The Möbius function of \mboxPSL(3,2p) for any prime p
Martino Borello, Francesca Dalla Volta, Giovanni Zini
Abstract.
Let G be the simple group PSL(3,2p), where p is a
prime number. For any subgroup H of G, we compute the Möbius
function of H in the subgroup lattice of G. To
this aim, we describe the intersections of maximal subgroups of G.
We point out some connections of the Möbius function with
other combinatorial objects, and, in this context, we compute the
reduced Euler characteristic of the order complex of the subposet of
r-subgroups of PGL(3,q), for any prime r and any prime
power q.
Keywords: Möbius function, subgroup lattice, Euler characteristic
2010 MSC: 05E15, 20D30, 20D06
1. Introduction
Let G be a finite group. The Möbius function of G is defined
recursively by μ(G)=1 and μ(H)=−∑K:H<K≤Gμ(K) for any H<G. It was introduced independently by Weisner [27] and Hall [11]; in
particular, Hall provides a formula to enumerate generating tuples of elements of G.
For any group K, let σn(K) and
ϕn(K) denote respectively the number of ordered n-tuples and generating n-tuples of
elements of K. Then σn(G)=∑H:H≤Gϕn(H), and by the Möbius
inversion formula ([26, Prop. 3.7.1]) one gets ϕn(G)=∑H:H≤Gσn(H)⋅μ(H).
Let ProbG(n) be the probability that n random elements of G generate G. Then
[TABLE]
It is possible to define the following complex function (see [19]):
[TABLE]
It satisfies PG(t)=ProbG(t) for t∈N.
In case of a profinite group G, the above function is generalized as follows:
[TABLE]
where H ranges over all open subgroups of G. Mann conjectured
that this sum is absolutely convergent in some half complex plane
for positively finitely generated (PFG) groups. This
conjecture is verified if the following two facts hold for any PFG
profinite group (see [19, 20]): ∣μ(H)∣ is bounded by a
polynomial function in [G:H]; the number of subgroups H of index
n satisfying ∣μ(H)∣=0 grows at most polynomially in n.
The conjecture was reduced by Lucchini [18] to the following one:
there exist c1,c2∈N such that, for any almost simple group G, ∣μ(H)∣≤[G:H]c1 for any H<G; and, for any n∈N, the number of subgroups H<G of index n in G satisfying G=Hsoc(G) and μ(H)=0 is upper bounded by nc2.
This was proved in [4] in the case of alternating and symmetric groups.
Not very much is known about the exact values of μ(H) when G is a
simple group; up to our knowledge, the only infinite families of
non-abelian simple groups for which the Möbius function is
completely known are the following.
The groups \mboxPSL(2,q); for q prime see [11], for any prime power q see [7], where also the groups \mboxPGL(2,q) are completely worked out.
The Suzuki groups Sz(q) for any non-square power q of 2; see [9].
The Ree groups Ree(q) for any non-square power q of 3; see [23].
The 3-dimensional unitary groups \mboxPSU(3,22n) for any n>0; see [28].
For any of these families Mann’s conjecture is verified. In this
paper we consider the case of 3-dimensional projective general
linear groups \mboxPGL(3,q); again, Table 1
confirms Mann’s conjecture when q=2p with prime p.
The main result of this paper, Theorem 3.1, provides the Möbius function of the simple group G=\mboxPSL(3,2p) for any odd prime p (note that G=\mboxPGL(3,2p)). The subgroups with non-zero
Möbius functions are summarized in Tables 1. In
the second column we specify the Aschbacher’s classes in which H
is contained, with an N when H is not maximal in G [2];
for example C1,C2 (N) means that H is
intersection of maximal groups in C1 and
C2 and H is not maximal.
For the sake of
completeness, Table 2 summarizes the case p=2, which can be easily computed using GAP, via the full table of marks of \mboxPSL(3,4).
We observe that, when considering the group \mboxPSL(3,2n) with a non-prime n, other maximal subgroups appear, such as subgroups isomorphic to \mboxPSL(3,2n′), or \mboxPGL(3,2n′), or \mboxPSU(3,2n′), or PG(3,2n′), for certain divisors n′ of n.
This makes the computations quite longer and not easily tractable.
Throughout the paper we will use the following group-theoretic
notation which is based on the ATLAS [5]: given two
subgroups H and K of G, H×K denotes their direct
product, H:K a split extension of H by K, and H.K a
(split or non-split) extension of H by K; Cn a cyclic group
of order n; Dn a dihedral group of order n; Epn an
elementary abelian group of order pn; Hm+n denotes the
extension Hm.Hn; \mboxSym(n) and \mboxAlt(n) denote
respectively the symmetric and the alternating group of degree n;
Sylr(G) denotes a Sylow r-subgroup of the group G under
consideration. For simplicity, we will use Sr for Sylr(G); when q is a power of 2, S2≅Eq1+2.
In order to prove Theorem 3.1, the intersections of
maximal subgroups of G, their conjugacy classes and normalizers
are carefully investigated and determined; for each subgroup, the
Möbius function is computed. To this aim, we apply geometric
arguments regarding the geometry of G and its subgroups in their
natural action on the plane PG(2,q) over Fq, and more
generally in their action on the plane
PG(2,Fq) over the algebraic closure of
Fq.
Finally, we point out that the Möbius function of a finite group
G has connections with different areas of
mathematics, in which the Möbius inversion formula turns out to be
applicable. We list some objects whose enumeration can be performed by means of the Möbius function of G.
- (1)
Epimorphisms from a free group of finite rank to the group G; see [9].
2. (2)
Graphs Γ~ which are a G-covering of a given graph Γ; see [17].
3. (3)
The structure of the group of units of the monoid of cellular automata over the configuration space AG, for a given finite set A; see [3, Section 4].
4. (4)
Reduced Euler characteristic of the order complex of posets P associated to G.
We will explicitly work out the computation of point (4) in Section 5, where we consider the order complex of the finite poset P=Lr of r-subgroups of \mboxPGL(3,q) ordered by inclusion, for any prime power q and any prime r.
The results are summarized in Table 3.
This paper is organized as follows. Section 2 contains preliminary results on the Möbius function of a finite group, and on the groups \mboxPGL(3,q). In Section 3 the main result on the Möbius function of \mboxPSL(3,2p) is stated, namely Theorem 3.1. Section 4 provides the proof of Theorem 3.1. Finally, Section 5 computes the reduced Euler characteristic of certain ordered complexes associated to \mboxPGL(3,q).
2. Preliminary results
For any locally finite poset (P,⪯), define the Möbius function
μP:P×P→Z by
[TABLE]
For x≺y, μP is equivalently defined by μP(x,y)=−∑z∈P:x⪯z≺yμP(x,z).
We will consider the poset P=L of subgroups of a finite group
G, ordered by inclusion; L is a lattice with greatest element
G and least element {1}. For simplicity, we denote by μ(H)
the Möbius function of H≤G. The function
μ:L→Z, H↦μ(H) will be called the
Möbius function of G.
Clearly, if H,K≤G are conjugated, then
μ(H)=μ(K). The following property restricts
the investigation to the intersections of maximal subgroups of G.
Theorem 2.1**.**
([11, Theorem 2.3])
If H≤G satisfies μ(H)=0, then H is the intersection of
maximal subgroups of G.
Let q be a prime power; we consider the group \mboxPGL(3,q).
Note that, when q=2p with p an odd prime,
\mboxPGL(3,q)=\mboxPSL(3,q); q=2p implies also
\mboxPSL(3,q)=SL(3,q), which allows us to use matrices to denote the
elements of \mboxPSL(3,q).
The classification of subgroups of \mboxPGL(3,q) goes back to Mitchell
[21] and Hartley [12] . We refer to [21, 12, 15] for the proof of the
following classical results, and to [14] for a general
reference on projective planes.
Theorem 2.2**.**
For any prime power q, the following are self-normalizing maximal subgroups of
\mboxPGL(3,q), and they are unique up to conjugation:
- (1)
the stabilizer Eq2:GL(2,q) of an Fq-rational point, of order q3(q−1)2(q+1);
2. (2)
the stabilizer Eq2:GL(2,q) of an Fq-rational line, of order q3(q−1)2(q+1);
3. (3)
the stabilizer (Cq−1)2:Sym(3) of an Fq-rational triangle, of order 6(q−1)2;
4. (4)
the stabilizer Cq2+q+1:C3 of an Fq3∖Fq-rational triangle, of order 3(q2+q+1).
If q=2p with p an odd prime, the only other maximal subgroup of
\mboxPGL(3,q) up to conjugation is the following:
the stabilizer of a subplane of order 2, of order 168 and isomorphic to \mboxPSL(3,2).
