J. geom. 70 (2001) 17 – 27 0047–2468/01/020017 – 11 $ 1.50 + 0.20/0 © Birkh¨auser Verlag, Basel, 2001

Two-transitive parabolic ovals* Mauro Biliotti, Vikram Jha and Norman L. Johnson

Abstract. We investigate finite affine planes π of even order possessing a parabolic oval  (| ∩ l∞ | = 1) and a collineation group G which leaves  invariant and acts 2-transitively on its affine points. The main attention is devoted to translation planes. The odd order case has already been considered by Enea and Korchmaros in [5]. Our main result shows that if π has even order n 6 = 26 and possesses two 2-transitive parabolic ovals which share at least two, but not all their affine points, then π is Desarguesian. Mathematics Subject Classification (2000): 51B10. Key words: Ovals, two-transitive.

1. Introduction Let π be a finite affine plane whose projective extension contains an oval  which intersects the line at infinity in a single point (∞). In this case,  is said to be a parabolic oval of π . If there is a collineation group G of π which leaves  invariant and acts 2-transitively on the affine points of , we shall call  a 2-transitive parabolic oval. When π is a translation plane admitting a 2-transitive parabolic oval and the order is odd, then the combined results of Ganley and Jha [8] , Cordero and Figueroa [4], Enea and Korchm´aros [5] and a remark in [2] give the following THEOREM 1.1. Let π be a finite translation plane of odd order admitting a 2-transitive parabolic oval. Then π is a Desarguesian or a commutative twisted field plane and the oval consists of the absolute points of an orthogonal polarity. Furthermore, in the above situation, π admits an autotopism group acting transitively on the non-fixed points of any side of the autotopism triangle. In this article we consider even order planes. Recently the question whether generalized twisted field planes could admit a 2-transitive parabolic oval was resolved by the authors. ∗ The research of the first author was supported by a grant from the M.U.R.S.T. The ideas for this article originated when the third author was visiting Caledonian University and The Ricco Institute during November of 1997 and when the second and third authors were visiting the Scuola Estiva delle Universit´a di Basilicata e Lecce in September of 1997. The authors gratefully acknowledge the support of the Universities of Basilicata, Lecce, Caledonian University and the Ricco Institute.

17

18

Mauro Biliotti, Vikram Jha and Norman L. Johnson

J. Geom.

THEOREM 1.2. (Biliotti, Jha and Johnson [3]) Let π denote a generalized twisted field plane of even order. If π admits a 2-transitive parabolic oval then π is Desarguesian. In addition, generalized Andr´e planes of order 2r generally admit translation ovals depending on the integer r (see [11]). However it has not been clear if a non-Desarguesian generalized Andr´e plane admits a 2-transitive parabolic oval. We are able to show that, in fact, this does not occur (see Theorem 3.5). Although there are certain things we can say about translation planes admitting a 2-transitive parabolic oval, we are primarily interested in what happens when there are two 2-transitive parabolic ovals. Of course, it is possible that there is a covering of the points not on two components by a set of parabolic ovals which share exactly one affine point [11]. For example this happens in generalized Andr´e planes. Clearly, any translation plane admits translations mapping a 2-transitive parabolic oval  into a disjoint oval and also kernel homologies mapping  into another oval with exactly one affine point in common with . So the assertion that the plane contains two 2-transitive parabolic ovals is significant only if one supposes that the two ovals share at least two distinct affine points. Our main result in this regard shows the plane to be Desarguesian in this situation (see Theorem 4.1). 2. Two-transitive parabolic ovals In this section we investigate the affine planes with a 2-transitive parabolic oval. THEOREM 2.1. Let π be a finite affine plane of even order n which admits a 2-transitive parabolic oval  with 2-transitive group G . Then the following hold (1)  is a translation oval, (2) n = 2r , G ≤ A0L(1, 2r ) and G contains the translation group of A0L(1, 2r ). Proof. Let (∞) denote the infinite point of . Let (0) denote the knot of . Let 8 be the pencil of the tangents to  through (0), different from l∞ . We have that G is faithful and 2-transitive on 8. Now we may apply the dual version of [10], Main Theorem. We then have that n = 2r , G has a regular normal subgroup N and one of the following holds for G,0 , where 0 denote any affine point of . (i) G,0 ≤ 0L(1, 2r ), n = 2r , (ii) SL(2, 2m ) ≤ G,0 ≤ 0L(2, 2m ), n = 22m , (iii) n = 24 .

