2-transitive abstract ovals of odd order

Aequatlones Mathematlcae 33 (1987) 208-219 Umverstty of Waterloo

0001 9054/87/002208-125150 + 0 20/0 © 1987 Blrkhauser Verlag, Basel

2-transitive abstract ovals of odd order

G. KORCHMAROS

I. Introduction Following Buekenhout [3,1, [4,1, an abstract oval (called also free oval, or Buekenhout oval, or for brevity B-oval) B = (M, ~ ) is defined as a set M of elements, called points, together with a sharply quasi 2-transmve set ~ of mvolutonal permutations of M, called mvoluttons. Here sharply quast 2-transitivity means that for any two pairs of points (a t, a2), (b~, b 2) with a, :# b~ 0,J = 1, 2) there exists a unique element f ~ ~- such that f ( a O = a2, f ( b t ) = b2. In this paper we shall be concerned with fintte abstract ovals We define the order n of a finite abstract oval by n = IMI- 1 The classwal abstract oval arises from the linear group PGL(2, q), q = p" and p prime, regarded m its 3-transmve representation over GF(q)" M = GF(q) w {~}, and ~ consists of all elements of order 2 of PGL(2, q) with the addmon, for p = 2, ~ts identity element It ~s interesting that there is a natural way of denwng abstract ovals from projective ovals, the classical one arises from the irreducible comc of PG(2, q). There are known, however, abstract ovals not obtainable m such a way. We do not discuss these here, the reader is referred to [5], [6-1, [7-1, [20,1. An automorphlsm g of an abstract oval is a permutation of M which reduces a map of ~ onto itself, I e. g ( f ) = gfg-t ~ ~ for e a c h f E ~ . The full automorphlsm group of the classmal abstract oval is PFL(2, q), and this property characterizes it. A weaker result is given m this paper. Our Theorem 1 shows that the classmal abstract oval is characterized as the only abstract oval w~th an automorphlsm group acting on M as PSL(2, q) m its usual 2-transitwe representahon Notice that for p = 2 this characterization was given by Buekenhout [3]. AMS (1980) subject classification Primary 51T20, 51T99 Secondary 12K05,20B10

Manuscrtpt recewedNovember 14 1985

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An abstract oval ts called 2-transmve ff tt has an automorphism group actmg 2transittvely on M. An involution f e ~ is called regular tf tt is an automorphism. Clearly, all involutions of the classical abstract oval are regular The study of regular involutions and automorphism groups generated by them is a powerful tool in the study of abstract ovals Buekenhout's paper [3] has been the basis for regular revolutions His stimulating and penetrating mvesttgations were followed by numerous authors. Details cannot be given here, the reader should refer to [7], [16] We mention only Theorem 8 8 of [3] which states that, if all involutions of an abstract oval of even order are regular, then it ts the classtcal one Thts result was extended to any odd order in [15] In a recent paper [8] we were concerned with regular involutions of 2-transitive abstract ovals of even order We have an almost complete classification in which only a few gaps remain It is in the case of even order that the picture is very rich and group theory is especmlly powerful For the case of odd order, the classtcal ones are the only known 2-transitive abstract ovals If no other example extsts, then all Involutions of any 2-transitive abstract oval of odd order are regular The author beheves that the classtcal one is the untque 2-transitive abstract oval of odd order Our Theorem 2 states this for n = 3 (mod 4) The case n = 1 (mod 4) is more complicated in our context because lnvolutorial automorphisms with x/~ + 1 fixed pomts can be involved In an effort to investigate this case, we give a proof under the assumption that the abstract oval admits some regular involutions, see Theorem 3. Moreover, a proof without any additional hypothesis is given by using the classification of the finite simple groups

2. Notation and preliminary results Fairly standard notation ts used A certam famtharlty with abstract ovals as well as with finite groups is assumed. The reader should refer to [3], [7], [13], [16], [18]. Throughout this paper, B = (M, ~ ) denotes an abstract oval of odd order n We shall assume the following elementary results on finite abstract ovals; for the proofs see [3]. consists ofn z involutions They are permutations of order two Each involution has either 2 or 0 fixed points, and it ts called a hyperbolic or elliptic mvolution, respectively. Let E denote the set of all elhptlc involutions Then Igl = n(n - 1)/2 For any two &stlnct points a, b of M, let E(a, b) denote the set of all eUlptlc Involutions which exchange a with b. Then IE(a, b)l = (n - 1)/2. For hyperbolic revolutions, let H and H(a,b) be defined similarly Then IHI = n(n + 1)/2,

