MAURO
HYPEROVALS
BILIOTTI
WITH
AND GABOR
KORCHMAROS
A TRANSITIVE
COLLINEATION
GROUP
ABSTRACT.We investigatethe structure of a collineation group G leaving invariant a hyperoval (n + 2 - arc) ~ of a finite projective plane n of even order n. The main result is that n = 2,4 or 16 when G acts transitively on £~ and 4] IGI. The case n = 16 is investigated in some details. In 1975 M. Hall [5] determined all hyperovals lying in the Desarguesian plane of order 16. There is exactly one class of hyperovals in that plane other than conics with their knots. That class was discovered by L. Lunelli and M. Sce [14] in 1958. As pointed out by M. Hall, each of its hyperovals has the very unusual property of being invariant under the action of a collineation group which operates transitively on its points. Only two other planes with the same property are known: namely, those of order 2 or 4. One might ask whether any finite projective plane n of even order n, containing a hyperoval fl which is left invariant by a collineation group G acting transitively on its points, must be Desarguesian of order 2, 4 or 16. Since n + 2 must divide the order of Aut n, it is not difficult to see that the answer is affirmative when n is Desarguesian, as pointed out by G. K o r c h m a r o s in [12]. V. Abatangelo [1] showed that n has order 2 or 4 when G acts 2-transitively on ft. In this paper we prove that the problem can essentially be solved if we assume that some involution in G fixes some point in ft.
M A I N T H E O R E M . Let n be a finite projective plane of even order, ~ a hyperoval o f n and G a collineation group of n which leaves ~) invariant and acts transitively on its points. I f 46 ] G ] , then n has order 2, 4 or 16. We remark that in a Desarguesian plane of order 2, 4 or 16 the whole collineation group leaving invariant a hyperoval always has order divisible by 4. Some techniques used here may be of independent interest when studying collineation groups which leave a hyperoval invariant. They are collected in Section 1. In Section 3 we determine the structure of a collineation group satisfying the assumptions of the Main Theorem when n is a plane of order 16. Comparing our results with those obtained by Hall [5], Payne and Conklin [15] and K o r c h m a r o s [12] about the Desarguesian case, it seems likely that no other possibility can occur. Geometriae Dedicata 24 (1987), 269-281. © 1987 by D. Reidel Publishing Company.
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.
We shall use standard notation. The reader is referred to [4] and [10] for the necessary background about projective planes and to [11] for information concerning finite groups. The internal structure of the groups SL(2,q), Sz(q), SU(3,q 2) and PSU(3,q 2) with q even is assumed to be known. In particular, see [17] for Sz(q)and [3J for PSU(3, q2 ). Throughout the paper, re denotes a finite projective plane of even order n and ~ a hyperoval of u - i.e. a set of n + 2 points of ze, no three of which are collinear. Each line of ~ intersects fl in 0 or 2 points and is called an external line or a secant of ~ respectively. There are (n + 2)/2 secants and n/2 external lines of fl through each point of g not in ft. The total number of external lines of is n(n - 1)/2. Let ~ be a collineation of z~. We denote by F(~) the structure consisting of all points and lines fixed by ~. If F(~) consists of a non-incident point-line pair (A,a) together with i points A 1 , . . . , A i on a and the lines A A i , 1 ~
2, then Aut(u) has odd order. Since here we are mainly interested in collineation groups of even order, we assume throughout the paper that has order n = 0(4). P R O P O S I T I O N 1.1. I f a is a involutorial Baer collineation of ~ which leaves ff~ invariant, then one of the following holds: (1) ~ c ~ F ( a ) = (7) and aCAltfl. F(tr) contains exactly (n + 2)/2 secants of~. (2) ff~c~F(tr) is a hyperoval of F(g) and ~ ~ Altfl /of n = 0(8), whereas tr¢ Alt ~ / f n - 4(8). Each line of F(g) is a secant of ~. I f tr is an elation of ~ which leaves ~ invariant, then tr is involutorial and one of the following holds:
(3) the axis of a is an external line of ff~ and ~ ¢: Alt ~. (4) the axis of ~ is a secant of ~ and a ~ Alt ft. I f n > 4 then any two distinct elations of zc, both mapping ~ onto itself, have different centres.