Vol. 70, 2001

Two-transitive parabolic ovals

19

Note that all involutions in N are conjugate under G . By [1], Prop. 5, either |N | = 2 or all involutions in N are elations with the axis l∞ , that is they are translations and  is a translation oval. Trivially, (1) and (2) holds in P G(2, 2). Let S be a Sylow 2-subgroup of G,0 . Any involution σ in S fixes (∞) and 0 on . Hence it is a Baer collineation by [1]. Again by [1], Prop. 5, S must be cyclic. Then the cases (ii) and (iii) cannot occur. Note that in the case (iii) G,0 ∼ = A6 or A7 . This completes the proof. ¨ THEOREM 2.2. Let π be a finite affine plane of even order n which admits a 2-transitive parabolic oval  with group G . Then either all the translations of Aut(π ) lie in N , or π is a translation plane. Proof. Let |N | = n = 2r . Furthermore, let h = |Aut(π )(Q∞ , l∞ )| for Q∞ 6 = (0) , (∞), k = |Aut(π)((0) , l∞ )| and s = |Aut(π )((∞) , l∞ )|. Clearly, |Aut(π )(l∞ , l∞ )| = ht where t | n. We then have (n − 1) (h − 1) + (k − 1) + (s − 1) + 1 = ht. But (n − 1) (h − 1) + (k − 1) + (s − 1) + 1 = nh − (n + h − k − s). Then (1) nh − (n + h − k − s) ≤ 21 nh, or (2) nh − (n + h − k − s) = nh. In the case (1), 21 nh ≤ (n + h − k − s) for h > 2 yields nu ≤ (n + h − k − s) for u ≥ 2. This is possible only for h = n, k = s = 0. We have a contradiction since k, s ≥ 1. So, h = 2. Then the translations of π in the directions different from (0) and (∞) are exactly those in N . We have n ≤ (n + h − k − s). Then k + s ≤ h. So k = s = 1 and every translation of π lies in N . In the case (2) n+h−k−s =0 Then, say k >

n 2.

So |Aut(π)((0) , l∞ ) · N | > n n2 . This yields |Aut(π )((0) , l∞ ) · N | = n2

and π is a translation plane. We call two 2-transitive parabolic ovals  and 9 distinct if they differ for some affine point. ¨

20

Mauro Biliotti, Vikram Jha and Norman L. Johnson

J. Geom.

THEOREM 2.3. Let π be a finite affine plane of even order n 6 = 26 admitting two distinct 2-transitive parabolic ovals  and 9. Then π is a translation plane and either (1) the associated hyperovals  and 9 share their infinite points (∞), (0) and the full collineation group of π leaves the pair {(∞), (0)} invariant or (2) π is a Desarguesian plane. Proof. By Theorem 2.2 we have to investigate the case N = N9 = N . So,  and 9 are two distinct orbits of N. Up to a change of (0) with (∞)—the role of (∞) and (0) is symmetric—we may suppose that both  and 9 contains (∞). Let r be any affine line through (∞) and let A = r ∩ , B = r ∩ 9. Clearly A 6 = B since A and B lie in different orbits of N. Since G,A is transitive on the points of r − {A} and G9,B is transitive on the points of r − {B} we have that F = hG,A , G9,B i is 2-transitive on r. Let K be the kernel of the representation of F on r. By [10], Main Theorem, F = F /K possesses an elementary abelian regular normal subgroup M. Note that K has odd order because an involution in K should fix the line r pointwise and the point (0) and hence it would be an homology, a contradiction. Thus, we may suppose that F contains an elementary abelian normal subgroup M acting regularly on r. Each non-trivial element of N does not fix r. Hence N ∩ M = h1i. Each involution in M fixes some point on l∞ , other than (0) and (∞), and it is not a Baer collineation since it fixes r, but no points on r. Hence M is a group of translations and π is a translation plane by Theorem 2.2. Suppose that the hyperovals  and 9 do not share their infinite points. If the infinite points of  and 9 are four distinct points then it is easily seen that G = hG , G9 i is transitive on l∞ . Hence the plane is flag-transitive. Since the group G0 is transitive on the points different from 0 on (x = 0), it follows that the stabilizer of 0 is transitive on the remaining affine points. That is, the plane admits a doubly transitive group on the affine points. By Jha-Johnson [13], the plane is Desarguesian. If  and 9 have a common infinite point (we may assume (∞)), then G = hG , G9 i acts doubly transitively on l∞ − {(∞)} since the infinite points of  and 9, different from (∞), must be distinct. Then (∞) is an elation center and π is a generalized twisted field plane by the results of Ganley-Jha [8] and Cordero-Figueroa [4]. However, generalized twisted field planes that admit a 2-transitive parabolic oval are Desarguesian by Biliotti-JhaJohnson [3]. ¨