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IH(a, b)l = (n - 1)/2. Moreover, for any point a of M, let H(a) denote the set of all revolutions fixing a. Then In(a)l = n. For any two distinct points a, b of M, I n ( a ) n n(b)l = 1. For any automorphlsm 0, let # denote the m a p f---) gfo-L Then # maps ~- onto ~tself. So, any automorphism group G mduces a permutation group (7 on ~ . G and (7 are ~somorphic. (7 ~s not transmve, as it maps E and H onto ~tself. Assume that g has order two, and denote by M ( g ) the set of all fixed points of # Then IM(g)l = 0, 2 or x/n + 1. If IM(g)l -- 2, then # is a regular revolution. This result is not true in general for IM(g)l = 0 If IM(o)I = ~ + 1, then ~(f) = f f e ~-, lmphes f ~ H Gwen a finite set M, let B, = (M, ~ , ) (t = 1, 2) be abstract ovals of odd order n such that for a n y f a ~ Ha,f2 ~ E2, the e q u a t l o n f l ( x ) = f2(x), x ~ M, has at most two solutmns Put ~ = H~ u E 2. Then also B = (M, ~-) is an abstract oval Moreover, ff there exists a permutation group G of M which is an automorphlsm group of both B, = (M, ~-,) such that (7 acts transitively on Ha, then the above can be weakened as follows" There exists a f l e Ha such that for any f2 e E2 the equation f l ( x ) = f2(x), x ~ M, has at most two solutions.

3. A characterization of the classical abstract oval Our aim is to prove the following THEOREM 1. L e t B = (M, :~r) denote an abstract oval o f order q = pr wtth odd prime p I f B = (M, ~r) admtts an automorphtsm group G actm9 on M as P S L ( 2 , q) m tts usual 2-transltwe representatton, then B = (M, :~r) ts the classwal abstract oval Under our hypothesis M m a y be identified with GF(q) w { ~ } m such a way that G coincides with PSL(2, q). Let B' = (M, ~ - ' ) denote the classical abstract oval over GF(q). We prove Theorem 1 by proving ~- = ~ ' . Case q = 1 (mod 4) Any a u t o m o r p h l s m g ~ G of order two has two fixed points Therefore g e H. Take any f e E, and let (7y denote the stabihzer o f f In (7. Then IGzl > q + 1 by IGI = IGI = q(q2 _ 1)/2. For any ~ e (7y the corresponding g ~ G is either without fixed points, or it has an even number of fixed points. Since each element of order p of G has a unique fixed pomt, p X 1(7sl follows. Let K = {1, gl, 02, 93} be any Klein subgroup of G Each g, has two fixed points, let a,~ (j = 1, 2) denote them. Then a,j ~ a,v when (t,j) ~ (u, v), 1 ~< t, u ~< 3, and 1 ~<j, v <~ 2, because the stablhzer of a point in G contains no commuting elements of order 2. Moreover. a~(a, .) = a, ~. a J a , . ) = a,~ for 1 ~< i. i ~< 2, t ¢ I. A s s u m e / ~

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f i x e s f T h e n f ( a x j ) = alk, f(a2j) = a2k with j, k as a b o v e T h e r e f o r e f = 93, which is impossible in o u r case a s f ~ E by hypothesis F r o m this fact we can infer t h a t no s u b g r o u p of G 1 is i s o m o r p h i c to $4, A4, o r As. By D l c k s o n ' s t h e o r e m ([13] 8.24 F a l l 2) it follows that t71 is a d i h e d r a l g r o u p of o r d e r q+l. Let S d e n o t e the dihedral s u b g r o u p of o r d e r q + 1 of G for which ~ = (7I . Since G = P S L ( 2 , q), there exists an lnvolutorlal p e r m u t a t m n s ~ P G L ( 2 , q) - P S L ( 2 , q) which centrahzes S. N o t e that also f centrahzes S Let k be a n y p e r m u t a t i o n of o r d e r 2 of M w i t h o u t fixed p o i n t which centrahzes S F o r a n y p o i n t a ~ M there is exactly one lnvolutorlal p e r m u t a t i o n d ~ S fixing a. Let b d e n o t e the s e c o n d fixed p o i n t of d. Then k d = dk implies k(a) = b T h u s k is uniquely d e t e r m i n e d This means that s = k = f Therefore f ~ P G L ( 2 , q). T h u s the t h e o r e m is p r o v e d for q - 1 (rood 4) Case q = 3 ( m o d 4) Step 1 Let ~* = E' w H

