J. Geom. 82 (2005) 001 – 024 0047–2468/05/0200001 – 24 © Birkh¨auser Verlag, Basel, 2005 DOI 10.1007/s00022-004-1645-2

Two-transitive orbits in finite projective planes Mauro Biliotti and Eliana Francot

Abstract. The problem of classifying finite projective planes
of order n with an automorphism group G and a point orbit O on which G acts two-transitively is investigated in considerable detail, under the assumption that O has length at last n. Combining old and new results a rather satisfying classification is obtained, even though some cases for orbit lengths n and n + 1 remain unsolved. Mathematics Subject Classification (2000): 51E15, 20B25. Key words: Projective plane, automorphism, orbit.

1. Introduction In a paper of about 30 years ago, Cofman [15] posed the following question: “Let
be a finite plane with a subset O of points admitting a collineation group G which maps O onto itself and induces a doubly transitive collineation group on the points of O. What can we say about the plane, the set O and the collineation group G?”. Here the term finite plane may be intended in the affine sense as well as in the projective sense. A survey about the knowledge on this subject at that time is given by Cofman in the same paper. Clearly, either the structure of a non-trivial 2-design is induced on O, or O is an arc or it is contained in a line. Hence a fundamental contribution to the problem is certainly the Kantor classification [39] of 2-designs with an automorphism group 2-transitive on the points, which is ultimately a consequence of the classification of finite simple groups. Nevertheless, after Cofman’s paper, conclusive results have been obtained for example when
is a projective plane and O is an oval [41] or a unital [38]. Relevant results are also been achieved when O consists of a line [42] or a line minus a point [33]. In a recent paper Ganley, Jha and Johnson [24] essentially classified the line-sized sets O in a translation plane for G non-solvable. Also the case when O is the line at infinity of a translation plane is now completely settled [37]. The main aim of this paper is to attempt a classification of the triples (
, O, G) under the assumptions given in Cofman’s question and under the additional assumption that the G-orbit O has length at least equal to the order n of
. For orbits of smaller lengths the question seems to become very difficult. Exhaustive results are obtained when O has length at least n + 1 for n odd and length at least n + 2 for n even. When O has length n + 1 and n is even, the case where O is a line or an oval remains essentially open in the case that G is 1

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a semilinear 1-dimensional affine group, despite several efforts Korchmaros and Bonisoli [13], [12] and some progress for the case of the line obtained in this paper. When O has length n the characterization of
seem to be difficult, but several results about O and G are given in our Theorem 3.13. 2. Preliminaries Let
= (P, L) be a finite projective plane of order n and let D be any subset of P. A line l of
is called an external line, a tangent or a secant of D, according to whether |D ∩ l| = 0, 1 or k, with k > 1. We may regard D as an incidence structure, where the lines are the secants of D. Any incidence structure represented in this manner is called embedded in
. Suppose that D admits a parallelism, that is the lines of D are partitioned into parallel classes such that each class is a partition of the points. If for each pencil  of parallel lines, the lines of  pass through a common point P of
and all P lie in the same line lD of
, then we say that the projective extension of D is embedded in
. In particular, any orbit O of a collineation group G of
may be regarded as an embedded structure. When G is transitive on the secants of O, then O is a 2 − (v, k, 1) design with v = |O|. A point-orbit O of G is called 2-transitive if G acts 2-transitively on its points. In this case O is a 2 − (v, k, 1) design with an automorphism group 2-transitive on the points. We have the following: THEOREM 2.1 (Kantor). Let D be a 2 − (v, k, 1) design with k ≥ 3 and at least two distinct blocks. If D possesses an automorphism group 2-transitive on the points, then D is one of the following designs: (i) PG(d, q), d ≥ 2, q = p r , p prime, (ii) AG(d, q), d ≥ 2, q = pr , p prime, (iii) the unitary designs with v = q 3 + 1 and k = q + 1 associated with PSU(3, q 2 ) or 2 G (q), 2 (iv) the nearfield plane of order 32 or the Hering plane of order 33 , or (v) one of the two Hering designs with v = 36 and k = 32 . For a proof see [39]. Thus, we have the following possibilities for a 2-transitive orbit O of
: – O is one of the designs given in Theorem 2.1, – O is a v-arc, – O consists of v points on a line. We shall investigate the three cases separately, under the assumption that |O| ≥ n. We also give some number theoretical results which will be useful in Section 5. PROPOSITION 2.2. (1) The Diophantine equation a m − b2 = 1 with m ≥ 2 has no integer solution,

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−1 (2) The unique integer solutions of the Diophantine equation xx−1 = y 2 with x > 1 and a > 2 are (x, a, y) = (7, 4, 20) or (3, 5, 11), a −1 (3) The Diophantine equation xx−1 = y 2 + 1 with x > 1 and a > 2 has no integer solution. a

For a proof of (1) and (2) see [48], (A3.1), (A7.1) and (A8.1). In case (3) the given equation is equivalent to x x x−1−1 = y 2 . This implies that both x and a−1

x a−1 −1 x−1

are square numbers and the result easily follows from (2).

3. The embedded designs Several designs in Theorem 2.1 with v ≥ n are not embeddable in
if one requires that they must be invariant under some particular, not necessarily 2-transitive, collineation groups of
. First we give some results in this direction. PROPOSITION 3.1. Let D ∼ = AG(d, q) or PG(d, q), d > 2, be embedded in
and let v ≥ n. If D is invariant under a non-identity collineation α of
fixing pointwise an s-dimensional subspace of D, then s < 21 d + 1. Proof. Suppose that s ≥ 21 d + 1. Then α is a planar collineation and through a point of Fix α ∩ D there are at least (q s − 1)/(q − 1) fixed lines. Thus we must have
√ (q s − 1)/(q − 1) ≤ n + 1 ≤ q d + 1 (1) when D ∼ = AG(d, q) and (q s − 1)/(q − 1) ≤

√

n+1≤




(q d+1 − 1)/(q − 1) + 1

(2)

when D ∼ = PG(d, q). Clearly, (1) is in contrast with s ≥ 21 d + 1, while (2) is in contrast with s ≥ 21 d + 1, except for d = 3 and q = 2 or 3. However, d = 3 and s ≥ 21 d + 1 implies s = 3 and (2) is unsatisfied for s = d = 3.


PROPOSITION 3.2. Let D be any design listed in Theorem 2.1. Suppose that D is embedded in
and let v ≥ n. If D is invariant under a non-identity collineation α of
such that α|D = I , then one of the following occurs: (1) n = q 2 , q 2 + q or q 2 + q + 1 and D ∼ = PG(2, q), (2) n = q 2 , D ∼ = AG(2, q) and the projective extension of D is be embedded in
, (3) n = 34 or 36 , D is one of the planes in the case (iv) of Theorem 2.1 and the projective extension of D is be embedded in
.