It follows immediately that every Sylow subgroup of
\mboxPGL(3,q) in contained in one of the maximal subgroups (1) to (4)
in Theorem 2.2.
For the reader’s convenience, we recall in Remark
2.3 which points, lines or triangles in
PG(2,Fq) are stabilized by any element
σ∈\mboxPGL(3,q), in terms of \mboxord(σ).
Remark 2.3**.**
Let q be a power of a prime r and
σ∈\mboxPGL(3,q)∖{1}. Then one of the following cases
holds.
\mboxord(σ)=r* and σ is an elation, i.e. σ stabilizes every line through an Fq-rational point C and every point of an Fq-rational line ℓ passing through C; C and ℓ are called the center and the axis of σ.*
\mboxord(σ)=r=2, or r=2 and ord(σ)=4. Also, σ stabilizes exactly one point P and one line ℓ; both P and ℓ are Fq-rational, and P∈ℓ.
\mboxord(σ)∣(q−1)* and σ is a homology, i.e. σ stabilizes every line through an Fq-rational point C and every point of an Fq-rational line not passing through C; C and ℓ are the center and the axis of σ.*
\mboxord(σ)=r⋅d* with 1=d∣(q−1); σ stabilizes two Fq-rational points C and P, the line CP, and another Fq-rational line passing through P.*
2=\mboxord(σ)∣(q−1)* and σ stabilizes three non-collinear Fq-rational points P,Q,R and the lines PQ,PR,QR.*
\mboxord(σ)∣(q2−1)* and \mboxord(σ)∤(q−1). Also, σ stabilizes an Fq-rational point P and two Fq2∖Fq-rational points Q,R which are conjugated under the Fq-Frobenius collineation: Qq=R, Rq=Q; σ stabilizes the Fq-rational line QR and the Fq2∖Fq-rational lines PQ and PR.*
\mboxord(σ)∣(q2+q+1)* and σ stabilizes three non-collinear Fq3∖Fq-rational points P,Q,R which are an orbit of the Fq-Frobenius collineation;
σ stabilizes the Fq3∖Fq-rational lines PQ, PR, QR.*
Remark 2.4 lists some of the
actions of \mboxPGL(3,q), which will be used in the proof of Theorem 3.1.
Remark 2.4**.**
Let q be a prime power and G=\mboxPGL(3,q).
G* is 2-transitive on the points of PG(2,q).*
G* is transitive on the points of PG(2,q2)∖PG(2,q); the stabilizer in G of a point P∈PG(2,q2)∖PG(2,q) stabilizes also its Frobenius conjugate Pq.*
G* has two orbits on the points of PG(2,q3)∖PG(2,q); namely, one is made by the
points on the Fq-rational lines, the other is made by the remaining points.*
G* is 2-transitive on the Fq-rational lines.*
G* is transitive on the Fq-rational point-line pairs (P,ℓ) with P∈ℓ.*
G* is transitive on the Fq-rational point-line pairs (P,ℓ) with P∈/ℓ.*
G* is transitive on the non-collinear triples (P,Q,R) of Fq-rational points.*
G* is transitive on the Fq3∖Fq-rational triangles {P,Pq,Pq2}⊂PG(2,q3)∖PG(2,q) left invariant by the Fq-Frobenius collineation.*
G* is transitive on the projective frames of PG(2,q), i.e. on the 4-tuples of Fq-rational points no three of which are collinear.*
3. The Möbius function of \mboxPSL(3,2p) for any odd prime p
We state the main result, Theorem 3.1, whose proof is worked out in Section 4.
We assume that p is an odd prime and q=2p,
so that G=\mboxPSL(3,q)=\mboxPGL(3,q).
The main argument in the proof is to find which subgroups of G are intersection of maximal subgroups.
Roughly speaking, we start with the intersection of two
maximal subgroups M1 and M2. Through the geometry of M1
and M2, we determine the structure of M1∩M2 and we are able to identify which other maximal subgroups
of G contain M1∩M2 ; also, we study whether the group
M1∩M2 is unique up to conjugation in G. Clearly, the group
M1∩M2 may vary when M1 and M2 run in their conjugacy
classes. For instance, if M1 is the stabilizer of a point
P∈PG(2,q) and M2 is the stabilizer of an Fq-rational
triangle T, then M1∩M2≅(Cq−1)2:C2 if P is a
vertex of T; M1∩M2≅C2(q−1) if P is on a side of
T but not a vertex; and M1∩M2≅\mboxSym(3) if P is not
on a side of T.
We continue by intersecting M1∩M2 with other maximal subgroups, stopping when the geometry of the chosen maximal subgroups forces their intersection to be trivial.
Theorem 3.1**.**
The subgroups H<G which are intersection of maximal subgroups of
G are exactly the groups in Table 4, where the
normalizer NG(H) and the Möbius function μ(H) are provided.
For any such H there is just one conjugacy class in G.
4. Proof of Theorem 3.1
We use the same notation as in Section 3. The
proof is divided into the following
steps: for any H in Table
4, we prove that G has exactly one conjugacy class, and we determine NG(H) (Proposition 4.1); we show that the intersections of maximal subgroups of G are exactly the groups in Table 4
(Propositions 4.2 and 4.3); for any H in Table 4, we determine μ(H) (Proposition 4.4).
Proposition 4.1**.**
*For any group H in Table 4 there is exactly
one conjugacy class in G, and NG(H) is as in Table
4.
*
Proof.
We consider the groups H according to their line in Table
4.
Lines 1 to 5. H is a maximal subgroup; the claim follows from Theorem 2.2.
Line 31, H={1}. The claim is trivial.
Line 30, H≅C2. The involution α of H is an elation, and hence is uniquely determined by its center P, its axis ℓ, and its action on a third point not on ℓ. Hence, there is just one conjugacy class for H by Lemma 2.4.
Also, NG(H) stabilizes P and ℓ. Thus, up to conjugation,
P=(1:0:0), ℓ:Z=0, and
[TABLE]
Line 29, H≅C3. Since 3∣(q+1), the fixed points of
a generator α of H are three non-collinear points
P∈PG(2,q), Q,R∈PG(2,q2)∖PG(2,q); note that
R=Qq and the line ℓ=QR is Fq-rational. By Lemma
2.4 P and ℓ are unique up to
conjugation; since GP,ℓ acts as GL(2,q) on ℓ,
{Q,R} is also unique up to conjugation. Hence, there is just one
conjugacy class for H.
The pointwise stabilizer of {P,Q,R} in G is cyclic of order
q2−1. Also, NG(H) stabilizes P and acts on {Q,R}. Thus,
∣NG(H)∣=2(q2−1) and NG(H)≅Cq2−1:C2.
Line 28, H≅E4≤GP1,P2,P3. Here, P1,P2,P3
are three collinear Fq-rational points. For i∈{1,2,3} let
αi∈G be the elation with center Pi and axis ℓ=P1P2P3, with α3=α1α2 and
H=⟨α1,α2⟩≅E4. For any
P∈PG(2,q)∖ℓ, the set F={P1,P2,P,α3(P)}
is a projective frame of PG(2,q). Also, H is uniquely
determined by F; in fact, α1(P)=PP1∩P2α3(P)
and α2(P)=PP2∩P1α3(P). Then there is just one
conjugacy class for H by Lemma 2.4.
The normalizer NG(H) acts on {P1,P2,P3}; for any
σ∈NG(H), σ(Pi)=Pj implies
σαiσ−1=αj. Thus, the pointwise stabilizer
S of {P1,P2,P3} in NG(H) is made by those σ∈NG(H) which commute with αi for any i and stabilize
ℓ pointwise. Since no homology with axis ℓ commutes with
an elation with axis ℓ, S is only made by elations with axis
ℓ. Since two elations commute if and only if they have the same
center or axis, this implies S≅Eq2. Now, NG(H)/Eq2
acts faithfully on {P1,P2,P3} and hence is a subgroup of
\mboxSym(3). Finally we can choose H up to conjugation and obtain
NG(H) as follows:
[TABLE]
By direct computation,
NG(H)={σ0,b,c,1:b,c∈Fq}:⟨τ,ω⟩≅Eq2:\mboxSym(3).
Line 27, H≅E4≤Gℓ1,ℓ2,ℓ3. Here,
ℓ1,ℓ2,ℓ3 are three Fq-rational lines, concurrent in
P. Let H={1,α1,α2,α3=α1α2}≅E4, where αi has center P and axis ℓi. A dual argument with respect to the one used in the previous point, shows that there is just one conjugacy class for
H, and NG(H)≅Eq2:\mboxSym(3).