3. Two-transitive parabolic ovals in translation planes In this section, we deal with translation planes possessing a 2-transitive parabolic oval . We denote by 0 the origin of π and we assume—without loss of generality—that 0 ∈ .

Vol. 70, 2001

Two-transitive parabolic ovals

21

3.1. The group Since G,0 ≤ 0L(1, 2r ) by Theorem 2.1, we recall the main results concerning the classification of doubly transitive subgroups of A0L(1, 2r ) as given by Foulser in [6], Section 15, and also by Woltermann [17]. THEOREM 3.1. (Foulser). Let the elements of A0L(1, 2r ) be represented by the mappings x 7 −→ gx µ + b over GF (2r ) with b ∈ GF (2r ), g ∈ GF (2r )∗ and µ ∈ Aut GF (2r ). Let w denote a primitive root of GF (2r ). Let w, α and τa denote the mappings x 7 −→ wx, x 7 −→ x 2 , x 7 −→ x + a, respectively. Let T = hτa : a ∈ Ki. Let F be a doubly transitive subgroup of A0L(1, 2r ). Then (1) T ⊂ F and F = T · H is the split extension of T by H = F0 . (2) H may be represented in the form hw d , we α s i where (i) (ii) (iii) (iv)

d > 0 and d | 2r − 1, s > 0 and r ≡ 0 (mod sd), 0 ≤ e < d and (d, e) = 1, the prime divisors of d divide 2s − 1,

and H1 = G0,1 = hα sd i. (3) G contains a minimal sharply 2-transitive subgroup. Furthermore, r = sd in a representation of a minimal 2-transitive group. e = T ·H e of A0L(1, 2r ) = (4) Two doubly-transitive solvable subgroups G = T ·H and G T ·hw, αi are isomorphic if and only if they are conjugate under an element of hw, αi. Proof. See [6], Lemma 15.2, Propositions 15.3 and 15.5 and Corollary 15.4 and also [17], Section 3, for parts (1), (2) and (3), except for the fact that r = sd for minimal 2-transitive groups. Assume that G is sharply 2-transitive. We have that H1 = hα sd i = h1i. This forces sd ≡ 0(mod r). Since r ≡ 0 (mod sd) by (2.ii) and (2.vi), we have r = sd. To see that (4) is valid, the arguments of Foulser [6] (13.1) thru (13.9) generalize to the more general setting. In particular, the generalization of (13.9) then shows that two isomorphic 2-transitive subgroups of T · hw, αi are conjugate under an element of hw, αi. Note that the exceptional case of (13.9) for 2r = 64 and groups hw7 , w0 α 6 i and hw21 , we α 2 i, where (e, 3) = 1 and 0 ≤ e < 21, cannot occur because these groups are not doubly-transitive. This completes the proof of (4). ¨ REMARK 3.2. From Foulser’s results, we infer that whenever G is doubly transitive on  − {(∞)}, there is a subgroup G of G which is sharply 2-transitive on  − {(∞)}. In this case G ∼ = hwd , we α s i with r = ds and {2s , d} is a Dickson pair (e.g. see [15]). We start by giving some general results.

