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ARS MATHEMATICA CONTEMPORANEA 6 (2013) A1–A21
On the rank two geometries of the groups
PSL(2, q): part II∗
Francis Buekenhout , Julie De Saedeleer , Dimitri Leemans †
Université Libre de Bruxelles, Département de Mathématiques - C.P.216
Boulevard du Triomphe, B-1050 Bruxelles
Received 30 November 2010, accepted 25 January 2013, published online 15 March 2013
Abstract
This document contains an appendix to the paper On the rank two geometries of the
groups PSL(2, q): part II, Ars Math. Contemp. 6 (2013), 365–388.
Appendix
Proof of Lemma 9
Proof. In order to determine all subgroups H of PSL(2, q) such that (H,D10) is a two-
transitive pair we scan the list of maximal subgroups of PSL(2, q). For each maximal
subgroup we analyse its subgroup lattice. There are six cases to consider.
1. The group Eq : q−1(2,q−1) contains a subgroup D10
∼= E5 : 2 if 5|q. In this situation and
in view of (1) in Proposition 7, H ∼= E5 : 4 which is not a subgroup of PSL(2, q),
under the given conditions.
2. Take D2d with d | q±12 . In view of (16)-(18) in Proposition 7, D2d acts two-
transitively on the cosets of D10 if and only if the index of D10 in D2d equals 2
or 3 (d = 10 or 15). Therefore (D20, D10) and (D30, D10) are two-transitive pairs.
3. A4 and S4 do not contain any subgroup of order 10.
4. In view of (6) in Proposition 7 (A5, D10) is a two-transitive pair.
5. In view of (6), (7), (8) and (10) in Proposition 7, PSL(2, q′) acts two-transitively on
the cosets of D10 ∼= Eq′ : q
′−1
2 only if q
′ = 5, therefore q = 5r for r an odd prime.
(PSL(2, 5), D10) is a two-transitive pair.
∗This paper is a part of SIGMAP’10 special issue Ars Math. Contemp. vol. 5, no. 2.
†Supported by the “Communauté Française de Belgique - Actions de Recherche Concertées”
E-mail addresses: fbueken@ulb.ac.be (Francis Buekenhout), judesaed@ulb.ac.be (Julie De Saedeleer),
dleemans@ulb.ac.be (Dimitri Leemans)
Copyright c© 2013 DMFA Slovenije
A2 Ars Math. Contemp. 6 (2013) 1–21
6. In view of (12) in Proposition 7, PSL(2, q′) acts two-transitively on the cosets of
D10 ∼= Eq′ : q
′−1
2 only if q
′ = 5 and q′ − 1 = 2, which leads to a contradiction.
Proof of Lemma 10
Proof. In order to determine all subgroups H of PSL(2, q) such that (H,A4) is a two-
transitive pair we scan the list of maximal subgroups of PSL(2, q). For each maximal
subgroup we analyse its subgroup lattice. There are six cases to consider.
1. If q = 5r, the group Eq : q−12 does not contain any subgroup isomorphic to A4
∼=
E4 :3 because 4 | q is in contradiction with the condition q = 5r.
If q = p = ±1(5), the group Eq : q−12 does not contain any subgroup isomorphic
to E4 : 3 because 4 | p implies that 4 = p, which is in contradiction with p an odd
prime, the same argument holds for q = p2 = −1(5).
If q = 4r with r prime, the (2T )1 condition, the maximality and the conditions
given on q imply that the only candidate of the form Eq : q−12 is E16 : 3. Now
(E16 :3, E4 :3) is a two-transitive pair.
2. Take D2d with d | q±1(2,q−1) . We know that dihedral groups only contain cyclic groups
and dihedral groups, they do not contain an A4.
3. If q = 4r with r prime, the group PSL(2, q) does not contain a subgroup isomorphic
to S4, because this is in contradiction with q = ±1(8). The same argument holds for
q = 5r with r an odd prime.
If q = p = ±1(5) or q = p2 = −1(5), in view of (11) in Proposition 7 (S4, A4) is a
two-transitive pair provided q = ±1(8).
4. In view of (6) in Proposition 7 (A5, A4) is a two-transitive pair.
5. If q = p = ±1(5), the group PSL(2, q) cannot contain any PSL(2, q′) with q′m = q,
m an odd prime, the same argument holds for q = p2 = −1(5).
If q = 5r with r an odd prime; orif q = 4r with r prime, the only candidates q′
for PSL(2, q′) are 4 and 5. In this situation we have PSL(2, q′) ∼= PSL(2, 4) ∼=
PSL(2, 5) ∼= A5. This situation has been treated in (4).
6. If q = p = ±1(5); or q = 5r with r an odd prime, the group PSL(2, q) cannot
contain any PGL(2, q′) with q′2 = q.
If q = p2 = −1(5) in view of (12) in Proposition 7 PGL(2, q′) with q′2 = q acts
two-transitively on the cosets of A4 if q′ = 4. In this situation we have PSL(2, q′) ∼=
PGL(2, 4) ∼= A5. This situation has been treated in (4).
If q = 4r with r prime, the group PSL(2, q) contains PGL(2, q′) if q′2 = q which
implies that q′ = 2r. In view of (12) in Proposition 7, (PGL(2, 4), E4 :3) is a two-
transitive pair provided q′ = 2r with r = 2.
Part of proof of Proposition 13.
Proof. Subcase 1: G01 = G0 ∩G1 ∼= D10.
By Lemma 9 the four possibilities for G1 are D20 provided 10 | q±1(2,q−1) , D30 provided
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A3
15 | q±1(2,q−1) , PSL(2, 5) ∼= A5 provided q = 5
r and A5.
1.1 We consider the case where G1 ∼= D20, provided 10 | q±1(2,q−1) .
The given conditions imply that either q = p = ±1(20) or q = p2 = −1(20). In both
situations there are two conjugacy classes of A5 in PSL(2, q). Since q±110 is even there are
two conjugacy classes of D10 in PSL(2, q). The index of D10 in D20 equals two, therefore
the D10 in a D20 are not all conjugate. The number of conjugacy classes of D20 depends on
whether q±120 is even or odd. In order to determine all geometries under the given conditions
we distinguish the cases where q±120 is even or odd.
• q±120 is even. This implies that NPSL(2,q)(D10) = D20 and NPSL(2,q)(D20) = D40,
with two conjugacy classes of D20. Therefore the number of D20 containing a given D10
is one.
There are two classes of A5 and D10 and the latter is contained in one D20; therefore
there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);A5, D20, D10) up to
conjugacy, provided q±120 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 8 the two
classes of D10, D20 and A5 are fused under the action of PGL(2, q) and thus also under
the action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);A5, D20, D10) up to isomorphism provided q±120 is even.
• q±120 is odd. In this situation there is one conjugacy class of D20 in PSL(2, q). The
condition on q implies that NPSL(2,q)(D10) = D20 and NPSL(2,q)(D20) = D20. Therefore
the number of D20 containing a given D10 is one.
Up to conjugacy, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);
A5, D20, D10) provided q±120 is odd.
Let us deal with the fusion of non-conjugate classes. Up to isomorphism there is exactly
one such geometry, since following Lemma 8 the two classes of D10 and A5 are fused under
the action of PGL(2, q) and thus also under the action of PΓL(2, q).
To summarize, up to conjugacy there exist exactly two RWPRI and (2T )1 geometries
Γ5 = Γ (PSL(2, q);A5, D20, D10) provided q = p = ±1(20). Up to isomorphism there
exists exactly one such geometry. Also, up to conjugacy there exist exactly two RWPRI
and (2T )1 geometries Γ12 = Γ (PSL(2, q);A5, D20, D10) provided q = p2 = −1(20).
Up to isomorphism there exists exactly one such geometry.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 19, 41, 61. For q = 19, it is also confirmed by [20].
1.2. We consider the case where G1 ∼= D30 , provided 15 | q±1(2,q−1) .
The condition 15 | q±1(2,q−1) implies that either q = 4
r with r prime, q = p = ±1(5) or
q = p2 = −1(5). Hence, there are three cases namely q = 4r = ±1(15) with r prime;
q = p = ±1(30); or q = p2 = −1(30). We distinguish the first case from the other two.