Then also B = ( M , ~ * )

is an abstract oval

By the final p r o p o s i t i o n of Section 2, it suffices to exhibit a n l n v o l u t ~ o n f ~ H such that for any g ~ E ' the e q u a t i o n f ( x ) = g(x), x ~ M, has at m o s t two solutions Let f be the h y p e r b o l i c involution of B = (M, ~ - ) fixing 0 a n d ~ T h e n there exists a p o l y n o m i a l P ( x ) ~ G F ( q ) [ x ] such that f ( x ) = P ( x ) for x ~ GF(q) Go,~ consists o f all p e r m u t a t i o n s a(x) = a x with a ~ [ ] where [ ] = {x2[x GF(q) - {0}} Since 6 ( f ) = f , P ( a x ) = a P ( x ) holds identically for any a ~ I-]. It is easy to see t h a t this lmphes P ( x ) = s x (q÷x)/2 + u x with s 2 - u 2 = 1 a n d t = s + u ~ A , w h e r e / ~ = GF(q) -- {[3 w {0}} This m e a n s

P(x)

ff'tx t-lx

for x e [ ] for x e A

(1)

W e are able to p r o v e that for a n y O e E ' the e q u a t l o n f ( x ) = 9(x), x e M, has at m o s t two solutions. N o t e that such a p o i n t x belongs necessarily to GF(q) - {0}. M o r e over, if there exist m o r e than two solutions, then 0 ( f ) = f h o l d s # ( f ) = f Implies g(0) = ~ , hence 9 ( x ) = d / x with d ~ / ~ F r o m this we can refer that if the e q u a t i o n f ( x ) = 9(x), x ~ M , has m o r e t h a n two solutions, then the s a m e x are s o l u t i o n s of one of the following equations: tx = d/x, x ~ tq, t - l x = d/x, x ~ A . Since b o t h of these have at m o s t one solutaon, o u r assertion IS p r o v e d Step2

H=

H'.

F o r any two dtstmct a, b ~ GF(q), let g(a,b) ~ G denote the p e r m u t a t i o n

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Clearly g(a'b)(O) = b, g(a,b)(~) = a. Hence H consists of all hyperbohc revolutions of the form

F o r a n y u, v ~ G F ( q ) -- {0}, such that 1 -

of B = ( M , ~ - * )

( u / v ) ~ l-q, let h' denote the i n v o l u t i o n for which h'(O) = u, a n d h ' ( ~ ) = v T h e n h ' ¢ E ' Hence h e o ~ ,

h(O) = u, h ( ~ ) = v imply

with statable a, b ~ G F ( q ) with a 4: b. Here a 4 : 0 4: b, as u, v ~ G F ( q ) by (4)

{0} T h e n

P u t x = b - a, m = b ( b - a ) / ( a x ) . T h e n m 4 : 1 a n d a = x / ( l - m ) . F r o m (5) we have

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213

hence u

[f(mx)

-

mx][f(x)

v

If(rex)

-

x][f(x)

- =

-

x]

(6)

mx]

Case x e [], m ~ []

W e h a v e f ( x ) = tx, f ( m x )

-t

1 + t 2) + 2t m + - t

m 2

--

I

--

V

= mtx

F r o m (6) it follows

= 0

(7)

/)

Therefore m ~ G F ( q ) If a n d only If (1 -

t) 2 -

(1 + t ) 2 - u ~ [ ]

(8)

V

W e have f ( x )

C a s e x e f-q, m ~ A

.urn2 + .[ 2( 1u ) .

= tx, f ( r n x )

]

t

(1 . + t 2) m + t 2 u =

V

= mt-lx

F r o m (6) It follows

0

(9)

t~

Therefore m ~ G F ( q ) if a n d only if

(1 -

t) 2 + 4 t u ~ [ ]

(10)

V

Case x ~ A, m ~ []

W e have f ( x )

= t-ix, f(mx)

= mt-lx.

