Regular balanced Cayley maps on ${\rm PSL}(2,p)$

0.15 MB·

by  Haimiao Chen

Regular balanced Cayley maps on PSL(2,p)
6
1
0 Haimiao Chen
∗
2
Beijing Technology and Business University, Beijing, China
l
u
J

5

Abstract
]
O
A regular balanced Cayley map (RBCMforshort)onafinite group
C
Γ is an embedding of a Cayley graph on Γ into a surface, with some
.
h special symmetric property. People have classified RBCM’s for cyclic,
t dihedral, generalized quaternion, dicyclic, and semi-dihedral groups.
a
m In this paper we classify RBCM’s on the group PSL(2,p) for each
prime number p>3.
[

2
v 1 Introduction
1
5
Let Γ be a finite group and let Ω be a generating set not containing the
2
5 identity and ω 1 Ω whenever ω Ω. The Cayley graph Cay(Γ,Ω) is
−
0 ∈ ∈
the graph having vertex set V = Γ and arc set Γ Ω, where (η,ω) means
. ×
1 the arc from the vertex η to ηω. If ρ is a cyclic permutation on Ω, then
0
it gives a cyclic order to the set of arcs starting from η for each η, via
6
1 (η,ω) (η,ρ(ω)). This determines a uniquecellular embeddingof the Cay-
: 7→
v ley graph Cay(Γ,Ω) into a closed oriented surface, where “cellular” means
i that each connected component of the complement of the embedded graph
X
is homeomorphic to a disk. Such an embedding is called a Cayley map and
r
a is denoted by (Γ,Ω,ρ).
CM
An isomorphism between two Cayley maps is an isomorphism of the
underlying graphs which is compatible with cyclic orders.
A Cayley map (Γ,Ω,ρ) is called regular if its automorphism group
CM
acts transitively on the arc set, and called balanced if ρ(ω 1) = ρ(ω) 1 for
− −
all ω Ω. From now on we abbreviate “regular balanced Cayley map”
∈
to “RBCM”. The following proposition collects some well-known facts, for
which one may refer to [10–12].
∗Email: [email protected]. This work is supported by NSFC-11401014.
1
Proposition 1.1. (a) A Cayley map (Γ,Ω,ρ) is a RBCM if and only
CM
if ρ extends to an isomorphism of Γ.
(b) For a RBCM (Γ,Ω = ω : 1 i m ,ρ) with ρ(ω ) = ω , all
i i i+1
CM { ≤ ≤ }
of the elements of Ω have the same order, and
(I) either m = 2n and ω = ω 1 for all i,
i+n i−
(II) or all the ω ’s are involutions, (i.e., have order 2).
i
A RBCM in the case (I) or (II) is said to be of type I or II, and denoted by
I-RBCM or II-RBCM, respectively.
(c) If (Γ,Ω,ρ) and (Γ,Ω,ρ) are two RBCM’s of the same type,
′ ′ ′
CM CM
then they are isomorphic if and only if there exists an isomorphism f :Γ
→
Γ such that f(Ω)= Ω and f ρ= ρ f.
′ ′ ′
◦ ◦
A RBCM on a group Γ is not only a combinatorial object with good
symmetry property, but also can be considered as an extra structure on
Γ. So far, RBCM’s have been classified for cyclic, dihedral, generalized
quaternion, dicyclic, and semi-dihedral groups; see [7,9,12]. Recently, the
first author [2] classified RBCM’s for several subclasses of abelian p-groups.
In this paper, we classify RBCM’s on PSL(2,p) with p > 3 a prime
number. Thisisthefirsttimetoobtainconcreteresultsforafamilyofsimple
groups. Twokeyingredientsareinvolvedinouridea: (i)eachautomorphism
of PSL(2,p) is the conjugation by a unique element of PGL(2,p), and (ii)
maximal subgroups of PSL(2,p) are known, as recalled in Proposition 2.2.