• Let us first assume that q = 4r = ±1(15) with r prime. In this situation there is only
one conjugacy class of A5. The number of classes of D30 and D10 in PSL(2, q) depends
on whether q±115 is even or odd. The even case cannot occur because of the condition
q = 4r given on q. If q±115 is odd there is only one conjugacy class of D30 and also
one of D10 in PSL(2, q). Then the index
|D30|
|D10| 6= 2, and therefore all D10 in D30 are
conjugate. And A5 contains one D10 up to conjugacy. The odd condition on q±115 implies
A4 Ars Math. Contemp. 6 (2013) 1–21
that NPSL(2,q)(D10) = D10 and NPSL(2,q)(D30) = D30. Therefore the number of D30
containing a given D10 is one.
To summarize, up to conjugacy there exists exactly one RWPRI and (2T )1 geometry
Γ1 = Γ (PSL(2, q);A5, D30, D10) and thus also exactly one up to isomorphism provided
either q = 4r with r prime; or q±115 odd. This geometry is new and the number of classes
up to conjugacy (resp. isomorphism) is confirmed by MAGMA for q = 16 and is also
confirmed by [20].
• The cases q = p = ±1(30) and q = p2 = −1(30) with p an odd prime can be
treated together. In this situation there are two conjugacy classes of A5, but the number of
conjugacy classes of D30 and D10 depends on whether q±130 is even or odd.
Assume q±130 is even. This implies that NPSL(2,q)(D10) = D20 and NPSL(2,q)(D30) =
D60, with two conjugacy classes of D10 and also two of D30. The number of subgroups
D30 containing a given subgroup D10 in PSL(2, q) is equal to
| PSL(2, q) |
| D60 |
· | D30 |
| D10 |
· | D20 |
| PSL(2, q) |
= 1.
Up to conjugacy, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);A5,
D30, D10) provided q±130 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 8 the two
classes of D10, D30 and A5 are fused under the action of PGL(2, q) and thus also under
the action of PΓL(2, q). Therefore there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);A5, D30, D10) up to isomorphism provided q±130 is even.
Assume q±130 is odd. This implies that NPSL(2,q)(D10) = D10 and NPSL(2,q)(D30) = D30,
with one conjugacy class of D10 and also one of D30. The number of subgroups D30
containing a given subgroup D10 in PSL(2, q) is equal to
| PSL(2, q) |
| D30 |
· | D30 |
| D10 |
· | D10 |
| PSL(2, q) |
= 1.
Up to conjugacy, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);A5,
D30, D10) provided q±130 is odd.
Let us deal with the fusion of non-conjugate classes. Following Lemma 8 the two
classes of A5 are fused under the action of PGL(2, q) and thus also under the action of
PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ(PSL(2, q);
A5, D30, D10) up to isomorphism provided q±130 is odd.
To summarize, there exist exactly two RWPRI and (2T )1 geometries Γ6 =Γ(PSL(2, q);
A5, D30, D10) up to conjugacy and exactly one up to isomorphism provided q = p =
±1(30). Also, up to conjugacy there exist exactly two RWPRI and (2T )1 geometries
Γ13 = Γ (PSL(2, q);A5, D30, D10) and exactly one up to isomorphism provided q =
p2 = −1(30). This geometry is new and the number of classes up to conjugacy (resp.
isomorphism) is confirmed by MAGMA for q = 29, 31, 61.
1.3. Consider the case G0 ∼= G1 ∼= A5.
There are four situations, which are q = 5r with r odd prime, q = p = ±1(5),
q = p2 = −1(5) with p an odd prime and q = 4r with r prime. Cases 2 and 3 can be
treated together. We distinguish them from the others in our discussion.
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A5
• Assume q = 5r with r an odd prime. Using that PSL(2, 5) ∼= A5, there is only
one conjugacy class of PSL(2, 5). We must check whether this geometry exists, that is
whether there are two subgroups isomorphic to A5 in PSL(2, 5r) that have a subgroup D10
in common. There is only one conjugacy class of E5 : 2. Since PSL(2, 5r) is simple and
A5 maximal, A5 is self-normalized. Also, since PSL(2, 5) is simple and E5 : 2 maximal,
E5 : 2 is self-normalized in PSL(2, 5) and also in PSL(2, 5r). Therefore the number of
subgroups PSL(2, 5) containing a given subgroup E5 :2 in PSL(2, 5r) is equal to
| PSL(2, 5r) |
| PSL(2, 5) |
· | PSL(2, 5) |
| E5 :2 |
· | E5 :2 |
| PSL(2, 5r) |
= 1
which implies that the geometry does not exist.
• Assume that either q = p = ±1(5) or q = p2 = −1(5) with p an odd prime. There
are two conjugacy classes of A5. The number of conjugacy classes of D10 depends on
whether q±110 is even or odd.
If q±110 is even there are two conjugacy classes of D10. Notice that all D10 in an A5 are
conjugate and NPSL(2,q)(D10) = D20. The number of subgroups A5 containing a given
subgroup D10 in PSL(2, q) is equal to
| PSL(2, q) |
| A5 |
· | A5 |
| D10 |
· | D20 |
| PSL(2, q) |
= 2.
Therefore there exist exactly two RWPRI and (2T )1 geometries
Γ7 = Γ (PSL(2, q);A5, A5, D10) up to conjugacy, provided q±110 is even with q an odd
prime and also exactly two RWPRI and (2T )1 geometries Γ14 = Γ(PSL(2, q);A5, A5,
D10) up to conjugacy, provided q+110 is even with q = p
2; one geometry for each class of
A5.
Let us deal with the fusion of non-conjugate classes. Following Lemma 8 the two
classes of A5 and D10 are fused under the action of PGL(2, q) and thus also under the
action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ7 = Γ (PSL(2, q);A5, A5, D10) up to isomorphism provided q±110 is even with q an odd
prime and also exactly one RWPRI and (2T )1 geometry Γ14 = Γ (PSL(2, q);A5, A5, D10)
up to isomorphism provided q+110 is even with q = p
2.
Assume that q±110 is odd. There is only one conjugacy class of D10 and
NPSL(2,q)(D10) = D10. The number of subgroups A5 containing a given subgroup D10 in
PSL(2, q) is equal to
| PSL(2, q) |
| A5 |
· | A5 |
| D10 |
· | D10 |
| PSL(2, q) |
= 1.
Since there are two conjugacy classes of A5 there exists exactly one RWPRI and (2T )1
geometry Γ8 = Γ (PSL(2, q);A5, A5, D10) up to conjugacy and thus also exactly one up
to isomorphism provided q±110 is odd with q an odd prime. Also, there exists exactly one
RWPRI and (2T )1 geometry Γ15 = Γ (PSL(2, q);A5, A5, D10) up to conjugacy and thus
also exactly one up to isomorphism provided q+110 is odd with q = p
2.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism) is
confirmed by MAGMA for q = 9, 11, 19, 29, 31, 41, 49. For q = 9, it is also confirmed by
A6 Ars Math. Contemp. 6 (2013) 1–21
[3] and for q = 11, 19 by [20].
• If q = 4r with r prime. We know that there is only one conjugacy class of A5. We must
check whether this geometry exists, that is whether there are two subgroups isomorphic to
A5 in PSL(2, 4r) that have a subgroup D10 in common. The condition given on q implies
that q±15 is odd, therefore there is only one class of D10 and NPSL(2,q)(D10) = D10. The
number of subgroups A5 containing a given subgroup D10 in PSL(2, q) is equal to
| PSL(2, q) |
| A5 |
· | A5 |
| D10 |
· | D10 |
| PSL(2, q) |
= 1.
In this situation there is only one conjugacy class of A5, therefore we may conclude that
there exists no such geometry.
Proof of Proposition 14
Proof. Let G0 ∼= A4 with q prime, q > 3 and either q = 3, 13, 27, 37(40) or q = 5.
In view of (5) in Proposition 7 the only possibility for G01 is the cyclic subgroup of order 3.
If H is a subgroup of G such that (H, 3) is a two-transitive pair then one of the following
holds: H ∼= Z6 provided 6 | q±12 , H ∼= D6 and H ∼= A4. They are the three only
G1-candidates.
Notice that q prime, q > 3 and so 3 divides either q+12 or
q−1
2 .
We review all possibilities for G1 as well as the number of classes of geometries with re-
spect to conjugacy (resp. isomorphism).