It can be s h o w n that

m e G F ( q ) if a n d o n l y if (8) lS true

W e have f ( x ) = t - i x , f ( m x ) = m t x It can be s h o w n t h a t m ~ G F ( q ) tf a n d o n l y if (10) is true W e are a b l e to p r o v e that t = - 1. It suffices to prove t h a t if t # - 1 then there Case x e A,

m e/~

exist u, v ~ G F ( q ) -

U

{ ~ } with 1 - - e [ ] such that neither (8) n o r (10) hold. P u t t i n g V

U

X 2 = 1 - - , we p r o v e that the system 13

(1 -- t) 2 (1

--

/)2

(1 + 0 2 ( 1 - X +

2) = - Y 2 ,

4t(1 -- X 2) = - Z

2,

has s o m e s o l u t i o n s X, Y, Z over G F ( q ) p r o v i d e d that t # - 1, t ~ A.

(i1) (12)

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Let w denote the n u m b e r of its solutions over GF(q). By a theorem of B Segre 1-19], then w = w 1 + w 2 ( m o d p) where Wl and w 2 denote the n u m b e r of solutions of (11) and (12) over GF(q), respectively. In o u r case wl = q + 1, w 2 = q + 1. Therefore w = 2 (mod p). Hence the above system has at least two solutions over GF(q). This proves our assertion. f ( x ) = - x implies f e PGL(2, q). By (4) each involution of H belongs to PGL(2, q). This means that H -- H ' .

Step 3. E = E'. F o r each k e E ' , set t T ( k ) = {#(k)l#e (7}. T h e n t T ( k ) = E', as m v o l u t o n a l elements of PSL(2, q) are always conjugated. If in a d d m o n k e E, then tT(k) is contamed in E By IEI = IE'I it follows that E c~ E ' ¢ 0 lmphes E = E'. Thus, to prove E = E', it suffices to exhibit an elliptic involution k e E ' which also belongs to E. Take k as k(x) = ( - 1 ) / x Clearly k(0) = o0. Since k e G, ~ leaves E(0, o0) invanant. As 2 ,t" IE(0, ~)1, ~ has some fixed elhptlc involution in E(0, o0). We want to prove that there exists just one such revolution. In order to do this, some preparation is needed. In particular, we want to solve the equation m ( x ) = f ( x ) , x e M , for any involution rn e #- such that m(0) = ~ . F o r m e H(0, oo), it has no solution because m(x) = a/x with a suitable a e [ ] F o r m e E(0, ~ ) , it has at most two solutions, as m ( x l ) = f ( x l ) , re(x2) = f ( x 2 ) with xl, x2 e M, imply x2 = f ( x l ) . Since IM - {0, oo}l = 21E(0, o0)1 and no two equations of this type have c o m m o n solutions, we can infer that there are exactly two solutions for each m e E(0, o0) Let e e H(0, ~ ) be any involution such that ~ ( e ) = e. T h e n e ( x l ) = f ( x l ) , e(x2) = f ( x 2 ) for two suitable xl, x2 e M - {0, ~ } . Here x2 = f ( x l ) . F r o m this we can refer k ( x l ) = f ( x l ) . In fact, by/~(f) = f w e have e(k(xl)) = k(e(xl)) = k ( f ( x l ) ) = f ( k ( x l ) ) , whence k ( x l ) = x2 follows because k has no fixed point. Also k(x2) = f(x2). If there were another c e E(0, o0) with /~(c)= c, then we would have other x], x~ e M - {0, o0} such that k(x'l) = f(x'l), k(x'2) = f(x'2). But this is not possible, a s f ¢ k. Therefore, there exists only one e e E(0, o0) such that/~(e) = e. We are able to prove k = e. Let a e M be any point, and let s denote the revolution such that s(0) = ~ , s(a) = k(a). Then/~(s) = s. Here s ¢ H(0, ~ ) , as the equation s(x) = f ( x ) has solutions. Thus s e E(0, oo). Therefore s = e, which imphes k(a) = e(a). Hence k e E. 4. 2-transitive abstract ovals of order n - 3 (mod 4) We begin with the following PROPOSITION 1. Let B = (M. ~ )