By Proposition 1.1, to classify 2n-valent I-RBCM’s on PSL(2,p), it suffices
to find all pairs (σ,ω) with σ PGL(2,p) and ω PSL(2,p) such that σ
∈ ∈
has order 2n, σnωσ n = ω 1 and σiωσ i,i = 1,...,n generate PSL(2,p),
− − −
(these conditions ensure that the conjugacy class of ω under σ has size 2n).
Note that the last condition is equivalent to that the subgroup generated by
σiωσ i,i = 1,...,n is not contained in any maximal subgroup of PSL(2,p).
−
The RBCM’s determined by two pairs (σ,ω) and (σ ,ω ) are isomorphic if
′ ′
andonlyifthereexistsτ PGL(2,p)suchthatσ = τστ 1 andω = τωτ 1.
′ − ′ −
∈
The method for classifying II-RBCM’s is similar.
The content is organized as follows. In Section 2 we recall some well-
known facts about PSL(2,p) and PGL(2,p). In Section 3 and 4 we classify
I-RBCM’s and II-RBCM’s on PSL(2,p), respectively; in each section we
separately deal with the cases p = 5 and p > 5, because PSL(2,5) = A
5
plays a special role in subgroup structure of PSL(2,p).
Notation 1.2. For a RBCM (PSL(2,p),Ω,ρ), if Ω = σiωσ i: 1 i
−
CM { ≤ ≤
m with σ PGL(2,p), then we denote (PSL(2,p),Ω,ρ) by σ, with
ω
} ∈ CM CM
theunderstandingthatthepermutationρisgivenbyσiωσ i σi+1ωσ (i+1).
− −
7→
2
For a set X, denote its cardinality by #X.
For an element µ of a group Γ, denote its order by µ . Given a set of
| |
elements µ ,...,µ Γ, denote the subgroup they generate by µ ,...,µ .
1 ℓ 1 ℓ
∈ h i
For two permutations ω and ψ, use ωψ to mean “first do ω, then do ψ”.
For instance, (12)(23) = (132).
Denote the 2 2 identity matrix by ε.
×
2 Preliminary
For a finitefield F, let F denote themultiplicative groupof units. Consider
×
F as a subfield of F . Fix a generator e of the cyclic group F and fix a
p p2 ×p
square root √e ∈ F×p2, then elements of Fp2 are linear combinations a+b√e
with a,b F . The norm
p
∈
N :F×p2 → F×p,
(a+b√e) =a2 eb2 = (a+b√e)p+1,
N −
is a surjective homomorphism (see Problem 1 on Page 87 of [8]).
Fix a generator w1 +w2√e of the cyclic group F×p2 and let w = w1/w2,
then w2 e has no square root in F .
p
−
Let
e 0 1 1 w e
α = , β = , γ = . (1)
0 1 0 1 1 w
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Itiswell-knownthat(onemayreferto[4],Page68)eachelementofPGL(2,p)
is conjugate to αk for some k, or β, or γℓ for some ℓ. Furthermore, αk′ is
conjugate to αk if and only if k = k, and γℓ′ is conjugate to γℓ if and only
′
±
if ℓ = ℓ.
′
±
Thefollowing enables ustoconveniently dealwiththeordersof elements
of PSL(2,p), and can be proved by repeatedly applying Hamilton-Cayley
Theorem η˜2 = tη˜ ε.
−
Proposition 2.1. Let η˜ SL(2,p) with tr(η˜) = t, and let η PSL(2,p) de-
∈ ∈
note the image of η˜under the quotienthomomorphism SL(2,p) PSL(2,p).
→
(a) η = 2 if and only if t = 0,
| |
(b) η = 3 if and only if t2 = 1,
| |
(c) η = 4 if and only if t2 = 2,
| |
3
(d) η = 5 if and only if (t2 1)2 =t2.
| | −
The following result is quoted from Proposition 2.1 of [5]; also see [3].