Line 26, H≅C4. Let α be a generator of H≅C4, with fixed point P∈PG(2,q) and fixed Fq-rational line
ℓ, where P∈ℓ. Then α is uniquely determined by
its action on F={P,Q,α(Q),R}, where Q∈PG(2,q)∖ℓ and R∈PG(2,q)∖Qα(Q). Since F is a projective frame of
PG(2,q), there is just one conjugacy class for H in G. Up to
conjugacy, we have
[TABLE]
Thus, NG(H)≤NG(⟨α2⟩) and
NG(⟨α2⟩) is in Equation
(1). By direct checking,
σa,b,c,λ∈NG(H) if and only if λ=1 and
either a=c+1 or a=c. Also,
Z(NG(H))={σ0,b,c,1:b∈Fq}. Therefore, NG(H)≅Eq.E2q.
Line 25, H≅\mboxSym(3). Let
H=⟨α⟩:⟨β⟩ with o(α)=3,
o(β)=2. Then ⟨α⟩ is uniquely determined by
its fixed points P∈PG(2,q),
Q,R∈PG(2,q2)∖PG(2,q), while β fixes P,
interchanges Q ad R, and is uniquely determined by the
projective frame F={P,Q,R,S}, where S is an Fq-rational
point on the axis of β different from P. Hence, there is
just one conjugacy class for H in G.
From the previous points about C2 and C3 follows that
NG(H) contains H×Cq−1, where Cq−1 is made by the
homologies with center P and axis QR. Also, NG(H) fixes P
and acts on the three intersection points between QR and the axes
of the three elations in H. Since Cq−1 is the whole subgroup
of GP acting trivially on QR, NG(H)/Cq−1≤\mboxSym(3) and hence NG(H)=H×Cq−1.
Line 24, H≅C7≤GT,Π. Here, Π is a subplane of order 2, and T is a triangle; we have
GT≅Cq2+q+1:C3 or GT≅(Cq−1)2:\mboxSym(3),
according to p>3 or p=3, respectively. Suppose p>3. There is
just one conjugacy class for H in G, because H is
characteristic in GT which is unique up to conjugation in G (Theorem 2.2); this also shows NG(H)=GT≅Cq2+q+1:C3. If p=3, the claim follows by direct inspection with Magma [1].
Line 23, H≅D8. As in Line 26, we can assume that an element α∈H of order 4 is
as in Equation (2). Let β∈H be an involution
with β=α2. Let τ∈NG(H)≤NG(⟨α⟩) be an involution with
τατ−1=α−1. By direct checking, either
τ=σ1,b,0,1 or τ=σ0,b,1,1 for some
b∈Fq; also, τ is conjugated by some σ0,b,0,1
either to β or to αβ. Thus, there is just one
conjugacy class for H in G. By direct checking, an element
σa,b,c,1∈NG(⟨α⟩) is in NG(H) if
and only if a,c∈{0,1}, and
Z(NG(H))={σ0,b,0,1:b∈Fq}. Therefore, NG(H)≅Eq.E4.
Line 22, H≅C7:C3. If p=3, then the claim follows using Magma [1]. Suppose p>3. Then
H is characteristic in M≅Cq2+q+1:C3, which is unique up to conjugation; hence, H is unique up to conjugation. NG(H)
stabilizes the triangle T~ fixed pointwise by C7≤H,
hence NG(H)≤M; thus NG(H)≅Cm:C3 with
7∣m∣(q2+q+1). Since G does not contain subgroup E9 (as Syl3(G) is cyclic), the action by conjugation of the 2m 3-elements of NG(H) on the 14 3-elements of H is fixed-point-free;
thus, m=7 and NG(H)=H.
Line 21, H≅\mboxSym(4)=Gℓ,Π. For any subplane
Π of order 2, the maximal subgroup
GΠ≅\mboxPSL(3,2) of G has two conjugacy classes of
subgroups \mboxSym(4), containing respectively the groups
Gℓ,Π and the groups GP,Π, where ℓ and P range over the 7 lines and points of Π.
As GΠ is unique up to conjugation in G, the same holds for H.
For any σ in the centralizer CG(H), σ commutes with all elations in H and hence stabilizes
their centers; hence, σ stabilizes Π pointwise, so that CG(H) is trivial. Thus NG(H)=H, because Aut(\mboxSym(4))≅\mboxSym(4).
Line 20, H≅\mboxSym(4)=GP,Π. The claim follows as in the previous point.
Line 19, H≅Cq−1=GP1,…,Pq+1,ℓ1,…,ℓq+1. Here,
P1,…,Pq+1∈PG(2,q) are distinct collinear points and
ℓ1,…,ℓq+1 are Fq-rational distinct lines,
concurrent in a point P and different from ℓ=P1P2⋯Pq+1. It is easily seen that there
is just one conjugacy class for H in G, determined by (P,ℓ);
H is the center of NG(H),
because the following holds up to conjugation:
[TABLE]
Line 18, H≅Eq=GP1,…,Pq+1,ℓ1,…,ℓq+1. Here,
P1,…,Pq+1∈PG(2,q) are distinct points, collinear in a
line ℓ, and ℓ1,…,ℓq+1 are Fq-rational
distinct lines, concurrent in a point P of ℓ. Then H is the
group of elations with center P and axis ℓ; H is uniquely
determined by (P,ℓ). Hence, there is just one conjugacy class
for H in G. Also, NG(H) fixes P and stabilizes ℓ
linewise. Thus, up to conjugation and by direct checking, H and
NG(H) are as follows:
[TABLE]
Line 17, H≅C2(q−1). By Remark 2.3,
H=⟨α⟩ stabilizes two distinct points
P,C∈PG(2,q), the line CP, and another Fq-rational line
ℓ through P. By Line 19,
⟨α2⟩ is unique up to conjugation in G. The involutions of NG(⟨α2⟩) form a unique
conjugacy class, as the same happens in
NG(⟨α2⟩)/⟨α2⟩≅\mboxPGL(2,q).
Hence, there is just one conjugacy class for H in G. Since the
normalizer NG(H) stabilizes {P,C,CP,ℓ} elementwise, we
have up to conjugation and by direct computation that P=(1:0:0),
C=(0:1:0), CP:Z=0, ℓ:Y=0, and
[TABLE]
where ϵ is a primitive element of Fq.
Line 16, H≅Eq:Cq−1=Gℓ1,…,ℓq+1,P.
Here P∈PG(2,q) and ℓ1,…,ℓq+1 are
Fq-rational lines concurrent in a point C=P. The elementwise
stabilizer of {ℓ1,…,ℓq+1,P} contains the
elations Eq of center C and axis CP, and the homologies with
center C and axis through P; such homologies form
subgroups Cq−1 which are conjugated under Eq, as Eq
acts regularly on the q Fq-rational lines through P different
from CP. Thus, H=Gℓ1,…,ℓq+1,P≅Eq:Cq−1 and no elation and homology in H commute.
As G is 2-transitive on PG(2,q), H is unique up to conjugation in G. Also, NG(H) stabilizes both P and C; hence ∣NG(H)∣ divides q2(q−1)2 by the
orbit-stabilizer theorem. Up to conjugacy and by direct checking, we
have C=(1:0:0), P=(0:1:0), and
[TABLE]
Line 15, H≅Eq:Cq−1=GP1,…,Pq+1,ℓ.
Here P1,…,Pq+1 are Fq-rational points collinear in a
line r, and ℓ=r is another Fq-rational line. A dual argument with respect to the one in the previous point yields the claim.
Line 14, H≅(Cq−1)2. H is the pointwise
stabilizer of an Fq-rational non-degenerate triangle T, and
hence is unique up to conjugation in G. NG(H) is
the stabilizer of T, i.e. a maximal subgroup (Cq−1)2:\mboxSym(3)
of G.
Line 13, H≅(Cq−1)2:C2. The characteristic subgroup
(Cq−1)2 of H is unique up to conjugation in G, as shown in
the previous point. By Schur-Zassenhaus theorem, the complement
C2 is unique up to conjugation in H. Hence, there is just one
conjugacy class for H in G. Let T be the triangle pointwise
stabilized by (Cq−1)2 and P be the vertex of T stabilized
by H. Then NG(H) stabilizes T, so that NG(H)≤GT.
Also, NG(H) stabilizes P, which implies NG(H)=H.
Line 12, H≅Eq:(Cq−1)2. Such a H≤G
exists, for instance as follows:
[TABLE]
Consider a group L:M≤G, with L≅Eq, M≅(Cq−1)2. The group L is made by elations with either the same center or the same axis.