22

Mauro Biliotti, Vikram Jha and Norman L. Johnson

J. Geom.

PROPOSITION 3.3. Let π be a non-Desarguesian translation plane of even order 2r 6 = 26 which admits a 2-transitive parabolic oval  with group G . Then there exists a unique Desarguesian plane 6 which admits G,0 as a collineation group. Proof. The order of G,0 is divisible by 2r − 1 and thus contains an element g of order a prime 2-primitive divisor u of 2r − 1. Furthermore, by Theorem 2.1, G,0 ≤ 0L(1, 2r ) and  is a subspace. We may assume without loss of generality that G,0 fixes (x = 0), (y = 0) and  . Clearly, hgi is characteristic in GL(1, 2r ) ∩ G,0 which is normal in G,0 . By Johnson [14] (2.3), there is a unique Desarguesian plane 6 of order 2r whose spread is the set of line-sized invariant g-subspaces. Furthermore, the normalizer of hgi in G,0 is a collineation group of 6. ¨ THEOREM 3.4. Let π be a translation plane of even order 2r 6 = 26 which admits a 2-transitive parabolic oval  with group G . If G,0 is abelian then π is Desarguesian. Proof. By Theorem 2.1, G,0 ≤ 0L(1, 2r ). Furthermore, under the assumptions, G0 = G,0 is a cyclic subgroup of order 2r − 1 acting on 6 (the associated Desarguesian plane) and fixes at least three components of 6. Hence, the group is the kernel homology group of order 2r − 1 of 6. We assert that any component L of 6 different from (x = 0), (y = 0) is an oval of π . Note that L is a subspace of π which is left invariant by G0 . Since G0 is transitive on l∞ − {(∞), (0)}, it follows that every component of π intersects L − {0} in exactly one point. Moreover, L admits a 2-transitive group. That is, it must be that L is an oval (the reader might like to refer to Jha-Johnson [11]). Now conversely, any component M of π different from (x = 0), (y = 0) is an oval of 6. To see this, note that M is a subspace, so there is a transitive group on M which is a translation group of 6. Furthermore, M meets any component L of 6 different from (x = 0), (y = 0) in exactly one point, other than 0, since L is an oval of π. We may represent M in the form y = f (x) within 6 since M is disjoint from x = 0 and y = 0. Since M is an GF (2)-subspace and an oval in 6, it is a translation oval so that f is an additive GF (2r )-transformation and hence has a representation in the form r−1 X

i

ai x 2 .

i=0 N

But, it follows directly from Payne [16] that then f (x) = aN x 2 where (N, r) = 1 for −1 ). Since this is a collineation of the aN ∈ GF (2r ). Change bases by (x, y) 7 −→ (x, yaN associated Desarguesian plane 6, we may represent a given component M of π within the

Vol. 70, 2001

Two-transitive parabolic ovals

23

N

Desarguesian plane 6 in the form y = x 2 without disturbing the form of the group G0 which is h(x, y) 7 −→ (xa, ya) for all a in GF (2r ) − {0}i. This group is transitive on the components different from (x = 0), (y = 0) of π. Hence, N N every component of π has the general form (x = 0), (y = 0), (y = x 2 a 1−2 ) for all a −1 N in GF (2r ). Change bases by (x, y) 7−→ (x, y σ ), where σ is x 7−→ x 2 , to represent r ∗ the components of π as (x = 0), (y = 0), (y = xb) for all b in GF (2 ) . Hence, π is Desarguesian. ¨ 3.2. Generalized Andr´e planes Assume a generalized Andr´e plane admits a 2-transitive parabolic oval  with group G . By Foulser [7], the group which fixes (x = 0) and (y = 0) (the standard coordinate axes) may be written in the form h(x, y) 7 −→ (x σ a, y ρ b)i for various elements a, b in GF (2r ) and where σ and ρ are automorphisms of GF (2r ), possibly depending on a and b. We may assume that there is a group G0 which fixes 0 and is regular on each axis (x = 0), (y = 0) and on . It follows that G0 | (x = 0) is isomorphic to G0 | (y = 0) and note that both groups must lie within the group hw, αi (the same 0L(1, 2r )). Since the groups are isomorphic then the two groups induced on the axes are conjugate by an element of hw, αi. Hence, the group G0 may be written as follows h(x, y) 7 −→ (xh, y0(h)) for all h in G0 | (x = 0)i where 0 is a group isomorphism which is induced by an element k of hw, αi by conjugation. Note, for example, in the Desarguesian case, if k = α then the group action on (y = 0) is y 7 −→ y 2

