ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 16 (2019) 377-401 https://doi.org/10.26493/1855-3974.1765.d73
(Also available at http://amc-journal.eu)
The Möbius function of PSU(3, 22")
Giovanni Zini *
Department of Mathematics and Applications, University of Milano-Bicocca, via Cozzi 55, 50125 Milano, Italy
Received 25 July 2018, accepted 10 November 2018, published online 29 January 2019

Abstract
Let G be the simple group PSU(3,22"), n > 0. For any subgroup H of G, we compute the Möbius function ^L(H, G) of H in the subgroup lattice L of G, and the Möbius function ([H], [G]) of [H] in the poset L of conjugacy classes of subgroups of G. For any prime p, we provide the Euler characteristic of the order complex of the poset of non-trivial p-subgroups of G.
Keywords: Unitary groups, Möbius function, subgroup lattice. Math. Subj. Class.: 20G40, 20D30, 05E15, 06A07
1 Introduction
The Mobius function ^(H,G) on the subgroups of a finite group G is defined recursively by ^(G, G) = 1 and J2K>H G) = 0 if H < G. This function was used in 1936 by Hall [12] to enumerate k-tuples of elements of G which generate G, for a given k.
The combinatorial and group-theoretic properties of the Mobius function were investigated by many authors; see the paper [14] by Hawkes, Isaacs, and Ozaydin. The Mobius function is defined more generally on a locally finite poset (P, <) by the recursive definition ^(x, x) = 1, ^(x, y) = 0 if x < y, and J2x<z<y m(z, y) = 0 if x < y; for instance, the poset taken into consideration may be the subgroup lattice L of a finite group G ordered by inclusion. Mann [19, 20] studied ^(H, G) in the broader context of profinite groups G and defined a probabilistic zeta function P (G, s) associated to G, related to the probability of generating G with s elements when G is positively finitely generated.
The Mobius function on a poset P also appears in the context of topological invariants of the order simplicial complex A(P) associated to P, see the works of Brown [2] and
* This research was partially supported by Ministry for Education, University and Research of Italy (MIUR) and by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA -INdAM). The author would like to thank Francesca Dalla Volta and Martino Borello for their useful comments and suggestions, and Emilio Pierro for pointing out a mistake in a previous version of this paper.
E-mail address: giovanni.zini@unimib.it (Giovanni Zini)
©® This work is licensed under https://creativecommons.org/licenses/by/4.0/
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Quillen [25]; if P is the subgroup lattice of a finite group G, then the reduced Euler characteristic of A(P) is equal to ^({1}, G). This motivates the search for ^({1}, G) independently of the knowledge of ^(H, G) for other subgroups H of G, see for instance [26, 27] and the references therein; ^({1}, G) is often called the Möbius number of G. Shareshian provided a formula in [26] for ^({1}, Sym(n)), and computed ^({1},G) in [27] when G € {PGL(2, q), PSL(2, q), PGL(3, q), PSL(3, q), PGU(3, q), PSU(3, q)} with q odd or G is a Suzuki group Sz(22h+1).
Consider the poset L of conjugacy classes [H] of subgroups H of a finite group G, ordered as follows: [H] < [K] if and only if H is contained in some conjugate of K in G. After Hawkes, Isaacs, and Özaydin [14], we denote by A(H, G) the Möbius function ^([H], [G]) in L, while ^(H, G) is the Mobius function in L. Some attempt was done to search relations between the Mobius functions ^(H, G) and A(H, G); Hawkes, Isaacs, and Özaydin [14] proved that, if G is solvable, then
M{1},G) = |G'| • A({1}, G).	(1.1)
The property (1.1), which we call A)-property, does not hold in general for non-solvable groups; see [1]. Pahlings [23] proved that, if G is solvable, then
^(H, G) = [Na (H) : H n G'] • A(H, G)	(1.2)
for any subgroup H of G. The analysis of the generalized A)-property (1.2), although false in general for non-solvable groups, is of interest since it relates the Mobius functions ^(H, G) and A(H, G).
A lot of work was done by several authors about probabilistic functions for groups; see for instance [6, 10, 19, 20]. In particular, Mann posed in [19] a conjecture, the validity of which would imply that the sum
V	MH, G)
V	[G : H]s
over all subgroups H < G of finite index of a positively finitely generated profinite group G is absolutely convergent for s in some right complex half-plane and, for s € N large enough, represents the probability of generating G with s elements. Lucchini [18] showed that this problem can be reduced so that Mann's conjecture is reformulated as follows: there exist two constants c1, c2 € N such that, for any finite monolithic group G with non-abelian socle,
1.	|m(H, G)| < [G : H]C1 for any H < G such that G = Hsoc(G), and
2.	the number of subgroups H < G of index n in G such that H soc(G) = G and ^(H, G) = 0 is upper bounded by nc2, for any n € N.
It seems natural to investigate this conjecture on finite monolithic groups starting by almost simple groups. Mann's conjecture has been shown to be satisfied by the alternating and symmetric groups [3], as well as by those families of groups G for which ^(H, G) has been computed for any subgroup H; namely, PSL(2, q) [8, 12], PGL(2, q) [8], the Suzuki groups Sz(22h+1) [9], and the Ree groups P(32h+1) [24].
In this paper, we take into consideration the three dimensional projective special unitary group G = PSU(3, q) over the field with q = 22" elements, for any positive n (note that PSU(3, q) = PGU(3, q) as 3 \ (q + 1)). In particular, the following results are obtained.
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379
(i)	We compute ^(H, G) for any subgroup H of G, as summarized in Table 1. This shows that the groups PSU(3, 22") satisfy Mann's conjecture.
(ii)	We compute A(H, G) for any subgroup H of G, as summarized in Table 1. This shows that the groups PSU(3,22") satisfy the A)-property, but do not satisfy the generalized A)-property.
(iii)	We compute the Euler characteristic x(A(Lp \ {1})) of the order complex of the poset Lp \ {1} of non-trivial p-subgroups of G, for any prime p, as summarized in Table 2.
For the subgroups listed in Table 1, the isomorphism type determines a unique conjugacy class in G.
Table 1: Subgroups H of G = PSU(3, q), q = 22", with ^(H) = 0 or A(H) = 0.
Isomorphism type of H	|H |	ng(h )	m(h,g)	A(H, G)
G	q3(q3 + 1)(q2 -1)	H	1	1
(Eq . Eq2 ) X Cq2 —1	q3(q2 -1)	H	-1	-1
PSL(2, q) X Cq+1	q(q2 - 1)(q + 1)	H	-1	-1
(Cq+1 X Cq+i) X Sym(3)	6(q + 1)2	H	-1	-1
Cq2—q+1 X C3	3(q2 - q + 1)	H	-1	-1
Eq X Cq2 — 1	q(q2 -1)	H	1	1
(Cq+1 X Cq+1) X C2	2(q + 1)2	H	1	1
Sym(3)	6	Sym(3) X Cq+1	q + 1	1
C3	3	Cq2 —1 X C2	2(q2 1) 3	1
C2	2	(Eq . Eq2 ) X Cq+1	q3(q+1) 2	-1
Table 2: Euler characteristic of the order complex of the poset of proper p-subgroups of G.
Prime p
p i |G| p = 2 p | (q +1) p | (q - 1) p | (q2 - q + 1)
X(A(Lp \{1}))
0 q3 + 1 - q6-2q5-q4 + 2q3-3q2

The paper is organized as follows. Section 2 contains preliminary results on the Mobius functions ^(H, G) and A(H, G) and the relation between the Mobius function and the Euler characteristic of the order complex; this section contains also preliminary results on the groups G = PSU(3, 22"), whose elements are described geometrically in their action on the Hermitian curve associated to G. Sections 3 and 4 are devoted to the determination of ^(H, G) and A(H, G), respectively, for any subgroup H of G. Section 5 provides the Euler characteristic of the order complex of the poset of proper p-subgroups of G, for any prime p.
q6+q
2
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2 Preliminary results
Let (P, <) be a finite poset. The Mobius function :	Z is defined recursively
as follows:
(x, y) = 0 if x < y;	(x, x) = 1;	(z, y) = 0 if x<y.
xKzKy
If x < y, then (x, y) can be equivalently defined by
(x,z) = 0.
x<z<y
To the poset P we can associate a simplicial complex A(P) whose vertices are the elements of P and whose ¿-dimensional faces are the chains a0 < ■ ■ ■ < a of length i in P; A(P) is called the order complex of P. Provided that P has a least element 0, the Euler characteristic of the order complex of P \ {0} is computed as follows (see [28, Proposition 3.8.6]):
X(A(P\{0})) = - £ (0, x).
xep\{0}
Given a finite group G, we will consider the following two Mobius functions associated to G.
(i)	The Mobius function on the subgroup lattice L of G, ordered by inclusion. We will denote ^L(H, G) simply by H).
(ii)	The Mobius function on the poset L of conjugacy classes [H] of subgroups H of G, ordered as follows: [H] < [K] if and only if H is contained in the conjugate gKg-1 for some g G G. We will denote ([H], [G]) simply by A(H).
Two facts will be used to compute ^(H). The first easy fact is that, if H and K are conjugate in G, then ^(H) = ^(K). The second fact is due to Hall [12, Theorem 2.3], and is stated in the following lemma.