Suppose that the elations of L have the same axis ℓ. Consider
an element β∈M of order q−1 with exactly three fixed
points P,Q,R; let R be the one out of ℓ. Then β does
not commute with any α∈L∗ (otherwise α stabilizes
R), and hence the action of
⟨β⟩ by conjugation on Eq∗ is fixed-point-free.
Also, βαβ−1 has the same center as α.
Therefore, the elations of L have the same center. Similarly,
if the elations of L have the same center, they also have the same axis. Thus, L is made by
the elations with the same center P and axis ℓ.
Being uniquely determined by (P,Q,R), H is unique up to conjugation in G. Being L characteristic in H, we have NG(H)≤NG(L); thus, by Equations (4) and (7), NG(H)=H.
Line 11, H≅Eq2:Cq−1=Gℓ1,…,ℓq+1. Here,
ℓ1,…,ℓq+1 are distinct Fq-rational lines
concurrent in a point P. Thus, the elementwise stabilizer of
{ℓ1,…,ℓq+1} is made by the elations and
homologies with center P. The elations with center P (resp. the homologies with center P and given axis ℓ) form a
subgroup L≅Eq2 (resp. M≅Cq−1). The action by
conjugation of the elements of L on the homologies with center P
is fixed-point-free (as no elation with center P stabilizes a line
ℓ not through P); hence, Gℓ1,…,ℓq+1=L:M.
Since H is uniquely determined by P and G is transitive on
PG(2,q), H is unique up to conjugation in G. As NG(H)=GP, Theorem 2.2 yields the claim.
Line 10, H≅Eq2:Cq−1=GP1,…,Pq+1.
Here, P1,…,Pq+1 are distinct collinear Fq-rational
points. The claim follows with a dual argument with respect to the
one used in the previous point.
Line 9, H≅Eq2:(Cq−1)2=Gℓ,r. Here, ℓ,r are distinct Fq-rational lines. Since G is
2-transitive on PG(2,q) and hence also in the dual plane, Gℓ,r is unique up to conjugation in G. Let
P=ℓ∩r, so that Gℓ,r≤GP. Then Gℓ,r
contains the group L≅Eq2 of elations with center P, and
by Remark 2.3 Gℓ,r does not contain any
other 2-element; in particular, L⊴Gℓ,r.
Also, Gℓ,r contains the pointwise stabilizer
N≅(Cq−1)2 of an Fq-rational triangle with two sides in
ℓ and r. Hence, Gℓ,r contains L:N. Every nontrivial
element whose order divides q+1 does not stabilize two
Fq-rational lines; since ∣GP∣=q3(q+1)(q−1)2, this implies
Gℓ,r=L:N≅Eq2:(Cq−1)2. NG(H) acts on {ℓ,r}, and H is the pointwise stabilizer of ℓ,r. Thus, NG(H)≅H:C2, where the complement C2 is given by a suitable elation; in fact, the
group of elations with center C=P and axis CP acts
transitively on the lines through P different from CP.
Line 8, H≅Eq2:(Cq−1)2=GP,Q. Here,
P,Q∈PG(2,q), P=Q. The claim follows with a dual
argument with respect to the one used in the previous point.
Line 7, H≅GL(2,q). Such a subgroup H≤G exists, as
shown in Equation (3), and H=NG(Z), where
Z≅Cq−1 is the center of H and is made by homologies with
a given center P and axis ℓ. H is uniquely
determined by (P,ℓ); hence, by Remark 2.3, H
is unique up to conjugation in G. Since Z is characteristic in
H, we have NG(H)≤NG(Z)=H and the claim follows.
Line 6, H≅Eq1+2:(Cq−1)2. Equation
(4) shows that such a subgroup H≤G exists, and
H=NG(K) where K≅Eq is made by the elations with a common
center P and axis ℓ. Since K is unique up to conjugation in
G (see the point above regarding Line 18), H is also unique up
to conjugation in G. NG(H) stabilizes P and
ℓ, and hence preserves the elations with center P and axis
ℓ. This implies NG(H)≤NG(K), so NG(H)=H.
∎
Proposition 4.2**.**
Every H in Table 4 is intersection
of maximal subgroups of G.
Proof.
For every group H, we use what has been already shown
in the proof of Proposition 4.1
and we give a summary description of each case.
Lines 1 to 5. In these cases, H is a maximal subgroup of
G.
Lines 6 and 7, H≅Eq1+2:(Cq−1)2 and H≅GL(2,q), respectively.
We have H≤GP,ℓ, where P and ℓ are an
Fq-rational point and line, with either P∈ℓ (Line 6) or
P∈/ℓ (Line7). From Remark 2.3 and the
orbit-stabilizer theorem follows H=GP,ℓ=GP∩Gℓ. No
other maximal subgroup contains H.
Line 8, H=GP,Q≅Eq2:(Cq−1)2=GP∩GQ for some distinct points P,Q∈PG(2,q). No other
maximal subgroup contains H.
Line 9, H=Gℓ,r≅Eq2:(Cq−1)2=Gℓ∩Gr for some distinct Fq-rational lines
ℓ,r. No other maximal subgroup contains H.
Line 10, H=GP1,…,Pq+1≅Eq2:Cq−1. We
have H=GP1∩⋯∩GPq+1∩Gℓ, where
ℓ is an Fq-rational line and P1,…,Pq+1 are the
q+1 distinct Fq-rational points of ℓ. No other maximal
subgroup of contains H.
Line 11, H=Gℓ1,…,ℓq+1≅Eq2:Cq−1. We have H=GP∩Gℓ1∩⋯∩Gℓq+1, where P∈PG(2,q) and ℓ1,…,ℓq+1
are the q+1 distinct Fq-rational lines through P. No other
maximal subgroup contains H.
Line 12, H≅Eq:(Cq−1)2. Up to conjugation, H is
as in Equation (7). Then H=GP,Q,ℓ,r, where P=(1:0:0), Q=(0:1:0), ℓ:Y=0, r:Z=0. Thus H=GP∩GQ∩Gℓ∩Gr. No other maximal subgroup contains H.
Line 13, H≅(Cq−1)2:C2. H stabilizes an
Fq-rat. triangle T={P,Q,R} and a vertex P. Then
H=GP∩GQR∩GT. No other max. subgroup
contains H.
Line 14, H≅(Cq−1)2. H is the pointwise
stabilizer in G of an Fq-rational triangle T={P,Q,R}. Then
H=GP∩GQ∩GR∩GPQ∩GPR∩GQR∩GT,
and no other maximal subgroup contains H.
Line 15, H=GP1,…,Pq+1,ℓ≅Eq:Cq−1.
H is the subgroup of G stabilizing the line r
through P1,…,Pq+1 pointwise, and the line ℓ
linewise. Hence, H=GP1∩⋯∩GPq+1∩Gr∩Gℓ. No other maximal subgroup contains H.
Line 16, H=Gℓ1,…,ℓq+1,P≅Eq:Cq−1. H is the subgroup of G stabilizing
all lines ℓ1,…,ℓq+1 through a point
C∈PG(2,q), and a point P∈PG(2,q)∖{C}. Hence,
H=GP∩GC∩Gℓ1∩⋯∩Gℓq+1. No
other max. subgroup contains H.
Line 17, H≅C2(q−1). We have H≤GP,C,CP,ℓ,
where P,C∈PG(2,q), P=C, ℓ is an Fq-rat. line,
P∈ℓ, ℓ=CP. No other Fq-rat. point or line is
stabilized by H. Also, H stabilizes exactly q
Fq-rat. triangles Ti={C,Qi,Ri}, where the {Qi,Ri}’s are the
orbits of H on ℓ(Fq)∖{P}. Thus, H=GP∩GC∩GCP∩Gℓ∩GT1∩⋯∩GTq, and
no other maximal subgroup contains H.
Line 18, H=GP1,…,Pq+1,ℓ1,…,ℓq+1≅Eq. Here,
P1,…,Pq+1 are the Fq-rat. points of a line ℓ,
and ℓ1,…,ℓq+1 are the Fq-rat. lines through
a point P of ℓ; H is the group of elations with
center P and axis ℓ. Thus, H=GP1∩⋯∩GPq+1∩Gℓ1∩⋯∩Gℓq+1. No other
maximal subgroup contains H.
Line 19,
H=GP1,…,Pq+1,ℓ1,…,ℓq+1≅Cq−1.