1. Consider the case where G1 ∼= Z6, provided 6 | q±12 .
The conditions on q prime are that q = ±1(12) and q = 3, 13, 27, 37(40). This implies
that q = 13, 37, 83, 107(120) with q prime. The group A4 contains one cyclic group of
order 3 up to conjugacy. The cyclic group of order 3 is contained in exactly one Z6 and
all Z6 in PSL(2, q) are conjugate. Since PSL(2, q) is simple and A4 maximal, A4 is self-
normalized. It is also the case for the cyclic subgroups of order 3 in A4. Now NZ6(3) = Z6
and NPSL(2,q)(3) = NPSL(2,q)(Z6) = Dq+1 provided 6 | q+12 and Dq−1 provided 6 |
q+1
2 .
The number of subgroups Z6 containing a given cyclic subgroup of order 3 in PSL(2, q) is
equal to
| PSL(2, q) |
| q ± 1 |
· 1 · | q ± 1 |
| PSL(2, q) |
= 1.
Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ1 = Γ(PSL(2, q);A4,
Z6, 3) up to conjugacy, and also exactly one up to isomorphism, provided q = 13, 37, 83,
107(120). This geometry is new and the number of classes up to conjugacy (resp. isomor-
phism) is confirmed by MAGMA for q = 13, 37, 83.
2. Consider the case where G1 ∼= D6.
All cyclic subgroups of order 3 are conjugate in PSL(2, q). The number of conjugacy
classes of D6 depends on whether q±16 is odd or even. We distinguish the cases
q±1
6 odd
or even.
The group A4 contains one cyclic group of order 3 up to conjugacy. We know that
the normalizer of D6 in PSL(2, q) is D6 provided q±16 is odd, and that it is D12 provided
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A7
q±1
6 is even. The normalizer of the cyclic group of order 3 in D6 is D6 and its normalizer
in PSL(2, q) is a dihedral group of order q ± 1. Therefore the number of subgroups D6
containing a given cyclic subgroup of order 3 in PSL(2, q) is equal to{ |PSL(2,q)|
|D6| · 1 ·
|q±1|
|PSL(2,q)| =
q±1
6 if
q±1
6 odd
|PSL(2,q)|
|D12| · 1 ·
|q±1|
|PSL(2,q)| =
q±1
12 if
q±1
6 even.
To get the number of geometries up to conjugacy we need to know whether the sub-
group A4 normalizes each of the D6, which is the case because
|NPSL(2,q)(3) ∩NPSL(2,q)(A4)| = 3.
In order to determine the number of classes of geometries up to conjugacy we distin-
guish the cases q±16 odd or even.
• Assume q±16 is odd. There is only one class of D6 and every given cyclic subgroup
of order 3 in PSL(2, q) is contained in exactly q±16 dihedral groups D6. Up to conjugacy
there exist exactly q±16 geometries.
• Assume q±16 is even. There are two classes of D6 and every given cyclic subgroup
of order 3 in PSL(2, q) is contained in exactly q±112 dihedral groups D6. Up to conjugacy
there exist exactly q±16 geometries.
To summarize, up to conjugacy there exist exactly q−16 RWPRI and (2T )1 geome-
tries Γ3 = Γ (PSL(2, q);A4, D6, 3) provided q−16 is odd and exactly
q−1
6 RWPRI and
(2T )1 geometries Γ5 = Γ (PSL(2, q);A4, D6, 3) up to conjugacy, provided q−16 is even.
Also, there exist exactly q+16 RWPRI and (2T )1 geometries Γ2 = Γ (PSL(2, q);A4, D6, 3)
up to conjugacy, provided q+16 is odd and exactly
q+1
6 RWPRI and (2T )1 geometries
Γ4 = Γ (PSL(2, q);A4, D6, 3) up to conjugacy, provided q+16 is even.
Let us deal with the fusion of non-conjugate classes. We remember that q is prime and
thus PΓL(2, q) ∼= PGL(2, q). We also find that NPGL(2,q)(A4) = S4, NPGL(2,q)(3) =
D2(q±1) and NPGL(2,q)(D6) = D12. In order to determine the number of classes of ge-
ometries up to isomorphism we distinguish the cases q±16 odd or even.
• Assume q±16 odd. There is only one conjugacy class of D6. If we fix A4 and the
cyclic group of order 3, there is one D6 which is fixed and the others are exchanged two
by two, because D6 in PSL(2, q) is its own normalizer. They merge two by two under
the action of PΓL(2, q). Therefore, the number of RWPRI and (2T )1 geometries Γ2 =
Γ (PSL(2, q);A4, D6, 3) up to isomorphism, provided q+16 odd, is exactly
q+1
6 −1
2 + 1,
and the number of RWPRI and (2T )1 geometries Γ3 = Γ (PSL(2, q);A4, D6, 3) up to
isomorphism, provided q−16 odd, is exactly
q−1
6 −1
2 + 1.
• Assume q±16 is even. There are two conjugacy classes of D6. They both merge under
the action of PGL(2, q) and thus also in PΓL(2, q) (see Lemma 11). If we fix A4 and
the cyclic group of order 3, we fix two D6, one of each conjugacy class and all others are
exchanged two by two. They merge two by two under the action of PΓL(2, q). Therefore,
the number of RWPRI and (2T )1 geometries Γ4 = Γ (PSL(2, q);A4, D6, 3) up to isomor-
phism, provided q+16 even, is exactly
( q+1
6 −2
2 + 1
)
= q+112 and the number of RWPRI and
(2T )1 geometries Γ5 = Γ (PSL(2, q);A4, D6, 3) up to isomorphism, provided q−16 even,
A8 Ars Math. Contemp. 6 (2013) 1–21
is exactly
( q−1
6 −2
2 + 1
)
= q−112 .
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 5, 13, 37, 43, 53, 67. For q = 5, it is also confirmed by
[3] and for q = 13 by [20].
3. Consider the case where G0 ∼= G1 ∼= A4.
We must check whether this geometry exists or not, that is whether there are two subgroups
isomorphic to A4 in PSL(2, q) that have a cyclic subgroup of order 3 in common. We
know that NPSL(2,q)(A4) = A4 and that NA4(3) = 3. Moreover, the group A4 contains
4 maximal cyclic subgroups of order 3, all conjugate. The normalizer of 3 in PSL(2, q) is
Dq−1 if 3 | q− 1 and Dq+1 if 3 | q + 1. Therefore the number of subgroups A4 containing
a given cyclic subgroup of order 3 in PSL(2, q) is equal to{ |PSL(2,q)|
|A4| · 4 ·
q−1
|PSL(2,q)| =
q−1
3 if 3 | q − 1
|PSL(2,q)|
|A4| · 4 ·
q+1
|PSL(2,q)| =
q+1
3 if 3 | q + 1.
Knowing that there exists only one conjugacy class of A4 and using the conditions on q we
know that this geometry exists. There exist exactly, up to conjugacy, q−13 − 1 RWPRI and
(2T )1 geometries Γ7 = Γ (PSL(2, q);A4, A4, 3), provided 3 | q − 1 and exactly q+13 − 1
RWPRI and (2T )1 geometries Γ6 = Γ (PSL(2, q);A4, A4, 3) up to conjugacy, provided
3 | q + 1.
Let us deal with the fusion of non-conjugate classes. We remember that q is prime and
thus PGL(2, q) ∼= PΓL(2, q). We find that NPGL(2,q)(A4) = S4 and NPGL(2,q)(3) =
D2(q±1). Therefore the number of subgroups A4 containing a given cyclic subgroup of
order 3 in PGL(2, q) is equal to q±13 . To count the geometries up to isomorphism we need
to know the action of PGL(2, q) on subgroups A4 containing a given cyclic subgroup of
order 3. If we fix A4 ∼= G0 and the cyclic subgroup of order 3 we know that |NPGL(2,q)(3)∩
NPGL(2,q)(A4)| = |D6| = 2|3|. This D6 is contained in two S4 in PGL(2, q), which
implies that there is one other A4 fixed and all others are exchanged two by two. Thus they
merge under the action of PGL(2, q). Hence, there exist exactly (
q−1
3 −2)
2 + 1 RWPRI and
(2T )1 geometries
Γ7 = Γ (PSL(2, q);A4, A4, 3) up to isomorphism, provided 3 | q − 1 and exactly
( q+13 −2)
2 + 1 RWPRI and (2T )1 geometries Γ6 = Γ (PSL(2, q);A4, A4, 3) up to isomor-
phism, provided 3 | q + 1.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 5, 13, 37, 43, 53, 67. For q = 13, it is also confirmed by
[20].