Lemma 2.1. If H < G satisfies ^(H) = 0, then H is the intersection of maximal subgroups of G.
For any prime p, let Lp be the subposet of L given by all p-subgroups of G, so that
x(A(Lp \{1})) = - £ ({1}, H).	(2.1)
h eLp\{ 1}
By a result of Brown [2], x(A(Lp \ {1})) is congruent to 1 modulo the order |G|p of a Sylow p-subgroup of G. In order to compute explicitly x(A(Lp \ {1})) we will use the following result of Hall [12, Equation (2.7)]:
Lemma 2.2. Let H be a p-group of order pr. If H is not elementary abelian, then MLp({1},H) = 0. If H is elementary abelian, then ^Lp ({1},H) = ( — 1)rp(2).
We describe now the group G which will be considered in the following sections. Let n be a positive integer, q = 22", Fq be the finite field with q element, and Fq be the algebraic
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closure of Fq. Let U be a non-degenerate unitary polarity of the plane PG(2, q2) over Fq2, and Hq C PG(2, Fq) be the Hermitian curve defined by U. The following homogeneous equations define models for Hq which are projectively equivalent over Fq2:
X q+! + Y q+! + Zq+1 = 0;	(2.2)
XqZ + XZq - Yq+1 =0.	(2.3)
The models (2.2) and (2.3) are called the Fermat and the Norm-Trace model of Hq, respectively. The set of Fq2 -rational points of Hq is denoted by Hq (Fq2), and consists of the q3 + 1 isotropic points of U. The full automorphism group Aut(Hq) of Hq is defined over Fq2, and coincides with the unitary subgroup PGU(3, q) of PGL(3, q2) stabilizing Hq(Fq2), of order | PGU(3, q)| = q3(q3 + 1)(q2 - 1).
The combinatorial properties of Hq (Fq2) can be found in [16]. In particular, any line I of PG(2, q2) has either 1 or q +1 common points with Hq(Fq2), that is, I is either a tangent line or a chord of Hq (Fq2); in the former case I contains its pole with respect to U, in the latter case I doesn't. Also, PGU(3, q) acts 2-transitively on Hq(Fq2) and transitively on PG(2, q2) \ Hq; PGU(3, q) acts transitively also on the non-degenerate self-polar triangles T = {P1, P2, P3} C PG(2, q2) \ Hq with respect to U. Recall that, if a G PGU(3, q) stabilizes a point P G PG(2, q2), then a stabilizes also the polar line of P with respect to U, and vice versa.
The curve Hq is non-singular and Fq2 -maximal of genus g = q(q2-1), that is, the size of Hq (Fq2) attains the Hasse-Weil upper bound q2 + 1 + 2gq. This implies that Hq is Fq4 -minimal and Fq6 -maximal, so that Hq(Fq4) \ Hq (Fq2) = 0 and |Hq (Fqa) \ Hq(Fq2 )| = q6 + q5 - q4 - q3. Let $ q2 be the Frobenius map (X, Y, Z) ^ (Xq , Yq , Zq ) over PG(2, Fq2); then the Fq6 \Fq2 -rational points of Hq split into q6+q5-q4-q3 non-degenerate triangles {P, $q2 (P), $^2 (P)}. The group PGU(3, q) is transitive on such triangles.
Since 3 f (q + 1), we have PGU(3, q) = PSU(3,q); henceforth, we denote by G the simple group PSU(3, q). The following classification of subgroups of G goes back to Hartley [13]; here we use that log2(q) has no odd divisors different from 1. The notation S2 stands for a Sylow 2-subgroup of G, which is a non-split extension Eq . Eq2 of its elementary abelian center of order q by an elementary abelian group of order q2.
Theorem 2.3. Let n > 0, q = 22", and G = PSU(3, q). Up the conjugation, the maximal subgroups of G are the following.
(i)	The stabilizer M1(P) = S2 x Cq2_1 of a point P G Hq (Fq2), of order q3(q2 - 1).
(ii)	The stabilizer M2(P) ^ PSL(2, q) x Cq+1 of a point P G PG(2,q2) \Hq (Fq2), of order q(q2 - 1)(q + 1).
(iii)	The stabilizer M3 (T) = (Cq+1 x Cq+1) x Sym(3) of a non-degenerate self-polar triangle T = {P, Q, R} C PG(2, q2) \ Hq with respect to U, of order 6(q + 1)2.
(iv)	The stabilizer M^T) = Cq2-q+1 x C3 of a triangle T = {P, $q2 (P), $2 (P)} C
Hq (Fq6 ) \ Hq (Fq2 ), of order 3(q2 - q + 1).
For a detailed description of the maximal subgroups of G, both from an algebraic and a geometric point of view, we refer to [11, 21, 22].
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In our investigation it is useful to know the geometry of the elements of PGU(3, q) on PG(2, Fq), and in particular on Hq(Fq2). This can be obtained as a corollary of Theorem 2.3, and is stated in Lemma 2.2 with the usual terminology of collineations of projective planes; see [16]. In particular, a linear collineation a of PG(2, Fq) is a (P, £)-perspectivity, if a preserves each line through the point P (the center of a), and fixes each point on the line £ (the axis of a). A (P, £)-perspectivity is either an elation or a homology according to P G £ or P / I. Lemma 2.4 was obtained in [21] in a more general form (i.e., for any prime power q).
Lemma 2.4. For a nontrivial element a G G = PSU(3, q), q = 22", one of the following cases holds.
(A)	ord(a) | (q +1) and a is a homology, with center P G PG(2, q2) \ Hq and axis £P which is a chord of Hq (Fq2); (P,£p) is a pole-polar pair with respect to U.
(B)	2 { ord(a) and a fixes the vertices Pi,P2,P3 of a non-degenerate triangle T c PG(2, q6).
(B1) ord(a) | (q + 1), Pi, P2, P3 G PG(2, q2) \ Hq, and the triangle T is self-polar with respect to U.
(B2) ord(a) | (q2 - 1) and ord(a) \ (q + 1); Pi G PG(2, q2) \ Hq and P2, P3 G
Hq (Fq2 ).
(B3) ord(a) | (q2 - q + 1) andPi,P2,Ps G Hq(Fq6) \ Hq(Fq2).
(C)	ord(a) = 2; a is an elation with center P G Hq (Fq2) and axis £P which is tangent to Hq at P, such that (P, £P) is a pole-polar pair with respect to U.
(D)	ord(a) = 4; a fixes a point P G Hq (Fq2) and a line £P which is tangent to Hq at P, such that (P,£p) is a pole-polar pair with respect to U.
(E)	ord(a) = 2d where d is a nontrivial divisor of q +1; a fixes two points P G Hq (Fq2) and Q G PG(2, q2) \ Hq, the polar line PQ of P, and the polar line of Q which passes through P.
For a detailed description of the elements and subgroups of G, both from an algebraic and a geometric point of view, we refer to [11, 21, 22], on which our geometric arguments are based.
Throughout the paper, a nontrivial element of G is said to be of type (A), (B), (B1), (B2), (B3), (C), (D), or (E), as given in Lemma 2.4. Also, the polar line to Hq at P G PG(2, q2) is denoted by £P. Note that, under our assumptions, any element of order 3 in G is of type (B2). We will denote a cyclic group of order d by Cd and an elementary abelian group of order d by Ed. The center Z(S2) of S2 is elementary abelian of order q, and any element in S2 \ Z(S2) has order 4; see [11, Section 3].
3 Determination of ) for any subgroup H of G
Let n > 0, q = 22", G = PSU(3, q). This section is devoted to the proof of the following theorem.
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Theorem 3.1. Let H be a proper subgroup of G. Then H is the intersection of maximal subgroups of G if and only if H is one of the following groups:
S X Cq2_i,
(Cq+i x Cq+1) X Sym(3),
Cq+1 x Cq+1,
PSL(2,q) x Cq+1,
Eq X Cq2 _ 1, Cq2- 1,
Cq
q2_q+1
X C3,
Cq+1 = Z(M2(P)) for some P, Eq,
C3,	C2,
(Cq+1 X Cq+1) X C2, C2(q+1),	(3.1)
Sym(3), {1}.
Given a type of groups in Equation (3.1), there is just one conjugacy class of subgroups of G of that isomorphism type.
The normalizer NG(H) of H in Gfor the groups H in Equation (3.1) are, respectively:
H, H,
H X Sym(3), PSL(2, q) x H,
Cq2 _1 X C2,
H,	H,
H,	H,
H X C2,	Eq X Cq+1,
S2 X Cq2 _ 1,	H X Cq+1,
S2 X Cq+1,	G.
(3.2)
The values ^(H) for the groups H in Equation (3.1) are, respectively:
-1, -1, 0, 0,
-1, 1, 0, 0,
1,
2(q2 - 1) q3(q + 1)

2
1, 0,
q +1, 0.
(3.3)
The proof of Theorem 3.1 is divided into several propositions.
Proposition 3.2. The group G contains exactly one conjugacy class for any group in Equation (3.1).
Proof. Case 1: The first four groups in Equation (3.1), i.e.,
S2 x Cq2_ 1, PSL(2, q) x Cq+1, Cq2_q+1 x C3, and (Cq+1 x Cq+1) x Sym(3),
are the maximal subgroups of G, for which there is just one conjugacy class by Theorem 2.3.