Here, P1,…,Pq+1 are the Fq-rational points of a line
ℓ, and ℓ1,…,ℓq+1 are the Fq-rational lines
through a point P∈/ℓ; H is the group of homologies with
center P and axis ℓ. Also, H stabilizes (pointwise) the
(2q+1) Fq-rational triangles with one vertex in P and
two vertices on ℓ. Thus, H=GP∩GP1∩⋯∩GPq+1∩Gℓ1∩⋯∩Gℓq+1∩Gℓ∩GT1∩⋯∩GT(2q+1). No other
maximal subgroup contains H.
Line 20, H=GP,Π≅\mboxSym(4). We have H=GP∩GΠ, for some subplane Π of order 2 and some point
P∈Π.
No other maximal subgroup contains H.
Line 21, H=Gℓ,Π≅\mboxSym(4). We have H=GP∩GΠ for some subplane Π⊂PG(2,q) of order 2 and some
Fq-rational line ℓ containing three points of Π. No
other maximal subgroup contains H.
Line 22, H≅C7:C3. By double counting arguments, H=GT~∩GΠ, with a subplane
Π of order 2 and a triangle T~.
No other max. subgroup contains H.
Line 23, H≅D8. Let P and ℓ be the unique point
and line stabilized by an element of order 4 in H; then
H≤GP,ℓ. By double counting arguments, H
is contained in exactly 2q max. subgroups
GΠ1,…,GΠq/2 isomorphic to \mboxPSL(3,2);
also, H is equal to the intersection of any two of them (see Lines 20 and 21).
Thus, H=GP∩Gℓ∩GΠ1∩⋯∩GΠq/2.
No other max subgroup contains H.
Line 24, H=C7≤GT,Π. If p>3, then H stabilizes
no Fq-rat. points or lines; if p=3, the Fq-rat.
points and lines stabilized by H are the vertices P,Q,R and the
sides of T.
The number of subplanes Πi of order 2
stabilized by H is either 7q2+q+1 or 7, according to
p>3 or p=3, respectively; H is equal to the intersection of
any two of them (see Line 22). Thus, either H=GT∩M1∩⋯∩M(q2+q+1)/7 or H=GP∩GQ∩GR∩GPQ∩GPR∩GQR∩GT∩M1∩⋯∩M7, according to p>3 or p=3, respectively.
No other max. subgroup contains H.
Line 25, H≅\mboxSym(3). Let P∈PG(2,q) and
Q,R∈PG(2,q2)∖PG(2,q) be the fixed points of the
3-elements of H. Then P,QR are the unique Fq-rat.
point and line fixed by H. By double counting arguments, H
stabilizes exactly q−1 Fq-rat. triangles Ti’s and q−1 subplanes Πi’s of order 2. For i=j, we have H=GTi∩GTj and H=GΠi∩GΠj. Thus H=GP∩GQR∩GT1∩⋯∩GTq−1∩M1∩⋯∩Mq−1. No other max. subgroup contains H.
Line 26, H=C4. The group H stabilizes exactly one
Fq-rat. point P and one Fq-rat. line ℓ.
By double counting arguments, H stabilizes exactly 4q2
subplanes Πi’s of order 2. Every overgroup of H isomorphic
to D8 stabilizes exactly 2q of such subplanes (see
Line 23); hence, GΠ1,…,Πq2/4=H. Thus,
H=GP∩Gℓ∩M1∩⋯∩Mq2/4. No other
max. subgroup contains H.
Line 27, H≅E4, H≤Gℓ1,ℓ2,ℓ3. Let
P the center of the elations in H, and
ℓ1,…,ℓq+1 be the Fq-rat. lines through P.
Then H stabilizes P,ℓ1,…,ℓq+1, and no other
Fq-rat. points or lines. Also, H
stabilizes exactly 4q2 subplanes Πi’s of order 2,
and GΠ1,…,Πq2/4=H (see Line 23). Thus, H=GP∩Gℓ1∩⋯∩Gℓq+1∩GΠ1∩⋯∩GΠq2/4, and no other max. subgroup contains H.
Line 28, H≅E4, H≤GP1,P2,P3. As in Line 27, H=GP1∩⋯∩GPq+1∩Gℓ∩GΠ1∩⋯∩GΠq2/4, where ℓ is a line through the Pi’s, and the Πi’s are subplanes of order 2. No other max. subgroup contains H.
Line 29, H≅C3. The group H stabilizes exactly one
Fq-rat. point P and one Fq-rat. line ℓ. By
double counting arguments,
[TABLE]
where the Ti’s are distinct Fq-rat. triangles, the
T~i’s are distinct Fq3∖Fq-rat. triangles, and the Πi’s are distinct subplanes of PG(2,q) of
order 2. No other max. subgroup contains H.
Line 30, H≅C2. Let α be the elation of H,
P1,…,Pq+1 be the Fq-rat. points of the axis of
α, and ℓ1,…,ℓq+1 be the Fq-rat. lines
through the center of α. By double counting argument,
[TABLE]
where the Ti’s are distinct Fq-rat. triangles and the
Πi’s are distinct subplanes of PG(2,q) of order 2. No
other max. subgroups contain H.
Line 31, H={1}. As G is simple, H is the Frattini
subgroup of G.
∎
Proposition 4.3**.**
Let H<G be the intersection of maximal subgroups of G. Then H
is one of the groups in Table 4.
Proof.
The claim is proved as follows: we consider every subgroup K<G in
Table 4, starting from the maximal subgroups of
G; for any maximal subgroup M of G satisfying K≤M, we show that H:=K∩M is one of the groups in
Table 4.
Line 1: K=GP with P∈PG(2,q).
If H=K∩GQ, Q∈PG(2,q)∖{P}, then H=GP,Q≅Eq2:(Cq−1)2.
If H=K∩Gℓ, ℓ an Fq-rat. line, then H≅Eq1+2:(Cq−1)2 if P∈ℓ, H≅GL(2,q) if P∈/ℓ.
If H=K∩GT, T an Fq-rat. triangle, then either P is a vertex of T, and H≅(Cq−1)2:C2; or P is not a vertex but on a side of T, and H≅C2(q−1); or P is not on the sides of T, and H≤\mboxSym(3).
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, H≤C3 by Lagrange’s theorem.
If H=K∩GΠ, Π a subplane of order 2, then either P∈/Π and H={1}; or P∈Π and H≅\mboxSym(4).
Line 2: K=Gℓ for some Fq-rational line ℓ.
If H=K∩Gr, r=ℓ an Fq-rat. line, then H=Gℓ,r=Eq2:(Cq−1)2.
If H=K∩GT, T an Fq-rat. triangle, then either ℓ is a side of T, and H≅(Cq−1)2:C2; or ℓ contains exactly one vertex of T, and H≅C2(q−1); or ℓ does not contain any vertex of T, and H≤\mboxSym(3).
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H≤C3.
If H=K∩GΠ, Π a subplane of order 2, then either ℓ is not a line of Π and H={1}; or ℓ is a line of Π and H≅\mboxSym(4).
Line 3: K=GT for some Fq-rational triangle T={A,B,C}.
If H=K∩GT′ for some Fq-rat. triangle T′={A′,B′,C′}=T and H≤\mboxSym(3), then some element of H∖{1} stabilizes T and T′ pointwise. Hence, T∪T′⊂{P}∪ℓ for some point P and line ℓ, so that T and T′ have a vertex in common.
If T and T′ have another vertex in common, then H is the
group Cq−1 of homologies. Otherwise, H≅C2(q−1).
If H=K∩GT~, T~ an Fq3∖Fq-rational triangle, then H≤C3.
If H=K∩GΠ, Π a subplane of order 2, and H≤\mboxSym(3), then q=8 and H≅C7:C3 by direct checking with Magma [1].
Line 4: K=GT~ for some Fq3∖Fq-rational triangle T~.
If H=K∩GT~′, for some Fq3∖Fq-rational triangle T~′=T~, then H≤C3.
If H=K∩GΠ, Π a subplane of order 2, and H≤C3, then H≅C7:C3.
Line 5: K=GΠ for some subplane Π of order 2. Let H=K∩GΠ′, Π′ a subplane of order 2.
Assume that H contains the only proper subgroup of \mboxPSL(3,2) not appearing in Table 4, namely \mboxAlt(4). Let Λ⊂PG(2,q) be the pointset containing the centers of the 3 elations in \mboxSym(4) and the Fq-rat. points stabilized by one of the 4 subgroups C3 of \mboxSym(4). Then Λ has 7 distinct Fq-rat. points and is a subplane of order 2; thus Λ=Π=Π′ and H=K.
Line 6: K=GP,ℓ≅Eq1+2:(Cq−1)2, where P and ℓ are Fq-rational, and P∈ℓ.