be an abstract oval o f odd order havma a

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regular automorphtsm 9roup S. I f S is an elementary Abehan 2-group then some but not all non-trtvtal elements are regular mvoluttons. Proof Since g.m d (IEI, ISI) = 1, ~ has some fixed elhptlc revolutions Let f e E be such an revolution Let Y be the set consisting of the (n + 1)/2 non ordered pairs {a,f(a)} of points. Clearly S leaves Flnvanant. Let S* denote the permutation group reduced by S on Y. As ISI = n + 1, IYI = (n + 1)/2, for any y ~ Y there exists a non-trivml automorph~sm s ~ S such that s*(y) = y From th~s we can mfer that s*(y) = y for each y E Y, as S* is a transmve group on Y. Since s is without fixed point, s = f f o l l o w s Assume that each non-tnvml element of S is a regular elliptic mvolut~on. For any s ~ S, let F(s-) denote the set of all fixed hyperbohc revolutions of g Then IF(s')l = (n + 1)/2 holds. Moreover, for each point a, H(a, s(a)) contains a fixed hyperbohc revolution of g, as I{H(a, s(a))la e M}I = (n - 1)/2 is an odd number by IMI = ISI = 2'. It follows that each H(a, s(a)) contains just one hyperbohc involution fixed by g. Next we prove that F(g 0 and F(g2) are disjoint when s 1 ~ s2. If there were a common fixed hyperbohc m v o l u t l o n f then both of s, would exchange the fixed points o f f Hence, sis 2 would have some fixed points: a contradiction. Since IHI = n(n + 1)/2, from these facts we can Infer that each hyperbohc revolution is fixed by exactly one non-tnwal element of S. Let f e H(a, s(a)) be such that g ( f ) - ~ f , and let s x ~ S be a non-trivial element such that g ~ ( f ) = f Set sl(a) = b. Clearly b v~ s(a), as S is regular ssl = sis lmphes s(b) = sl(s(a)). Also, s l f = fsl implies &(f(a)) = f(b). By s(a) = f ( a ) we get s(b) = f(b). Therefore s = f But this Is impossible as s e E and f e H Our mare result is the following THEOREM 2. The classwal abstract oval ts the only 2-transttive abstract oval o f order n --- 3 (mod 4).

Proof Let B = (M, ~ ) be an abstract oval of order n --- 3 (mod 4), having a 2transmve automorph~sm group G. By Theorem 1 it suffices to prove that G contains a subgroup K isomorphic to PSL(2, q) and acts on M as PSL(2, q) in its usual 2transitive representation Assume that a normal minimal subgroup S of G Is solvable. Then S is an elementary Abehan 2-group acting regularly on M. Moreover, any two non-triwal elements of S are conjugated under G. But this is lmposs~ble in our situation because of Proposition 1. Assume that no element of order two of G has fixed point. Then Bender's theorem [2] can be applied, and the minimal normal subgroup of G has the desired property.

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If there ~s an element of order two m G w~th some fixed points, then the maximal number of fixed points of elements of order two m G is 2. Therefore, Hermg's result [11] can be apphed Since n # 5, the minimal normal subgroup of G has the desired property.

5. 2-transitive abstract ovals of order n --- 1 (mod 4)

In preparation for the proof of Theorem 3 we state and prove the following techmcal PROPOSITION 2. Let B = (M, ,~) be an abstract oval of order n - 1 (mod 4) havm9 a 2-group o f automorphtsms S whtch contams a regular mvolutton. I f ISI >- 4, then S contains a regular hyperbohc mvolutton

Proof Put U = S n Alt M By ISI ~ 4, U is non-trivial The central elements of order two of U together w~th the identity automorphlsm form a non-trivial subgroup V of U L e t f ~ S be any regular involution. T h e n f c o m m u t e s w~th an element v e V of order two Here v ~s either a hyperbohc regular mvolut~on, or ~t has w/n + 1 fixed points In the latter ease vf = fv implies f ~ H This proves our assertion. We are able to prove the following THEOREM 3 I f a 2-transtttve abstract oval o f order n = 1 (mod 4) contams a regular mvolutton, then tt ts the classwal abstract ovaL