Case 2: Let a1, a2 G G have order 3, so that they are of type (B2) and a fixes two distinct points Pj, Qj G Hq (Fq2). The group G is 2-transitive on Hq (Fq2), and the pointwise stabilizer of {Pj, Qj} is cyclic of order q2 - 1. Hence, (a1) and (a2) are conjugated in G.
Case 3: Let a1, a2 G G have order 2, so that they are of type (C) and aj fixes exactly one point Pj on Hq(Fq2). Up to conjugation P1 = P2, as G is transitive on Hq(Fq2). The involutions fixing P1 in G, together with the identity, form an elementary abelian group Eq, which is normalized by a cyclic group Cq_1; no nontrivial element of Cq_1 commutes
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with any nontrivial element of Eq (see [11, Section 4]). Hence, a1 and a2 are conjugated under an element of Cq_ i.
Case 4: Let a1 , a2,p^p2 € G satisfy o(oj) = 3, = 2, and H := (a^P^ = Sym(3). As shown in the previous point, we can assume a1 = a2 up to conjugation. Let P,Q € Hq (Fq2) and R € PG(2, q2)\Hq be the fixed points of «1. Since A a^r1 = a-1, we have that pj fixes R and interchanges P and Q; P is then uniquely determined from the Fq2 -rational point of PQ fixed by P (namely, the intersection between PQ and the axis of P). Since the pointwise stabilizer Cq2_1 of {P, Q} acts transitively on PQ(Fq2) \ Hq, P1 and p2 are conjugated, and the same holds for H1 and H2.
Case 5: Any two groups isomorphic to Cq2_1 are conjugated in G, because they are generated by elements of type (B2) and G is 2-transitive on Hq(Fq2).
Case 6: Any two groups isomorphic to Eq are conjugated in G, because any such group fixes exactly one point P € Hq (Fq2), G is transitive on Hq (Fq2), and the stabilizer GP = M1 (P) contains just one subgroup Eq.
Case 7: Any two groups H1 , H2 = Eq x Cq2_1 are conjugated in G. In fact, their Sylow 2-subgroups Eq coincide up to conjugation, as shown in the previous point. The normalizer NG(Eq) fixes the fixed point P € Hq(Fq2) of Eq, and hence NG(Eq) = M1(P) = S x Cq2_ 1. The complements Cq2_1 are conjugated by Schur-Zassenhaus Theorem; hence, H1 and H2 are conjugated.
Case 8: Any two groups isomorphic to C2(q+1) are conjugated in G, because they are generated by elements of type (E) and two elements a1 , a2 of type (E) of the same order are conjugated in G. In fact, aj is uniquely determined by its fixed points Pj € Hq (Fq2) and Qj € ¿p(Fq2) \ Hq; here, ¿p is the polar line of Pj. Up to conjugation P1 = P2, from the transitivity of G on Hq(Fq2). Also, S2 has order q3 and acts on the q2 points of ¿p (Fq2) \ Hq with kernel Eq, hence transitively. We can then assume Q1 = Q2.
Case 9: Let ZPi be the center of M2 (Pj), i = 1, 2. As shown in [5, Section 4], ZPi = Gq+1 and ZPi is made by the homologies with center Pj, together with the identity. Since G is transitive on PG(2,q2) \ Hq, we have up to conjugation that M2(P1) = M2(P2) and
ZPl = ZP2 .
Case 10: Any two groups H1, H2 = Cq+1 x Gq+1 are conjugated in G. In fact, H is the pointwise stabilizer of a self-polar triangle Tj = {Pj, Qj, Rj} c PG(2, q2) \ Hq (see [5, Section 3]), and the stabilizers of T\ and T2 are conjugated by Theorem 2.3.
Case 11: Any two groups H1, H2 = (Cq+1 x Cq+1) x C2 are conjugated in G. In fact, their subgroups Cq+1 x Cq+1 coincide up to conjugation as shown above, and fix pointwise a self-polar triangle T = {P, Q, R} c PG(2, q2) \ Hq. Let Pj € Hj have order 2, i = 1, 2. Then pj fixes one vertex of T and interchanges the other two vertexes. Up to conjugation in M3 (T) we have p1 (P) = p2 (P) = P. Then H1 = H2, as they coincide with the stabilizer of P in M3(T).	□
Proposition 3.3. The normalizers NG(H) of the groups H in Equation (3.1) are given in Equation (3.2).
Proof. Casel: Clearly NG(H) = H for any H from the first four groups of Equation (3.1) as H is maximal in G.
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Case 2: Let H = Eq x C^^. Then H < M2(P), where P is the unique fixed point of in PG(2, q2) \ Hq. The group H has a unique cyclic subgroup Cq+1 of order q +1; namely, Cq+1 is the center of M2(P) and is made by the homologies with center P; since q is even, H is a split extension Cq+1 x (Eq x Cq-1). Hence, NG(H) < NG(Cq+1) = M2(P). The group H/Cq+1 = Eq x Cq-1 is maximal and hence self-normalizing in M2(p)/Cg+1 = PSL(2, q); thus, NG(Eg x Cq_1) = H and Ng(H) = H.
Case 3: Let H = Cq+1 x Cq+1. Then NG(H) < Ms(T), where T is the self-polar triangle fixed pointwise by H. Since H is the kernel of M3(T) in its action on T, we have
Ng(H) = M3(T) and |Ng(H)| = 6|H|.
Case 4: Let H = (Cq+1 x Cq+1) x C2. Then Cq+1 x Cq+1 is normal in NG(H), being the unique subgroup of index 2 in H. Hence NG(H) < M3(T), where T is the self-polar triangle fixed pointwise by H. Also, NG(H) fixes the vertex P of T fixed by H, so that
Ng(H) = M3 (T). This implies Ng(H) = H.
Case 5: Let H = Cq2_1. Then H is generated by an element a of type (B2) with fixed points P, Q G Hq(Fq2) and R G PG(2, q2) \ Hq. Let P be an involution satisfying P(R) = R, P(P) = Q, and P(Q) = P; then P g Ng(H), because H coincides with the pointwise stabilizer of {P, Q} in G. An explicit description is the following: given with equation (2.3), we can assume up to conjugation that a = diag(aq+1, a, 1) where a is a generator if F*2 (see [11]); then take
/0 0 1N
P = 10 1 0 | .	(3.4)
V1 0 0y
Since Ng(H) acts on {P, Q} and P G NG(H), the pointwise stabilizer H of {P, Q} has index 2 in Ng(H). This implies Ng(H) = Cg2_1 x G2 and |Ng(H)| = 2|H|.
Case 6: Let H = C2(q+1), so that H is generated by an element a of type (E) fixing exactly two points P G (Fq2) and Q G (Fq2) \ . Then NG(H) fixes P and Q. The subgroup of M1(P) commutes with H elementwise, while any 2-element in M1 (P) \ has order 4 and does not fix Q; hence, the Sylow 2-subgroup of NG (H) is . Also, Ng(H) = x Cd, where Cd is a subgroup of Cq2_1 containing the subgroup Gq+1 of H. Let C2 be the subgroup of H of order 2; the quotient group (C2 x Cd)/Cq+1 = C2 x Cacts faithfully as a subgroup of PGL(2, q) on the q +1 points of ^q n . By the classification of subgroups of PGL(2, q) ([7]; see [17, Hauptsatz 8.27]), this implies d =1; that is, Ng(H) = Eq x Cg+1 and |Ng(H)| = f |H|.
Case 7: Let H = Cq+1 = Z(Mf(P)). Since H is the center of Mf(P), Mf(P) < Ng(H). Conversely, H is made by homologies with center P, and hence NG (H) fixes P. Thus, Ng(H) = Mf(P) and |Ng(H)| = q(q2 - 1)|H|.
Case 8: Let H = Eq. Since Eq has a unique fixed point P on Hq (Fq2) and Eq = Z(M1(P)), we have Ng(H) < M1(P) and M1(P) < Ng(H), so that Ng(H) = M1(P) and |Ng(H)| = q2(q2 - 1)|H|.
Case 9: Let H = Sym(3) = (a, P), with o(a) =3 and o(P) = 2. Let P, Q G Hq(Fq2) and R G PG(2, q2)\Hq be the fixed points of a; P fixes R, interchanges P and Q, and fixes another point on ¿R nHq. The group NG(H) acts on {P, Q} and on {A^, , }.
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The pointwise stabilizer Cq2 _ 1 has a subgroup Cq+i which is the center of M2 (P) and fixes PQ pointwise, while any element in Cq2_1 \ Cq+1 acts semiregularly on PQ \ {P, Q}; hence, Cq2_1 nNG(H) = C3(q+1). If an element 7 G NG(H) fixes {P, Q} pointwise, then Y fixes a point in {A^, Aap, A^}, and hence 7 G {ft, aft, a2 ft}. Therefore, NG(H) = C3(q+1) X C2 = H x Cq+1 and |Ng(H)| = (q + 1)|H|.
Case 10: Let H = C3 and a be a generator of H, with fixed points P, Q G Hq (Fq2) and R G PG(2, q2) \ Hq. The normalizer NG(H) fixes R and acts on {P, Q}. There exists an involution ft G G normalizing H and interchanging P and Q (see Equation (3.4)). Then the pointwise stabilizer of {P, Q} has index 2 in NG(H). Also, the pointwise stabilizer
of {P, Q} in G is cyclic of order q2 - 1. Then NG(H) = Cq2_1 x G2 and |NG(H)| =
|H|.