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∼ AG(d, q) or PG(d, q). By Proposition 3.1 we must have d = 2. FurtherProof. Let D = more, (2) yields q 2 ≤ n ≤ q 2 + q + 1 when D ∼ = PG(2, q), while (1) yields n = q 2 ∼ when D = AG(d, q). We have (1) and (2) by Bruck’s theorem – note that α must be a Baer collineation when (2) occurs. A similar argument excludes the case (v) of Theorem 2.1, while in the case (iv) it leads to n = 34 or 36 and the projective extension of D must be embedded in
since α is a Baer collineation. For unitary designs the same argument leads to q = 2, n = 4, D ∼
= U(2) and hence α = I , a contradiction. PROPOSITION 3.3. Let D ∼ = PG(d, q) be embedded in
and let v ≥ n. If there is a collineation α of
such that α|D is induced by a transvection, then d = 2. Proof. Assume that d ≥ 3. Clearly, α is planar and there is a point C of D such that α|D fixes all the secants of D through C. The same argument of Proposition 3.2 yields α = I , a contradiction.
PROPOSITION 3.4. Let D ∼ = AG(d, q) be embedded in
and let v ≥ n. (1) If there is a collineation α of
such that α|D is a transvection, then d ≤ 3. (2) If d = 3, then there cannot exist collineations α and β of
such that α|D and β|D are transvections with the same hyperplane, but different directions. Proof. (1) follows from Proposition 3.1. Suppose that d = 3 and assume there exist collineations α and β of
such that α|D and β|D are transvections with the same fixed plane H, but different directions. Then S = Fix α ∩ Fix β is a subplane of order at least q since it contains H. If Fix α  = Fix β, then
must have order at least q 4 , in contrast with q 3 ≥ n. If Fix α = Fix β, then two different pencils of parallel lines of D lie in S, but then D ⊂ S. We have a contradiction.
PROPOSITION 3.5. Let D ∼ = AG(d, q), d ≥ 2, q odd, be embedded in
and let v ≥ n. If D is left invariant by a collineation group G of
such that G|D ∼ = T α , where T is the translation group of D and α is an involutory dilatation of D, then one of the following holds (a) v = n. If d ≥ 3 then the projective extension of D is embedded in
, T is a group of elations with axis lD and α is a homology with axis lD , (b) v = n2 , D =
l for some line l of
, (c)
∼ = PG(2, 4) and D consists of absolute points and non-absolute lines of a Hermitian polarity. (d)
∼ = PG(2, 7) and |D| = 32 . Proof. Suppose that G does not act faithfully on D. Then d = 2 and v = n by Proposition 3.2 and we have (a). Now, suppose that G acts faithfully on D. We note

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that α fixes at least (q d − 1)/(q − 1) lines through a point. Using (1) with s = d we have that either α is a perspectivity or d = 2 and v = n. We may assume that α is a perspectivity. 1) If G fixes a line l then the cases (a) or (b) occur. Suppose that n is even. Each point of D is the centre of exactly one elation in G and all the centres lie on l. Then D ⊆ l, a contradiction. Suppose that n is odd. If l is the axis of any involutory homology in G, then l is the axis of all homologies in G, since all homologies in G are conjugate. Then by [35], Theorem 4.25, T is a group of elations of
with axis l. Furthermore T contains q elations for exactly (q d − 1)/(q − 1) centres corresponding to the different directions of the secants of D. Thus the projective extension of D is embedded in
. By [19] either (q d − 1)/(q − 1) = n + 1 or v = n. In the first case D =
l by [35], Theorem 4.26. In the other case we have (a). If l is not the axis of any homology in G, then the centres of all homologies in G lie on l and D ⊆ l, a contradiction. 2) If G fixes a point P then the cases (a) or (b) occur. Clearly P ∈ / D, so that no involutory perspectivity in G has centre P . Hence the axis of every involutory perspectivity contains P . Suppose that n is even. The axis of an involutory elation in G meets D in exactly one point, namely the centre of the elation. Thus q d ≤ n + 1. By our assumptions either q d = n + 1 or q d = n. In the first case T is transitive on the lines of the pencil   with centre P . Let s and r be two distinct parallel lines of D and let Q = s ∩ r. Then TQ  ≥ q and TQ ≤ TPQ . Since no secant of D passes through P , we have that P = Q. But then the group T cannot be transitive on the lines of , a contradiction. In the second case T fixes exactly one line l through P . Then α also fixes l. We obtain a contradiction as in 1). Suppose that n is odd. Assume there are two distinct involutory homologies α and β in G with the same axis r. The centres of α and β are distinct, being the centres of two distinct involutory dilatations in D. By [35], Theorem 4.25, T contains a non-identical subgroup T0 of translations with axis r. Thus T fixes r, being an abelian group. Since α fixes r, then G fixes r and we argue as in the case 1). Hence, no two involutory homologies in G have the same axis. Since all axes pass through P then q d ≤ n + 1. For q d = n + 1 each line through P is an axis meeting D in exactly one point, which will be the centre of another homology. Let Q be any point of D and let PQ be the axis of the homology γ . The point Q is the centre of a homology δ which fixes PQ. Thus, δ centralizes γ and we have a contradiction since G does not contain two commuting involutions. If q d = n, then T fixes exactly one line l through P , which is not the axis of any involutory homology in T α . Then α also fixes l. We obtain a contradiction since P cannot be the centre of α.

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3) If G does not fix any point or line, but G fixes a triangle , then one of the cases (c) or (d) occurs. Let  = {A, B, C}. Suppose that n is even. Then any involutory elation in G has the axis through a vertex of  and the centre on the opposite side. Thus there are three disjoint lines of D containing all the points of D. This yields q d = 32 . The case (c) occurs by [7], Lemma 2.9. Suppose that n is odd. Let α be an involutory homology in G. If the centre of α is a vertex of  then |O| = 3. This is impossible. Thus the axis of α passes through a vertex of  and the centre lies on the opposite side. As before, q d = 32 . Note that if two involutory homologies in G have the same axis l, then G fixes l, as we have already seen. A side AB of  contains two vertices, the centres of three involutory homologies and the intersection points of the three corresponding axes. A centre and an intersection point of an axis cannot coincide since G does not contain distinct commuting involutions. So n ≥ 7. Actually n = 7 since n + 1 ≤ 32 and the case (d) occurs. 4) G cannot be irreducible. Suppose that G does not fix any point, line or triangle. Then the centres and the axes of the perspectivities in G generate a subplane
0 by [28], Lemma 3.3. Furthermore, D ⊆
0 since each point of D is the centre of an involutory perspectivity in G. Thus G acts faithfully on
0 . By [28], Theorem 5.5, G possesses a unique minimal normal subgroup which is either elementary abelian of order 9 or nonabelian simple. We have a contradiction.
Now we return to the 2-transitive case. The following two lemmas investigate the case when D is an affine plane of order q embedded in a projective plane of order q 2 . LEMMA 3.6. Let D be an affine plane of order q = p r embedded in a projective plane
of order q 2 . If D is left invariant by a collineation group G of
such that G acts 2-transitively on D, then the projective extension of D is embedded in
. Proof. If G does not act faithfully on D, then the assertion follows from Proposition 3.2, (2). Assume that G acts faithfully on D. Then by [35], Theorem 14.11, G = T GO , where T is the translation group of D and O is any point of D. Note that the group T must fix a line l disjoint from D and a point L on l. Suppose that a pencil  of parallel lines of D is not a pencil of
with centre on l. Then the subgroup T of T leaving invariant each line of  fixes the vertices of a triangle . Since T is a p-group, T is planar. If some line s through L meets D in exactly one point, then T acts regularly on the lines through L different from l. Since some of these lines must lie in Fix T we have a contradiction. Thus we may assume that a line s through L meets D in at least two points. Then sT is a pencil  of q parallel lines of D through L. Since G is transitive on the pencils of parallel lines of D, it is easily seen that there exist exactly q + 1 fixed points of T in
. Since G is transitive on Fix T , then either Fix T is a subplane or it is contained in a line – we may