Proposition 2.2. Suppose p 5. Theneachmaximal subgroup of PSL(2,p)
≥
has one of the following forms:
(i) the stabilizer of a point on the projective line P1(F );
p
(ii) D , the dihedral group of order p 1;
p 1
± ±
(iii) A , S or A .
4 4 5
Remark 2.3. A subgroup of form (i) is the same as one whose elements
have a common eigenvector.
A subgroup of form (ii) means one isomorphic to D , and similarly
p 1
±
for (iii). Subgroups in (ii) or (iii) do not always exist, and even when they
exist, they may not be maximal.
Finally, recall some facts about S , PSL(2,5) = A and PGL(2,5) = S :
4 5 5
Proposition 2.4. (a) Nontrivial conjugacy classes of S are listed below
5
(using [µ] to denote the conjugacy class containing µ):
[(12345)], [(123)], [(12)(34)], [(12)(345)], [(1234)], [(12)],
where the first three classes are contained in A .
5
(b) S has a presentation X,Y X2,Y3,(XY)4 , so any group generated
4
h | i
by two elements µ,η with µ = 2, η = 3 and µη = 4 is a quotient
| | | | | |
of S .
4
(c) A has a presentation X,Y X2,Y3,(XY)5 , so any nontrivial group
5
h | i
generated by two elements µ,η with µ = 2, η = 3 and µη = 5 is
| | | | | |
isomorphic to A .
5
(d) Each non-abelian proper subgroup of A is isomorphic to D , D or
5 6 10
A .
4
We explain (d). Let Γ A be a non-abelian proper subgroup, then
5
≤
#Γ 6,10,12,15,20 . Clearly Γ is dihedral if #Γ 6,10 . By [1]
∈ { } ∈ { }
Theorem 7.8.1 which classifies groups of order 12, Γ = A if #Γ = 12.
4
By [1] Theorem 7.7.7 (a), each group of order 15 is cyclic, so #Γ never
equals 15. Finally, if #Γ = 20, then by Sylow’s Theorem, Γ has exactly one
subgroup of order 5, so all the other 15 elements are involutions, which are
exactly allthe15involutions inA , buttheproductof(12)(34) and(12)(35)
5
is (345), whose order cannot divide 20.
4
3 Type I regular balanced Cayley maps
3.1 I-RBCM’s on PSL(2,5) = A
5
Suppose σ is a I-RBCM on PSL(2,5) = A , with ω A , ω > 2,
ω 5 5
CM ∈ | |
σ S , σ = 2n. Clearly 2 < 2n 6, hence n = 2 or n = 3. We may
5
∈ | | ≤
assumethatω isoneofrepresentatives ofconjugacyclassesasinProposition
2.4 (a), namely, ω = (123) or ω = (12345). Denote ω = σiωσ i, 1 i 2n.
i −
≤ ≤
There are four possibilities.
If n= 2 and ω = (123), then it follows from ω = ω 1 that σ = (k4ℓ5)
2 −
•
with k,ℓ 1,2,3 . We may find τ (1),ω,ω2 such that
τστ 1{= (}142⊂5){and } τστ−1 = σ. Just∈ass{umeσ = (}1425). Now
− ω ∼ ω
CM CM
ω = (543), ωω ωω 1 = (14)(23) has order 2, and (ωω ωω 1)ω =
1 1 1− 1 1− 1
ωω ω = (13254) has order 5, thus ωω ωω 1,ω = A and also
1 h 1 1− 1i 5
ω ,ω = A .
1 2 5
h i
If n = 2 and ω = (12345), then by replacing σ by ωkσω k for some
−
•
k if necessary, we may assume σ fixes the letter 5. Hence σ = (1243)
or σ = (1342), due to the condition ω = ω 1. If σ = (1243), then
2 −
ω = (31425) = ω 2; if σ = (1342), then ω = (24135) = ω2. In
1 − 1
neither case ω,σωσ 1 = A , as A is not cyclic.