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Proof. The proof is straightforward. See also [2], [13]. As in [2, Prop. 2], we can also prove LEMMA 1.2. If ~ and fl are two distinct Baer involutions of ~z mapping onto itself, then F(oO ~ F(fl). The following proposition is essential for our purpose. PROPOSITION 1.3. Let K be an elementary abelian collineation group of order 4 of ~z which leaves ~) invariant. Then K contains an involutorial elation belonging to Alt ft. Proof. The assertion is trivial when n - 4(8), since in this case Alt fl does not contain any Baer involution by Proposition 1.1. Therefore, we may assume that n - 0(8). We divide the proof into two steps. 1. If K ~
+ 1 lines. In case (II) the
common pencil of F(a) and F(z) with centre A must contain (x/~ + 2)/2 secants of ¢2~ and (x/~ + 2)/2 secants of fl,. Again, these lines must be pairwise distinct and we have a contradiction, because the common pencil of F(a) and F(z) contains just x/~ + 1 lines. If fl~ n ~, ~ O then the assertion follows from [2, Prop. 3]. 2. If K - ~ Alt r , then K ~ Alt ~) contains an involutorial elation. Let K = {1, a, r, ~fl }, with a, ~fl ¢ K n Alt Y~, and assume that fl ~ K n Alt~ is a Baer involution. If a is an elation, then its axis lies in F(fl) and hence it is a secant of fl by Proposition 1.!. This is a contradiction, since a e A l t ~ . Therefore, we may assume that a is a Baer involution. Denote by ~ and the collineations induced by ~ in F(fl) and by fl in F(~) respectively. First,
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suppose that both ~ and fl are elations. Their common axis r is a secant of since it lies in F(fl). Moreover, r c~ F(~) = r c~ F(fl) is disjoint from ~, since F(~) c~ fl = (D by Proposition 1.1. It follows that both ~ and fl interchange the two common points of r and f2 which are thus fixed by ~fl; but this is again a contradiction, because ~fl¢Alt f~. Therefore, both ~ and fl are Baer involutions and fix F(~)r~F(fl) pointwise. Clearly, F(~)c~ F(fl) is a Baer subplane of both F(~) and F(fl). Since F(~) is disjoint from ~, F(~)c~ F(fl) is disjoint from f2p = F(fl)c~ f~ and so there exists a line s of F(~)c~ F(fl) which is external to fl~ in F(fl) (see Proposition 1.1). Since fleAltfl, the line s intersects fl-fl~ in two points which are interchanged by both ~ and ft. It follows that ~fl fixes some point in D~ which is a contradiction. This completes the proof. Moreover, we have L E M M A 1.4. Let ~ be a collineation of ~z mapping f2 onto itself such that ~2 is a Baer involution (and hence ~4 = 1). Then we have: (1) ~)~2= F(~ 2) c~ ~ is a hyperoval ofF(~ 2) (2) If ~ denotes the collineation induced by ~ on F(~ 2), then F(~) is a Baer subplane of F(~ 2) and F(~t)c~~ 2 is a hyperoval of F(~).
Proof. (1) The assertion follows from Proposition 1.1, since ~2e Alt f2. (2) If s is a line of F(~), then s must be a secant of f2 because it lies in F(~t2). It follows that s cannot be an external line of ~ 2 since otherwise ~2 would fix two points of ~ - ~ 2 and hence it would be the identity, in contradiction to our assumptions. The result can now be obtained by using Proposition 1.1. Using Proposition 1.3 and Lemma 1.4 we may argue as in [2, Prop. 5], to show the following P R O P O S I T I O N 1.5. Let E be a 2-group of collineations of ~ which leaves invariant and does not contain any involutorial elation. Then E is a cyclic group. In addition, we have P R O P O S I T I O N 1.6. Let E be a cyclic 2-group of collineations of 7t which leaves ~ invariant and does not contain any involutorial elation. If E c~ Alt f~ 4 (1), then E is planar and F(E) c~ ~2 is a hyperoval of F(E). Proof. By induction on the order of E and using Lemma 1.4. We now shall obtain some information about p-groups that have no fixed points on f~
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L E M M A 1.7. Assume that P is a collineation group of rt satisfying the following conditions: (1) P is a p-group with p odd. (2) P leaves ~ invariant and acts fp.f. on its points. Then one of the following occurs: (a) p # 3 and F(P) is a subplane. (b) p = 3 and either F(P) = (Q, ~ ), or F(P) is of type Di with i >~2, or F(P) is a subplane. Moreover, each line of F(P) is external to ~.