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suppose q > 2. If Fix T is a subplane then we must have q = m2 + m, which is impossible. This completes the proof.
LEMMA 3.7. Let D be an affine plane of order q = p r embedded in a projective plane
of order q 2 . If D is left invariant by a collineation group G of
such that G acts faithfully and 2-transitively on D, then G = T GO where O is any point of D and T is a translation group of
of order q 2 . Proof. We have that G = T GO for O a point of D and T the whole translation group of D of order q 2 . Furthermore, by Lemma 3.6
contains the projective extension of D with respect to the line lD . Suppose that an element τ ∈ T fixes a point A of
− lD . Clearly A∈ / D. There is exactly one line m of D through A. Hence mτ = m. Among the q 2 − q points of m − (D ∪ lD ) there must be a point fixed by τ and different from A. As τ already fixes q + 1 point on lD , τ is a Baer collineation. Hence all non-trivial elements of T are Baer collineations being conjugate under GO . Note that there are q 2 − q points of lD not in the projective extension of D and if N is one of these points, then TN  = 1 . We have a contradiction. We may assume that each element of T acts f.p.f. on
− lD . Let M ∈ lD in the projective

of D. Then T acts on the pencil consisting of the q 2 − q lines through M not extension D in D and different from lD . If |Ts | > q for some line s ∈ and T denotes the subgroup of T leaving invariant each line of a pencil  of parallel lines of D with direction different from M, then Ts ∩ T  = 1 and there is τ ∈ Ts ∩ T , τ  = 1, such that τ fixes a point in
− lD , in contrast with our assumption. Hence |Ts | ≤ q for each line s ∈ . Let  be the

set of lines external to D and different from lD , passing through some point L ∈ lD ∩ D. 2 Clearly || = (q − q)(q + 1). Each non-trivial element of T fixes the same number k of lines of  since the elements of T are conjugate under G and  is G-invariant. Thus, if T has t orbits on , then the following relation holds tq 2 = k(q 2 − 1) + (q 2 − q)(q + 1). Hence q 2 − 1|t. As |Ts | ≤ q for each s ∈ , each T -orbit has length at least q. This implies that there are q 2 − 1 orbits, each exactly of length q. In particular, |Ts | = q for each line s ∈ . As before we must have Ts ∩ T = 1 for each pencil  of direction different from M. Hence the subgroups T , together with Ts , make a partition of T . This implies that Ts = T for all s ∈ , where  denotes the pencil of direction M. Thus T fixes all the lines through M and hence it consists of translations of
. Thus T is a translation group of
.
∼ AG(d, q), q even, be embedded in
and let v ≥ n. If D PROPOSITION 3.8. Let D = is left invariant by a collineation group G of
such that G|D is 2-transitive on the points of D and contains the translation group T of D, then one of the following holds:

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(a) v = n. If d ≥ 3 then the projective extension of D is embedded in
and T is a group of elations with axis lD , (b) v = n2 , D =
l for some line l of
. (c)
∼ = PG(2, 3) and D is a conic. Proof. When G does not act faithfully on D, then d = 2 and v = n by Proposition 3.2, (2) and we have (a). Suppose that G acts faithfully on D. Since
has an odd number of points, then T fixes a point P with P ∈ / D. Suppose that an involution in T is a Baer involution. Then all involutions in T are Baer since they are conjugate under G. Assume there are not secants of D through P . As q d ≥ n, there are at least n lines through P which are tangents of D. Furthermore, T is transitive on the tangents of D through P since it is transitive on the points of D. If τ ∈ T , τ  = 1, then the lines through P of Fix τ are T -invariant since T is abelian and we have a contradiction. Thus, there is a secant s of D through P . Hence, if  is the pencil of parallel lines of D containing s, then all the lines of  pass through P . A secant of D, not in , cannot pass through P , since it already meets some line of  in a point of D. Since T
G and G is d −1 transitive on the pencils of parallel lines of D, it is easily seen that T fixes exactly qq−1 points. Suppose that Fix τ = Fix τ  for each τ, τ  ∈ T − {1}. If l is a line of Fix τ , then √ √ T acts semiregularly on the points of l − Fix τ ∩ l. Thus n ≥ |T | + n = q d + n, in contrast with our assumptions. Hence we may suppose there exist τ, τ  ∈ T − {1} with Fix τ  = Fix τ  . Clearly Fix T ⊆ Fix τ ∩ Fix τ  . Since the number of points in Fix τ ∩ Fix τ    d −1 √ √ √ √ ≤ n + 4 n + 1 ≤ q d + 4 q d + 1. This is at most n + 4 n + 1 we must have qq−1 yields d = 2. By Lemma 3.7 the elements of T cannot be Baer. Thus we may assume that all involutions in T are perspectivities. Any pencil  of parallel lines in D has a centre P which is the common centre of the perspectivities of T , the group fixing each line of . Clearly P  = P for   = . As T contains two commuting involutions with different centres we have a contradiction when n is odd and (d, q)  = (2, 2) by [38]. When n is even all the (q d − 1)/(q − 1) centres of involutions in T lie in the same line l, which is the common axis and we obtain (a) or (b) arguing as in Proposition 3.5, (1). If (d, q) = (2, 2) and n = 3, then D is a conic of PG(2, 3).
THEOREM 3.9. Let D ∼ = PG(2, q) be embedded in
and let v ≥ n. If D is left invariant by a collineation group G of
which is 2-transitive on the points of D, then one of the following occurs (a)
is a Desarguesian or a generalized Hughes plane of order q 2 , D is a Baer subplane and G contains PSL(3, q),

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(b)
is the generalized Hughes plane over the exceptional nearfield of order 72 , D is a Baer subplane and G contains SL(3, 7). Proof. Since v ≥ n, then either n = q 2 and D is a Baer subplane, or n = q 2 + q or q 2 + q + 1 by Bruck’s theorem [35]. When D is a Baer subplane we have (a) or (b) by [45], Theorem 2 and Corollary 6. Suppose that n = q 2 +q or q 2 +q +1. There is a subgroup F of G inducing PSL(3, q) in D by Ostrom-Wagner Theorem [35] and we may assume that F is minimal with respect to this property. Let K be the kernel of the representation of F on D. Since by questions of order < Fix α = <
, α ∈ K, then K acts semiregularly there cannot exist any chain of subplanes D = on the points of l − (l ∩ D) for any secant l of D. This yields that either |K| | q 2 or |K| | q 2 + 1. Hence, if α ∈ K, α  = 1, then [F : C] ≤ q 2 where C = CF (α). Suppose that CK/K is properly contained in F /K ∼ = PSL(3, q). By [17], Table 1, the index of a proper subgroup of PSL(3, q) is at least q 2 + q + 1. Since [F : C] ≥ [F /K : CK/K] we have a contradiction. Thus CK/K = F /K. Nevertheless, CK/K ∼ = C/(C ∩K) ∼ = PSL(3, q). By the minimality of F , we must have F = C. Thus K ≤ Z(F ). Furthermore, F  = F again by the minimality of F . Using Shur’s multipliers we have that F ∼ = PSL(3, q) or SL(3, q), except possibly when q = 2 or 4 (e.g. see [26]). Note that both q 2 + q and q 2 + q + 1 are non-squares by Proposition 2.2, (2). Hence the involutions in F are perspectivities. Thus K must have odd order. This yields F ∼ = PSL(3, q) or SL(3, q) also for q = 2 or 4. Suppose that q is even. Let l be a secant of D. There is a subgroup T of order q 2 of F consisting of elations of D with axis l. Clearly the elements of T are also elations of
. This yields n = q 2 + q. Let P ∈ l − (l ∩ D). T must act semiregularly on the q 2 + q lines through P , different from l. This is impossible. Suppose that q is odd. An involution in F is a perspectivity of
inducing an homology in D. Hence it must be a homology of
. So, n = q 2 + q + 1. Let α be an elation of D with axis the secant l and with centre L. Let s by any secant of D through L. Since |α| = p for q = pr , p prime, and |s − (s ∩ D)| = q 2 + 1, then α fixes at least one point on s − (s ∩ D). Hence α is planar and Fix α has order at least q + 1 since α fixes at least q + 2 points on l. This yields (q + 1)2 ≤ q 2 + q + 1, a contradiction. This completes the proof.
THEOREM 3.10. Let D ∼ = H (q), q > 2, be embedded in
and let v ≥ n. If D is left invariant by a collineation group G of
which is 2-transitive on the points of D, then
is Desarguesian of order q 2 and G contains PSU(3, q). Proof. By Theorem 2.1, G|D contains F ∼ = PSU(3, q). Furthermore, by Proposition 3.2 G acts faithfully on D. Let σ be an involution in F . Assume that q is odd. As it is well known (e.g. see [34]) σ fixes exactly q + 1 points of D lying on a secant l. Furthermore, σ leaves invariant exactly q 2 −q secants, other than l. Denote by Sσ this set of secants. Then Sσ ∪{l}