Case 11: Let H = C2 and a be a generator of H, with fixed point P G Hq(Fq2). Then Ng(H) fixes P, i.e. Ng(H) < M1(P) = S2 x Cq2_1. Since any involution of M1(P) is in the center of S2, the Sylow 2-subgroup of Ng(H) has order q3. Let ft G Cq2_1. If o(ft) | (q + 1), then ft commutes with any involution of S2. If o(ft) f (q + 1), then ft does not commute with any element of S2. This implies that NG(H) = S2 x Cq+1, and |Ng(H)| =	|H |.	□
Lemma 3.4. Let a G G be an involution, and hence an elation, with center P and axis £P. Then there exist exactly q3/2 self-polar triangles Tjj = {Pi, Qi,j, R^j}, i = 1,..., q2, j = 1,..., §, such that a stabilizes Tj,j. Also, Pi G and P G Qj,j Rj,j for any i and j.
Proof. The number of involutions in G is (q3 + 1)(q — 1), since for any of the q3 + 1 Fq2 -rational points P of Hq the involutions fixing P form a group Eq. The number of self-
polar triangles T c PG(2, q2) \ Hq is [G : M3(T)] = ^{q^^^. For any self-polar triangle T = {A1, A2, A3} c PG(2, q2) \ Hq, the number of involutions in G stabilizing T is 3(q +1). In fact, for any of the 3 vertexes of T there are exactly q +1 involutions a1,..., aq+1 fixing that vertex, say A1, and interchanging A2 and A3; ai is uniquely determined by its center A2A3 n Hq. Then, by double counting the size of
{(ft, T) | ft G G, o(ft) = 2, T c PG(2, q2) \ Hq is a self-polar triangle,
ft stabilizes T},
q3
a stabilizes exactly ^ self-polar triangles T. For any such T, one vertex Pi of T lies on the axis of a, because a is an elation, and the other two vertexes {Qi,j, Rj,j} of T lie on the polar line £Pi of Pi. Since M1(P) is transitive on the q2 points P1,..., Pq2 of £P (Fq2) \ {P}, any point Pi is contained in the same number | of self-polar triangles Ti,j
stabilized by a.	□
q2 1
Lemma 3.5. Let a G G have order 3. Then there are exactly ^-3— self-polar triangles
Tj C PG(2, q2) \Hq, i = 1,..., ^r, which are stabilized by a. Also, there are exactly 2(q3_1) triangles
3
Tj = {Pj, $q2 (Pj), ^2 (Pj )} C Hq(Fq6 ) \ Hq(Fq2 ), j = 1, . . . , ^^ which are stabilized by a.
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Proof. By Proposition 3.2, any two subgroups of G of order 3 are conjugated in G. Also, any element of order 3 is conjugated to its inverse by an involution of G. Hence, any two element of order 3 are conjugated in G.
Now the claim follows by double counting the size of
{(P, T) I P e G, o(P) = 3, T c PG(2, q2) \ Hq is a self-polar triangle,
P stabilizes T},
and
{(P,T) | P e G, o(P) = 3, T = {P, $q2(P), (P)} with
P e Hq (Fq6) \ Hq (Fq2), p stabilizes T},
using the following facts. The number of elements of order 3 in G is (q +1) • 2. The number of self-polar triangles T c PG(2,q2) \ Hq is [G : M3(T)]. The number of elements of order 3 stabilizing a fixed self-polar triangle T is 2(q+1)2, because any element acting as a 3-cycle on the vertexes of T has order 3 (see [5, Section 3]). The number of triangles T = {P, $q2 (P), $2 (P)} C Hq(Fq6) \ Hq(Fq2) is [G : M4(T)]. The number
of elements of order 3 stabilizing a fixed triangle T is 2(q2 - q + 1), because any element
in M4(T) \ Cq2-q+1 has order 3 (see [4, Section 4]).	□
Lemma 3.6. Let H < G be isomorphic to Sym(3), H = (a) x (P). Then there are exactly q +1 self-polar triangles
Ti = {Pi, Qi, Ri} C PG(2, q2) \ Hq, i = 1,... , q +1,
which are stabilized by H. Up to relabeling the vertexes, we have that P^ ..., Pq+1 lie on the axis of the elation P, Q1,..., Qq+1 lie on the axis of the elation ap and R1,..., Rq+1 lie on the axis of the elation a2 P.
Proof. By Proposition 3.2, any two subgroups K1, K2 < G with Ki = Sym(3) are conjugated, and |NG(Ki)| = 6(q + 1); hence, the number of subgroups of G isomorphic
to Sym(3) is [G : NG(Ki)] = te^1^"^. The number of self-polar triangles T is
[G : M3(T)] = (q2-q+16)q3(q-1). Then the claim on the number of self-polar triangles follows by double counting the size of
{(K, T) | K < G, K = Sym(3), T c PG(2, q2) \ Hq is a self-polar triangle,
K stabilizes T},
once we show that, for any self-polar triangle T = {A, B, C}, there are in G exactly (q + 1)2 subgroups isomorphic to Sym(3) which stabilize T.
Let K < M3(T), K ^ Sym(3), K = (a, P) with o(a) = 3, o(P) = 2. Let P, Q, R be the fixed points of a, with P e PG(2,q2) \ Hq, Q, R e Hq(Fq2). By Proposition 3.3, Ng(K) = K x Cq+1 where Cq+1 is made by homologies with center P; this implies Ng(k) n M3(T) = K. Hence, there are at least [M3(T) : Sym(3)] = (q + 1)2 distinct groups Sym(3) stabilizing T, namely the conjugates of K through elements of M3(T). On the other side, M3(T) contains exactly (q + 1)2 subgroups K of order 3, with fixed points P e PG(2, q2) \ Hq, Q, R e Hq(Fq2). Any involution P of M3(T) normalizing
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K is uniquely determined by the vertex of T that ft fixes, because ,0(P) = P, ,0(Q) = R, and ^(R) = Q. Thus, K is contained in exactly one subgroup of M3(T) isomorphic to Sym(3). Therefore the number of subgroups isomorphic to Sym(3) which stabilize T
is (q + 1)2.
Finally, the configuration of the vertexes of Ti,..., Tq+1 on the axes of the involutions of H follows from Lemma 2.4 and the fact that every involution fixes a different vertex of Ti.	□
Proposition 3.7. Any group H in Equation (3.1) is the intersection of maximal subgroups of G.
Proof. Case 1: The first four groups of Equation (3.1) are exactly the maximal subgroups of G.
Case 2: Let H = Eq x Cq2_1. Let P G Hq(Fq2) be the unique point of Hq fixed by Eq; Eq fixes £P pointwise. Also, the fixed points of Cq2_1 are P, Q G Hq (Fq2) and R G PG(2, q2) \ Hq, where R G £P and PQ = £R. Then H < M1(P) n M2(R). Conversely, from M1(P) n M2(R) < M1(P) follows M1(P) n M2(R) = K x Cd with K < S2 and Cd < Cq2_1. From M1(P) n M2(R) < M2(R) follows that K does not contain any element of type (D), so that K < Eq. Thus, M1(P) n M2(R) < H, and H = M1(P) n M2(R).
Case 3: Let H = (Cq+1 x Cq+1) x C2. Let T = {P, Q, R} c PG(2, q2) \ Hq be the self-polar triangle fixed pointwise by Cq+1 x Cq+1, and let P be the vertex of T fixed by C2. Then H < M3(T) n M2(P). Conversely, since M3(T) n M2(P) fixes P and acts on { Q, R}, the pointwise stabilizer Cq+1 x Cq+1 of T has index at most 2 in M3 (T) n M2 (P), so that M3(T) n M2(P) < H. Thus, H = M3(T) n M2 (P).
Case 4: Let H = Cq+1 x Cq+1. Let T = {P, Q, R} c PG(2, q2) \ Hq be the self-polar triangle fixed pointwise by Cq+1 x Cq+1. Since H is the whole pointwise stabilizer of T in G, we have H = M2(P) n M2(Q) n M2(R).
Case 5: Let H = Cq2_1 and let a be a generator of H, with fixed points P, Q G Hq (Fq2) and R G PG(2, q2) \ Hq. The pointwise stabilizer of {P, Q} in G is exactly H; thus, H = M1(P) n M2(Q).
Case 6: Let H = C2(q+1) and let a be a generator of H, of type (E), with fixed points P G Hq(Fq2) and Q G £p(Fq2) \ Hq. By Lemma 3.4 there are 2 self-polar triangles stabilized by the involution aq+1 having one vertex in Q and two vertexes on £q; let T =
{Q, R1, R2} be one of these triangles. Then H < M1(P) n M2(Q) n M3(T).
Conversely, let a G (M1(P) n M2(Q) n M3(T)) \ {1}. If a fixes {R1, R2} pointwise, then from a G M1(P) follows that a is in the kernel Cq+1 < H of the action of M2(Q) on £q. The quotient (M1(P) n M2(Q) n M3(T))/Cq+1 acts on £q as a subgroup of PSL(2, q) fixing P and interchanging R1 and R2. From [17, Hauptsatz 8.27] follows (M1(P) n M2(Q) n M3(T))/Cq+1 = C2, and hence H = M1(P) n M2(Q) n M3(T).