Proof First note that if a line l of F(P) is a secant of f~, then P fixes I n f~ pointwise, in contradiction to its f.p.f, action of f~. By condition (2) we have that p [ n + 2. I f p ~ 3, then p divides neither n(n - 1)/2 nor [n(n - 1)/2] - 1 and hence P fixes at least two external fines of f~. Moreover, p~n + 1,n, n - 1, so that P has at least three fixed points on each fixed line. F r o m this we infer that F(P) is a subplane. N o w assume p = 3 and F(P) # ((7), Q). If P fixes a line, then it fixes some point on it; so we m a y suppose that P fixes a point Q. Since p Xn + 1 and p Xn there are at least two distinct lines through Q which are fixed by P. Likewise, there exist at least two fixed points on each fixed line. N o w the assertion follows easily. L E M M A 1.8. Let P be a collineation group of ~ satisfying conditions (1) and (2) of Lemma 1.7. In the case p ~ 3, no involutorial collineation of rc exists which normalizes P, leaves f~ invariant and acts on fl as an even permutation. In the case p = 3 involutorial collineations of that kind do not exist when P is planar, and Baer involutions of that kind do not exist when F(P) v~ ( ~ , 0). Proof Assume that there exists a Baer involution ~ which leaves both f~ and F(P) invariant and acts as an even permutation on f~. If p # 3 or p = 3 and F(P) ~ (Q, Q), then, using L e m m a 1.7, it is easy to see that cr fixes some line l of F(P). By Proposition 1.1, I must be a secant of fL and that is in contrast with L e m m a 1.7. N o w let r/ be an involutorial elation with the same properties as a. If F(P) is a subplane, then it contains the axis of q. By Proposition 1.1, this axis is a secant off~ and again we have a contradiction.
. We recall that only three classes o f h y p e r o v a l s are k n o w n which are left invariant by a collineation group acting transitively on their points; namely, Lunelli--Sce hyperovals in the Desarguesian plane of order 16 and conics with their k n o t s in the planes of order 2 or 4. F o r further details see [9]. T h r o u g h o u t this section we assume that f~ is a hyperoval of a projective
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plane r~ of order n - 0(4), where n > 4, and G a collineation group of rt satisfying the following conditions: (1) G leaves fl invariant and acts transitively on its points, (ii) 4 11al. P R O P O S I T I O N 2.1. G contains involutorial elations whose axes are secants
off~. Proof Let E be a Sylow 2-subgroup of G. If E contains two commuting involutions, then the assertion follows from Proposition 1.3. Otherwise, E contains exactly one involution a and aeAltfL since 41 [El. If a is a Baer involution, then, by Propositions 1.5 and 1.6, E fixes some point on ~ and hence G cannot be transitive on ~, since I~L is even. Therefore a is an involutorial elation. We shall denote by S the subgroup of G generated by all elations whose axes are secants of f~. As usual O(S) denotes the maximal normal subgroup of odd order of S. We notice that two distinct elations of G must have different centres because of the assumption n > 4. P R O P O S I T I O N 2.2. S satisfies the followino conditions: (1) I f l is the axis of an elation in S, then St does not contain an elation with axis different from I. (2) There is no centre or axis of elations in S which is fixed by S. Proof (1) Assume that Sl contains an (M,m)-elation with m ¢ l and M e I. Since l is the axis of some elation in S, then by (10, Lemma 4.15], S must also contain an (M, /)-elation; but this is a contradicti'on, since there would be two distinct elations with the same centre. (2) Let a be an (L,l)-elation of S. Suppose that S fixes L. If P ~ f ~ - ( f ~ l } , then, by the transitivity of G on f~, S must contain an elation with axis PL. This implies the existence of ar/(L, PL)-elation, and, again, we have a contradiction. Now suppose that S fixes 1. Clearly S must contain another elation whose axis is different from I and whose centre belongs to l; but, as we have seen, this is impossible. Now we are able to prove the following proposition: P R O P O S I T I O N 2.3. The order of S is not divisible by 4. To prove this we need a group-theoretical lemma. L E M M A 2.4. Assume G ~ SL(2,q), Sz(q), SU(3,q 2) or PSU(3,q2), where q = 2 h, h > 1, and let E be a Sylow 2-suboroup of G. Then NG(E ) does not contain any subgroup of index 2.