−1

−1

7−→ y 2 wi 7 −→ yw2i

and the global group action is (x, y) 7−→ (xwi , yw2i ). This implies that the oval is (y = x 2 ). Hence, 0(h) = k −1 hk for k in hw, αi. Note that we have the oval as {(1h, 1k −1 hk) for all h in G0 | (x = 0)}.

24

Mauro Biliotti, Vikram Jha and Norman L. Johnson

J. Geom.

Let xk = x ρ b and xh = x σa a. Then −1

−1

xk −1 hk = (x ρ b−ρ )hk = (x ρ

−1 σ

a

b−ρ

−1 σ a

a)k = x σa a ρ b1−σa .

Hence, the group has the following form: h(x, y) 7 −→ (x σa a, y σa a ρ b1−σa ) for all a in GF (2r )∗ i and where ρ is in Aut GF (2r ) and b in GF (2r ) are fixed. Hence, the oval is  = {(a, a ρ b1−σa ) for all a in GF (2r )}. However, we note that (y = x) maps to (y = x(a −1 a ρ b1−σa )) and since the group is transitive on l∞ − {(∞), (0)}, it follows that the plane is Desarguesian and the oval has the N form (y = cx 2 ) for c ∈ K ∗ . Hence, we have shown: THEOREM 3.5. If π is a generalized Andr´e plane of even order which admits a 2 -transitive parabolic oval then π is Desarguesian. 4. Two 2-transitive parabolic ovals THEOREM 4.1. Let π be a finite affine plane of even order n 6 = 26 which admits two distinct 2-transitive parabolic ovals which share at least two affine points. Then π is Desarguesian. Proof. Let  and 9 denote the two distinct 2-transitive ovals which share at least two affine points. The plane is a translation plane by Theorem 2.3 and there exist two groups G,0 = G0 and G9,0 = H0 acting transitively on  − {0} and 9 − {0} respectively, where we assume that the two ovals share the zero vector 0. By Theorem 2.3, we may suppose that the hyperovals associated to  and 9 have identical infinite points {(∞), (0)}. We note that the action of G0 is isomorphic on the three fixed subspaces (x = 0), (y = 0) and . Moreover, by Foulser [6], we may assume that there is a solvable sharply transitive subgroup. Thus, we may assume that the indicated groups are sharply transitive on (x = 0), (y = 0) and  and on (x = 0), (y = 0), 9, respectively. Let u be a prime 2-primitive divisor of 2r − 1 for 2r = n and suppose that G0 ∩ H0 contains an element g of order u. It follows that g must fix both infinite points of the hyperovals and permute the affine points of  ∩ 9. Since both affine ovals are subspaces,  ∩ 9 is a subspace and it is different from h0i by our assumptions. It follows that g is planar.