Case 7: Let H = Cq+1 = Z(M2(P)). Then H is made by the homologies of G with center P, together with the identity. Thus, H = M1(P1) n M1(P2) n M1(P3), where P1, P2, P3 are distinct point in n Hq.
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Case 8: Let H = Eq and let P be the unique point of Hq (Fq2) fixed by any element in H. Then H = M2(Pi) n M2(P2) n M2(P3), where Pi, P2, P3 are distinct points in
¿p (Fq2 ) \{P}.
Case 9: Let H = C2, a be a generator of H with fixed point P G Hq (Fq2), and Pi, P2,P3 G ¿p(Fq2) \ {P}. Let T = {Pi,Qi,i,Ri,i} be a self-polar triangle stabilized by a. Then H < M2(Pi) n M2(P2) n M2'(P3) n M3(T). Since the elation a is uniquely determined by the image of one point not on its axis ¿P, H < M3(T) implies H = M2(Pi) n M2(P2) n M2(P3) n M3(T).
Case 10: Let H = C3. By Lemma 3.5, H stabilizes 2(q23-i) triangles T c Hq(Fq6) \ Hq (Fq2); let Ti and T2 be two of them. Then H < M4(Ti) n ^4(2^2). If H < M4(Tfi) n M4(T2), then there exist a nontrivial a G G stabilizing pointwise both Tfi and T2, a contradiction to Lemma 2.4. Thus, H = M4(Ti) n M4^).
Case 11: Let H = Sym(3). By Lemma 3.6, H stabilizes q +1 self-polar triangles Ti,... ,Tq+i, so that H < M3(Ti) n • • • n M3(Tq+i). Suppose by contradiction that H = M3(Ti) n • • • n M3(Tq+i). Then M3(Ti) n • • • n M3(Tq+i) contains a nontrivial element a fixing every triangle Ti pointwise. Since the triangles Tj's do not have vertexes in common, this is a contradiction to Lemma 2.4. Thus, H = M3(Ti) n • • • n M3(Tq+i).
Case 12: Let H = {1}. Since G is simple, H is the Frattini subgroup of G.	□
Proposition 3.8. If H < G is the intersection of maximal subgroups, then H is one of the groups in Equation (3.1).
Proof. We proceed as follows: we take every subgroup K < G in Equation (3.1), starting from the maximal subgroups Mj of G; we consider the intersections H = K n Mj of K with the maximal subgroups of G; here, we assume that K < Mj. We show that H is again one of the groups in Equation (3.1).
Case 1: Let K = S2 x Cq2_i = Mi(P) for some P G Hq(Fq2).
Let H = K n Mi(Q), Q = P. Then H is the pointwise stabilizer of {P, Q} c Hq(Fq2), which is cyclic of order q2 - 1, i.e. H = Cq2_i.
Let H = K n M2(Q). Suppose Q G ¿p. Then H = Eq2 x Cq2_i, where Eq2 is made by the elations with axis PQ and Cq2_i is generated by an element of type (B2) with fixed points Q, P, and another point R G ¿q. Now suppose Q G ¿P. Then H stabilizes ¿q and hence also the point R = ¿P n ¿q. Then H stabilizes QR and hence also the pole A of QR; by reciprocity, A G PQ. Thus, H fixes three collinear point A, P, Q, and hence every point on AP. Then H = Gq+i = Z(M2(R)).
Let H = K n M3(T), T = {A, B, G}, with P on a side of T, say P G AB. Then H fixes G and acts on {A, B}. Thus, H is generated by an element of type (E) with fixed points P, G and fixed lines PG, AB; hence, H = C2(q+i).
Let H = K n M3(T), T = {A, B, G}, with P out of the sides of T. By reciprocity, no vertex of T lies on ¿P. This implies that no elation acts on T, so that 2 \ |H |; this also implies that no homology in M3(T) fixes P, so that H has no nontrivial elements fixing T pointwise. Thus H < G3.
Let H = K n M4(T). By Lagrange's theorem, H < G3.
Case 2: Let K = PSL(2, q) x Gq+i = M2(P) for some P G PG(2, q2) \ H,
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Let H = K n M2(Q), Q = P, and R be the pole of PQ. If R e PQ, then H is the pointwise stabilizer of PQ and is made by the elations with center R; thus, H = Eq. If R e PQ, then H is the pointwise stabilizer of T = {P, Q, R}; thus, H = Cq+1 x Cq+1.
Let H = K n M3(T) with T = {A, B, C}. If P is a vertex of T, then H = (Cq+1 x Cq+1) x C2. If P is on a side of T but is not a vertex, say P e AB, then H fixes the pole D e AB of C. Then H fixes pointwise T' = {P, C, D} and acts on {A, B}. This implies that H fixes AB pointwise and H = Cq+1 = Z(M2(C)). If P is out of the sides of T, then no nontrivial element of H fixes T pointwise; thus, H < Sym(3). Let H = K n M4(T). By Lagrange's theorem, H < C3.
Case 3: Let K = (Cq+1 x Cq+1) x Sym(3) = M3(T) for some self-polar triangle
T = {A, B, C}.
Let H = K n M3(T') with T' = {A', B', C'} = T. If T and T' have one vertex A = A' in common, then H = C2(q+1) is generated by an element of type (E) fixing A and a point D e BC = B'C'. If A' e AC \ {A, C}, then H stabilizes B'C', because B'C' is the only line containing 4 points of {A, B, C, A', B', C'}. Then H fixes A', A, and C; hence also B. Since H acts on {B', C'}, H cannot be made by nontrivial homologies of center B; thus, H = {1}.
Let H = K n M4(T'). By Lagrange's theorem, H < C3.
Case 4: Let K = Cq2-q+1 x C3 = M4(T) for some T c Hq(Fq6). Let H = Kn M^T') with T' = T. Since 3 does not divide the order of the pointwise stabilized Cq2_q+1 of T, H contains no nontrivial elements fixing T or T' pointwise. Thus, H < C3.
Case 5: Let K = Eq x Cq2-1 and P e Hq(Fq2), Q e tP \ {P} be the fixed points of K.
Let H = K n M1(R) with R = P .If R e ¿Q, then H = Cq2_1. If R e then H fixes the pole S of PR; by reciprocity S e PQ, so that H fixes PQ pointwise and also R ePQ. Thus, H = {1}.
Let H = K n M2(r) with R = Q. If R e ¿P, then H is the pointwise stabilizer Eq of PQ. If R e ¿P, then H fixes pointwise the self-polar triangle {Q, R, S} where S is the pole of QR. Hence, either H = Cq+1 = Z(M2(Q)) or H = {1} according to P e RS or P e RS, respectively.
Let H = K n M3(T) with T = {A, B, C}. If P is on a side of T, say P e BC, then either H = {1} or H = Cq+1 = Z(M2(A)). If P is out of the sides of T, then no nontrivial element of H can fix T pointwise; thus, H < Sym(3). Let H = K n M4(T). By Lagrange's theorem, H < C3.
Case 6: Let K = (Cq+1 x Cq+1) x C2 = M3(T) n M2(A), where T = {A, B, C}.
Let H = K n M1 (P). If P e BC, then H = C2(q+1) is generated by an element of type (E). If P e BC, then H = {1}.
Let H = K n M2(P), P = A. If P e {B, C}, then H is the pointwise stabilizer Cq+1 x Cq+1 of T .If P e AB \ {A, B} or P e AC \ {A, C}, then H = Cq+1 = Z(M2(C)) or H = Cq+1 = Z(M2(b)), respectively. If P e BC \ {B, C}, then H fixes A, P, the pole of AP, and acts on {B, C}; thus, H = Cq+1 = Z(M2(A)). If P is not on the sides of T, then no nontrivial element of H can fix T pointwise; thus, H < C2.
Let H = K n M3(T') with T' = {A', B', C'} = T. Since 3 f |H|, H fixes a vertex of T', say A'. If A' = A, then H = C2(q+1). If A' e {B, C}, then H fixes T pointwise and acts on {B', C'}; thus, H = Cq+1 = Z(M2 (A')). If A' e (AB U AC) \ {A, B, C}, then H fixes AB or AC pointwise and acts on {B', C'}; thus, H = {1}. If A' e BC, then H
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fixes A, A', and the pole D of AA'; as H acts on {B, C}, this implies H = {1}. If A' is not on the sides of T, then no nontrivial element of H fixes T pointwise and H < C2. Let H = K n M4(T'). By Lagrange's theorem, H < C3.
Case 7: Let K = Cq+1 x Cq+1 = M3(T) n M2(A) n M2(B) n M2(C) with T =
{A, B, C}.
Let H = K n M1 (P) or H = K n M2 (P). If P is not on the sides of T, then H = {1}; if P is on a side of T, say P G BC, then H = Cq+1 = Z(M2 (A)).
Let H = K n M3(T') with T' = {A', B', C'}. Since K is not divisible by 2 or 3, H = {1} only if H fixes T' pointwise. Up to relabeling, this implies A' = A, B', C' G BC, and H = Cq+1 = Z (M2(A)).
Let H = K n M4(T'). By Lagrange's theorem, H = {1}.