− 5 5
h i
If n = 3 and ω = (123), then σ = (k k )(ℓ ℓ ℓ ) with k ,k
1 2 1 2 3 1 2
• { } ⊂
1,2,3 due to ω = ω 1. By replacing σ by ωkσω k for some k if
3 − −
{ }
necessary, we may assume σ = (12)(345), hence ω = (215), ω =
1 2
(124). Now ωω 1 = (15)(23), (ωω 1)ω = (15234), so ωω 1,ω =
1− 1− 2 h 1− 2i
A and also ω ,ω ,ω = A .
5 1 2 3 5
h i
If n = 3 and ω = (12345), then it is impossible that σ3ωσ 3 = ω 1,
− −
•
since σ3 is a transposition.
(1425)
Theorem 3.1. Each 4-valent I-RBCM on A is isomorphic to ,
5 CM(123)
(12)(345)
each 6-valent type I RBCM on A is isomorphic to , and there
5 CM(123)
does not exist a 2n-valent I-RBCM for n > 3.
3.2 I-RBCM’s on PSL(2,p) for p > 5
Suppose σ is a 2n-valent I-RBCM with σ = 2n 4, noting that
ω
CM | | ≥
PSL(2,p) is not cyclic. Now that σ is conjugate to αk or γℓ, we may just
assume σ = αk with 1 k (p 1)/2, or σ = γℓ with 1 ℓ (p+1)/2.
≤ ≤ − ≤ ≤
5
Setting
ε, if σ = αk,
τ = √e e (2)
 − , if σ = γℓ,
√e e

(cid:18) (cid:19)
one has 
(√e)k, if σ = αk,
s 0
τστ−1 = 0 1/s ∈ PGL(2,p2), with s = ( w−√e −ℓ, if σ = γℓ;
(cid:18) (cid:19) √w2 e
−
(cid:16) (cid:17) (3)
note that s has order 4n as an element of F×p2, hence s2n = −1.
Suppose
a b
τωτ 1 = PSL(2,p2), with a2 bc= 1. (4)
−
c d ∈ −
(cid:18) (cid:19)
a b d b
Theconditionσnωσ−n =ω−1 isequivalentto c −d = c −a ,
(cid:18) − (cid:19) (cid:18) − (cid:19)
implying a = d.
Lemma 3.2. The elements σiωσ i,i = 1,...,n generate PSL(2,p) if and
−
only if abc = 0 and 2a2 = 1 when n =2.
6 6
Proof. Let Γ = ψ ,...,ψ with ψ = τσiωσ iτ 1. The task is to show
1 n i − −
h i
τ 1Γτ = PSL(2,p). We have
−
a s2ib a2+s2(i j)bc (s2i +s2j)ab
ψ = , ψ ψ = − .
i s 2ic a i j (s 2i +s 2j)ac a2+s2(j i)bc
− − − −
(cid:18) (cid:19) (cid:18) (cid:19)
(5)
If a = 0, then ψ is counter-diagonal, so each element of Γ is either
i
diagonal or counter-diagonal, hence
#Γ 2(p2 1) < p(p2 1)/2 = #PSL(2,p).
≤ − −
If bc = 0, then ψ is unipotent, hence also #Γ < #PSL(2,p).
i
Thus a necessary condition for τ 1Γτ = PSL(2,p) is abc = 0. We show
−
6
that this is also sufficient except for the case when n = 2 and 2a2 = 1.
Suppose abc = 0. Then ψ > 2, as tr(ψ )= 2a = 0.
i i
6 | | 6
(a) If Γ D , then ψ Z/mZ, but by Eq.(5) Γ is not abelian.