Proof Let B = (M, ~,~) be an abstract oval of order n --- 1 (mod 4) having a 2transmve automorphlsm group G which contains a regular involution. By Theorem 1, it suffices to prove that then G contains a subgroup N ~ PSL(2, q) acting on M as PSL(2, q) m its usual 2-transmve representation Step 1. G contams no regular normal subgroup. Any regular normal subgroup of G is an elementary Abehan 2-group Thus n + 1 Is a power of 2: a contra&ctlon with n -- 1 (rood 4). Step 2. Each hyperbolic mvolutton ts regular. Let S be a 2-Sylow subgroup of G which contains a regular involution. By Step 1 G is non-solvable. Thus ISI > 4. Applying Proposition 2, we have that G contams a regular hyperbohc involution. Since G is 2-transitive, each hyperbolic involution is regular.

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Step 3. G ts 5/2-transttlve Given any two dtstlnct pomts a, b ~ M, we prove that A = {fo}lf, 9 ~ H(a, b)} Is either transltwe or has two orbits of length (n - 1)/2 on M - {a, b} F o r any c e M - {a, b}, there is a unique hyperbohc m v o l u n o n m H(a, b) fixmg c, and two distract hyperbolic m v o l u n o n s m H(a, b) cannot send c in the same pomt. Thus A(c) has at least (n - 1)/2 points F o r IA(c)l > (n - 1)/2, A(c) meets any orbit A(d) with d e M - {a, b, c} Thus A is transltwe on M - {a, b} Since A is a subset of G, our assertmn ~s proved

Step 4. G~ ts primitive on M - {a} Let • be a set of lmpnmttlvtty of M - {a} under G a Let b, c ~ • any two &stmct pomts. As we have seen m Step 3, IGa b(C)l /> (n -- 1)/2 Thus 141 ~ Ia~,b(c)l + 1 > (n + 1)/2. Since i~J > (n + 1)/2 tmphes q) = M - {a}, our assertion follows.

Step 5. For any two dtstmct hyperbohc lnvoluttons f, 9 ~ H(a), f9 has odd order H(a) n Z(Ga) = O. Each power of )cO belongs to E = {hklh, k ~ H(a)}. Thus no power of f9 distinct from the identity a u t o m o r p h l s m has fixed point on M - {a} Since M - {a} consists of an odd n u m b e r of points, also f9 has odd order In particular, no two &stlnct hyperbohc revolutions of H(a) commute, this proves the second assertion

Step 6 Ga contams a normal subgroup whtch ts reoular on M - {a} G w e n any f e H(a), let $2 be a Sylow 2-subgroup of Ga containing f By Step 5, we have H(a) c~ $2 = {f} In particular {gfg-ll9 ~ Ga} c~ $2 = {f} Since by Step f ¢ Z(Ga), from G l a u b e r m a n ' s theorem ([10], [13] V Satz 22 10) it follows that G~ contains a normal s u b g r o u p of odd order. By the F e l t - T h o m p s o n theorem ([19], [13] p. 128), the mimmal normal subgroup of G, is an elementary Abehan p - g r o u p It is regular on M - {a}, by Step 4

Step 7. G contains a normal subgroup N ~ PSL(2, q) actm9 on M as PSL(2, q) m tts usual 2-transttwe representatton By Steps 1 and 6 we can apply the H e n n g - K a n t o r - S e l t z theorem ([11], [13] XI. T h e o r e m 13 8). Since n = 1 (mod 4), a normal subgroup N o f G is asomorphlc to one of the simple groups PSL(2, q), q = n, PSU(3, q2), q3 = n, such that N acts on M as the corresponding g r o u p in its usual 2-transitive representation The latter case cannot occur m our situanon, as each element of order two of PSU(3, q2) has q + 1 fixed points while q + 1 :~ 2, x/~ + 1 Finally we prove

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THEOREM 4. A n y 2-transmve abstract oval o f order n = 1 ( m o d 4) has a 2transtttve stmple group o f automorphtsms. I f tt ts one o f the known stmple groups, then zt ts isomorphic to P S L ( 2 , q) and acts on M as P S L ( 2 , q) m tts usual 2-transtttve representatton.