Proof of Proposition 18
Proof. Let G0 ∼= S4.
We subdivide our discussion in three cases, namely the three G01-candidates given by (11),
(12) and (13) in Proposition 7 which are: D6, D8 and A4. In each of these three cases
we review all possibilities for G1 given in the previous Lemmas as well as the number of
classes of geometries with respect to conjugacy (resp. isomorphism).
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A9
Subcase 1: G01 = G0 ∩G1 ∼= D6.
By Lemma 15 the three possibilities for G1 are D12 provided 6 | q±12 , D18 provided
9 | q±12 and S4.
The number of conjugacy classes of D6 depends on whether q±16 is odd or even. In
order to determine all geometries under the given conditions we distinguish the cases q±16
odd or even.
Recall that when q > 2 is a prime and q = ±1(8) there are two conjugacy classes of
S4 in PSL(2, q).
1.1. Consider the case where G1 ∼= D12, provided 6 | q±12 .
Since q±16 is even, following Lemma 4 there are two conjugacy classes of D6 in PSL(2, q).
The number of conjugacy classes of D12 depends on whether q±112 is even or odd. The
conditions on q are that q = ±1(8) and q = ±1(12). Which implies that q±112 even. In
this situation there are two classes of D12 in PSL(2, q). Now the index of
|D12|
|D6| = 2,
therefore the D6 in a D12 are not all conjugate. Also, every D12 contains two D6 which
are not conjugate. And S4 contains one D6 up to conjugacy. Since q±112 is even we have
NPSL(2,q)(D6) = D12 = ND12(D6) and NPSL(2,q)(D12) = D24. Therefore the number
of D12 containing a given D6 is one. Since there are two classes of S4, D6 and D12, there
exist exactly two RWPRI and (2T )1 geometries Γ1 = Γ (PSL(2, q);S4, D12, D6) up to
conjugacy when q±112 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of D6, D12 and S4 are fused under the action of PGL(2, q) and thus also under
the action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ1 = Γ (PSL(2, q);S4, D12, D6) up to isomorphism, provided q±112 is even.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 23.
1.2. Consider the case where G1 ∼= D18, provided 9 | q±12 .
The number of conjugacy classes of D18 and D6 depends on whether q±118 is even or odd.
The conditions on q are that q = ±1(8) and q = ±1(18). Which implies that q±118 .
Now the index |D18||D6| 6= 2, therefore all D6 in a D18 are conjugate. And S4 contains
one D6 up to conjugacy.
• Assume q±118 is even. This implies that NPSL(2,q)(D6) = D12 and NPSL(2,q)(D18) =
D36. In this situation there are two conjugacy classes of D6 and also two of D18. The
number of subgroups D18 containing a given subgroup D6 in PSL(2, q) is equal to
| PSL(2, q) |
| D36 |
· | D18 |
| D6 |
· | D12 |
| PSL(2, q) |
= 1.
Since there are two conjugacy classes of S4 there exist exactly two RWPRI and (2T )1
geometries Γ(PSL(2, q);S4, D18, D6) up to conjugacy, provided q±118 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of D6, D18 and S4 are fused under the action of PGL(2, q) and thus also under
the action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);S4, D18, D6) up to isomorphism, provided q±118 is even.
• Assume q±118 is odd. This implies that NPSL(2,q)(D6) = D6 and NPSL(2,q)(D18) =
D18. In this situation there is one conjugacy class of D6 and also one of D18. The number
A10 Ars Math. Contemp. 6 (2013) 1–21
of subgroups D18 containing a given subgroup D6 in PSL(2, q) is equal to
| PSL(2, q) |
| D18 |
· | D18 |
| D6 |
· | D6 |
| PSL(2, q) |
= 1.
Since there are two conjugacy classes of S4 there exist exactly two RWPRI and (2T )1
geometries Γ(PSL(2, q);S4, D18, D6) up to conjugacy, provided q±118 is odd.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of S4 are fused under the action of PGL(2, q) and thus also under the action of
PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ(PSL(2, q);
S4, D18, D6) up to isomorphism, provided q±118 is odd.
To summarize, there exist exactly two RWPRI and (2T )1 geometries
Γ2 = Γ (PSL(2, q);S4, D18, D6) up to conjugacy and one up to isomorphism, provided
q = ±1(72) or q = ±17(72). This geometry is new and the number of classes up to con-
jugacy (resp. isomorphism) is confirmed by MAGMA for q = 17 and is also confirmed by
[20].
1.3. Finally we consider the case where G0 ∼= G1 ∼= S4.
• Assume q±16 is even. There are two conjugacy classes of D6. Now all the D6 are
contained in a S4 and all D6 in a S4 are conjugate. The normalizer of D6 in PSL(2, q) is
D12. The number of subgroups S4 containing a given subgroup D6 in PSL(2, q) is equal
to
| PSL(2, q) |
| S4 |
· | S4 |
| D6 |
· | D12 |
| PSL(2, q) |
= 2.
Therefore, there exist exactly two RWPRI and (2T )1 geometries Γ3 = Γ(PSL(2, q);S4,
S4, D6) up to conjugacy, provided q±16 is even, one for each class of S4.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of S4 are fused under the action of PGL(2, q) and thus also under the action
of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ3 =
Γ (PSL(2, q);S4, S4, D6) up to isomorphism, provided q±16 is even.
• Assume q±16 is odd. There is one conjugacy class of D6. This implies that normalizer
NPSL(2,q)(D6) = D6. The number of subgroups S4 containing a given subgroup D6 in
PSL(2, q) is equal to
| PSL(2, q) |
| S4 |
· | S4 |
| D6 |
· | D6 |
| PSL(2, q) |
= 1.
Since there are two conjugacy classes of S4, there exists exactly one RWPRI and (2T )1
geometry Γ4 = Γ (PSL(2, q);S4, S4, D6) up to conjugacy and thus also one up to isomor-
phism, provided q±16 is odd.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 7, 17, 23, 31, 41. For q = 17, it is also confirmed by [20].
Subcase 2: G01 = G0 ∩G1 ∼= D8.
By Lemma 16 the three possibilities for G1 are D16 provided 8 | q±12 , D24 provided
12 | q±12 and S4. Observe that under the hypothesis there are two conjugacy classes of S4
in PSL(2, q).
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A11
2.1. Consider the case where G1 ∼= D16, provided 8 | q±12 .
Since q±18 is even there are two conjugacy classes of D8. The conditions on q are that
q ± 1(8) and q ± 1(16). Which implies that q = ±1(16). The index of D8 in D16 equals
two, therefore the D8 in a D16 are not all conjugate. And also, every D16 contains two D8
which are not conjugate. Moreover S4 contains one D8 up to conjugacy. The number of
conjugacy classes of D16 depends on whether q±116 is even or odd. In order to determine all
geometries under the given conditions we distinguish the cases q±116 odd or even.
• Assume q±116 is even. This implies that NPSL(2,q)(D8) = D16 = ND16(D8) and
NPSL(2,q)(D16) = D32, with two conjugacy classes of D16. Therefore the number of D16
containing a given D8 is one.
Since there are two classes of S4, D8 and D16, there exist exactly two RWPRI and
(2T )1 geometries Γ(PSL(2, q);S4, D16, D8) up to conjugacy, provided q±116 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of D8, D16 and S4 are fused under the action of PGL(2, q) and thus also under
the action of PΓL(2, q). Therefore there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);S4, D16, D8) up to isomorphism provided q±116 is even.
• Assume q±116 is odd. This implies that NPSL(2,q)(D8) = D16 and NPSL(2,q)(D16) =
D16, with one conjugacy class of D16. Therefore the number of D16 containing a given
D8 is one.
Hence, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);S4, D16,
D8) up to conjugacy, provided q±116 is odd.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of D8 and S4 are fused under the action of PGL(2, q) and thus also under the
action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);S4, D16, D8) up to isomorphism, provided q±116 is odd.