Case 8: Let K = Cq2_1 = (a), with a of type (B2) fixing the points P G PG(2, q2) \ Hq and Q,R G Hq(Fq2).
Let H = K n M1(A) or H = K n M2(A). Since the nontrivial elements of H are either of type (B2) or of type (A) with axis QR, we have H = {1} unless A g QR; in this case, H = Cq+1 = Z(M2(P)).
Let H = K n M3(t) or H = K n M4(T). By Lagrange's theorem, H < C3.
Case 9: Let K = C2(q+1) = (a) with a of type (E) fixing the points P G Hq(Fq2) and Q G PG(2, q2) \Hq.
Let H = K n M1(R) or H = K n M2(R). If R G ¿Q, then H = Cq+1 = Z(M2(Q)). If R G then H = {1}.
Let H = K n M3(T); recall that H < K .If Q is a vertex of T, then H = Cq+1 = Z(M2(Q)). If Q is not a vertex of T, then no homology in K acts on T; hence, H < C2. Let H = K n M4(T). By Lagrange's theorem, H = {1}.
Case 10: Let K = Cq+1 = Z(M2(P)) for some P G PG(2, q2) \ Hq and a G K \ {1}. Then a fixes no points out of {P} U ; also, the triangles fixed by a have one vertex in P and two vertexes on . Thus, K n Mj = {1} for any maximal subgroup Mj of G not containing K.
Case 11: Let K = Eq and a G Eq \ {1}. Recall that K fixes one point P G Hq(Fq2) and the line pointwise. Also, a fixes no points out of . If a fixes a triangle T = {A,B,C}, then one vertex of T lies on (Fq2), say A, and a is uniquely determined by a(B) = C. Thus, K n M1(Q) = K n M2(Q) = K n M4(T') = {1} and K n M3(T) < C2.
Case 12: Let K g {Sym(3), C3, C2, {1}}. Then every subgroup of K is in Equation (3.1).
□
Proposition 3.9. The values ^(H) for the groups in Equation (3.1) are given in Equation (3.3).
Proof. Let H be one of the groups in Equation (3.1). By Lemma 2.1 and Proposition 3.8, ^(H) only depends on the subgroups K of G such that H < K and K is in Equation (3.1).
Case 1: If H is one of the first four groups in Equation (3.1), then H is maximal in G, and hence ^(H) = -1.
Case 2: Let H = Eq x Cq2_1. Let P G Hq(Fq2) and Q G PG(2, q2) \ Hq be the fixed points of H. Then H = M1(P) n M2(Q) and H is not contained in any other maximal
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subgroup of G. Thus, ^(H) = -{^(G) +	)) + ^(M2(Q))} = 1.
Case 3: Let H = (Gq+1 x Gq+1) x C2. Let T = {P, Q, R} be the self-polar triangle stabilized by H, with H(P) = P. No point different from P is fixed by H. Also, if a triangle T' = {P', Q'} = T is fixed by H, then P is a vertex of T', say P = P', and {Q', R'} c QR; but Cq+1 x Cq+1 has orbits of length q + 1 > |{Q', R'}|, so that H cannot fix T'. Then H = M2(P) n M3(T) and H is not contained in any other maximal subgroup of G. Thus, ^(H) = 1.
Case 4: Let H = Cq+1 x Cq+1 and T = {P, Q,R} be the self-polar triangle fixed pointwise by H. The vertexes of T are the unique fixed points of the elements of type (B1) in H. Also, any triangle T' = T fixed by an element of type (A) in H has two vertexes on a side I of T; but H has orbits of length q + 1 > 2 on ¿, so that H does not fix T'. Then H = M3(T) n M2(P) n M2(Q) n M2(R) and H is not contained in any other maximal subgroup of G.
If K is one of the groups M3(T) n M2(P), M3(T) n M2(P), M3(T) n M2(P), then K contains H properly, and ^(K) = 1 as shown in the previous point. The intersection of three groups between M3(T), M2(P), M2(Q), and M2(R) is equal to H. Thus, by direct computation, ^(H) = 0.
Case 5: Let H = Cq2-1 with fixed points P G PG(2, q2) \ Hq and Q, R G Hq(Fq2). Then H = Mx (Q) n M- (R) = Mx (Q) n Mx (R) n M2 (P). We already know ^(M1 (Q) n M2(P)) = m(M1(R) n M2(P)) = 1. Moreover, Cq2-1 has no fixed triangles, by Lagrange's theorem, and no other fixed points. Thus, by direct computation, ^(H) = 0.
Case 6: Let H = C2(q+1) = (a); a is of type (E), fixes the points P G Hq(Fq2) and Q G PG(2, q2) \ Hq, and fixes the lines and ¿Q. Since a2 is a homology with center Q, the orbits on ¿Q of H coincide with the orbits on ¿Q of the elation aq+1. By Lemma 3.4, the self-polar triangles Ti stabilized by H have a vertex in Q and two vertexes on ¿Q; there are exactly f such triangles T,..., Tq .No other triangle and no other point different from P and Q is fixed by H, so that H = M1(P) n M2(Q) n M3(T1) n • • • n M3(Tq) and H is not contained in any other maximal subgroup of G.	2
If K is the intersection of M2(Q) with one of the groups M1(P), M3(T1),..., M3(Tq), then K = Eq x Cq2-1 or K = (Cq+1 x Gq+1) x C2; hence, K contains H properly and ^(K) = 1 as shown above. The intersection of K with a third maximal subgroup of G containing H coincides with H. Finally, the intersection of any two groups in {M1(P),M3(T1),..., M3(Tq)} coincides with H. Thus, by direct computation, ^(H )=0.	2
Case 7: Let H = Cq+1 = Z(M2(P)). Denote ¿p n Hq = {P1,..., Pq+1} and ¿(Fq2) \ Hq = {Q1,..., Qq2—q} such that, for i = 1,..., ', T = {P, Qi, Q.+q2-q} are the
i+ 2
self-polar triangles with a vertex in P. Then
q+1	q2—q	(q2—q)/2
H = p M1(Pi) n M2(P) n p M2(Qi) n p M3(Ti)
i=1 i=1 i=1
and H is not contained in any other maximal subgroup of G. By direct inspection, the intersections K of some (at least two) maximal subgroups of G such that H < K < G are exactly the following.
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(i)	K = Ml (Pi) n Mi(Pj) for some i = j; in this case, K = C^ and ^(K) = 0.
(ii)	K = Mi (Pi) n M2(P) with i g {1,..., q + 1}; in this case, K = Eq x Cq2-1 and ^(K) = 1. These q +1 groups are pairwise distinct.
(iii)	K = M1(Pi) n M3(Tj) for some i, j; in this case, K = C2(q+1) and ^(K) = 0.
(iv)	K = M2(P) n M2(Qi) for some i; in this case, K = Cq+i x Cq+i and ^(K) = 0.
2
(v)	K = M2(P) n M3(Ti) with i G {1,..., ^}; in this case, K = (Cq+i x Cq+i) x
2 _
C2 and ^(K) = 1. These groups are pairwise distinct.
(vi)	K = M2(Qi) n Ms(Ti) or K = M2(Q.+^) n Ms(Ti), with i G {1,..., };
i+ 2	2
in this case, K = (Cq+i x Cq+i) x C2 and ^(K) = 0. These q2 - q groups are pairwise distinct.
To sum up, the only subgroups K with H < K < G and ^(K) =0 are the maximal subgroups, q +1 distinct groups of type Eq x Gq2_i, and 3(q2_q) distinct groups of type
(Cq+i x Gq+i) x C2. Thus, m(H) = 0.
Case 8: Let H = Eq. Let P be the point of Hq (Fq2) fixed by H; H fixes pointwise. We have H = Mi(P) n M2(Qi) n • • • n M2(Qq2), where Qi,..., Qq2 are the Fq2-rational points of \ {P}; H is not contained in any other maximal subgroup of G. The intersections K of at least two maximal subgroups of G such that H < K < G are exactly the q2 groups Mi(P) n M2(Qi) = Eq x Cq2_i, with ^(K) = 1. Thus, by direct computation, ^(H) = 0.
Case 9: Let H = Sym(3) = (a,fi) with o(a) = 3 and o(fi) = 2. Let P G PG(2,q2) \Hq and Q, R G Hq be the fixed points of a, and A g QR be the fixed point of fi on Hq, so that fi fixes ¿A = AP. By Lemma 3.6 and its proof, H = M2(P) n M3(Ti) n • • • n M3(Tq+i), where Ti has one vertex on ¿A \ {P, A} and the other two vertexes are collinear with A; H is not contained in any other maximal subgroup of G.
For any i, j G {1,..., q +1} with i = j, no vertex of Tj is on a side of Ti; hence, no nontrivial element of M3(Ti) n M3(Tj) fixes Ti pointwise. This implies M3(Ti) n M3(Tj) = H. Analogously, no nontrivial element in M3(Ti) n M2(P) fixes Ti pointwise, and this implies M3(Ti) n M2(P) = H. Thus, by direct computation, ^(H) = q +1.
Case 10: Let H = C3 = (a) with fixed points P G PG(2, q2) \ Hq and Q, R G Hq. By Lemma 3.5,
(q2_i)/3	2(q2 _ i)/3
H = Mi(Q) n Mi(R) n M2(P) n f M3(Ti) n f] M4(Ti)
i=i i=i
and H is not contained in any other maximal subgroup of G. By direct inspection, the intersections K of at least two maximal subgroups of G such that H < K < G are exactly the following.
