a_13^2*a_17*a_25 ],
[ 25, -t_3*s_2^2*a_13^2*a_17*a_24 ]];;
evalResCom(cenToEval);
#-t_3*s_2^2*a_13^2*a_17*a_24*a_25+4*t_3*s_2^2*a_17^2*a_18*a_19*a_24
Value(evalResCom(cenToEval), e_old, e_new);
#0
#Then we simply have
#\lambda([x_{2, 4, 7, 10, 11, 16}(t), x_3(s_3)])=\phi(-c_{8, 9, 12, 14, 15, 20}s_3t).
#If a_{8, 9, 12, 14, 15, 20}:= c_{8, 9, 12, 14, 15, 20} \ne 0, then we can
#apply the Reduction Lemma. We get in this case a family F_2
#\chi^{a_{8, 9, 12, 14, 15, 20}, a_{13}, a_{17}, a_{18}, a_{19}, a_{24}}
#of (q-1)^6 characters of degree q^6.
#Finally, if c_{8, 9, 12, 14, 15, 20} = 0, then X_3 and X_{2, 4, 7, 10, 11, 16}
#commute. We get a last family F_3, that consists of the
#\chi_{b_{2, 4, 7, 10, 11, 16}, b_3}^{a_{13}, a_{17}, a_{18}, a_{19}, a_{24}}
#of q^2(q-1)^5 characters of degree q^5.