To summarize, there exist exactly two RWPRI and (2T )1 geometries
Γ5 = Γ (PSL(2, q);S4, D16, D8) up to conjugacy and exactly one up to isomorphism, pro-
vided q = ±1(16). This geometry is new and the number of classes up to conjugacy (resp.
isomorphism) is confirmed by MAGMA for q = 17, 31. For q = 17, it is also confirmed by
[20].
2.2. We now consider the case G1 ∼= D24, provided 12 | q±12 .
The index |D24||D8| 6= 2, therefore all D8 in a D24 are conjugate. And S4 contains one D8 up
to conjugacy. The number of conjugacy classes of D8 and D24 depends on whether q±124 is
even or odd. In order to determine all geometries under the given conditions we distinguish
the cases q±124 odd or even.
• Assume q±124 is even. This implies that NPSL(2,q)(D8) = D16 and NPSL(2,q)(D24) =
D48. In this situation there are two conjugacy classes of D8 and also two of D24. The
number of subgroups D24 containing a given subgroup D8 in PSL(2, q) is equal to
| PSL(2, q) |
| D48 |
· | D24 |
| D8 |
· | D16 |
| PSL(2, q) |
= 1.
Therefore, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);S4, D24,
D8) up to conjugacy, provided q±124 is even.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of D8, D24 and S4 are fused under the action of PGL(2, q) and thus also under
A12 Ars Math. Contemp. 6 (2013) 1–21
the action of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ(PSL(2, q);S4, D24, D8) up to isomorphism provided q±124 is even.
• Assume q±124 is odd. This implies that NPSL(2,q)(D8) = D8 and NPSL(2,q)(D24) =
D24. In this situation there is one conjugacy class of D8 and also one of D24. The number
of subgroups D18 containing a given subgroup D8 in PSL(2, q) is equal to
| PSL(2, q) |
| D24 |
· | D24 |
| D8 |
· | D8 |
| PSL(2, q) |
= 1.
To summarize, there exist exactly two RWPRI and (2T )1 geometries Γ(PSL(2, q);S4, D24,
D8) up to conjugacy.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of S4 are fused under the action of PGL(2, q) and thus also under the action of
PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ(PSL(2, q);
S4, D24, D8) up to isomorphism, provided q±124 is odd.
To summarize, there exist exactly two RWPRI and (2T )1 geometries
Γ6 = Γ (PSL(2, q);S4, D24, D8) up to conjugacy and exactly one up to isomorphism pro-
vided q = ±1(24). This geometry is new and the number of classes up to conjugacy (resp.
isomorphism) is confirmed by MAGMA for q = 23.
2.3. At last, consider the case G0 ∼= G1 ∼= S4.
The number of conjugacy classes of D8 depends on whether q±18 is even or odd. In order
to determine all geometries under the given conditions we distinguish the cases q±18 odd or
even.
• Assume q±18 is even. There are two conjugacy classes of D8. In S4 all D8 are
conjugate and the normalizer of D8 in PSL(2, q) is D16. The number of subgroups S4
containing a given subgroup D8 in PSL(2, q) is equal to
| PSL(2, q) |
| S4 |
· | S4 |
| D8 |
· | D16 |
| PSL(2, q) |
= 2.
Therefore, up to conjugacy there exist exactly two RWPRI and (2T )1 geometries
Γ7 = Γ (PSL(2, q);S4, S4, D8) provided q±18 is even, one for each class of S4.
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of S4 are fused under the action of PGL(2, q) and thus also under the action
of PΓL(2, q). Therefore, there exists exactly one RWPRI and (2T )1 geometry Γ7 =
Γ (PSL(2, q);S4, S4, D8) up to isomorphism, provided q±18 is even.
• Assume q±18 is odd. There is one conjugacy class of D8. This implies that normalizer
NPSL(2,q)(D8) = D8. The number of subgroups S4 containing a given subgroup D8 in
PSL(2, q) is equal to
| PSL(2, q) |
| S4 |
· | S4 |
| D8 |
· | D8 |
| PSL(2, q) |
= 1.
Since there are two conjugacy classes of S4, there exists exactly one RWPRI and (2T )1
geometry Γ8 = Γ (PSL(2, q);S4, S4, D8) up to conjugacy and thus also exactly one up to
isomorphism provided q±18 is odd.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 7, 17, 23, 31, 41. For q = 7, it is also confirmed by [3]
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A13
and for q = 17 by [20].
Subcase 3: G01 = G0 ∩G1 ∼= A4.
By Lemma 17 the possibilities for G1 are S4 and A5 provided q = ±1(5). In the latter
situation there are two conjugacy classes of A5.
3.1. Consider the case where G0 ∼= G1 ∼= S4.
We have q = ±1(8) which implies that there are two conjugacy classes of S4 and also two
of A4. Now all A4 in a S4 are conjugate and every given A4 is contained in exactly one
S4, which implies that there exists no geometry in this situation.
3.2. Consider the case where G1 ∼= A5.
If q = p = ±1(5) = ±1(8) with p prime, this case has already been dealt with in
Proposition 13. Therefore, there exist exactly two RWPRI and (2T )1 geometries Γ9 =
Γ (PSL(2, q);S4, A5, A4) up to conjugacy and exactly one up to isomorphism for q = p =
±1(40) and for q = p = ±9(40) with p an odd prime.
Proof of Proposition 20
Proof. Let G0 ∼= PSL(2, 2n).
We subdivide our discussion in three cases according to the three G01-candidates given by
(3), (4), (6) and (10) in Proposition 7 namely: the case of the cyclic subgroup of order 3
provided q′ = 2; the case of D10 provided q′ = 4 and the case of E2n : (2n − 1).
In each of these three cases we review all possibilities for G1 given in the previous
Lemmas as well as the number of classes of geometries with respect to conjugacy (resp.
isomorphism). In order to determine all geometries under the given conditions we subdi-
vide our discussion in a particular case and a general one depending on whether n = 1 or
not.
Particular case: n = 1 and m = 2.
In this situation q′ = 2 and q = 4. In view of (3) and (4) in Proposition 7 there are two
cases to consider: the cyclic group of order 3 and the cyclic group of order 2.
Subcase 1: G01 = G0 ∩G1 ∼= 2.
Since G ∼= PSL(2, 4), (PSL(2, 2), 2) and (22, 2) are the only two-transitive pairs. We
obtain the following geometries
Γ2 = Γ (PSL(2, 4); PSL(2, 2),PSL(2, 2), 2) and Γ3 = Γ
(
PSL(2, 4); PSL(2, 2), 22, 2
)
.
They are indeed RWPRI and (2T )1 geometries as we need because we already met them
in [5], Proposition 15. Since PSL(2, 4) ∼= PSL(2, 5) and PSL(2, 2) ∼= S3, these are the
RWPRI and (2T )1 geometries corresponding to the Petersen graph and the Desargues’
configuration.
Subcase 2: G01 = G0 ∩G1 ∼= 3.
Since G ∼= PSL(2, 4) ∼= A5, (PSL(2, 2), 3) and (A4, 3) are the only two-transitive pairs.
The geometry Γ (PSL(2, 4); PSL(2, 2),PSL(2, 2), 3) has been treated in [5] Proposi-
tion 15 since PSL(2, 2) ∼= D6 and it does not exist. We obtain the following geometry
Γ4 = Γ (PSL(2, 4); PSL(2, 2), A4, 3), which is indeed a RWPRI and (2T )1 geometry
as we need because we already met it in Proposition 14 since PSL(2, 4) ∼= PSL(2, 5).
A14 Ars Math. Contemp. 6 (2013) 1–21
General case: n 6= 1 and m is a prime.
In view of (10) in Proposition 7 there are two cases to consider: E2n : (2n − 1) and D10
provided q′ = 4.
Subcase 1: G01 = G0 ∩G1 ∼= D10, provided q′ = 4.
This situation has been treated in Proposition 13, Subcase 1. We obtained the following
RWPRI and (2T )1 geometry Γ5 = Γ (PSL(2, 4m); PSL(2, 4), D30, D10), provided q±115 is
odd.
Subcase 2: G01 = G0 ∩G1 ∼= E2n : (2n − 1).
By Lemma 19 the possibilities for G1 are E22n : (2n−1) provided m = 2, and PSL(2, 2n).
Notice that if n = 2, PSL(2, 2n) ∼= A5.