(i)	K = Mi(Q) n M2(P) or K = Mi(R) n M2(P); in this case, K = Eq x Cq2_i and ^(K) = 1.
(ii)	K = Mi(Q) n Mi(R); in this case, K = Cq2_i and ^(K) = 0.
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(iii) There are exactly groups K containing H with K = Sym(3), and hence ^(K) = q +1. In fact, any involution P G G satisfying (H, P) = Sym(3) interchanges Q and R and fixes a point of (QR n Hq) \ {P, Q}; conversely, any of the q-1 points A1,..., Aq—1 of (QRnHq) \{P, Q} determines uniquely the involution Pi G G such that P(A>), Pi(Q) = R, Pi(R) = Q, and hence (H,Pi) ^ Sym(3). The involutions Pi, «Pi, and a2Pi, together with H, generate the same group; thus, there are exactly 1 groups Sym(3) containing H.
Thus, by direct computation, ^(H) = 2(q3 1).
Case 11: Let H = C2 = (a), where a has center P. Let (Fq2) \ {P} = {P1,..., Pq2}. By Lemma 3.4,
q2	q2 q/2
H = M1(P) n f] M2(Pi) n f f M3(Ti,j),
i=1 i=1j=1
where the triangles Ti,j are described in Lemma 3.4; H is not contained in any other maximal subgroup of G. By direct inspection, the intersections K of at least two maximal subgroups of G such that H < K < G are exactly the following.
(i)	K = M1(P) n M2(Pi) for i = 1,... ,q2; in this case, K = Eq x Cq2 —1 and ^(K) = 0.
(ii)	K = M2(Pi) n M2(Pj) with i = j; in this case, K = Eq and ^(K) = 0.
(iii)	K = M1(P) n M3(Ti,j); in this case, K = Eq x C2(q+1) and ^(K) = 0.
3
(iv)	K = M2(Qi) nM3(Ti,j) with i g {1,..., q2} and j G {1,..., §}; these qr distinct groups are of type (Cq+1 x Cq+1) x C2, so that ^(K) = 1.
(v)	There are exactly N = groups K containing H such that K = Sym(3), and hence ^(K) = q +1. This follows by double counting the size of
I = {(H, K) | H, K < G, H = C2, K = Sym(3), H < K}.
Arguing as in the proof of Lemma 3.4, |11 = (q3 + 1)(q - 1)N; arguing as in the proof of Lemma 3.6, |11 = q3(q3+61)(q—1) • 3. Hence, N = ^.
Thus, by direct computation, ^(H) = - q ('2+1).
Case 12: Let H = {1}. Then ^(H) = - E{1}<k<g m(K,G). By the values ^(K) computed in the previous cases, Propositions 3.2, and Proposition 3.3, only the following groups K have to be considered:
(i)	1 group G;
(ii)	q3 + 1 groups S2 x Cq2 — 1;
(iii)	q2(q2 - q +1) groups PSL(2, q) X Cq+1;
(iv)	q3(q—1)(6q2—q+1) groups (Cq+1 x Cq+1) x Sym(3);
(v)	q3(q+132(q—1) groups Cq2—q+1 x C3;
(vi)	(q3 + 1)q2 groups Eq x Cq2 — 1;
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(vii) q (q_1)(2q -q+1) groups (Cq+1 x Cq+l) x C2;
(viii) q3(q3+61)(q-1) groups Sym(3);
(ix)	q3(q2+1) groups C3;
(x)	(q3 + 1)(q — 1) groups C2.
Thus, by direct computation, ^(H) = 0.	□
4 Determination of A(H) for any subgroup H of G
Let n > 0, q = 22", G = PSU(3, q). This section is devoted to the proof of the following theorem.
Theorem 4.1. Let H be a proper subgroup of G. Then A(H) = 0 if and only H is one of the following groups:
Eq x Cq2-1,	(Cq+1 x Cq+1) x C2, Sym(3),
C3,	S x Cq2_1,	PSL(2,q) x Cq+1, (4.1)
(Cq+1 x Cq+1) x Sym(3), Cq2_q+1 x C3,	C2.
For any isomorphism type in Equation (4.1) there is just one conjugacy class of subgroups of G.
If H is in the first row of Equation (4.1), then A(H) = —1; if H is in the second row of Equation (4.1), then A(H) = 1.
Proof. By Proposition 3.2, for any isomorphism type in Equation (4.1) there is just one conjugacy class of subgroups of G of that type. Hence, we can use the notation [M1], [M2], [M3] and [M4] for the conjugacy classes of M1(P), M2(P), M3(T) and M4(T), respectively. If H = G, then A(H) = 1; if H is one of the groups in the second row of Equation (4.1) and H = C2, then A(H) = —1 as H is maximal in G.
Case 1: Firstly, we assume that H is not a subgroup of Sym(3), and that H is not a group of homologies, i.e. H < Cq+1 = Z(M2(Q)) for any point Q.
(i)	Let H < M4 (T) for some T. From H = C3 follows that some nontrivial element in H fixes T pointwise; hence, H is not contained in any maximal subgroup of G other than M4(T). Thus, inductively, A(H) = —{A(G) + A(M4(T))} = 0.
(ii)	Let H < M1(P) for some P; we assume in addition that gcd(|H|,q — 1) > 1. Here, the assumption H < Sym(3) reads H G {{1}, C2, C3}. If H contains an element of order 4, then H is not contained in any maximal subgroup of G other than M1 (P). Thus, inductively, A(H) = 0.
We can then assume that the 2-elements of H are involutions, so that H — E22 x Cd with 0 < r < 2n and d | (q2 — 1) (see [15, Theorem 11.49]). This implies that H < M1(P) n M2(Q) for some Q G ; the eventual nontrivial elements in H whose order divides q +1 are homologies with center Q. Then we have [H] < [M1], [H] < [M2]; by Lagrange's theorem, [H] < [M4]. From the assumptions gcd(|H|, q — 1) > 1 and H < Sym(3) follows [H] < [M3].
If H = Eq x Cq2_1, then no proper subgroup of M1(P) or M2(Q) contains H properly; thus, A(H) = 1. If H = Eq x Cq2_1, then H < Eq x Cq2_1 = M1(P) n
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M2(Q) up to conjugation. Thus, inductively, the only classes [K] with [H] < [K] and A(K) = 0 are [K] g {[G], [Mi], [M2], [Eq x Cq2_i]}. This implies A(H) = 0.
(iii)	Let H < M2 (Q) for some Q, and assume also H < Mi (P) for any P .As H < C3, we have [H] < [M4]. The group H := H/(H n Z(M2(Q))) acts as a subgroup of PSL(2, q) on £q n Hq; we assume in this point that H is one of the following groups (see [17, Hauptsatz 8.27]): PSL(2, 22h) with 0 < h < n; a dihedral group of order 2d where d is a divisor of q - 1 greater than 3; Alt(5). Then, by Lagrange's theorem, [H] < [M3]. Thus, inductively, G and M2(Q) are the only groups K with H < K and A(K) = 0, so that A(H) = 0.
Note that, since we are under the assumptions H < Mi (P) for any P, H < Sym(3), and H < Cq+i = Z(M2(Q)), we have that the only subgroups H of PSL(2, q) for which A(H) still has not been computed are the cyclic or dihedral groups of order d or 2d (respectively), where d is a nontrivial divisor of q +1.
(iv)	Let H < M3(T) for some T, and assume also H < Mi (P) for any P .As H < C3, we have [H] < [M4]. Here, the assumption H < Sym(3) means that some nontrivial element of H fixes T pointwise. Hence, the assumption H < Cq+i = Z(M2(Q)) for any vertex Q of T, together with H < Mi(P), implies that H contains some element of type (B1). Write H = L x K, with K < Sym(3) and L < Cq+i x Cq+i.
If K = C3 or K = Sym(3), then [H] < [M2]; thus, inductively, G and M3(T) are the only groups K with H < K and A(K) = 0, so that A(H) = 0.
If K = C2 and L = Cq+i x Cq+i, then H < M2(Q) for some vertex Q of T. Since H := H/(H n Z(M2(Q))) is dihedral of order 2(q + 1), [17, Haptsatz 8.27] implies the non-existence of groups K with H < K < M2(Q) (except for q = 4 and K = Alt(5); in this case, A(K) =0 by the previous point). Thus, A(H) = -{A(G) + A(M2(Q)) + A(M3(T))} = 1.
If K = C2 and L < Cq+i x Cq+i, then again H < M2(Q) with Q vertex of T. The group H is dihedral of order 2d, where d | (q +1); d > 1 because L contains elements of type (B1). By the previous point and [17, Hauptsatz 8.27], the only groups K with H < K < M2(Q) are such that K is dihedral of order dividing q +1. Thus, inductively, A(H) = 0.
If K = {1}, then H G M2(Q) for any vertex Q of T. The group H < PSL(2, q) on the line £q n Hq is cyclic of order d | (q +1); d > 1 because H has elements of type (B1). By [17, Hauptsatz 8.27], the groups K with H < K < M2(Q) are such that either K is cyclic of order dividing q + 1, or we have already proved that A(K) = 0. Thus, inductively, A(K) = 0.