2m i
≤ ∈
6
(b) If Γ is contained in a subgroup of form (i) in Proposition 2.2, then
ψ and ψ have a common eigenvector (x,y)t F2 , hence both (x,y) and
1 2 ∈ p2
(sx,y)t are eigenvectors of ψ ; this implies x = 0or y = 0, whichcontradicts
1
the assumption that bc= 0.
6
(c) If Γ S , then Γ = A or S according to (a). It is well-known that
4 4 4
≤
each automorphism of A or S is the conjugation by some element in S ,
4 4 4
whose order belongs to 2,3,4 , hence n = 2, s4 = 1 and tr(ψ ψ ) = 2a2,
1 2
{ } −
using Eq.(5). If ψ = 3, then Γ = A , and 4a2 = (tr(ψ ))2 = 1, so
i 4 1
| |
tr(ψ ψ ) = 1/2, implying ψ ψ = 2,3, but this contradicts ψ ψ A .
1 2 1 2 1 2 4
| | 6 ∈
Thus ψ = 4, and 2a2 1= tr(ψ2) = 0, i.e., 2a2 = 1.
| i| − 2
Conversely, if n = 2 and 2a2 = 1, then (denoting τστ 1 by ς)
−
(tr(ςψ ))2 = (s+s 1)2a2 = 1, tr(ς2ψ ) = (s2+s 2)a2 = 0,
4 − 4 −
hence ςψ = 3and ψ 1ς 2 = ς2ψ = 2;thistogetherwith (ψ 1ς 2)(ςψ ) =
| 4| | 4− − | | 4| | 4− − 4 |
ς = 4 implies that ς,ψ = ψ 1ς 2,ς̟ is a quotient of S . Thus
| | h 4i h 4− − i 4
#Γ 24.
≤
(d) If Γ A , then Γ = A by (a) and Proposition 2.4 (d). By Theorem
5 5
≤
3.1, there are two possibilities; in both cases ψ = 3 hence 4a2 = 1.
i
| |
(i) n= 2(sothats4 = 1),thentr(ψ ψ )= 2a2 = 0, 1andisnotaroot
1 2
− 6 ±
of (t2 1)2 = t2, hence ψ ψ / 2,3,5 , contradicting ψ ψ A .
1 2 1 2 5
− | | ∈ { } ∈
(ii) n= 3(sothats4 s2+1 = 0),thenthereexistsanisomorphismA = Γ
− 5 ∼
sending(215) to ψ and (124) to ψ , henceit sends (154) = (215)(124)
1 2
to ψ ψ . But tr(ψ ψ )= 3a2 1 = 1/4 = 1, a contradiction.
1 2 1 2
− − 6 ±
Ifσ = αk,then2n p 1andk = (p 1)u/2nforsomeuwith(u,2n) = 1,
| − −
b 1 0
1 u < n. Now ϕ:= − commutes with αk and
≤ 0 1
(cid:18) (cid:19)
a 1
ϕωϕ 1 = =:ω(i,a), (6)
− a2 1 a
(cid:18) − (cid:19)
thus αk = αk . Furthermore, for a = a, ω(i,a) is not conjugate to
CMω ∼ CMω(i,a) 6 ′
ω(i,a′), as tr(ω(i,a)) 6= tr(ω(i,a′)), so CMαωk(i,a) ≇ CMαωk(i,a′).