Vol. 70, 2001

Two-transitive parabolic ovals

25

However, G0 is sharply transitive on (x = 0). So, there can be no planar elements in G0 . Hence, G0 ∩ H0 has no 2-primitive elements. If the group hG0 , H0 i | (x = 0) is non-solvable then by Ganley, Jha, Johnson [9], the plane must be Hall. However, by the orbit structure the order can only be 9 which cannot occur. Suppose that hG0 , H0 i | (x = 0) is solvable. Let G0 | (x = 0) = hwd1 , we1 α s1 i, where hwi = GF (2r )∗ , α : x 7 −→ x 2 , d1 s1 = r and conditions (i) thru (iv) of Theorem 3.1 are satisfied for (d1 , e1 , s1 ). Similarly, we let H0 | (x = 0) = hwd2 , we2 α s2 i with d2 s2 = r. We have that R = hG0 , H0 i | (x = 0) = hw d , we α s i, which has properties (i) thru (iv) of Theorem 3.1 as well. Let u be a 2-primitive divisor of 2r − 1 and let ub be the largest u-power divisor of 2r − 1. It is easily seen that R | (x = 0) has a unique Sylow u-subgroup Su , namely those generated r b r b by w (2 −1)/u . Su is linear and cyclic of order ub . Note that hw(2 −1)/u i ≤ hwd i since (d, u) = 1 by the condition (iv) of Theorem 3.1. Let Su1 and Su2 denote the Sylow u-subgroups of G0 and H0 respectively. Note that, as before, both are unique and cyclic of order ub . Since Su1 | (x = 0) is Su2 | (x = 0) and    i  A 0 A 0 then Su2 = Su1 | (y = 0) is Su2 | (y = 0), it follows that if Su1 = 0 Bj 0 B as both groups are cyclic. So, Su1 and Su2 commute and hSu1 , Su2 i = Su1 × Su2 has order u2b . Let H[x=0] be the full homology group with axis (x = 0) and let H1 = R ∩ H[x=0] . Then R/H1 ∼ = R | (x = 0). Hence H1 has order divisible by ub , because R has order divisible by u2b as Su1 × Su2 ≤ R. Nevertheless, the order of H[x=0] is (2r − 1)/t for some integer t since any homology group acts semi-regularly. So, ub is the largest u-power divisor of |H1 | and a Sylow u-subgroup of R has order u2b . Note that H1 acts faithfully on (y = 0). So H1 | (y = 0) is isomorphic to a subgroup of R | (y = 0) which possesses a unique Sylow u -subgroup (of order ub ). Hence H1 has a unique Sylow u-subgroup Zu1 of order ub which is normal in R. Similarly, H 2 has a unique

Sylow u-subgroup Zu2 of order ub which is normal in R. Then Z = Zu1 , Zu2 = Zu1 × Zu2 = Su1 × Su2 is normal in R. Hence it is the unique Sylow u-subgroup of R. Furthermore, the group R is transitive on l∞ − {(∞), (0)}. Let L be any component of π different from (x = 0) and (y = 0). Then RL has order divisible by ub , so that ZL has order ub since Z is the unique Sylow u-subgroup of R. Moreover a non-trivial element of Z does not fix any point different from 0. So, ZL induces an irreducible group on L. Since Z is abelian, π is a generalized Andr´e plane by L¨uneburg [15], Corollary 9.3. From the previous section, we have that any generalized Andr´e plane which admits a 2-transitive parabolic oval must be Desarguesian. Hence, we have the proof to the theorem. ¨

26

Mauro Biliotti, Vikram Jha and Norman L. Johnson

J. Geom.

5. Final conclusions and coverings We have seen that finite affine planes of even order n 6 = 26 containing parabolic ovals  and 9 are Desarguesian in two main situations (see Theorems 2.3 and 4.1). Actually, only one case remains open—up to a translation. Precisely that when  and 9 have exactly one affine point in common and do not correspond in a kernel homology with centre the common point. In this regard, we have the following PROPOSITION 5.1. Let π be a finite translation plane of order 2r 6 = 26 which admits two 2-transitive parabolic ovals  and 9 that share exactly one affine point and share both infinite points (∞) and (0) of their hyperovals. Then either π is Desarguesian or there is a 2-primitive divisor u such that groups G,0 and G9,0 which fix  and 9 respectively share a Sylow u-subgroup and there is a Desarguesian affine plane 6 with components (x = 0), (y = 0),  and 9 (the affine parts). Proof. We may assume that the group hG,0 , G9,0 i is solvable and if the two groups do not share their unique Sylow u-subgroup, where u is a 2-primitive divisor of 2r − 1 , then the previous arguments show that the plane is generalized Andr´e and hence Desarguesian. Hence, there is a u-element g which fixes both ovals and (x = 0), (y = 0). It follows that there is a Desarguesian plane 6 with components (x = 0), (y = 0) and the affine parts of  and 9. ¨ THEOREM 5.2. Let π be a finite translation plane of order 2r 6 = 26 . If there is a covering of the affine points not on (x = 0) and (y = 0) by 2-transitive parabolic ovals sharing 0, then π is Desarguesian. Proof. If there is a covering of the affine points as indicated, then there is a set of parabolic ovals whose associated hyperovals share the same infinite points and the ovals mutually share exactly one affine point, namely the zero vector. Thus, the Desarguesian affine plane 6 of Proposition 5.1 has components (x = 0), (y = 0) and this set of 2-transitive parabolic ovals. It follows that each component of π other than (x = 0), (y = 0) is a translation oval in 6. Hence, π is a plane which is ovally derived from a Desarguesian affine plane 6. Thus it is a generalized Andr´e plane by Jha-Johnson [12] and hence it is Desarguesian by previous results. ¨ References [1]