P r o o f By Burnslde's theorem ([13], XI. T h e o r e m 7.12) the minimal normal s u b g r o u p N of any 2-transitive a u t o m o r p h l s m g r o u p of B = (M, ~-) is simple. Smce n = 1 (mod 4), also N is 2-transltwe by Ascbacher's theorem [1] Since N ~< Alt M,

then the elements of order two of N have either 2 or x//~ + 1 fixed points In the case where no element of order two of N has x / n + 1 fixed points, Hermg's result [11] can be apphed. Therefore, m general N _~ P S L ( 2 , q) holds; an exception is only possible if n = 5, and N ~ A6. O n the other hand, the unique abstract oval of order 5 is the classical one (cf. [6]) whose full a u t o m o r p h l s m g r o u p P G L ( 2 , 5) does not contain A 6 We m a y assume that N admits some elements of order two with x/~ + 1 fixed points. In particular n ts a square number. All the possible 2-transltwe representauons of the k n o w n simple groups are given m [17] Table I. N o t e that only three of them have degree n + 1 with square n: P S L ( d , q ) , Alt(,+l), P S U ( 3 , q2) m their usual 2transltwe representation. Clearly N 4: Alt(.+l) for n >= 4. Also N v~ P S U ( 3 , q2) by Step 7 of the p r o o f of T h e o r e m 3. Finally, N = P S L ( d , q) lmphes d = 2.

REFERENCES

[1] ASCHBACHER,M, On doubly transmvepermutauon 9roups ofdeoree n -- 2 (mod 4) Ilhnols J Math 16 (1969), 276-279 [2] BENDER,H, Endhche zwetfaeh transtttve Permutattonsgruppen deren lnvoluttonen keme Ftxpunkte haben Math Z 104 (1968), 175-204 [3] Bt~KENHOUT,F, Etude mtrmseque des ovales Rend Mat 25 (1966), 333-393 [4] BUEKENHOUT,F, Ovales et ovales projecttfs Rend Accad Naz Lmcel (8) 40 (1966), 46-49 [5] FAINA,G, A new class of2-transmve mvolutorypermutatton sets Aequatlones Math 24 (1982), 175-178 [6] FAIr'A,G, The B-ovals of order <8 J Comb Theory Ser A 36 (1984), 307-314 [7] FAINA,G, The geometry of buekenhout ovals A Survey Preprmt [8] FAINA,G and KORCHMAROS,G, ll sottooruppo 9enerato dalle mvoluzlom regolarl dt un B-ovale transmvo, to appear in Rend Sere Mat Umv Padova [9] FEIT,W and THOMPSON,J G, Solvablhty of groups ofoddorder Pacific J Math 11 (1963), 755 1029 [10] GLAUBERMAN,G, Central elements of core-free groups J Algebra 4 (1966), 403-420 [11] HER•NG• C • Zwetfach transttwe Permutatt•ns•ruppen m denen zwet dte maxtmale Anzah• v•n Ftxpunten stud Math Z 104 (1968), 150-174 [12] HER1NG,C, KANTOR,W M and SEITZ,G M, Fmtte groups with a spht BN-patrs of rank 1 J Algebra 20 (1970), 435-475 [13] HUePERT,B, Endhche Gruppen I Springer, Berhn-New York, 1967 [14] HUPPEgT,B and BLACKBURN,Fmtte groups II1 Spnnger, Berhn-New York, 1982 [15] KORCHMAROS,G, Su una class~ficazlone delle ovah dotate dt automorfismt Rend Accad Naz XL (5) I 2 (1975-76), 77-86 [16] KORCHMAROS,G, Questwm relattve adovah astratte Conf Sere Mat Umv Ban no 171 (1979) 1-11

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[17] SHULT,E, Permutatton groups wtth f e w fixed pomts In Geometry-yon Staudt s Points o f Vten NATO Adv Study Inst Ser C-7D Reldel, Dordrecht, 1980, pp 275-311 [18] SEGRE, B, lntroductton to Galots geometrle~ Mem Accad Naz Lmcel, (8) (1967), 137 236 [19] SEGRE,B, St~temt dl equaztont net campl dt Galots In Attt Convegno sulla teorta det gruppt fimtt Flrenze, 1960, pp 66-80 [20] CHEROWlTZO, W, Harmontc oval9 o f even order, In Fmtte Geometrws, Lecture Notes in Pure and Apphed Mathematics 103 M Dekker, New York, 1985, pp 65-81 Inslttulo dt Matematwa Unwerstta degh Studt della Baslhcata 1-85100 Potenza ltal k