Ifσ = γℓ,then2n p+1andℓ =(p+1)v/2n forsomev with(v,2n) = 1,
|
1 v < n. Note that, if n = 2, then 2a2 never equals 1 since the Legendre
≤
7
symbol (2/p) = 1 by Theorem 1 (b) on Page 53 of [6]. By Eq.(2) and (4),
−
a+(b+c)/2 √e(b c)/2 x ez
ω = − = − PSL(2,p), (7)
(c b)/2√e a (b+c)/2 z 2a x ∈
(cid:18) − − (cid:19) (cid:18) − (cid:19)
where x = a+(b+c)/2 and z = (c b)/2√e are elements of F , hence
p
−
(x a)2 ez2 = bc = a2 1, (8)
− − −
i.e., (x a +z√e) = a2 1. When a is fixed, this equation has p + 1
N − −
solutions, since the homomorphism N : F×p2 → F×p is surjective. Choose and
fix a solution (x ,z ), and put
i,a i,a
x ez
ω˜(i,a) = i,a − i,a . (9)
z 2a x
i,a i,a
(cid:18) − (cid:19)
Noticing
w √e 0 a b
τγτ−1 = − , and τω˜(i,a)τ−1 = ′
0 w+√e c a
′
(cid:18) (cid:19) (cid:18) − (cid:19)
for some b,c with bc = a2 1, we easily see that the p+1 elements
′ ′ ′ ′
−
γhω˜(i,a)γ h, h = 1,...,p+1
−
are distinct from each other. Thus for the present ω, there exist (a unique)
h 1,...,p+1 such that γhω˜(i,a)γ h = ω, and hence γℓ = γℓ .
∈ { } − CMω ∼ CMω˜(i,a)
Theorem 3.3. Suppose is a 2n-valent I-RBCM on PSL(2,p) with p > 5,
M
then 2n p 1 or 2n p+1.
| − |
(i) If 2n p 1, then = α(p−1)u/2n for a unique pair (a,u) such that
| − M∼ CMω(i,a)
a / 1,0 , (u,2n) =1, 1 u < n, and moreover, 2a2 = 1 if n =2.
∈ {± } ≤ 6
γ(p+1)v/2n
(ii) If 2n p+1, then = for a unique pair (a,v) such that
| M ∼ CMω˜(i,a)
a / 1,0 , (v,2n) = 1 and 1 v < n.
∈ {± } ≤
4 Type II regular balanced Cayley maps
4.1 II-RBCM’s on PSL(2,5) = A
5
Suppose σ is an n-valent II-RBCM on PSL(2,5) = A , with ω A ,
ω 5 5
CM ∈
ω = 2, and σ S , σ = n 6. Since a nonabelian group generated by
5
| | ∈ | | ≤
8
two involutions must be dihedral, we have n > 2. Also note that the action
of the conjugation by σ on the set of involutions of A has at most one fixed
5
element, hence n 15 or n 14 which implies n = 3 or n = 5.
| |
Denote ω = σiωσ i,i = 1,...,n, and denote Γ = ω ,...,ω .
i − 1 n
h i
Ifn = 3,wemayassumeσ = (123)andωfixestheletter1. Thecondition
ω ,ω ,ω = A requiresω (24)(35),(25)(34) . Theconjugation by(45)
1 2 3 5
h i ∈{ }
fixes σ and takes (25)(34) to (24)(35), so let us just assume ω = (24)(35).
Then ω = (14)(25), ω = (15)(34), ω ω ω ω = (153), ω (ω ω ω ω ) =
1 2 1 2 1 3 1 1 2 1 3
ω ω ω = (14523), thus ω ,ω ω ω ω = A , and also Γ = A .
2 1 3 1 1 2 1 3 5 5
h i
If n = 5, we may assume σ = (12345). There are three possibilities.
(i) If ω = (12)(34), then ω = (15)(23), ω = (12)(45), ω = (15)(34),
5 1 2 3
ω = (23)(45), soω ω = (152),ω ω = (13542), ω ω ω ω = (13)(45),
4 3 5 5 4 5 4 5 3
hence ω ω ω ω ,ω ω = A and also Γ = A .
5 4 5 3 3 5 5 5
h i
(ii) If ω = (13)(24), then ω = (13)(25), ω = (14)(25), ω = (14)(35),
5 1 2 3
ω = (24)(35), soω ω = (135),ω ω = (13452), ω ω ω ω = (25)(34),
4 4 5 5 2 5 2 5 4
hence ω ω ω ω ,ω ω = A and also Γ = A .