2.1. Consider the case where G1 ∼= E22n : (2n − 1) provided m = 2.
In this situation there is only one conjugacy class of PSL(2, 2n) and also one of E22n :
(2n− 1) in PSL(2, 22n). There is one conjugacy class of E2n : (2n− 1) in PSL(2, 2n) and
also one in PSL(2, 22n). Notice that there are 2n + 1 conjugacy classes of E2n : (2n − 1)
in E22n : (2n − 1). Since PSL(2, 22n) is simple and both PSL(2, 2n) and E2n : (2n − 1)
are maximal, PSL(2, 2n) and E2n : (2n − 1) are self-normalized. Moreover the normalizer
of E2n : (2n− 1) in PSL(2, 22n) is itself. We also find that NPSL(2,22n)(E22n : (2n− 1)) =
E22n : (2
2n − 1). Therefore the number of subgroups E22n : (2n − 1) containing a given
subgroup E2n : (2n − 1) in PSL(2, 22n) is equal to
| PSL(2, 22n) |
| E22n : (22n − 1) |
· | E2
2n : (2n − 1) |
| E2n : (2n − 1) |
· (2n + 1) · | E2
n : (2n − 1) |
| PSL(2, 22n) |
= 1.
Hence, the RWPRI and (2T )1 geometry Γ1 = Γ(PSL(2, 22n); PSL(2, 2n), E22n : (2n−1),
E2n : (2
n − 1)) provided n 6= 1 exists and is unique up to conjugacy and also up to
isomorphism.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 16, 64. For q = 16, it is also confirmed by [20].
The particular situation where n = 2, has also been dealt with in Proposition 13, which
showed that Γ(PSL(2, 42);A5, E16 : 3, A4) exists and is unique up to conjugacy, and also
up to isomorphism.
2.2. Consider the case where G0 ∼= G1 ∼= PSL(2, 2n).
In this situation there is only one conjugacy class of PSL(2, 2n) in PSL(2, 2nm). We must
check whether this geometry exists, that is whether there are two subgroups isomorphic
to PSL(2, 2n) in PSL(2, 2nm) that have the subgroup E2n : (2n − 1) in common. Since
PSL(2, 2nm) is simple and PSL(2, 2n) maximal, PSL(2, 2n) is self-normalized. More-
over, the group PSL(2, 2n) contains 2n+1 maximal subgroups E2n : (2n−1) all conjugate.
The normalizer of E2n : (2n − 1) in PSL(2, q) is the group itself. Therefore the number of
subgroups PSL(2, 2n) containing a given subgroup E2n : (2n − 1) in PSL(2, q) is equal to
| PSL(2, 2mn) |
| PSL(2, 2n) |
· | PSL(2, 2
n) |
| E2n : (2n − 1) |
· | E2
n : (2n − 1) |
| PSL(2, 2mn) |
= 1
which implies that the geometry does not exist.
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A15
The particular situation where n = 2, has also been treated in Proposition 13 since
PSL(2, 4) ∼= A5, which showed that Γ(PSL(2, 4m); PSL(2, 4),PSL(2, 4), A4) does not
exist.
Proof of Proposition 24
Proof. Let G0 ∼= PSL(2, pn).
We subdivide our discussion in three cases according to the four G01-candidates given by
(5)-(10) in Proposition 7 namely: A4 provided q′ = 5, S4 provided q′ = 7, A5 provided
q′ = 9, 11 and Eq′ : q
′−1
2 .
In each of these four cases we review all possibilities for G1 given in the previous Lem-
mas as well as the number of classes of such geometries with respect to conjugacy (resp.
isomorphism). In order to determine all geometries under the given conditions we subdi-
vide our discussion in a particular case and a general one depending on whether n = 1 or
not.
Particular case: n = 1.
In this situation q′ = p. The candidates for G01 are Ep : p−12 , A4 provided q
′ = 5, S4
provided q′ = 7, A5 provided q′ = 11.
Subcase 1: G01 = G0 ∩G1 ∼= Ep : p−12 .
By Lemma 21 the only possibility for G1 is PSL(2, p). We distinguish two particular situ-
ations, namely PSL(2, 3) ∼= A4 (provided p = 3) and PSL(2, 5) ∼= A5 (provided p = 5).
All other situations will be treated in the general case, where n can take any value.
1.1 Consider the case where G0 ∼= PSL(2, 3) ∼= A4 ∼= G1.
In this situation G01 is the cyclic group of order 3. There is only one conjugacy class of
A4 in PSL(2, 3m). We must check whether this geometry exists, that is whether there exist
two subgroups isomorphic to A4 in PSL(2, 3m) that have the cyclic subgroup of order 3 in
common. Since PSL(2, 3m) is simple and A4 maximal, A4 is self-normalized. The cyclic
subgroup of order 3 is self-normalized in A4. Moreover A4 contains four cyclic subgroups
of order 3 which are all conjugate. The normalizer of 3 in PSL(2, 3m) is an elementary
abelian subgroup of order 3m. Therefore the number of subgroups A4 containing a given
subgroup 3 in PSL(2, 3m) is equal to
| PSL(2, 3m) |
| A4 |
.
| A4 |
| 3 |
.
| 3m |
| PSL(2, 3m) |
= 3m−1
and thus the geometry exists. There exist exactly 3m−1 − 1 RWPRI and (2T )1 geometries
Γ1 = Γ (PSL(2, 3
m);A4, A4, 3) up to conjugacy when m 6= 3. There exist exactly 8
RWPRI and (2T )1 geometries Γ2 = Γ
(
PSL(2, 33);A4, A4, 3
)
up to conjugacy when m =
3.
Let us deal with the fusion of non-conjugate classes. We find that NPΓL(2,q)(A4) =
(S4 : Cm) and NPΓL(2,q)(3) = (3m.2 : Cm). Therefore the number of subgroups A4
containing a given cyclic subgroup of order 3 in PΓL(2, 3m) is equal to
| PΓL(2, 3m) |
| S4.m |
.
| A4 |
| 3 |
.
| 3m.2.m |
| PΓL(2, 3m) |
= 3m−1.
A16 Ars Math. Contemp. 6 (2013) 1–21
To count the geometries up to isomorphism we need to know the action of PΓL(2, 3m) on
the subgroups A4 containing a given cyclic subgroup of order 3. If we fix A4 ∼= G0 and the
cyclic subgroup of order 3 we know that |NPΓL(2,3m)(A4)∩NPΓL(2,3m)(3)| = |D6|.|Cm|.
We distinguish the cases m = 3 and m 6= 3:
• Let us first assume that m = 3. In this situation there are three subgroups A4 fixed
and the others are exchanged 6 by 6. Thus they merge under the action of PΓL(2, 3m).
Therefore, there exist exactly 3
3−1−3
6 + 1 = 2 RWPRI and (2T )1 geometries Γ2 =
Γ
(
PSL(2, 33);A4, A4, 3
)
up to isomorphism for m = 3.
• Now we assume m 6= 3. Using Fermat’s Last Theorem for m an odd prime we
know that m | 3m−1 − 1. In this situation there is only one A4 ∼= G0 fixed. All others
are exchanged 2m by 2m. Therefore, there exist exactly 3
m−1−1
2m RWPRI and (2T )1
geometries Γ1 = Γ (PSL(2, 3m);A4, A4, 3) up to isomorphism, provided m 6= 3 is an odd
prime.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 27.
1.2 Consider the case where G0 ∼= PSL(2, 5) ∼= A5 ∼= G1.
This RWPRI and (2T )1 geometry Γ (PSL(2, 5m),PSL(2, 5), A5, E5 :2) has already been
dealt with in Proposition 13 and it does not exist.
Subcase 2: G01 = G0 ∩G1 ∼= A4, provided q = 5m with m an odd prime.
This RWPRI and (2T )1 geometry Γ (PSL(2, 5m),PSL(2, 5), A5, A4) has already been
dealt with in Proposition 13, Subcase 2.3 and it does not exist.
Subcase 3: G01 = G0 ∩G1 ∼= S4, when q = 7m with m an odd prime.