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makes a partition of the points of D, C = CF (σ ) induces on Sσ a transitive permutation group C¯ ∼ = PGL(2, q) and C¯ r is cyclic of order q + 1 for each r ∈ Sσ . Suppose that σ is a Baer involution. By [34] we must have n > q 2 . Hence there is a point P ∈ Fix σ ∩ l, but P ∈ / D. If there are two distinct lines r and s of Sσ through P , then all the lines of Sσ pass through P because C¯ r , C¯ s ∼ = PGL(2, q), as it is easily seen. Thus Fix σ has order at least 2 q − q. If there is at most one line of Sσ through P , then l contains at least q 2 − q points in Fix σ − (l ∩ D). In any case Fix σ has order at least q 2 − q. Since (q 2 − q)2 > q 3 + 1 for q ≥ 3, we have a contradiction. Assume that q is even. As it is well known [34] σ fixes a point P of D and all the q 2 secants through P . Since n ≤ q 3 + 1, σ cannot be a Baer involution. Thus, we may assume that all involutions in F are perspectivities. Note that each secant of D is the axis of exactly one involutory perspectivity in F when q is odd, while each point of D is the centre of q − 1 involutory perspectivities generating a subgroup of order q in F when q is even. Furthermore, if n is even, then q is even. On the contrary, every involution in CF (σ ) would be an elation with the same centre of σ and hence the involutions in CF (σ ) would generate an abelian group, a contradiction. 1) F does not fix any point of
. Let P be any point of
and suppose that F = FP . Clearly P ∈ / D. Suppose that n is even. Hence q is even. Since two involutory elations with distinct centres must have different axes, there are at least q 3 + 1 distinct axes through P . Hence n + 1 ≥ q 3 + 1 ≥ n. Let n + 1 = q 3 + 1. Then each line through P is an elation axis and F acts on the pencil with centre P as PSU(3, q) in its usual 2-transitive representation. Suppose that F fixes a line l. Then l must contains all the elation centres, which is clearly impossible. Hence in the dual plane
∗ of
the group F acts 2-transitively on the line P and fixes no point in
∗P . This contradicts [5] and [6]. The case n = q 3 + 1 cannot occur since q would be odd. Suppose that n is odd. If P is the centre of some involution in F , then all involutions in F have centre P . If q is even we have a contradiction. If q is odd, then by [35], Theorem 4.25, F would have an elation group of order q 4 + q 2 − q 3 and of index two, again a contradiction. Hence the axes of involutory homologies in F pass through P and again we have a contradiction. 2) F does not fix any line of
. Suppose that F = Fl for some line l of
. Clearly l∩D = ∅. Suppose that l contains all the centres of involutory perspectivities in F . Then q must be odd. Take two involutory perspectivities σ and φ in F with axis a and b, respectively, and such that Sσ ∩ Sφ consists of a secant m. The secants in Sσ and in Sφ meet l in the centre of σ and φ, respectively. Both centres coincide with l ∩ m. Then σ fixes Sσ ∪ Sφ , a contradiction. Suppose that l is the axis of all involutions in F . If q is odd each secant of D is an axis, while if both q and n are even, the centres do not lie on a line. Let n be

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odd and let q be even. Then by [35], Theorem 4.25, F contains an elation group of order q 3 + 1 and the homologies in F with the same centre make a Frobenius complement. This yields q = 2. Clearly, if F leaves invariant a triangle, then F fixes the triangle pointwise for q > 2. Furthermore the centres and the axes of the perspectivities in F generate a subplane containing all the q 4 +q 2 −q 3 secants of D. By our assumption v ≥ n, the subplane must coincide with
. Thus F is strongly irreducible on
[28]. By [32] and by [8], Lemma 5.4,
is Desarguesian of order q 2 .
THEOREM 3.11. Let D ∼ = R(q) be embedded in
and let v ≥ n. If D is left invariant by a collineation group G of
which is 2-transitive on the points of D, then
∼ = PG(2, 8), G∼ = P L(2, 8), G leaves invariant a line oval  in
and D consists of the = 2 G2 (q) ∼ external points of . Proof. By Theorem 2.1, G|D contains F ∼ = 2 G2 (q). Furthermore, by Proposition 3.2. G acts faithfully on D. There is only one class of involutions in F and if σ is any involution, then C = CF (σ ) ∼ = σ × PSL(2, q). Furthermore, σ fixes exactly q + 1 points of D lying on a secant l, σ is the unique involution in F fixing l ∩ D pointwise and C induces on l ∩ D the group PSL(2, q) in its usual 2-transitive representation [44]. Let P ∈ D − l. / l, s is the unique secant through P fixed by Then s = PPσ is fixed by σ . Since P ∈ σ . Furthermore, s ∩ l = ∅ since otherwise σ fixes some point on s − (s ∩ l). Thus σ leaves invariant exactly q 2 − q secants, other than l, making a set Sσ such that Sσ ∪ {l} is a partition of the points of D. If τ , τ  = σ , is an involution in C fixing s ∩ D pointwise, then σ fixes s. So s ∈ Sσ . Thus C acts on Sσ as C acts on its two classes of non-central involutions. Therefore, C¯ ∼ = PSL(2, q) splits Sσ into two orbits of length (q 2 − q)/2 and C¯ s is dihedral of order q + 1 for each s ∈ Sσ . Assume that the involutions in F are Baer involutions. We may suppose there is a point P ∈ l ∩ Fix σ , but P ∈ / D, since otherwise n = q 2 and we have a contradiction by [44]. Let q > 3. Assume there are two distinct lines r and s of Sσ through P in the same orbit ¯ Since as it is easily seen C¯ r , C¯ s ∼ of C. = PSL(2, q) for q > 3, then all the lines of Sσ in the same orbit under C¯ pass through P and as in Theorem 3.10. Fix σ has order at least (q 2 − q)/2. Since [(q 2 − q)/2]2 > q 3 + 1 for q ≥ 27, we must have v < n. Now let q = 3. As v ≥ n, then n ≤ 28. If P is the only point in (l ∩ Fix σ ) − D, the 6 lines in Sσ passes through P . Thus Fix σ has order at least 6, a contradiction. The only possibility is that ¯ n = 25 and Fix σ there are exactly 2 points P1 and P2 in (l ∩ Fix σ ) − D , both fixed by C, has order 5. The plane Fix σ ∼ PG(2, 5) contains the collineation group C¯ ∼ = = A4 fixing a point. We have a contradiction. If all involutions in F are homologies we may use the same argument of [44] to obtain a contradiction. Suppose that n is even. As in Theorem 3.10 every involution in C = CF (σ )

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would be an elation with the same centre of σ and hence the involutions in C would generate an abelian group, a contradiction for q > 3. Now suppose that q = 3. Then G∼ = P L(2, 8) . A Sylow 2-subgroup S of G is elementary abelian of order = 2 G2 (q) ∼ 8 and consists of elations with the same centre CS . If S  ∈ Syl2 (G) with S  = S  , then CS   = CS since it is easily seen that G cannot fix a point. Thus GCS ∼ = NG (S) and G acts on the elation centres as P L(2, 8) in its usual 2-transitive representation of degree 9. Note that the line s = CS  CS is fixed by G. However, s cannot be an elation axis since any two involutions in G have different axes. Thus s contains the 9 elation centres. If s contains some other point U , then GU must have odd order, so that |U G | ≥ 56, in contrast with the assumption v ≥ n. Thus n = 8 and the assertion follows.
LEMMA 3.12. Let H ≤ L(1, F ), F ∼ = GF(pm ), p odd. If H is transitive on F ∗ , then the map x → −x lies in H .   Proof. We have that H = ωk , σ v ωj , where ω : x −→ ωx, ω = F ∗ , σ : x −→ x p ,