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Proof. If G ~ SL(2,q), then NG(E ) is a Frobenius group of order q(q - 1), with Frobenius kernel of order q, and hence it cannot contain a subgroup of index 2. Likewise, the assertion holds when G-~ Sz(q), since in this case NG(E) is a Frobenius group of order q2(q _ 1), with Frobenius kernel of order q2. Assume G - PSU(3,q 2) and let M < NG(E) with [NG(E):M] = 2. We have that Z(E) < M c~ E, since each element of Z(E) is the square of some element in E. Therefore M/Z(E) must be a Frobenius group of order (q2 _ 1)q2/2d, d = (3, q + 1), with Frobenius kernel of order q2/2, according to the structure of N6(E)/Z(E) (see [3, Prop 1.2(d)]). Hence ( q 2 _ 1)/d divides q2/2 - 1, which implies q = 2 in contradiction to our assumptions. A similar argument may be used when G -~ SU(3,q2).
Proof of Proposition 2.3. By Proposition 2.2, S satisfies the hypotheses of [6, Th. 5.1], and hence one of the following holds: (a) 4 X ISI, (b) S = SL(2,q), Sz(q), PSU(3,q 2) or SU(3,q2), where q = 2h, h > 1. Suppose that case (b) occurs. Since each point of f2 belongs to the axis of some elation in S, the centre Z(S) fixes f~ pointwise and hence Z(S) = (1). So we cannot have S ~ SU(3,q2), with 3 I q + 1. Note that in our situation two elations in S have the same axis if and only if they commute. Therefore, two elations in S have the same axis if and only if they lie in the centre of the same Sylow 2-subgroup of S. Let E be a Sylow 2-subgroup of S and let l be the common axis of the elations lying in Z(E). Clearly Ns(E) fixes I. Moreover, if l ~ f~ = {A,B}, then Ns(E ) fixes both A and B since, by Lemma 2.4, it does not contain any subgroup of index 2. Since Ns(E) is a maximal subgroup of S and S acts f.p.f, on f~ because of the transitivity of G on fL we have that there is exactly one elation axis through each point of f~ and S splits f~ into two orbits of length q + 1, q2 + 1 o r q3 ÷ 1 according to whether S --- SL(2,q), Sz(q) or PSU(3,q2). Moreover, S acts on both orbits in its usual doubly transitive representation. Now we shall prove that if S ~ SL(2,8), then S contains a subgroup C which satisfies the following conditions: (1) C is normalized by some involution in S; (2) C acts f.p.f, on f~; (3) the order of C is an odd prime different from 3. Then by Lemma 1.8 we have a contradiction. It is well known that S contains cyclic subgroups of order m = q + 1, q +__x/2q + 1 or q2 _ q + 1 according as S -~ SL(2, q), Sz(q) or PSU(3, q2). Each of these cyclic subgroups acts semiregularly on f~ and its normalizer in S has even order. Therefore, a subgroup C satisfying (1), (2) and (3) exists whenever there exists a prime divisor of m, different from 3. This is obvious for Sz(q) since (q + x/2q + 1, q - ~/2q + 1)= 1. If m = q + 1, then m = 3simplies q = 8 for q > 2. Let m = q2 _ q + 1 and sup-
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pose that 3lm. Then q ~ 2(3) and hence m = 3(9). So, the required prime divisor of m exists provided that q2 _ q + 1 > 3 and this occurs for q > 2. It remains to exclude that S ~ SL(2,8). In such a case, by [7,Th. 2.8], the order n of n must be of the form 8(7c + 1) for some integer c >/0. But n = 16, since S splits f2 into two orbits of length 9, and we have again a contradiction. The proof is complete. We shall say that G is minimal if G contains no proper subgroup satisfying both conditions (i) and (ii). P R O P O S I T I O N 2.5. S splits ~ into two orbits of length (n + 2)/2. If G is