By Lemma 23 the possibility for G1 ∼= PSL(2, 7) ∼= G0. In this situation there is only one
conjugacy class of PSL(2, 7) in PSL(2, 11m) and two conjugacy classes of S4. We must
check whether this geometry exists, that is whether there are two subgroups isomorphic
to PSL(2, 7) in PSL(2, 7m) which have the subgroup S4 in common. Since PSL(2, 7m)
is simple and PSL(2, 7) maximal, PSL(2, 7) is self-normalized. The normalizer of S4 in
PSL(2, 7m) and in PSL(2, 7) is the group S4 itself. Therefore the number of subgroups
PSL(2, 7) containing a given subgroup S4 in PSL(2, 7m) is equal to
| PSL(2, 7m) |
| PSL(2, 7) |
· | PSL(2, 7) |
| S4 |
· | S4 |
| PSL(2, 7m) |
= 1
which implies that the geometry does not exist.
Subcase 4: G01 = G0 ∩G1 ∼= A5, when q = 11m with m an odd prime.
By Lemma 25 the possibility for G1 ∼= PSL(2, 11) ∼= G0. In this situation there is only one
conjugacy class of PSL(2, 11) in PSL(2, 11m) and two conjugacy classes of A5. We must
check whether this geometry exists, that is whether there are two subgroups isomorphic to
PSL(2, 11) in PSL(2, 11m) which have the subgroup A5 in common. Since PSL(2, 11m)
is simple and PSL(2, 11) maximal, PSL(2, 11) is self-normalized. The normalizer of A5 in
PSL(2, 11m) and in PSL(2, 11) is the group A5 itself. Therefore the number of subgroups
PSL(2, 11) containing a given subgroup A5 in PSL(2, 11m) is equal to
| PSL(2, 11m) |
| PSL(2, 11) |
· | PSL(2, 11) |
| A5 |
· | A5 |
| PSL(2, 11m) |
= 1
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A17
which implies that the geometry does not exist.
General case:
Let us now discuss the general case, where n can take any value and pn is different from 3
and 5 because these two cases have been discussed in the particular case. The two candi-
dates for G01 are Eq′ : q
′−1
2 and A5 provided q
′ = 32.
Subcase 1: G01 = G0 ∩G1 ∼= Epn : p
n−1
2 .
By Lemma 21 the only possibility for G1 is PSL(2, pn) ∼= G0. In this situation there is only
one conjugacy class of PSL(2, pn) in PSL(2, pnm). We must check whether this geometry
exists, that is whether there are two subgroups isomorphic to PSL(2, pn) in PSL(2, pnm)
that have the subgroup Epn : p
n−1
2 in common. Since PSL(2, q) is simple and PSL(2, p
n)
maximal, PSL(2, pn) is self-normalized. Moreover, the group PSL(2, pn) contains 2n + 1
maximal subgroups Epn :
(
pn−1
2
)
all conjugate. There is only one conjugacy class of Epn :(
pn−1
2
)
in PSL(2, pmn). The normalizer of Epn : p
n−1
2 in PSL(2, q) is the group itself.
Therefore the number of subgroups PSL(2, pn) containing a given subgroup Epn :
(
pn−1
2
)
in PSL(2, q) is equal to
| PSL(2, pmn) |
| PSL(2, pn) |
· | PSL(2, p
n) |
| Epn : p
n−1
2 |
·
| Epn : p
n−1
2 |
| PSL(2, pmn) |
= 1
which implies that the geometry does not exist.
Subcase 2: G01 = G0 ∩G1 ∼= A5, when q = 9m with m an odd prime.
By Lemma 22 the possibility for G1 ∼= PSL(2, 9) ∼= G0. In this situation there is only one
conjugacy class of PSL(2, 9) in PSL(2, 9m) and two conjugacy classes of A5. We must
check whether this geometry exists, that is whether there are two subgroups isomorphic
to PSL(2, 9) in PSL(2, 9m) which have the subgroup A5 in common. Since PSL(2, 9m)
is simple and PSL(2, 9) maximal, PSL(2, 9) is self-normalized. The normalizer of A5 in
PSL(2, 9m) and in PSL(2, 9) is the group A5 itself. Therefore the number of subgroups
PSL(2, 9) containing a given subgroup A5 in PSL(2, 9m) is equal to
| PSL(2, 9m) |
| PSL(2, 9) |
· | PSL(2, 9) |
| A5 |
· | A5 |
| PSL(2, 9m) |
= 1
which implies that the geometry does not exist.
Proof of Proposition 29
Proof. Let G0 ∼= PGL(2, pn).
We subdivide our discussion in four cases, namely the four G01-candidates given by (11),
(12), (13) and (20) in Proposition 7 namely: Epn : (pn − 1), PSL(2, pn), D8 for pn = 3
and the case of S4 provided q = 52. In each of these four cases we review all possibilities
for G1 given in the previous Lemmas as well as the number of classes of such geometries
with respect to conjugacy (resp. isomorphism).
Subcase 1: G01 = G0 ∩G1 ∼= D8, provided q = 9.
A18 Ars Math. Contemp. 6 (2013) 1–21
By Lemma 25 the only case to consider is G0 ∼= G1 ∼= PGL(2, 3).
Since q = 9, there is only one conjugacy class of D8 and D8 is self-normalized in
PSL(2, 9). Therefore the number of subgroups PGL(2, 3) containing a given subgroup
D8 in PSL(2, 9) is equal to
| PSL(2, 9) |
| PGL(2, 3) |
· | PGL(2, 3) |
| D8 |
· | D8 |
| PSL(2, 9) |
= 1.
There are 2 conjugacy classes of PGL(2, 3) in PSL(2, 9). Hence, up to conjugacy and also
up to isomorphism there exists exactly one RWPRI and (2T )1 geometry
Γ3 = Γ (PSL(2, 9); PGL(2, 3); PGL(2, 3);D8). This is confirmed by [3].
Subcase 2: G01 = G0 ∩G1 ∼= Epn : (pn − 1).
By Lemma 26 the possibilities for G1 are Ep2n : (pn − 1) and PGL(2, pn). Notice that S4
is a particular case of PGL(2, pn) provided pn = 3.
2.1. Consider the case where G1 ∼= Ep2n : (pn − 1).
In this situation there is only one conjugacy class of Ep2n : (pn − 1) and two conjugacy
classes of PGL(2, pn) in PSL(2, p2n). Each PGL(2, pn) contains one conjugacy class of
Epn : (p
n − 1) and there are two conjugacy classes of Epn : (pn − 1) in PSL(2, p2n).
Notice that there are pn + 1 conjugacy classes of Epn : (pn − 1) in Ep2n : (pn − 1). Since
PSL(2, p2n) is simple and both PGL(2, pn) and Epn : (pn − 1) maximal, PGL(2, pn)
and Epn : (pn − 1) are self-normalized. Moreover the normalizer of Epn : (pn − 1) in
PSL(2, p2n) is itself. We also find that NPSL(2,p2n)(Ep2n : (pn − 1)) = Ep2n : p
2n−1
2 .
Therefore the number of subgroups PGL(2, pn) containing a given subgroup Epn : (pn−1)
in PSL(2, p2n) is equal to
| PSL(2, p2n) |
| PGL(2, pn) |
· | PGL(2, p
n) |
| Epn : (pn − 1) |
· | Ep
n : (pn − 1) |
| PSL(2, 22n) |
= 1.
Therefore, up to conjugacy, there exist exactly two RWPRI and (2T )1 geometries
Γ1 = Γ
(
PSL(2, p2n); PGL(2, pn);Ep2n : (p
n − 1);Epn : (pn − 1)
)
, corresponding to the
two conjugacy classes of subgroups isomorphic to Epn : (pn − 1).
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of PGL(2, pn) are fused under the action of PGL(2, p2n) and thus also under the
action of PΓL(2, p2n). This is also the case for the two classes of Epn : (pn−1). Therefore,
up to isomorphism there exists exactly one RWPRI and (2T )1 geometry
Γ1 = Γ
(
PSL(2, p2n); PGL(2, pn);Ep2n : (p
n − 1);Epn : (pn − 1)
)
.
This geometry is new and the number of classes up to conjugacy (resp. isomorphism)
is confirmed by MAGMA for q = 9, 25, 49.