∀x ∈ F ,

and the integers p, m, v, k and j satisfy suitable conditions (e.g. see [50]). In particular v | m and k | (pm −1)/(p v −1). Hence (p v −1) | (pm −1)/k, so that the order (p m −1)/k m
of ωk is even. Thus ω(p −1)/2 : x −→ −x lies in H . THEOREM 3.13. Let
be a projective plane of order n, G a collineation group of
and O a point-orbit of G of length v on which G acts 2-transitively. If v ≥ n and |l ∩ O| ≥ 3 for a secant l of O, that is O is one of the designs listed in Theorem 2.1, then one of the following holds (i) v = n and for n  = 34 , 36 either

(ii) (iii) (iv) (v)

(1) O is a Desarguesian affine plane and the projective extension of O is embedded in
, or (2) O ∼ = AG(d, q), d > 2, q > 2, the projective extension of O is embedded in
, G = T GO and GO ≤ L(1, q d ), √ v = n + n + 1, n square, O is a Baer subplane and
is one of the planes listed in Theorem 3.9, √ v = n n + 1, n square,
is Desarguesian, O is an Hermitian unital and √ G
PSU(3, n), v = n2 and O ∼ =
l∞ is a Desarguesian affine plane, the nearfield plane of order 2 3 or the Hering plane of order 33 v = 28, n = 8, G ∼ = 2 G2 (8) ∼ = P L(2, 8) leaves invariant a line oval  in
and O consists of the external points of .

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∼ PG(d, q), d ≥ 2. By [14] either PSL(d + 1, q) ≤ G, ¯ where Proof. Suppose that O = ¯ G is the group induced by G on O, or d = 3, q = 2, the action of G on O is faithful by Proposition 3.2 and G ∼ = A7 . In the first case we have d = 2 by Proposition 3.3 and (ii) holds by Theorem 3.9. In the other case let P be any point of O. Then GP ∼ = PSL(2, 7) ∼ = PSL(3, 2) (see [20]) and an involution σ in GP fixes exactly three secants of O through P . Furthermore, σ fixes s ∩ O pointwise for eactly one secant s through P (see [46]). Hence σ is a Baer collineation. As v ≥ n and v = 15, then n = 9. There are exactly 3 tangents to O through P and GP must fixes each of these tangents. Thus σ fixes exactly 6 lines through P , a contradiction. Suppose that O ∼ = AG(d, q), d ≥ 2, q > 2 and let O be the origin of O. It is easily seen that for v = n2 we must have d = 2 and (iv) holds by Theorem 2.1. Thus we may assume ¯ where G ¯ is the group induced by G on O, contains the translation that v  = n2 . By [40] G, group T of AG(d, q) for q > 2. ¯ then we have (i.1) by Proposition 3.2, (2). Assume that Suppose that d = 2. If G  = G, ¯ G = G. When q is even we have (i.1) by Proposition 3.8 and Lemma 3.6. Let q be odd. Then the structure of the stabilizer GO of the origin O of O has been determined in [22], Theorem 4. Either SL(2, q) ≤ GO or GO ≤ L(1, q 2 ) or GO is of some exceptional types listed in [22], Table II. An inspection of this table and Lemma 3.12 show that in any case the involutory kernel homology of O is in GO . Thus we have (i.1) or (iii) with q = 3, n = 4, by Proposition 3.5 and Lemma 3.6. Note that the case (d) of Proposition 3.5 cannot occur since G is not 2-transitive on O in that case. Let d ≥ 3. By [31], Theorem 6.7, a structure of d ∗ -dimensional vector space V over a field L∼ = GF(q h ), h|d, d = hd ∗ , may be defined on O in such a way that G ≤ AL(V ) and if O is identified with the zero-vector of V , then one of the following possibilities occurs for GO : (1) (2) (3) (4) (5) (6) (7)

GO ≤ AL(1, q h ), d ∗ = 1, GO  SL(d ∗ , q h ), d ∗ ≥ 2, GO  Sp(2m, q h ), d ∗ = 2m, m ≥ 2, GO  G2 (q h ) , d ∗ = 6, q even, GO  D8 ◦ Q8 , d = d ∗ = 4, q = 3 and GO /(D8 ◦ Q8 ) ≤ S5 , GO  SL(2, 5), d = 4, d ∗ = 2, q = 3, GO ∼ = SL(2, 13), d = d ∗ = 6, q = 3.

Suppose that the case (1) occurs. If q is even we have (i.2) by Proposition 3.8. If q is odd, the involutory dilatation with centre O lies in GO by Lemma 3.12 and we may use Proposition 3.5 to obtain (i.2). Now consider the case (2). Let τ ∈ GO induce a transvection in V . Then Fix τ ∩ O is ∗ a subspace of both V and O of order q h(d −1) . Thus it is a (d − h)-dimensional subspace

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of O. Hence d − d/d ∗ = d − h <

1 d +1 2

J. Geom.

(3)

by Proposition 3.1. Let d ∗ ≥ 3. If h ≥ 2, then d ≥ 6 and (3) does not hold. If h = 1, then (3) yields d = 3. Now Proposition 3.4, (2), may be used to exclude this case. When d ∗ = 2, by Proposition 3.5 for q odd and by Proposition 3.8 for q even we have that v = n = q d and the projective extension of O is embedded in
. As d > 2 and d ∗ | d we may assume in this case that d ≥ 4. Note that a Sylow p-subgroup S of GO ∼ = SL(2, q d/2 ), q = pt , p prime, is a Baer group. Indeed, OS = Fix S ∩ O is a d/2 -dimensional subspace of O, d/2 so that Fix S is a subplane of
of order u. By [19], Proposition 2.1, either u = q q−1−1 − 1 or u = q d/2 or u ≥ q d/2 + q. Since n = q d the third case cannot occur. In the first case OS is a maximal arc in Fix S (see [19]) and NG (S) acts 2-transitively on OS . By [18] we have a contradiction. Suppose that u = q d/2 and let lD be the line defining the projective extension of O. The extension of OS has exactly (q d/2 − 1)/(q − 1) points on lD which lie in Fix S. Let R be one of these points. There are exactly q d−1 lines of O and q d/2−1 lines of OS through R. Thus the lines through R in O − OS are q d/2−1 (q d/2 − 1). Since S has order q d/2 , the stabilizer in S of at least one of these lines, say a, must have order at least q. Then a is a line of Fix S and a ∩ O is a secant of O fixed by S. Since a ∩ O is disjoint from OS , S must act f.p.f. on a ∩ O. This implies |S| ≤ q, in contrast with d ≥ 4. So, the case (2) cannot occur. The case (3) may be excluded by the same argument of the case (2) since Sp(d ∗ , q h ) contains transvections – here d ∗ ≥ 4. In the case (4) G2 (q h ) contains an involution σ fixing a 4-dimensional subspace of V pointwise (e.g. see [33]). Thus Fix σ ∩ O is a 4h-dimensional subspace of O. By Proposition 3.1 we must have 4h < 3h + 1. This is impossible and the case (4) cannot occur. In cases (5)–(7) GO contains the involutory O-dilatation (see [3], [27]). Hence we have v = n by Proposition 3.5, but we supposed n = 34 , 36 . The same occurs when O is the nearfield plane of order 32 , the Hering plane of order 33 or one of the two Hering designs with v = 36 and k = 32 . Indeed in all cases the translation group T and the involutory O-dilatation of the underlying affine space over GF(3) lie in G and the same arguments of Proposition 3.5 work. ∼ AG(2, 3). Then we have (iii) by Theorem 3.10. Let O ∼ = H (q) for q > 2 – note that H (2) = Let O ∼ R(q). Then we have (v) by Theorem 3.11.
= PROPOSITION 3.14. Under the assumptions of Theorem 3.13 when n = 34 or 36 the following additional cases can occur for v = n:

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(1) O is the nearfield plane of order 32 or the Hering plane of order 33 , or (2) O ∼ = AG(d, q), G = TGO and either (a) d = 4, q = 3 and GO  SL(2, 5) or GO  D8 ◦ Q8 with GO /(D8 ◦ Q8 ) ≤ S5 , or (b) d = 6, q = 3 and GO ∼ = SL(2, 13), or (3) O is one of the two Hering designs with v = 36 and k = 32 . In all cases the projective extension of O is embedded in
. The proof easily follows from that of Theorem 3.13. Note that the case (i.1) of Theorem 3.13 occurs in Desarguesian planes of square order and also in the Hughes planes. The case (i.2) typically occurs in generalized Andr´e planes (see [47] and [25]). The case (i.1) of Theorem 3.13 occurs also when
is a Prohaska plane of order 34 . In this case GO is the central product of SL(2, 5) and a cyclic group of order 8 acting transitively on the non-zero points of a Desarguesian affine Baer subplane of order 32 [24]. 4. The v-arc The main aim of this section is to prove that a n-arc which is invariant under a collineation group 2-transitive on its points may be extended to an oval. PROPOSITION 4.1. Let G be a 2-transitive permutation group of degree n and assume that the minimal normal subgroup T of G is non-abelian simple. If n = m2 is a square, then one of the following holds (1) (2) (3) (4)

∼ = Am2 , m > 2, ∼ = PSL(4, 7), m = 20, ∼ PSL(5, 3), m = 11, = ∼ = PSp(6, 2), m = 6.

T T T T

Proof. We refer to the possibilities for T listed in [39]. We have to consider the Diophantine equations −1 (a) qq−1 = m2 , q ≥ 2, d ≥ 2, (d, q)  = (2, 2), (2, 3), (b) q 2 + 1 = m2 , q 3 + 1 = m2 , q > 2, (c) 22d−1 + 2d−1 = m2 , 22d−1 − 2d−1 = m2 , d ≥ 3. d

In the case (a) by Proposition 2.2, (2) the only admissible integer solutions (q, d, m) are (7, 4, 20) and (3, 5, 11) which lead to (2) and (3). In the case (b) there are not admissible solutions. In the case (c), the equations admit integer solutions if and only if d ≥ 3 is odd

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and 2d + 1 or 2d − 1 are, respectively, squares. 2d + 1 = s 2 yields d = s = 3 and we have (4), while 2d − 1 = s 2 has no integer solutions by Proposition 2.2, (1). For none of the sporadic cases n is a square.
PROPOSITION 4.2. Let  be a n-arc, n > 3, in a projective plane
of order n and let σ be an involutory collineation of
leaving  invariant. Then one of the following holds (1) n is odd, σ is a (L, l)-homology, L ∈ /  and |l ∩ | = 1, (2) n is even, σ is a (L, l)-elation, L ∈ /  and |l ∩ | = 0 or 2, (3) σ is a Baer collineation and either Fix σ ∩  = ∅ or √ (a) n is odd and |Fix σ ∩ | = n, or √ √ (b) n is even and |Fix σ ∩ | = n or n + 2. Proof. Suppose that σ is a perspectivity. If L ∈ , then σ fixes  pointwise. This implies n ≤ 3. Clearly |l ∩ | ≤ 2 and 2 | n − |l ∩ |. We have (1) and (2). Now suppose that σ is a Baer collineation and Fix σ ∩   = ∅. If P ∈ Fix σ ∩ , then Fix σ ∩  lies in √ the n + 1 lines of Fix σ through P . Since there are at most two tangents at  through √ √ √ P , then n ≤ |Fix σ ∩ | ≤ n + 2, where the case |Fix σ ∩ | = n + 2 may occur only when Fix σ ∩  is an hyperoval in Fix σ , that is n is even. As 2 | n − |Fix σ ∩ | we have (3).
THEOREM 4.3. Let  be a n-arc in a projective plane
of order n. If  is invariant under a collineation group G of
acting 2-transitively on its points, then  is not complete. Proof. 1) The socle T of G is elementary abelian. Suppose that T is non-abelian simple. Assume that n is a square. We use Proposition 4.1. Let T ∼ = An , n ≥ 5. There is an √ involution σ ∈ T fixing n − 4 points. If σ is a Baer involution then n − 4 = n or √ n − 4 = n + 2 for n even. This is impossible. Let T ∼ = PSL(d, q) with (d, q) = (4, 7) or q d−1 −1 (5, 3). There is an involution σ ∈ T fixing q−1 + 1 = 41 points for (d, q) = (5, 3) and q d−2 −1 q−1

+ q + 1 = 16 points for (d, q) = (4, 7). The order of Fix σ ∩  does not agree with ∼ PSp(6, 2). There is an involution any of the possibilities listed in Proposition 4.2. Let T = σ ∈ T fixing 16 points in the representation of degree 36 (e.g. see [20], p. 159). Again, Proposition 4.2 yields a contradiction. Therefore, we may assume that all involutions in T are perspectivities. If n is odd, then by Proposition 4.2 each involution in T fixes exactly one point of . By [30], Theorem 2 and Lemma 3, T ∼ = PSL(2, q), Sz(q) or P SU (3, q) with q even and TP , P ∈ , contains a Sylow 2-subgroup of T . It is easily seen that two commuting involutory homologies cannot have the same axis. Thus, there are q − 1 distinct axes through P . We have a contradiction since there are at most two tangents at  through P . Let n be even. By Proposition 4.2 an involution in T fixes 0 or 2 points on . Furthermore, T is 2-transitive on , except when T ∼ = PSL(2, 8) and G ∼ = P L(2, 8) ∼ = 2 G2 (3). In this

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last case an involution fixes 4 points, a contradiction [44]. Then by [4] and [29], either T ∼ = PSL(2, q) with q odd and || = q + 1 or T ∼ = A6 and || = 6. The last case does not occur since
cannot have order 6. It is easily seen that two distinct commuting involutory elations cannot have the same centre. So, they have the same axis. Let σ be an involution in T ∼ = PSL(2, q), q odd. The involutions in CT (σ ) have the same axis of σ . Hence they generate an elementary abelian group. This yields |CT (σ )| = 4, that is q = 3 or 5. For q = 5, we have n = 6, which is impossible. 2) If T is elementary abelian, then  is not complete. Suppose that n is odd. As n = p h for some odd prime p, the group T fixes a point P of
and P ∈ /  since T is regular on . Since n is odd, at least one line through P is a tangent of . But then, by the transitivity of T on , there are exactly n tangents and one external line through P at . Hence  is not complete. Suppose that n is even. As before T fixes a point-line incident pair P , l, where P ∈ /  and l is external to . If all the lines through P distinct from l are tangent to , then  is not complete. Suppose there is a secant m through P at . Then Tm = σ for some involution σ since T is transitive on . Furthermore, T is abelian and so all the n/2 secants through P at  are fixed by σ . Since we may assume n > 4, then σ is an elation. By the 2-transitive action of G, all involutions in T are elations. Since the centres of these involutions must be pairwise distinct, then all involutions in T must have axis l. There are exactly two points M and N of l which are not elation centres. Both these points are fixed by T . Thus the lines through M, different from l, are in a unique orbit under T . This implies that the lines through M, different from l, are tangent to  and again  is not complete.
The known results about 2-transitive v-arcs with v ≥ n are summarized in the following. THEOREM 4.4. Let
be a projective plane of order n, G a collineation group of
and O a point-orbit of G on which G acts 2-transitively. If O is a v-arc with v ≥ n, then one of the following holds (1) v = n, n = p h , p prime, O is extendable to an oval and either G ≤ Al(1, n) or n ∈ {52 , 72 , 112 , 232 , 292 , 592 , 34 , 36 }, (2) v = n + 1 and either (a) n is odd,
is Desarguesian, O is a conic and G  PSL(2, n), or (b) n is even and one of the following holds (1)
is Desarguesian, O is a conic and G  PSL(2, n), (2) n = 22r , r ≥ 3 odd, and G  Sz(2r ), (3) n ∈ {2, 4} or n = p 2h − 1 and either p is a prime with p ≡ 3 mod 4, h is odd and G ≤ Al(1, p2h ) or p 2h = 72 , 112 , 232 , 292 or 592 , (3) v = n + 2, O is an hyperoval and either n = 2 and A4 ≤ G ≤ S4 , or n = 4 and A5 ≤ G ≤ S6 .