(v)	Let H < M2(Q) for some Q. Let H = {1} be the induced subgroup of PSL(2, q) acting on £q n Hq. If H is cyclic or dihedral of order d or 2d (respectively) with d | (q + 1), then H < M3(T) for some T. Hence, A(H) = 0, as already computed in the previous point in the case K = {1} if H is cyclic, or in the case K = C2 if H is dihedral.
(vi)	Under the assumptions that H < Sym(3) and H is not a group of homologies, the only remaining case is H < Mi(P) for some P with gcd(|H|, q — 1) = 1. In this case H = E2r x Cd, where Cd is cyclic of order d | (q +1) and made by homologies, whose axis passes through P and whose center Q lies on £P. We have r > 0, because H < Z(M2(Q)).
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If r =1, then H is cyclic of order 2d generated by an element of type (E). By Lemma 3.4, H < M3(T), where T has a vertex in Q and two vertexes on ¿Q. Hence, [H] < [Mi], [H] < [M2], [H] < [M3], and [H] < [M4]. Let K be such that H < K < G and K is not of the same type of H, i.e. K is not cyclic of order 2d' with d' | (q +1). As shown in the previous points, A(K) = 0 if and only if [K] e {[G], [Mi], [M2], [M3], [E, x C,2-i], [(C,+ i x C,+ i) x C2]}. Thus, inductively, A(H) = 0.
Case 2: Let H < Cq+i = Z(M2(Q)) for some Q and K be a subgroup of G properly containing H. As shown above, A(K) =0 if and only if
[K] e {[G], [Mi], [M2], [M3], [E, x C,2_i], [(Cq+i x Cq+i) x C2]}.
Thus A(Z(M2(Q))) = 0 and, inductively, A(H) = 0.
Case 3: Let H = Sym(3) = (a) x (£) with o(a) = 3 and o(£) = 2. Let P e PG(2,q2) \ H, and Q, R e H, (F,2) be the fixed point of a, so that fi fixes P and interchanges Q and R. This implies [H] < [M2]. By Lemma 3.6, [H] < [M3]. From the computations above and Lagrange's theorem, no class [K] with K < G other than [G], [M2] and [M3] satisfies [H] < [K] and A(H) = 0. Thus, A(H) = 1.
Case 4: Let H = C3. By Lagrange's theorem and Proposition 3.2, H < K < G and A(K) = 0 if and only if
[K] e {[G], [Mi], [M2], [M3], [M4], [E, x C,2-i], [Sym(3)]}.
Thus, A(H) = 1.
Case 5: Let H = C2. By Lagrange's theorem and Proposition 3.2, H < K < G and A(K) = 0 if and only if
[K] e {[G], [Mi], [M2], [M3], [E, x C,2-i], [(C,+i x C,+ i) x C2], [Sym(3)]}.
Thus, A(H) = -1.
Case 6: Let H = {1}. Collecting all the classes [K] with A(K) = 0, we have by direct computation A(H) =0.	□
5 Determination of x(A (Lp \ {1})) for any prime p
Let n > 0, q = 22", G = PSU(3, q). If p is a prime number, we denote by Lp the poset of p-subgroups of G ordered by inclusion, by Lp \ {1} its subposet of proper p-subgroups of G, and by A(Lp \ {1}) the order complex of Lp \ {1}. In this section we determine the Euler characteristic x(A(Lp \ {1})) of A(Lp \ {1}) for any prime p, using Equation (2.1) and Lemma 2.2. The results are stated in Theorem 5.1 and in Table 2.
Theorem 5.1. For any prime number p one of the following cases holds:
(i)	p { |G| and x(A(Lp \ {1})) = 0;
(ii)	p = 2 and x(A(L2 \ {1})) = q3 + 1;
6 o 5 4,r>3 o 2
(iii)	p | (q + 1) and x(A(Lp \ {1})) = - " -2? -,3+2? -3? ;
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(iv)	p | (q - 1) andx(A(Lp \ {1})) = -;
~	6,54-
(v)	p | (q2 - q +1) and x(A(Lp \ {1})) = - q +q -q -q .
Proof. Since |G| = q3(q + 1)2(q - 1)(q2 - q +1), q is even, and 3 | (q - 1), the cases p f |G|, p = 2, p | (q + 1), p | (q - 1), andp | (q2 - q + 1) are exhaustive and pairwise incompatible. We denote by Sp a Sylow p-subgroup of G.
Case 1: Letp f |G|. Then A(Lp \ {1}) = 0, and hence x(A(Lp \ {1})) = x(0) = 0.
Case 2: Let p = 2. The group G has q3 + 1 Sylow 2-subgroups, and any two of them intersect trivially; see [15, Theorem 11.133]. Any nontrivial element a of S2 fixes exactly one point P on Hq (Fq2) which is the same for any a G S2; S2 is uniquely determined among the Sylow 2-subgroups of G by P. Hence, Equation (2.1) reads
x(A(L2 \{1})) = -(q3 + 1) £	({1},H),
HeL2\{1}, H(P) = P
where P is a given point of Hq (Fq2). By Lemma 2.2, we only consider those 2-groups in M1 (P) which are elementary abelian. Then we consider all nontrivial subgroups H of an elementary abelian 2-group Eq of order q. For any such group H = E2r of order 2r, with
1 < r < 2n, we have ^L2 ({1}, H) = (-1)r • 2(2) by Lemma 2.2. Thus,
X(A(L2 \ {1})) = -(q3 + 1) £(-1)r 2(2) f2")
r=1	V r / 2
where the Gaussian coefficient (2r ) 2 counts the subgroups of Eq of order 2r. Using the property
[2n\ (2n - A r(2" - 1' +2
V r / 2 Vr - 1 / 2
we obtain
2
2 Vr 1 / 2	/2
S("1)r 2(2,C") 2
=|(-Dr (T-1)+2(2)+r (2n -
= EC-Dr+i 2(f) (2"- 1)=+ £ (-1)r 2('f) (2"- ^
=(-U<2>(2n-^ + (-r(-:^=2
Thus, x(A(L2 \{1})) = q3 + 1.
Case 3: Letp | (q + 1). Then Sp < Cq+1 x Cq+1, and hence Sp = Cps x Cps, where ps | (q + 1) andps+1 f (q + 1). Let H be a subgroup of Sp. By Lemma 2.2, ^ ({1},H) = 0 only if H is elementary abelian of order p or p2; in this cases, ^Lp ({1}, Cp) = -1 and MLp({1}, Cp x Cp) = r. Now we count the number of elementary abelian subgroups of order p or p2 in G.
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(i)	A subgroup Ep2 of G of type Cp x Cp is uniquely determined by the maximal subgroup M3(T) such that Ep2 is the Sylow p-subgroup of M3(T). Hence, G contains
exactly [G : Ng(M3(T))] = q3(q2_q+1)(q_1) elementary abelian subgroups of order p2.
(ii)	A subgroup Cp made by homologies is uniquely determined by its center P € PG(2, q2) \ Hq of homology, because the group of homologies with center P is cyclic. Hence, G contains exactly | PG(2, q2) \ Hq | = q2(q2 - q +1) cyclic subgroups of order p made by homologies.
(iii)	A subgroup Cp which is not made by homologies is made by elements of type (B1), and fixes pointwise a unique self-polar triangle T. The Sylow p-subgroup Cp x Cp of M3(T) contains exactly 3 subgroups Cp made by homologies, namely the groups of homologies with center one of the vertexes of T. Since Cp x Cp contains p +1 subgroups Cp altogether, Cp x Cp contains exactly p — 2 subgroups Cp not made by homologies. Thus, the number of subgroups Cp of G not made by homologies is
(p — 2) • [G : Ng(M3(T))] = q3(q2_q+1)s(q_1)(p_2). Thus, by direct computation,
X(A(Lp \{1}))
q3(q2 — q +1)(q — 1)(p — 2)
- • r

+
6
q2 (q'2 — q +1)+ "V — q + !)(, — 1)(p — 2)
q6 — 2q5 — q4 + 2q3 — 3q2

3
Case 4: Let p | (q — 1). By Lemma 2.4, Sp is a subgroup of the cyclic group Cq 2 _ 1 fixing two points P, Q on Hq(Fq2); then a proper p-subgroup H of G satisfies ({1}) = 0 if and only if H has order p; in this case, ({1},H) = —1. Also, by Lemma 2.4, any two Sylow p-subgroups of G have trivial intersection. Then the number of subgroups Cp of G is equal to the number (q +1) of couples of points in Hq(Fq2); equivalently, this number is equal to [G : Ng(C^)], where |NG(Cq2_1 )| = 2(q2 — 1) by Proposition 3.3. Thus,
6 , 3
x(A(Lp \{1})) = — ^^.
Case 5: Let p | (q2 — q + 1). Then Sp < Cq2_q+1, and hence a proper p-subgroup H of G satisfies ({1}, H) =0 if and only if H has order p; in this case, ({1},H) = —1. The number of subgroups Cp of G is equal to the number of subgroups Cq2_q+1, and hence to the number [G : Ng(M^T))] = q3(q+1,)2(q_1) of maximal subgroups of type M4(T)
in g. Thus, x(A(Lp \ {1})) = — q3(q+1^2(q_1) = — q6+q5_ q4_q3.	□
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