If H=K∩GQ, Q∈PG(2,q)∖{P}, then either Q∈ℓ and H=GP,Q≅Eq2:(Cq−1)2; or Q∈/ℓ and H≅Eq:(Cq−1)2.
If H=K∩Gr, r=ℓ an Fq-rat. line, then either P∈r and H=Gℓ,r≅Eq2:(Cq−1)2; or P∈/r and H≅Eq:(Cq−1)2.
If H=K∩GT, T an Fq-rat. triangle, then one of the following holds:
P is a vertex and ℓ is a side of T. Then H≅(Cq−1)2.
P is a vertex of T and ℓ is not a side of T; or P is not a vertex of T and ℓ is a side of T. Then H≅C2(q−1).
P is not a vertex but is on a side of T, and ℓ is not a side of T. Then either H≅C2(q−1), or H≅C2.
P is not on a side of T and ℓ is not a side of T. Then H≤\mboxSym(3).
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H={1}.
If H=K∩GΠ, Π a subplane of order 2, and H={1}, then P and ℓ are a point and a line of Π, and H≅D8.
Line 7: K=GP,ℓ≅GL(2,q), where P and ℓ are Fq-rational, and P∈/ℓ.
If H=K∩GQ, Q∈PG(2,q)∖{P}, then either Q∈ℓ and H≅Eq:(Cq−1)2; or Q∈/ℓ and H≅Eq:Cq−1 is as in Line 15.
If H=K∩Gr, r=ℓ an Fq-rat. line, then either P∈r and H≅Eq:(Cq−1)2; or P∈/r and H≅Eq:Cq−1 is as in Line 16.
If H=K∩GT, T an Fq-rat. triangle, then one of the following holds:
P is a vertex of T and GP,T≅(Cq−1)2:C2. Then either H≅(Cq−1)2:C2 or H≤C2.
P is not a vertex but on a side of T and GP,T≅C2(q−1). Then either H={1}, or H≅Cq−1 is made by homologies.
P is not on a side of T and GP,T≤\mboxSym(3). Then H≤\mboxSym(3).
If H=K∩GT~, T~ an Fq3∖Fq-rational triangle, then H≤C3.
If H=K∩GΠ, Π a subplane of order 2, and H={1}, then P is a point and ℓ a line of Π, and H≅\mboxSym(3).
Line 8: K=GP,Q≅Eq2:(Cq−1)2 for some P,Q∈PG(2,q) with P=Q.
If H=K∩GR, R∈PG(2,q)∖{P,Q}, then either H≅Eq2:Cq−1 is as in Line 10, or H≅(Cq−1)2.
If H=K∩Gℓ with ℓ=PQ, and R=ℓ∩PQ, then either R∈{P,Q} and H≅Eq:(Cq−1)2; or R∈/{P,Q} and H≅Eq:Cq−1 is as in Line 15.
If H=K∩GT, T an Fq-rat. triangle, then either P,Q are vertices of T and H≅(Cq−1)2; or P,Q are not both vertices but still on the same side of T, and H is a group Cq−1 of homologies; or H≤\mboxSym(3).
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H={1}.
If H=K∩GΠ, Π a subplane of order 2, then either {P,Q}⊂Π and H={1}, or {P,Q}⊂Π and H≅E4 is as in line 28.
Line 9: K=Gℓ,r≅Eq2:(Cq−1)2 for some Fq-rat. lines ℓ,r with ℓ=r.
Dual arguments with respect to the ones used in the previous point
prove the claim.
Line 10: K=GP1,…,Pq+1≅Eq2:Cq−1 where P1,…,Pq+1 are the distinct Fq-rat. points of a line ℓ.
If H=K∩GP, P∈PG(2,q)∖ℓ, then H is a group Cq−1 of homologies.
If H=K∩Gr, r=ℓ an Fq-rat. line, then H≅Eq:(Cq−1)2.
If H=K∩GT, T an Fq-rat. triangle, then either H is a group Cq−1 of homologies, or H≤C2.
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H={1}.
If H=K∩GΠ, Π a subplane of order 2, then either ℓ is not a line of Π and H={1}; or ℓ is a line of Π and H≅E4 is as in Line 28.
Line 11: K=Gℓ1,…,ℓq+1≅Eq2:Cq−1 where ℓ1,…,ℓq+1 are concurrent
Fq-rat. lines. Dual arguments with respect to Line 10 prove the claim.
Line 12: K=GP,Q,ℓ,r≅Eq:(Cq−1)2, where
P,Q∈PG(2,q), P=Q, r=PQ, and ℓ=r is another
Fq-rational line with P∈ℓ.
If H=K∩GR, R∈PG(2,q)∖{P,Q}, then either R∈r and H≅Eq:Cq−1; or R∈ℓ and H≅(Cq−1)2; or H is a group Cq−1 of homologies.
If H=K∩Gs, s∈/{ℓ,r} an Fq-rat. line, then dual arguments yield either H≅Eq:Cq−1, or H≅(Cq−1)2, or H≅Cq−1 made by homologies.
If H=K∩GT, T an Fq-rat. triangle, and H≤\mboxSym(3), then some σ∈H∖{1} stabilizes T pointwise.
Since σ cannot stabilize a projective frame pointwise, the
points P,Q are on some side of T and the lines ℓ,r pass
through some vertex of T. If P,Q are vertices of T and
r,ℓ are sides of T, then H≅(Cq−1)2; otherwise, H
is a group Cq−1 of homologies.
If H=K∩GT~, T~ an Fq3∖Fq-rational triangle, then H={1}.
If H=K∩GΠ, Π a subplane of order 2, then H≤C2.
Line 13: K=GP,T≅(Cq−1)2:C2, P a vertex of an Fq-rational triangle T.
If H=K∩GQ, Q∈PG(2,q)∖{P}, then either Q is a vertex of T and H≅(Cq−1)2; or Q is on the side not through P and H≅C2(q−1); or Q is on a side thorugh P and H is a group Cq−1 of homologies; or H≅C2.
If H=K∩Gℓ, ℓ an Fq-rat. line, then dual arguments prove the claim.
If H=K∩GT′, T′=T an Fq-rat. triangle, and H≤\mboxSym(3), then either H≅C2(q−1) or H is a group Cq−1 of homologies, according to ∣T∩T′∣=1 or ∣T∩T′∣=2, respectively.
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H={1}.
If H=K∩GΠ={1}, Π a subplane of order 2, then either H≅C2, or p=3 and H≅C7.
Line 14: K=GT≅(Cq−1)2 for some Fq-rational triangle T. Let H=K∩M for some maximal subgroup M of G such that {1}<H<K.
Then either p>3 and H is a group Cq−1 of homologies; or
p=3 and H≅C7 is as in Line 24.
Line 15: K=GP1,…,Pq+1,ℓ≅Eq:Cq−1, where P1,…,Pq+1 are the Fq-rat. points of an Fq-rat. line r, and ℓ=r is an Fq-rat. line meeting r in P.
If H=K∩GQ, Q∈PG(2,q)∖r, then either Q∈ℓ and H is a group Cq−1 of homologies, or Q∈/ℓ and H={1}.
If H=K∩Gs, s∈/{r,ℓ} an Fq-rat. line, then either P∈s and H is a group Eq of elations, or P∈/s and H is a group Cq−1 of homologies.
If H=K∩GT, T an Fq-rat. triangle, then either H is a group Cq−1 of homologies, or H≤C2.
If H=K∩GT~, T~ an Fq3∖Fq-rat. triangle, then H={1}.
If H=K∩GΠ, Π a subplane of order 2, then H=≤C2.
Line 16: K=Gℓ1,…,ℓq+1,P≅Eq:Cq−1.
Dual arguments with respect to Line 15 prove the claim.
Line 17: K≅C2(q−1). Let H=K∩M, for a maximal subgroup M of G with {1}<H<K. Then either H≅C2 or H is a group Cq−1 of homologies.
Line 18: K≅Eq is the group of elations with a given axis and center. Let H=K∩M, for a maximal subgroup M such that {1}<H<K. Then H≅C2.
Line 19: K≅Cq−1 is the group of homologies with a given axis and center. Let H=K∩M, for a maximal subgroup M such that H<K. Then H={1}.
Lines 20 to 31: if K is in one of the Lines 20 to 31, then every subgroup of K is in Table 4, apart from the subgroup \mboxAlt(4), which has already been shown not to be intersection of maximal subgroups of G (see Line 5).
∎
Proposition 4.4**.**
For any H<G in Table 4, μ(H) is given in Table 4.
Proof.