minimal, then G = O(S) ), E, where E is a Sylow 2-subgroup of G and O(S) is a 3-group. Proof. Let a be an involutorial elation in S with axis l and let l ~ f ~ = {A1,A2}. Moreover, denote by O1 and 02 the orbits of O(S) on f~ containing A1 and A2, respectively. By Proposition 2.3 we have that 4 X ISI and hence any other elation in S is conjugated to a by an element of O(S). So its axis intersects f~ in two points belonging to O1 w 02. Then the transitivity of G on f~ implies that O1 w 02 = ft. Clearly an elation of S fixes both O1 and Oz whereas a Sylow 2-subgroup E of G interchanges O1 and 02. Now assume that G is minimal. Clearly G = O(S) )x E. Let P be a Sylow p-subgroup of O(S) with p :~ 3. By the Frattini argument N~(P) contains a Sylow 2-subgroup of G and hence P must fix some point on fL by Lemma 1.8. It follows that the stabilizer of a point of I2 in O(S) contains a Hall 3'subgroup of O(S). Therefore, if F is a Sylow 3-subgroup of O(S) which is normalized by E, then F k E satisfies both conditions (i) and (ii). By the minimality of G we must have O(S) = F and the proof is complete. Now we can prove P R O P O S I T I O N 2.6. S does not fix any point or line of n. Proof By Proposition 2.5 we may assume that G = O(S) ), E, where E is a Sylow 2-subgroup of G and O(S) is a 3-group. First, suppose that S fixes a point A of n. Clearly the axes of the elations in S are exactly the secants of Q through A. Each of these secants is the axis of exactly one elation and there is not any elation with centre A. By Lemmas 1.7 and 1.8, F(O(S)) is of type Di with i >~ 2 since A is fixed by O(S). A line of F(O(S)) through A is external to f~ and hence it cannot be fixed by any involutorial elation of S. Thus each line of F(O(S)) through A must contain exactly one other point of F(O(S)). so the points Ax . . . . . Ai of F(O(S)) lie on a line a, which is
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not incident to A and contains all the centres of involutions in S. Without any loss of generality we may assume that A~ = A2 for some involution tr ~ S. Then all involutions in S interchange A1 and A2. Let P be a point of AA1 with A # P ~ A~. It is very easy to see that the images of P under all involutorial elations in S consist of (n + 2)/2 distinct points of AA2. N o w let t/eZ(O(S)), t/4: 1, and assume P" = P. If a is any involutorial elation of S, then t/~= r/" since tleZ(O(S)) and hence (P~)"~= ( p ~ ) e = p , ~ = p~. So t/" fixes at least (n + 2)/2 distinct points of AA2 and hence it fixes at least (n + 2)/2 distinct lines through A~. Since at least one of these lines must be a secant of fl, we have that r/" fixes some point on ft. But then r/" fixes at least (n + 2)/2 points of ~ because tI'eZ(O(S)), which is impossible. Therefore, each element of Z(O(S))acts f.p.f, on AAI-{A, A1} and hence IZ(O(S))l [ n + 1. By Proposition 2.5, n + 2 = If~l = 2"3h, with h > 1. Then 3 I n - l , but 9 X n - 1 . So Iz(o(s))l = 3 and O(S) and Z(O(S)) have a c o m m o n orbit of length 3 on AAI-{A, A1}. Since O ( S ) c a n n o t split over Z(O(S)), this implies O(S) = Z(O(S)), which is a contradiction for n > 4. If S fixes a line, then F(O(S)) is again of type Di, with i >~2, and S leaves F(O(S)) invariant. N o w it is very easy to see that S must fix some point of F(O(S)). This completes the proof. P R O P O S I T I O N 2.7. If S leaves invariant some triangle of re, then n = 16. Proof. Again, we may assume that G = O(S) )x E, where E is a Sylow 2subgroup of G and S is a 3-group. Let A = {A1,A2,A3} be the triangle which is left invariant by S. By Proposition 2.6, we have only to consider the case that F does not fix any element of A. In such a situation, the axis of any elation in S must be incident to a vertex A~ of A and the centre lies on the opposite side AjAk. Moreover, by taking into account that S splits f~ into two orbits of length (n + 2)/2, it is easy to see that the sides of A are external to fL N o w let P be any point of f~ and assume that there exist two distinct involutorial elations al,tr2 in S whose axes pass through P; then ( a l , a 2 > is a dihedral group which must contain at least another involutorial elation with axis through P. So, it is easily seen that the number of elation axes through P is either 1 or 3. Since G is transitive on fl, it follows that there are exactly (n + 2)/6 or (n + 2)/'2 elation axes through each vertex Ai of A according to whether the number of elation axes through P is 1 or 3. Assume that there are (n + 2)/2 axes through A1 and let 1 be a line through A~ which is external to f~ and distinct from A~Az and AIA3. Then there must be a point Q e l, Q ~ A1, such that QA2 and QA3 are axes of two elations a l , a2~S. This is impossible because of the existence of a third elation in ( a l , tr2 > whose axis would be QA1. Therefore,
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there are exactly (n + 2)/6 elation axes through each vertex of A. Now let O0 be the subgroup of O(S) fixing A pointwise. Since [O(S): O0] = 3 and n > 4, then Oo 4:(1 ). First, assume that we have t/~ = t/- 1 for each ~/eZ(Oo), with o(~/) = 3, and for each involution tre S. In this case each element of order 3 of Z(Oo) lies in Z(O(S)), since an element of O(S) is the product of an even number of involutions in S. Let ~/EZ(Oo), with o(~/)= 3, and suppose that t/ fixes some point P on the line A1A2 with A1 4= P ¢ A2. Since t/belongs to Z(O(S)) and O(S) permutes the sides of A, we have that (t/) is planar. By our assumptions (t/) is normalized by all involutions in S and hence it fixes some point on f~ by Lemma 1.8. But then t/, being in Z(O(S)), fixes at least (n + 2)/2 points on f~, and this is impossible. Therefore, Z(Oo) acts semiregularly on A 1 A 2 - { A I , A 2 } and hence [/(Oo)lln-1. As in the proof of Proposition 2.6, we have that IO01 = IZ(Oo)l = 3 and hence [O(S)l = 9 and n = 16. Now assume that there exists an element t/eZ(Oo) with o(t/) = 3, and an involution a e S such that t/~¢ r/-~. Then either t/~ = t/ or (r/~) 4= (~/). If (t/~) ~ (7), then (t/,t/~) is an elementary abelian group of order 9 and centralizes a non-identical element of (t/,t/~). Therefore Z(Oo) contains an element v which centralizes a. But then v fixes the axis of a and hence it fixes two points of fl lying in two distinct orbits of O0 on f~. Since each orbit of Oo on f~ has length at least (n + 2)/6, it follows that v fixes at least (n + 2)/3 points of f] and hence F(v) is a subplane of order m >1 (n + 2)/3 - 2. This implies n = 16 by Bruck's theorem. According to Hering [8], a collineation group G of a projective plane P is called irreducible over P when G does not fix any point, line or triangle of P, and strongly irreducible when G is irreducible and does not leave invariant any proper subplane of P. Now we can prove P R O P O S I T I O N 2.8. S cannot be irreducible over ~z. The following lemma, which is needed in the proof, is probably well known, but no reference is available to us. So we sketch the proof. L E M M A 2.9. Let F be a collineation group of a projective plane P. Assume that (1) F is a Frobenius group of order 18, (2) F does not fix any point or line, (3) the involutions in F are elations. Then S acts faithfully on an F-invariant subplane Po of order 4 of P and centralizes a unitary polarity of Po. Proof. F contains exactly nine involutorial elations ei, 1--
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Ai and Aj, with A~ ~ Aj. Since F does not fix any line, there is exactly one other involutorial elation in F whose centre belongs to a,~ - precisely the one lying on ( ~ , a j ) . Likewisel denote by Ai~ the common point of the axes ai and a t. It is straightforward to show that the points A~ and A~, together with the lines a~ and a~j, form a subplane Po of order 4 of P. The unitary polarity of Po which is centralized by F is defined as follows: A i ~ ai,
A i j ~ , aij.