5 2 5 4 4 5 5 5
h i
(iii) If ω = (14)(23), then ω = (12)(35), ω = (15)(24), ω = (13)(45),
5 1 2 3
ω = (25)(34), soω ω = (14253). Onecancheck thatω = ω (ω ω )i,
4 4 5 i 5 4 5
i= 1,...,5, hence Γ = ω ,ω = D , which is impossible.
h 4 5i ∼ 10
(123)
Theorem 4.1. Each 3-valent II-RBCM on A is isomorphic to ,
5 CM(24)(35)
(12345) (12345)
each 5-valent II-RBCM on A is isomorphic to or ,
5 CM(12)(34) CM(13)(24)
and there does not exist an n-valent II-RBCM on A for n = 3,5.
5
6
4.2 II-RBCM’s on PSL(2,p) for p > 5
Let σ bean n-valent II-RBCM on PSL(2,p), with ω = 2 and σ = n>
ω
CM | | | |
2 (as PSL(2,p) is not dihedral). We may assume σ is equal to αk, β or γℓ.
Suppose
x y
ω = with x2+yz = 1. (10)
z x −
(cid:18) − (cid:19)
If n = p, then σ = β, and
x+iz y 2ix i2z
βiωβ−i = − − , 1 i p. (11)
z x iz ≤ ≤
(cid:18) − − (cid:19)
Lemma 4.2. The elements βiωβ i,i = 1,...,p generate PSL(2,p) if and
−
only if z = 0.
6
9
Proof. Let Γ = ω ,...,ω with ω = βiωβ i. If z = 0, then each ω is
1 p i − i
h i
upper-triangular, and so is each element of Γ, hence Γ = PSL(2,p).
6
Suppose z = 0, we shall prove that Γ = PSL(2,p).
6
Firstly, Γ is not contained in the stabilizer of any point of P1(F ): if
p
ξ F2 is a common eigenvector of ω and ω , then ξ and βξ are both
p 1 2
∈
eigenvectorsofω ,henceξandβξarelinearlydependent,whichisimpossible
1
when z = 0.
6
Secondly, we show that Γ is not contained in any other maximal sub-
group, by counting involutions. For a group ∆, let I(∆) denote the set of
involutions of ∆. It is obvious that βiφβ i = φ for any φ PSL(2,p) unless
−
6 ∈
φ is upper-unitriangular, in which case φ / I(Γ), so β acts freely on I(Γ)
∈ h i
by conjugation. Thus
(⋆) p #I(Γ) and #I(Γ) p.
| ≥
It is impossible that Γ D , since #I(D ) = p 1< p.
p 1 p 1
• ≤ − − −
If Γ D , then Γ = D since any p involutions generate D ;
p+1 p+1 p+1
≤
but #I(D )= p+1 is not a multiple of p, contradicting (⋆).
p+1
It is impossible that Γ A , since #I(A ) = 3< p.
4 4
• ≤
If Γ S , then #I(Γ) = p = 7 since #I(S ) = 9, but no subgroup of
4 4
• ≤
S contains exactly 7 involutions.
4
If Γ A , then #I(Γ) = p 7,11,13 since #I(A ) = 15, but A
5 5 5
• ≤ ∈ { }
does not have such a subgroup Γ, as can be checked.
Let m be an integer whose residue class modulo p is x/z, then
−
0 1/z
βmωβ−m = − =:̟(z), (12)
z 0
(cid:18) (cid:19)
hence β = β . Furthermore, for z = z , there does not exist τ with
CMω ∼ CM̟(z) 6 ′
τβτ 1 = β and τ̟(z)τ 1 = ̟(z ), so β ≇ β .
− − ′ CM̟(z) CM̟(z′)
If n = p, then take τ as in Eq.(2), so that
6
s 0
ς := τστ 1 = PGL(2,p2)
−
0 1/s ∈
(cid:18) (cid:19)
10