We make implicit use of what has been shown in the previous
propositions. In particular, the proof of Proposition
4.2 contains all the maximal subgroups of G
containing H. We denote by n(H) the number of such maximal subgroups.
Lines 1 to 5. Since H is a maximal subgroup of G, μ(H)=−1.
Line 6, H≅Eq1+2:(Cq−1)2. Since n(H)=2, μ(H)=1.
Line 7, H≅GL(2,q). Arguing as in Line 6, μ(H)=1.
Line 8, H=GP,Q≅Eq2:(Cq−1)2.
We have n(H)=3, H=GP∩GQ, H=GP∩GPQ, H=GQ∩GPQ. Thus, μ(H)=0.
Line 9, H=Gℓ,r≅Eq2:(Cq−1)2. Arguing as in Line 8, μ(H)=0.
Line 10, H=GP1,…,Pq+1≅Eq2:Cq−1.
Let ℓ be the line through P1,…,Pq+1. Then n(H)=q+2; the overgroups of H
in Table 4 are the following:
GP1,…,GPq+1,Gℓ, q+1 groups GPi,ℓ,
and groups GPi,Pj. Thus, μ(H)=0.
Line 11, H=Gℓ1,…,ℓq+1≅Eq2:Cq−1. Arguing as in Line 10, μ(H)=0.
Line 12, H=GP,Q,r,ℓ≅Eq:(Cq−1)2.
Here, r=PQ, P∈ℓ, Q∈/ℓ, and P,Q,r,ℓ are
Fq-rational. We have n(H)=4, and the overgroups of H in Table
4 are GP, GQ, Gr, Gℓ,
GQ,r, GP,r, GP,ℓ, GQ,ℓ, GP,Q,
Gr,ℓ. Thus, μ(H)=−1.
Line 13, H≅(Cq−1)2:C2. Then n(H)=3 and H=GP∩GQR∩GT for some Fq-rational triangle T={P,Q,R}.
The overgroups of H in Table 4 are GP,
GQR, GT, GP,QR. Thus, μ(H)=1.
Line 14, H≅(Cq−1)2.
Then the overgroups of H in Table
4 are the following: 7 maximal subgroups of
G; 6 groups of type Eq1+2:(Cq−1)2; 3 groups
GL(2,q); 3 groups (Cq−1)2:C2; 6 groups
Eq:(Cq−1)2. Thus, μ(H)=0.
Line 15, H=GP1,…,Pq+1,ℓ≅Eq:Cq−1.
Here, P1,…,Pq+1 are collinear in r, and ℓ=r;
let Q=r∩ℓ. Then the overgroups of H in
Table 4 are the following: q+3 maximal
subgroups of G; 2q+2 groups Eq1+2:(Cq−1)2 or
GL(2,q); groups GPi,Pj, Gr,ℓ,
GP1,…,Pq+1; q groups Eq:(Cq−1)2. Thus,
μ(H)=0.
Line 16, H=Gℓ1,…,ℓq+1,P≅Eq:Cq−1.
Arguing as in Line 15, μ(H)=0.
Line 17, H≅C2(q−1).
Then the overgroups K of H in Table
4 with μ(K)=0 are the following: q+4
maximal subgroups of G; 4 groups Eq1+2:(Cq−1)2 or
GL(2,q); 1 group Eq:(Cq−1)2; q groups (Cq−1)2:C2. Thus, μ(H)=0.
Line 18, H≅Eq.
Here, H is the group of elations with a given center and axis.
Then the overgroups K of H in Table 4 with
μ(K)=0 are the following: 2q+2 maximal subgroups of G;
(q+1)2 groups Eq1+2:(Cq−1)2 or GL(2,q); q2 groups
Eq:(Cq−1)2. Thus, μ(H)=0.
Line 19, H≅Cq−1.
Here, H is the group of homologies with a given center and axis.
Then the overgroups K of H in Table 4 with
μ(K)=0 are the following: n(H)=2q+4+(2q+1) maximal
subgroups of G; (q+2)2 groups Eq1+2:(Cq−1)2 or
GL(2,q); (q+1)2+(q+1)q groups Eq:(Cq−1)2;
(2q+1)+(q+1)q groups (Cq−1)2:C2. Thus, μ(H)=0.
Lines 20 to 22: we have n(H)=2 and hence μ(H)=1.
Line 23, H≅D8.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: 2+2q maximal subgroups of
G; 1 group Eq1+2:(Cq−1)2; 2q groups
GP,Π≅\mboxSym(4); 2q groups
Gℓ,Π≅\mboxSym(4). Thus, μ(H)=−2q.
Line 24, H≅C7.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: if p>3, 1+7q2+q+1
maximal subgroups, q2+q+1 groups C7:C3; if p=3,
14 maximal subgroups, 6 groups Eq1+2:(Cq−1)2,
3 groups GL(2,q), 6 groups Eq:(Cq−1)2, 3 groups
(Cq−1)2:C2, 7 groups C7:C3. Thus, μ(H)=0.
Line 25, H≅\mboxSym(3).
The overgroups K of H in Table 4 with
μ(K)=0 are the following: 2q maximal subgroups of G; 1
group GL(2,q); q+1 groups GP,Π≅\mboxSym(4); q+1 groups
Gℓ,Π≅\mboxSym(4). Thus, μ(H)=0.
Line 26, H≅C4.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: 4q2+8 maximal subgroups
of G; 1 group Eq1+2:(Cq−1)2; 4q2 groups
GP,Π≅\mboxSym(4); 4q2 groups
Gℓ,Π≅\mboxSym(4); 2q groups D8. Thus,
μ(H)=0.
Line 27, H≅E4, H≤Gℓ1,ℓ2,ℓ3.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: 4q2+4q+8 maximal
subgroups of G; q+1 groups Eq1+2:(Cq−1)2;
4q2 groups GP,Π≅\mboxSym(4);
3⋅4q2 groups Gℓi,Π≅\mboxSym(4);
23q groups D8. Thus, μ(H)=0.
Line 28, H≅E4, H≤GP1,P2,P3.
Arguing as in Line 27, μ(H)=0.
Line 29, H≅C3.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: 34q2+2 maximal subgroups; 1 group GL(2,q); 3q2−1 groups
GP,Π≅\mboxSym(4); 3q2−1 groups
Gℓ,Π≅\mboxSym(4); 32(q2−1) groups C7:C3.
Thus, μ(H)=0.
Line 30, H≅C2.
The overgroups K of H Table 4 with
μ(K)=0 are the following:
2q+2+2q3(q−1)+8q3(q−1) maximal subgroups of
G; 2q+1 groups Eq1+2:(Cq−1)2; q2 groups GL(2,q);
q2 groups Eq:(Cq−1)2; 2q3(q−1) groups
(Cq−1)2:C2; 38q3(q−1) groups
GP,Π≅\mboxSym(4); 38q3(q−1) groups
Gℓ,Π≅\mboxSym(4); 45q2(q−1) groups D8.
Thus, μ(H)=0.
Line 31, H={1}.
The overgroups K of H in Table 4 with
μ(K)=0 are the following: q2+q+1 groups GP≅Eq2:GL(2,q); q2+q+1 groups Gℓ≅Eq2:GL(2,q);
6q3(q+1)(q2+q+1) groups (Cq−1)2:\mboxSym(3);
3q3(q−1)2(q+1) groups Cq2+q+1:C3;
168q3(q3−1)(q2−1) groups \mboxPSL(3,2); (q2+q+1)(q+1)
groups Eq1+2:(Cq−1)2; (q2+q+1)q2 groups GL(2,q);
(q2+q+1)(q2+q)q groups Eq:(Cq−1)2;
2(q2+q+1)q3(q+1) groups (Cq−1)2:C2;
24q3(q3−1)(q2−1) groups GP,Π≅\mboxSym(4);
24q3(q3−1)(q2−1) groups Gℓ,Π≅\mboxSym(4);
21q3(q3−1)(q2−1) groups C7:C3;
4q2(q3−1)(q2−1) groups D8. Thus, μ(H)=0.
∎
5. The Möbius function and other combinatorial objects
There are several situations in which the knowledge of the Möbius
function of a group may be of help. We mention an application of in
topological graph theory; see [17] for the details. Given a
finite group G, define dk:=∣\mboxAut(G)∣1∑H≤Gμ(H)∣H∣k. Let Γ be a connected simple graph with vertex
set V(Γ) and edge set E(Γ), and let
β=∣E(Γ)∣−∣V(Γ)∣+1 be the first Betti number of
Γ. Let Isoc(Γ;G) be the number of isomorphism
classes of connected simple graphs Γ~ such that
Γ~ admits G as an automorphism group acting
semiregularly on Γ~ and the quotient graph
Γ~/G is isomorphic to Γ. Then Isoc(Γ;G)=dβ(G).