Biliotti, M. and Korchmaros, G., Collineation groups strongly irreducible on an oval, Combinatorics ’84, eds. A. Barlotti, M. Biliotti, A. Cossu, G. Korchmaros and G. Tallini, Ann. Discrete Math., Vol. 30 (NorthHolland Math. Stud., no. 123), North-Holland, Amsterdam-New York, 1986, 85–97.

Vol. 70, 2001 [2] [3] [4]

[5] [6] [7] [8] [9] [10] [11]

[12] [13] [14] [15] [16] [17]

Two-transitive parabolic ovals

27

Biliotti, M., Jha, V. and Johnson, N. L., The collineation groups of generalized twisted field planes, Geom. Dedicata 76 (1999), 97–126. Biliotti, M., Jha, V. and Johnson, N. L., Two-transitive ovals in generalized twisted field planes, Arch. Math. (to appear). Cordero, M. and Figueroa, R., Transitive autotopism groups and the generalized twisted field planes, Mostly Finite Geometries, ed. Norman L. Johnson, Lecture Notes in Pure and Appl. Math., vol. 190, Marcel Dekker, New York, 1997, 191–196. Enea, M. and Korchm´aros, G., Ovals in commutative semifield planes, Arch. Math. 69 (1997), 259–264. Foulser, D. A., The flag-transitive collineation groups of the finite Desarguesian affine planes, Canad. J. Math. 16 (1964), 443–472. Foulser, D. A., Collineation groups of generalized Andr´e planes, Canad. J. Math. 21 (1969), 358–369. Ganley, M. J. and Jha, V., On translation planes with a 2-transitive orbit on the line at infinity, Arch. Math. 47 (1986), 379–384. Ganley, M. J., Jha, V. and Johnson, N. L., The translation planes admitting a nonsolvable doubly transitive line-sized orbit, J. Geom. 69 (2000), 88–109. Hiramine, Y., On finite affine planes with a 2-transitive orbit on l∞ , J. Algebra 162 (1993), 392–409. Jha, V. and Johnson, N. L., On the ubiquity of Denniston-type translation ovals in generalized Andr´e planes, Combinatorics ’90, eds. A. Barlotti, A. Bichara, P.V. Ceccherini, and G. Tallini, Ann. Discrete Math., vol. 52, North-Holland, Amsterdam, 1992, 279–296. Jha, V. and Johnson, N. L., A characterisation of spreads ovally-derived from Desarguesian spreads, Combinatorica 14 (1994), 51–61. Jha, V. and Johnson, N. L., Affine transitive planes, J. Combin. Theory Ser. A 83 (1998), 165–168. Johnson, N. L., Translation planes of order q 2 that admit q + 1 elations, Geom. Dedicata 15 (1984), 329–337. L¨uneburg, H., Translation Planes, Springer-Verlag, Berlin–Heidelberg–New York, 1980. Payne, S. E., A complete determination of translation ovoids in finite Desarguesian planes, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 51 (1971), 328–331. Woltermann, M., A note on doubly transitive solvable permutation groups, Comm. Algebra 7 (1979), 1877–1883.

Mauro Biliotti Dipartimento di Matematica Universit`a di Lecce Via Arnesano I-73100 Lecce Italy e-mail: [email protected] Norman L. Johnson Department of Mathematics The University of Iowa Iowa City, Iowa 52242 U.S.A. e-mail: [email protected]

Received 20 July 1998.

Vikram Jha Mathematics Department Caledonian University Cowcaddens Road Glasgow Scotland e-mail: [email protected]