Referring to Lemma 2.9, let U be the substructure of Po consisting of absolute points and non-absolute lines of the unitary polarity centralized by F. It is easily seen that U is an affine plane of order 3 and F acts on U in the following way: (1) The centres of the elations in F are exactly the points of U, whereas the axes are the tangents at U; O(F) acts on U as the translation group of U and each involution in S induces a dilatation on U. (2) F splits the points of Po, not on U, into four triangular orbits.
Proof of Proposition 2.8. Suppose that S is irreducible over n. By [8, Lemmas 3.3 and 3.5], the centres and axes of the elations in S generate a subplane no on which S induces a strongly irreducible collineation group. It is very easy to see that S acts faithfully on no. Since S is solvable, we have by [8, Th. 5.5] that F(S)= O(S) is an elementary abelian group of order 9 and S is a Frobenius group of order 18 (see the proof of Theorem 5.5 of [8] for the last assertion). Since S does not fix any point or line of n, we can apply Lemma 2.9. Thus S contains triangular orbits, and that is a contradiction. Combing the results of Propositions 2.1, 2.6 and 2.8, we obtain the proof of the Main Theorem.
.
In this section we investigate the structure of G when n has order 16. P R O P O S I T I O N 3.1. Assume that n has order 16. Then S is a Frobenius group of order 18 acting on n as described in Lemma 2.9. The Sinvariant subplane no is disjoint from fL
Proof By what we have seen in Section 2, we may suppose that S leaves a triangle A = {AI, A2, A3 } invariant and does not fix any point or line of n. Moreover, as we have shown in the proof of Proposition 2.7, each vertex of A lies on exactly three axes of elations in S. Let Oo be the subgroup of O(S) fixing A pointwise. If 1Ool > 3, then there exists an element t/zOo, t t ~ 1, and a vertex Ai of A such that q fixes each of the three axes through Ai.
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Hence r/fixes six points of ~. Thus F(q) is a Baer subplane of zc and no line of F(~/) is external to ~. In particular, each side of A is a secant of f2, and that is impossible. Therefore, ]Ool = 3, LO(S)i = 9 and S is a Frobenius group of order 18 since it contains exactly nine involutions. Thus Lemma 2.9 can be applied. Let no be the S-invariant subplane of order 4 of n. A line l of no containing the centres of two distinct elations al,a2 ~ S cannot intersect ~, since otherwise the cyclic stem of ( a l , a2) would fix some point on fl, in contradiction to the semiregular action of O(S) on ft. Since each point of no lies on such a line, no and f2 must be disjoint. Clearly the group G must leave invariant the subplane no. We can prove: 1. The subgroup K of G fixing no pointwise has order at most 2. Moreover, K ~ Alt ~. An element of odd order of K fixes ~ pointwise, since it fixes all axes of elations in S, and hence it is the identity. Therefore, by Proposition 1.5, either K = ( 1 ) or K is a cyclic 2-group. In the latter case, IKI = 2 by the same argument as before. Since no is disjoint from fl, K ,g Alt f2 by Proposition 1.1. 2. One of the following holds: (2a) K = ( 1 ) and G = O(S) )~E, where E is some subgroup of a Sylow 2subgroup of PFU(3, 4) with IE I >~4; (2b) SKi = 2. I f Go = G ~ A l t ~ ) , then G = Go x K and Go = O(S) ),E, where E is either a cyclic group or order 2, 4 or 8 or a quaternion group.