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Using Theorem 4.3, (1) follows from Francot [23] for n odd and from Biliotti, Jha and Johnson [10] for n even. (3) is due to Abatangelo [2]. (2.a) is due to Korchmaros [41], while (2.b) follows from several results of Biliotti, Bonisoli and Korchmaros (see [12] and the references in it). Note that in [12] Bonisoli actually proves that either G is contained in the affine semilinear group or p 2h = 72 , 112 , 232 , 292 or 592 . Nevertheless by [13] the Sylow 2-subgroups of G must be generalized quaternion groups. So (2.b.3) follows from the following lemma. LEMMA 4.5. Suppose that L(1, q), q = p n , contains a generalized quaternion group Q. Then n = 2h with h odd and p ≡ 3 mod 4. Proof. Let L = GL(1, q). Then Q/Q ∩ L ∼ = QL/L is cyclic as well as Q ∩ L. By the structure of the generalized quaternion groups the only possibility is that Q ∩ L is cyclic of order 2s and Q has order 2s+1 , s ≥ 2. Hence there is an element η of order 4 of Q, not in n

n

Q ∩ L, such that η2 : x → −x. Then η : x → bx p 2 with bp 2 +1 = −1. Let Q ∩ L = α , n

n

α : x → ax. Then η−1 αη : x → a p 2 x and η−1 αη = α −1 implies a p 2 +1 = 1. As n n 4 | p 2 + 1 we must have p 2 ≡ 3 mod 4, that is n/2 is odd and p ≡ 3 mod 4.
Note that the case (1) occurs in the Desarguesian planes and also in the twisted field planes of odd order [21]. The case (2.b.2) occurs in the dual L¨uneburg planes, while the only known planes in which the case (2.b.3) occurs are those of order 2 and 4. 5. The line Suppose that the point-orbit of G on the projective plane
of order n is contained in a line. As we assumed the orbit has length at least n, we have to consider two cases: ¯ on l − {L}, (I) G fixes a flag (L, l) of
and induces a 2-transitive group G l (II) G fixes a line l of
and induces a 2-transitive group G on the points of l. The case (I) has been investigated by Hiramine [33]. ¯ admits an elementary THEOREM 5.1 (Hiramine). In the case (I) n is a prime power, G ¯ ¯ abelian subgroup H of order n and G0 satisfies one of the following conditions (i) (ii) (iii)

¯ 0 ≤ L(1, p m ), n = pm , G ¯ 0 ≤ L(2, pm ), n = p2m , SL(2, pm ) ≤ G 4 2 4 n ∈ {2 , 3 , 3 , 36 , 52 , 72 , 112 , 192 , 232 , 292 , 592 }.

Suppose that
is a translation plane. If the flag is of type (L∞ , l∞ ), then a complete classification is given in [9], Theorem 8.1 – except for n = 26 . The situation (i) of Hiramine’s theorem occurs in all cases. If the flag is of type (L∞ , l), with l an affine line, a

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classification is not available, but there are several examples. All the so called “weak rank 3 affine planes” admits such a flag [11] and all the situations of Hiramine’s theorem occur. In particular the situation (ii) occurs in the Hall planes while in the examples referring to the ¯ 0 in many cases. A characterization has been obtained situation (iii), A5 is involved in G for self-dual translation planes [9]. When n = 2m , m ≥ 1, the case (II) has been investigated by Korchmaros [42]. A subsequent improvement is given in [5] and [6]. For translation planes the case is completely settled (e.g. see [37]). Here we give a further improvement to [42] using some connections between this problem and a recent paper of Bonisoli [12] on 2-transitive ovals. THEOREM 5.2. In case (II) with n even one of the following holds (1) (2) (3) (4) (5)


is a Desarguesian plane and SL(2, n)
G, n = q 2 , q = 22s+1 , s ≥ 1, and Sz(q)
G, n = q 3 , q = 22s , s ≥ 1, PSU(3, q)
G and G fixes a point of
l , n = ph − 1, p an odd prime, and G ≤ AL(1, p h ), n = ph − 1, p h = 52 , 72 , 112 , 232 , 292 or 592 and Gl is sharply 2-transitive on l except possibly for ph = 52 or 292 .

We need some preliminary results. PROPOSITION 5.3. Let
be a projective plane of even order n possessing a collineation group F which leaves invariant a line l and satisfies the following conditions (1) F has order 2(n + 1), (2) F contains an involutory elation with axis different from l, (3) F acts transitively on the points of l. Then we have (i) F is a Frobenius group with abelian Frobenius kernel H of order n + 1, (ii) F fixes exactly one point N of
− l and all involutory elations in F have the axis through N , (iii) the orbits of F on
− {l ∪ N } are ovals with common knot N and the permutation action of F on each orbit is the same as on l. For a proof see [1]. PROPOSITION 5.4. A projective plane
of order 8 does not admit a collineation group G which fixes a line l and induces on l a 2-transitive group possessing a normal regular abelian subgroup.

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Proof. Since
is Desarguesian, then the full collineation group of
induces P L(2, 8) on l. An elementary abelian group transitive on l must be generated by an element α of order 3 in PSL(2, 8) and an element β of order 3 outside PSL(2, 8). Since these elements cannot be conjugate in P L(2, 8) we have the assertion.
Proof of Theorem 5.2. Let K be the kernel of the representation of G on l. As in [42] we may assume that if K = 1 , then K consists of (N, l)-homologies with the same centre N and that GO , O ∈ l, contains a non-trivial normal subgroup Q of even order consisting of the elations with centre O. Arguing as in [42] and using a result of Hering [30] we obtain l (1), (2) or (3) (see also [5] and [6])  the socle of G is non-abelian simple. If the socle  when G is elementary abelian, then S = Q = O(S)Q, where S acts transitively on l and Q has the structure of a Frobenius complement by [30], Theorem 2. Since O(S)  is characteristic in S and S
G, then O(S)l contains the socle T l of Gl , K ≤ T . Let T l  = ph , p odd. If n = m2 is a square, then the Diophantine equation p h − m2 = 1 must have integer solutions. Thus h = 1 by Proposition 2.1, (1) and so Gl ≤ AGL(1, p) by [36], Satz II.3.6. Since |K| ≡ 1 mod 2 we may infer that in any case an involution in Gl fixes exactly one point, being induced by an involutory elation in G. Note that a Sylow 2-subgroup R of G lies in GP for some point P ∈ l. Thus R contains a unique involution by the structure of Q. Hence R is either a cyclic or a generalized quaternion group. By [31], Theorem 6.7, a structure of k-dimensional vector space V over a field L ∼ = GF(q) may be defined on l in such a way that Gl ≤ AL(V ), and if O is identified with the zero-vector of V , then one of the following possibilities occurs for GlO , (we assume K ≤ GO ): (a) (b) (c) (d) (e)

k k k k k

= 1, = 2 and GlO  SL(2, q), q odd, = 2, GlO  SL(2, 5) and q ∈ {9, 11, 19, 29, 59}, = 2, GlO  SL(2, 3) and q ∈ {5, 7, 11, 23}, = 6, GlO ∼ = SL(2, 13) and q = 3.