We now give another topological application of the Möbius function.
Let (P,⪯) be a finite poset, and P^ be
the poset obtained from P by adjoining a least element 0^
and a greatest element 1^. Let Δ(P) be the
order complex of P, i.e. the simplicial complex whose
vertices are the element of P and whose k-dimensional faces
are the s a0≺a1≺⋯≺ak of distinct
elements a0,…,ak∈P. Let χ(Δ(P)) be
the Euler characteristic and χ~(Δ(P)) be
the reduced Euler characteristic of Δ(P). The
Möbius function of P^ is related to
χ~(Δ(P)) as stated in Proposition 5.1, which essentially restates a result by Hall [11] on the computation of μP^(0^,1^) by means of the chains of even and odd length between 0^ and 1^.
Proposition 5.1**.**
(see [26, Proposition 3.8.6])*
Let P be a finite poset. Then*
[TABLE]
Let r be a prime number and P=Lr be the poset of nontrivial r-subgroups of a finite group G ordered by inclusion.
Lemma 5.2**.**
([24, Prop. 2.1])
Let H∈Lr. If H is not elementary abelian, then
μL^r(0^,H)=0. If H is elementary abelian of
order rs, then μL^r(0^,H)=(−1)sr(2s).
For the rest of this section, q is any prime power and G is the
group \mboxPGL(3,q). Using Proposition 5.1 and Lemma
5.2, we determine χ~(Δ(Lr)) for any
prime r.
Proposition 5.3**.**
For any prime number r, exactly one of the following cases holds:
r∤∣G∣* and χ~(Δ(Lr))=0;*
r∣q* and χ~(Δ(Lr))=−(q3−1);*
r∣(q2+q+1), r=3, and χ~(Δ(Lr))=3q3(q−1)2(q+1);
r∣(q+1), r=2, and χ~(Δ(Lr))=2q3(q3−1);
r∣(q−1), r∈/{2,3}, and χ~(Δ(Lr))=−3q2(q2+q+1)(q2+q−3):
r=2, q is odd, and χ~(Δ(L2))=−3q2(q2+q+1)(q2+q−3);
r=3, 3∣(q−1), and χ~(Δ(L3))=−8q2(q6−q4+7q3−7q−8).
Proof.
From gcd(q−1,q2+q+1)∈{1,3} and the order of G follows that the divisibility conditions in the claim are exhaustive and pairwise incompatible.
Suppose r∤∣G∣. Then χ~(Δ(Lr))=χ(∅)=0.
Suppose r∣q, i.e. r is the characteristic of Fq. Then H≤G is an elementary abelian r-subgroup if and only if H is made by elations with the same center or axis.
Let Nac(i) be the number of subgroups Eri whose elements
have both the same axis and center. Let Na(i) (resp. Nc(i)) be
the number of subgroups Eri whose elements have a common axis
(resp. center) but not a common center (resp. axis). By duality,
Na(i)=Nc(i) for any i. The subgroup of all elations with
both the same axis and center is a Eq; the subgroup
of all elations with a given axis (resp. center) is a
Eq2. Using the Gaussian coefficient, this implies
[TABLE]
[TABLE]
[TABLE]
Using the property (kn)r=rk(kn−1)r+(k−1n−1)r, the claim follows.
Suppose r∣(q2+q+1) and r=3. Then Sr is contained in a maximal subgroup Cq2+q+1:C3 of G and determines it uniquely. Thus G has exactly [G:(Cq2+q+1:C3)] subgroups Cr, and
χ~(Δ(Lr))=−3q3(q−1)2(q+1)⋅(−1).
Suppose r∣(q+1) and r=2. Then Sr is contained in a Cq+1 and hence is cyclic. Cq+1≤G is uniquely determined by its fixed points P∈PG(2,q) and Q,R∈PG(2,q2)∖PG(2,q), where R=Qq. Hence, we have q2+q+1 choices for P, and then q2 choices for the line QR and 2q2−q choices for {Q,R} on QR.
Thus,
χ~(Δ(Lr))=−2(q2+q+1)q2(q2−q)⋅(−1).
Suppose r∣(q−1) and r∈/{2,3}. Then Sr is contained in a maximal subgroup GT≅(Cq−1)2:\mboxSym(3); hence, the elementary abelian r-subgroups of G have size r or r2.
A subgroup Er2 is contained in exactly one group GT; thus,
G has exactly [G:GT] subgroups
Er2. A subgroup Cr made by homologies is uniquely
determined by its center and axis; hence, G has exactly
(q2+q+1)q2 subgroups Cr of homologies. A subgroup Cr
not made by homologies stabilizes pointwise an Fq-rat.
triangle T; GT has exactly 3 subgroups Cr of
homologies and r−1r2−1−3=r−2 subgroups Cr not of
homologies. Then G has exactly
[G:GT]⋅(r−2) subgroups
Cr not of homologies. Altogether,
[TABLE]
Suppose r=2 and q odd. Every involution of G is a homology (see [21]); hence a subgroup C2≤G is uniquely determined by the choice of the center and axis, and there are (q2+q+1)q2 such choices.
A subgroup E4≤G is uniquely determined by the
Fq-rat. triangle fixed pointwise by E4; hence G has exactly
[G:GT] subgroups E4. No
more than 3 homologies of the same order in PSL(3,q) can
commute pairwise (as they stabilize the center and axis of each
other, and the homology of a given order, center and axis is unique).
Hence, G has no subgroups E8. Thus,
χ~(Δ(L2))=−(q2(q2+q+1)⋅(−1)+6q3(q+1)(q2+q+1)⋅2).
Suppose r=3∣(q−1). Then r∣(q2+q+1), and the elements of order 3 are either homologies, or stabilize pointwise a triangle.
G has exactly q2(q2+q+1) homologies of order 3, [G:GT] elements of order 3 stabilizing an Fq-rat.
triangle T,
and [G:GT~] elements of order 3 stabilizing an
Fq3∖Fq-rat. triangle
T~.
Altogether, G has
q2(q2+q+1)+6q3(q+1)(q2+q+1)+3q3(q−1)2(q+1) elements of order 3.
Let H≤G with H≅E9. Since 9∤(q2+q+1), H is
contained in a max. subgroup GT≅(Cq−1)2:\mboxSym(3). This implies in particular that G has no subgroups E27.
Suppose that H contains a homology, with center C. Then C is a
vertex of T, and H is the unique E9≤G
stabilizing T pointwise. Thus, the number of E9≤G
containing a homology is [G:GT].
Suppose that H does not contain any homology. It is easily seen
that there exists σ∈H∖{1} which stabilizes T pointwise.
Hence, H=⟨σ,τ⟩ where σ and τ act
respectively trivially and with a 3-cycle on T. Up to
conjugation, T is the fundamental triangle and
σ=\mboxdiag(λ,λ2,1), where λ∈Fq
has order 3; up to replacing τ with τ2, τ:(X:Y:Z)↦(μY:ρZ:X) with
μ,ρ∈Fq∗.
The short orbits of H in PG(2,Fq) are
exactly the four distinct triangles T,T1,T2,T3 stabilized
pointwise by some C3≤H. By direct checking,
T1,T2,T3 are either Fq-rat. or
Fq3∖Fq-rat., according to
3∣o(μρ)q−1 or 3∤o(μρ)q−1,
respectively. Therefore, ⟨σ⟩ is contained in
exactly 31⋅3(q−1)2 subgroups E9≤G
which stabilize four Fq-rat. triangles; and
⟨σ⟩ is contained in exactly
31⋅32(q−1)2 subgroups E9≤G which
stabilize one Fq- and three
Fq3∖Fq-rat. triangles.
Viceversa, any subgroup E9≤G stabilizing only
Fq-rat. triangle (resp. one Fq- and
three Fq3∖Fq-rat. triangles)
contains exactly 4 subgroups (resp. 1 subgroup) C3
stabilizing pointwise an Fq-rat. triangle. Then, a
double counting argument shows that G contains exactly:
[G:GT]⋅9(q−1)2⋅41
subgroups E9 with no homology and only Fq-rat.
fixed triangles;
[G:GT]⋅92(q−1)2 subgroups E9
stabilizing an Fq3∖Fq-rat.
triangle.
The claim now follows by direct computation.
∎
Acknowledgments
The research of F. Dalla Volta and G. Zini was supported by the
Italian National Group for Algebraic and Geometric Structures and
their Applications (GNSAGA - INdAM).