(2a) Let F be a Sylow 3-subgroup of G containing O(S). F fixes at least one of the four triangular orbits of S on no. If F ¢ O(S), we obtain a contradiction by arguing as in the proof of Proposition 3.1. Therefore, F = O(S). Moreover, since G acts faithfully on no it is isomorphic to a subgroup of PFU(3,4). Then it is easily seen that G = O(S) ), E, where E is isomorphic to a subgroup of a Sylow 2-subgroup D of PFU(3, 4). (2b) As in (2a) we have that Go = O(S) )~ E, where E is isomorphic to a subgroup of D. Let a be an involution of E and suppose that a ¢ S. Then is a Baer involution and each line of F(a) is a secant of if2 since E < An. Clearly ~ acts on no as a Baer involution and, as is well known by [16], it fixes some line of U. This is a contradiction, since each line of U is external to ~. Therefore, a s S and hence E contains a unique involution. Since D has order 16 and contains more than one involution, we have the assertion. As we have previously remarked, the Desarguesian plane of order 16 contains exactly one class of hyperovals with a transitive collineation group. If ~ is a hyperoval of this class, then [K] = 2 and Go = O(S) ), E, where E is cyclic of order 8 (see [12], [15]),
HYPEROVALS WITH A TRANSITIVE COLLINEATION G R O U P
281
REFERENCES 1. Abatangelo, V., 'Doubly Transitive (n+2)-arcs in Projective Planes of Even Order n', J. Comb. Th., Ser. A.42 (1986), 1-8. 2. Biliotti, M. and Korchmaros, G., 'Collineation Groups Strongly Irreducible on an Oval', Ann. Discr. Math. 30 (1986), 85-98. 3. Burkhardt, R., 'Uber die Zerlegunszahlen der unitiiren Gruppen PSU(3, 22f ), J. Algebra 61 (1979), 548-581. 4. Dembowski, P., Finite Geometries, Springer-Verlag, Berlin, Heidelberg, New York, 1968. 5. Hall, M., Jr, 'Ovals in the Desarguesian Plane of Order 16', Ann. Mat. Pura Appl. 102 (1975), 159-176. 6. Hering, C., °On Involutorial Elations of Projective Planes', Math. Z. 132 (1973), 91-97. 7. Hering, C., 'On Projective Planes of Type VI', Colloq. Int. Teorie Comb., Roma 1973; Atti Conv. Lincei 17, Tomo II (1976), 29-53. 8. Hering, C., 'On the Structure of Finite Collineation Groups of Projective Planes', Abh. Math. Sere. Hamburg 49 (1979), 155-182. 9. Hirschfeld, J. W. P., Projective Geometries over Finite Fields, Clarendon Press, Oxford, 1979. 10. Hughes, D. and Piper, F., Projective Planes, Springer-Vertag, Berlin, Heidelberg, New York, 1973. 11. Huppert, B., Endliche Gruppen I, Springer-Verlag, Berlin, Heidelberg, New York, 1979. 12. Korchmaros, G., ~Gruppi di collineazioni transitivi sui punti di una ovale [(q + 2)-arco] di S2.q, q pari', Atti Sere. Mat. Fis. Univ. Modena 27 (1978), 89-105. 13. Korchmaros, G., 'Una proprietS~ gruppale delle involuzioni planari che mutano in s+ un'ovale di un piano proiettivo finito', Ann. Mat. Pura Appl. 116 (1978), 189-206. 14. Lunelli, L. and Sce, M., 'k-Archi completi dei piani desarguesiani di rango 8 e 16, Pubbl. Centro Calc. Num., Politecnico di Milano, 1958. 15. Payne, S. E. and Conklin, J. E., 'An Unusual Generalized Quadrangle of Order Sixteen', J. Comb. Th., Ser. A, 24 (1978), 50-74. 16. Seib, M., 'Unitiire Polaritiiten endlicher projektiver Ebenen', Arch. Math. 21 (1970), 103112. 17. Suzuki, M., 'On a Class of Doubly Transitive Groups', Ann. Math. 75 (1962), 105 145.
Authors' addresses:
Mauro Biliotti, Dipartimento di Matematica, Universit~ di Lecce, Via Arnesano, 73100 Lecce, Italy Gabor Korchmaros, Istituto di Matematica, Universit~ degli Studi della Basilicata, Via N. Sauro, 85, 85100 Potenza, Italy (Received, June 12, 1986; revised version, December 2, 1986)