2.2 Let us now consider the case where G1 ∼= G0 ∼= PGL(2, pn).
In this situation there are two conjugacy classes of PGL(2, pn) and also two conjugacy
classes of Epn : (pn−1) in PSL(2, p2n). We must check whether this geometry exists, that
is whether there are two subgroups isomorphic to PGL(2, pn) in PSL(2, p2n) that have the
subgroup Epn : (pn−1) in common. Since PSL(2, p2n) is simple and PGL(2, pn) is maxi-
mal, PGL(2, pn) is self-normalized. The subgroup Epn : (pn−1) is also its own normalizer
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A19
in PGL(2, pn) and in PSL(2, p2n). Therefore the number of subgroups PGL(2, pn) con-
taining a given subgroup Epn : (pn − 1) in PSL(2, p2n) is equal to
| PSL(2, p2n) |
| PGL(2, pn) |
· | PGL(2, p
n) |
| Epn : (pn − 1) |
· | Ep
n : (pn − 1) |
| PSL(2, 22n) |
= 1.
Now all Epn : (pn − 1) in PGL(2, pn) are conjugate. This implies that the RWPRI and
(2T )1 geometry Γ
(
PSL(2, p2n); PGL(2, pn); PGL(2, pn);Epn : (p
n − 1)
)
does not exist.
Notice that in the particular case where pn = 3 and thus G1 ∼= S4 ∼= PGL(2, 3) the
geometry does not exist.
Subcase 3: G01 = G0 ∩G1 ∼= PSL(2, pn).
By Lemma 27 the possibilities for G1 are A5 provided pn = 3, PGL(2, pn). Notice
that S4 is a particular case of PGL(2, pn) provided pn = 3.
3.1. Consider the case where G1 ∼= A5 when pn = 3.
There are two conjugacy classes of PGL(2, 3) ∼= S4, of A4 and of A5 in PSL(2, 9). All
A4 in A5 are conjugate, it is also the case for all A4 in S4. Since PSL(2, 9) is simple
and both S4 and A5 are maximal, S4 and A5 are self-normalized. The normalizer of A4
in PSL(2, 9) and in S4 is S4. A4 is self-normalized in A5. The number of subgroups A5
containing a given subgroup A4 in PSL(2, q) is equal to
| PSL(2, q) |
| A5 |
· | A5 |
| A4 |
· | S4 |
| PSL(2, q) |
= 2.
To count the geometries up to conjugacy we need to know if the S4 normalizes each of
the A5 which is not the case because |NPSL(2,q)(A4) ∩ NPSL(2,q)(S4)| = |S4| = 2|A4|.
Therefore, up to conjugacy there exist exactly two RWPRI and (2T )1 geometries Γ2 =
Γ (PSL(2, 9);S4, A5, A4).
Let us deal with the fusion of non-conjugate classes. Following Lemma 11 the two
classes of A4, S4 and A5 are fused under the action of PGL(2, 9) and thus also under
the action of PΓL(2, 9). Therefore, there exists exactly one RWPRI and (2T )1 geometry
Γ2 = Γ (PSL(2, 9);S4, A5, A4) up to isomorphism. This is confirmed by [3].
3.2 Consider the case where G1 ∼= G0 ∼= PGL(2, pn).
In this situation there are two conjugacy classes of PGL(2, pn) and also two conjugacy
classes of PSL(2, pn) in PSL(2, p2n). We must check whether this geometry exists, that is
whether there are two subgroups isomorphic to PGL(2, pn) in PSL(2, p2n) that have the
subgroup PSL(2, pn) in common. Since PSL(2, p2n) is simple and PGL(2, pn) maximal,
PGL(2, pn) is self-normalized. The normalizer of the subgroup PSL(2, pn) in PGL(2, pn)
and in PSL(2, p2n) is PGL(2, pn). Therefore the number of PGL(2, pn) containing a
given PSL(2, pn) is one.
Now all PSL(2, pn) in PGL(2, pn) are conjugate, which implies that the RWPRI and
(2T )1 geometry Γ
(
PSL(2, p2n); PGL(2, pn); PGL(2, pn); PSL(2, pn)
)
does not exist.
Notice that in the particular case where pn = 3 we get G1 ∼= S4 ∼= PGL(2, 3).
Subcase 4: G01 = G0 ∩G1 ∼= S4, provided q = 52.
By Lemma 28 the only case to consider is G0 ∼= G1 ∼= PGL(2, 5).
A20 Ars Math. Contemp. 6 (2013) 1–21
In this situation where q = 25, there are two conjugacy classes of PGL(2, 5) and also
two conjugacy classes of S4 in PSL(2, 52). Since PSL(2, 52) is simple and PGL(2, 5) is
maximal, PGL(2, 5) is self-normalized and S4 is self-normalized in PGL(2, 5) and also
in PSL(2, 52). Therefore the number of PGL(2, 5) containing a given S4 is one. Now
all S4 in PGL(2, 5) are conjugate, which implies that the RWPRI and (2T )1 geometry
Γ
(
PSL(2, 52); PGL(2, 5); PGL(2, 5);S4
)
does not exist.
Case of Table 3, geometry Γ2
We know that s ≥ 2. Consider a path (a, b, c) as in the preceding case. Here, Gabc =
Z3. This acts on the three 1-elements d1, d2, d3 other than b in c⊥. The action is transitive
since otherwise Z3 would be in the kernel of the action of Gc on c⊥. This kernel for the
action of S4 on the cosets of D6 is reduced to the identity, a contradiction. This provides
s ≥ 3 for paths starting at a 0− element.
Next consider a path (h, i, j) as in the preceding case. Here, Ghij = Z2. This acts on
the two 0-elements k1, k2 other than i in j⊥. The action is transitive since otherwise Z2
would be in the kernel of the action of Gj on j⊥. This kernel for the action D18 on the
cosets of D6 is a group Z3, a contradiction. Hence s ≥ 3.
Applying Leemans’ method we get s = 2 or 3. Thus s = 3.
Case of Table 3, geometry Γ5
This geometry Γ (PSL(2, q);D16, S4, D8) is known as a locally 7-arc-transitive graph
due to Wong [22], hence s = 7.
Case of Table 3, geometry Γ7 and Γ8.
This geometry Γ (PSL(2, q);S4, S4, D8) is known as a locally 4-arc-transitive graph
due to Biggs-Hoare [1], hence s = 4 in this case.
Case of Table 4, geometry Γ1
We know that s ≥ 2. Consider a path (a, b, c) as in the preceding case. Here, Gabc =
2n. This acts on the 2n elements of type 1, d1, ..., d2n other than b in c⊥. The action is
transitive since otherwise a subgroup of order 2 would be in the kernel of the action of Gc
on c⊥. This kernel for the action of PSL(2, 2n) on the cosets of 2n : (2n − 1) is reduced
to the identity, a contradiction. This provides s ≥ 3 for paths starting at a 0− element.
Next consider a path (h, i, j) as in the preceding case. Here, Ghij = Z2n−1. This acts
on the 2n−1 elements of type 0, k1, k2n−1 other than i in j⊥. The action is transitive since
otherwise Zt with t prime and dividing 2n − 1 would be in the kernel of the action of Gj
on j⊥. This kernel for the action of 22n : (2n − 1) on the cosets of 2n : (2n − 1) is not
determined but its order divides 2n, a contradiction. Hence s ≥ 3.
Applying Leemans’ method we get s = 2 or 3. Thus s = 3.
Case of Table 6, geometry Γ1
We know that s ≥ 2. Consider a path (a, b, c) as in the preceding case. Here, Gabc =
pn. This acts on the pn elements of type 1, d1, ..., dpn other than b in c⊥. The action is
F. Buekenhout et. al.: On the rank two geometries of the groups PSL(2, q): part II A21
transitive since otherwise a subgroup of order p would be in the kernel of the action of Gc
on c⊥ . This kernel for the action of PGL(2, 2n) on the cosets of pn : (pn − 1) is reduced
to the identity, a contradiction. This provides s ≥ 3 for paths starting at a 0− element.
Next consider a path (h, i, j) as in the preceding case. Here, Ghij = Zpn−1. This acts
on the pn−1 elements of type 0, k1, kpn−1 other than i in j⊥. The action is transitive since
otherwise Zt with t prime and dividing pn − 1 would be in the kernel of the action of Gj
on j⊥. This kernel for the action of p2n : (pn − 1) on the cosets of pn : (pn − 1) is not
determined but its order divides pn, a contradiction. Hence s ≥ 3.
Applying Leemans’ method we get s = 2 or 3. Thus s = 3.