The case (b) cannot occur for q  = 5. We have that T l HOl with HOl ∼ = SL(2, q), q = ph , l l p an odd prime, K ≤ HO , is 2-transitive on l. Furthermore, |T HO | = q 3 (q 2 − 1) and |K| | q 2 − 2. By [36], Hauptsatz I,18.1, there is a complement H of K in T HO acting faithfully on l as T l HOl . Let H = T ∗ HO∗ . If σ is the central involution in HO∗ , then F = T ∗ σ satisfies the assumptions of Proposition 5.3. Let N be the unique fixed point of F on
− l. If P is a Sylow p-subgroup of HO∗ , then P fixes exactly q points on l. Note that
has order n = q 2 − 1. So, P fixes N and also it fixes some other point on ON − {O, N} since (p, q 2 − 2) = 1. Thus Fix P is a subplane of order q − 1 and P fixes exactly q − 2 points on ON − {O, N}. Let P1 , P2 be two distinct Sylow p-subgroup of HO∗ . If Fix P1 and Fix P2 have some common point on ON − {O, N}, then HO∗ = P1 , P2 fixes a point C on ON − {O, N} and hence it fixes the point-orbit  of T containing C. Since  is an oval, this cannot occur by [12]. Otherwise the set of points of ON − {O, N} which are fixed by some Sylow p-subgroup of HO∗ has order exactly (q − 2)(q + 1) and it is HO∗ -invariant. So,

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HO∗ acts transitively on the set of the remaining q points of ON − {O, N} – a p-Sylow of HO∗ acts regularly on the set. Then q = 3, 5, 7 or 11 (e.g. see [36], Satz II, 8.28). The case q = 3 cannot occur by Proposition 5.4. If q = 7, then Fix P would have order 6, which is impossible. Let q = 11. The central involutory elation in HO∗ leaves invariant Fix P . Hence Fix P admits collineations of even order. Thus 4 must divide the order of Fix P by [35], Theorem 13.18, a contradiction. In cases (c), (d) and (e), a trivial calculation shows that (|T l GlO |, |K|) = 1. Again, we may assume there is a group T ∗ G∗O acting faithfully on l and either |T ∗ | = q 2 with q ∈ {5, 7, 9, 11, 19, 23, 29, 59} or |T ∗ | = 36 . Let consider the first case. Suppose that G∗O does not act regularly on l − {O}. Then there is an element α ∈ G∗O of prime order r fixing exactly q points on l – Fix α must be a 1-dimensional subspace. Therefore, r | q 2 − q. Then α fixes at least one point on ON − {O, N} as |ON − {O, N}| = q 2 − 2. Hence Fix α is a subplane of order q − 1. As before, the central involutory elation in G∗O leaves invariant Fix α. Thus 4 must divide the order of Fix α. This occurs only for q ∈ {5, 9, 29}. Note that for q = 19 the group G∗O cannot be regular on l − {O}. So the case q = 19 cannot occur. When q = 9, G∗O contains a Sylow 3-subgroup U fixing exactly 9 points on l. Furthermore, U fixes at least one point on ON − {O, N} as |ON − {O, N}| = 79. Thus Fix U is a subplane of order 8. By [49], Theorem 9.4, the normalizer of U in T ∗ G∗O is 2-transitive on Fix U ∩ l, but this contradicts Proposition 5.4. Now let |T ∗ | = 36 . As in [12], Section 7, we have that if U is a Sylow 3-subgroup of G∗O , then Fix U is a subplane of order 8 and the normalizer of U in T ∗ G∗O is 2-transitive on Fix U ∩ l, but again this contradicts Proposition 5.4. THEOREM 5.5. In case (II) with n odd
is a Desarguesian plane and SL(2, n)
G. Proof. This result was proved by Cofman [16] under the additional assumption n  ≡ 1 mod 8 and it is probably known as folklore. So we only sketch a proof. Assume that n ≡ 1 mod 8. Let K be the kernel of the representation of G on l and let G∗ = G/K. Furthermore, let T ∗ be the socle of G∗ . T ∗ is non-abelian simple since n+1 = 2s yields n ≡ 3 mod 4 for s ≥ 2. Using the classification of 2-transitive groups [39], under the assumption n ≡ 1 mod 8 we have one of the following possibilities for T ∗ : (1) An+1 , d −1 − 1, (2) PSL(d, q), q odd, and n = qq−1 2 3 (3) PSU(3, q ), q odd, and n = q . Furthermore d = 2 in the case (2) when n is a square, as follows from the solutions of the d −1 Diophantine equation qq−1 = n + 1 by Proposition 2.2(3). √ Clearly any involution in G − K fixes either 2 or n + 1 points on l according to whether it is a homology or a Baer collineation. If K = 1 , then T ∗ contains involutions with a √ number of fixed points different from both 2 and n+1, except for T ∗ ∼ = PSL(2, q), which

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of course contradicts [16], Result D. Thus K  = 1 . It is then not difficult to prove that if G does not fix any point of
l , then
is a translation plane. So, we may assume that G fixes a point P of
l . Also we may assume that no proper subgroup of G acts 2-transitively on l, so that G∗ = T ∗ . Now we argue exactly as in Theorem 3.9 and we use [17], Table 1, to show that K is central in G – the case G∗ ∼ = PSL(2, 9) requires some ad hoc arguments. Then G is quasisimple by the minimality condition. Using Shur’s multipliers we have that one of the following occurs (1’) (2’) (2”) (3’)

n+1 , n ≡ 1 mod 8, G∼ =A q d −1 G∼ = SL(d, q), q odd, and n = q−1 − 1, G is a Shur extension of PSL(2, 9) and n = 9, G∼ = SU (3, q), q ≡ −1 mod 3, and n = q 3 .

In the case (1’), take in G∗ the involution σ ∗ = (A, A )(B, B  ), A, A , B, B  ∈ l. Let σ ∈ G such that σ → σ ∗ in the homomorphism G → G∗ . If σ is an involution, then the order of Fix σ ∩ l leads to a contradiction. Let o(σ ) = 4, C ∈ l − {A, A , B, B  } and τ ∗ = (B, B  )(C, C  ). We may assume that o(τ ) = 4. Now σ ∗ τ ∗ = τ ∗ σ ∗ , so that either σ, τ ∼ = Q8 or σ τ = τ σ and σ τ is an involution such that σ ∗ τ ∗ = (A, A )(C, C  ), a contradiction. When σ, τ ∼ = Q8 , take D ∈ l − {A, A , B, B  , C, C  } and let η∗ = (B, B  )(D, D  ). We have that σ ∗ η∗ = η∗ σ ∗ and η∗ τ ∗ = τ ∗ η∗ . Again, we may assume that o(η) = 4. Furthermore, η normalizes σ, τ . The group σ, τ, η has order 16 and σ, τ, η  ∼ = Q16 , since σ, τ, η /ρ ∼ = E8 , where ρ is the central involution in G. Hence σ, τ, η contains an involution ν fixing all the points of l not in {A, A , B, B  , C, C  , D, D  }. Since n ≡ 1 mod 8, the only possibility is n = 9. G∗ ∼ = A10 contains H ∗ ∼ = PSL(2, 9) in its action of degree ∼ 10 and H = PSL(2, 9) or SL(2, 9). As already noted, H ∼ = PSL(2, 9) cannot occur, while H ∼ = SL(2, 9) forces
to be Desarguesian by [43], Satz 2, a contradiction. If the case (2’) occurs and d = 2, then
is Desarguesian by [43], Satz 2. Let d > 2. So
has non-square order and the involutions in G are homologies. Since G contains involutions fixing more than 2 point, we have a contradiction. In the case (2”), using Shur’s multipliers and the fact that |Z(G)| | n − 1 = 8 we have that G∼ = SL(2, 9) and again
is Desarguesian. In the case (3’), |Z(G)| = 3 and we may use the argument on the number of fixed points of an involution to exclude this case.
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Received 8 May 2002, revised 25 September 2003.