JOURNAL OF ALGEBRA 208, 533᎐550Ž. 1998 ARTICLE NO. JA987514

Finite Projective Planes with Abelian Transitive Collineation Groups

Chat Yin Ho

Department of Mathematics, Uni¨ersity of Florida, Gaines¨ille, Florida 32611-8105 E-mail: [email protected] View metadata, citation and similar papers at core.ac.uk brought to you by CORE Communicated by Walter Feit provided by Elsevier - Publisher Connector

Received January 5, 1997

DEDICATED TO PROFESSOR J. G. THOMPSON ON THE OCCASION OF HIS 65TH BIRTHDAY

1. INTRODUCTION

In 1959, Ostrom and Wagnerwx OW proved that if a collineation group acts 2-transitively on the points of a finite projective plane, then the plane is Desarguesian. This celebrated result inspired the study of projective planes with transitive collineation groups. A collineation group acting regularlyŽ. i.e., sharply transitively on the points of a projective plane is called a Singer group. Thus abelian transitive collineation groups are Singer groups. In 1975, Ottwx O proved that a finite projective plane admitting more than one cyclic Singer group is Desarguesian. Recently Muller¨ wx M , using the classification of finite simple groups, gave a new proof of Ott’s result. It has been conjectured that the cyclic assumption in Ott’s result can be replaced by an abelian assumption. In this paper, we give an affirmative answer to this conjecture in Theorem 1 below. Since our argument does not use Ott’s result or the classification of finite simple groups, Ott’s theorem is a consequence of our result. THEOREM 1. A finite projecti¨e plane admitting more than one abelian Singer group is Desarguesian. Note that the collineation group generated by these Singer groups contains the little projective group of the plane. An abelian Singer group of a Desarguesian plane is cyclic. Our result gives support for the long standing conjecture that an abelian Singer group is cyclic. The situation for

533

0021-8693r98 $25.00 Copyright ᮊ 1998 by Academic Press All rights of reproduction in any form reserved. 534 CHAT YIN HO an infinite plane is very different. In 1964, KarzelŽwx K. proved that an infinite projective plane admitting a cyclic Singer group is non-Desargue- sian. Involutory collineations play an important role in the study of collineation groups. When an involution is not a perspectivity, i.e., a Baer involution, Theorem 2.3 below provides a tightly embedded subgroup. With an additional condition a strongly embedded subgroup can be con- structed. The starting point of the proof of Theorem 1 is the following result, which is of independent interest.

THEOREM 2. An abelian Singer group contained in a sol¨able collineation group of a finite projecti¨e plane is always normal. We now make some comments on the rest of the proof of Theorem 1. In analyzing a collineation group G containing more than one abelian Singer group, each involution of G may be assumed to be a Baer involution. So the plane has square order. This implies, by Hall’s multiplier theorem, that an abelian Singer subgroup has an involutory multiplier. Consider the subplane fixed pointwise by this multiplier and the subgroup of G leaving invariant this subplane. If the restriction of this subgroup to this subplane contains more than one abelian Singer subgroup, then induction applies. From this the plane is proved to be Desarguesian. Otherwise, we prove that the generalized Fitting subgroup of G has only one component, and it contains a strongly embedded subgroup. This enables us to use the Bender᎐Suzuki theorem. A final contradiction is reached by some combi- natorial counting arguments.

2. DEFINITIONS, NOTATIONS, AND SOME GENERAL RESULTS

2 For any number r, let ¨Ž.r s r q r q 1. We use the term s-group to mean planar Singer group in the rest of this paper. For an s-group S, the multiplier group of S, which is the stabilizer of a point of the normalizer of S in the full collineation group of ⌸, is denoted by MSŽ.. The planar order of S is the order of the projective plane on which S acts. A subset D of a group S is a planar difference set if for each nonidentity element z of S there exist a unique pair x, y g D such that 1 1 z s xyy and a unique pair c, d g D such that z s cy d. In this paper we write difference set to mean planar difference set. The study of a differ- ence set of a finite group is equivalent to the study of a finite projective plane ⌸ admitting an s-group isomorphic to S. A difference set of S corresponds to a line of ⌸ Žsee, e.g.,wx HP. . FINITE PROJECTIVE PLANES 535

For a set of collineations H, let PŽ.ŽH respectively, L Ž..H be the set of common fixed pointsŽ. respectively, lines of elements of H, and let FixŽ.H be the set of fixed-point-line substructure ŽŽ.P H , L Ž..H of H. A collineation ␣ is planar if FixŽ.␣ is a subplane. Our terminology in group theory is taken fromwx G , that of the projective plane is taken fromwx HP , and that of the difference set is taken from wx B orwx L .

LEMMA 2.1. Let G s KH be a finite group such that K s K1 = иии = K r is a normal subgroup of G and K1,...,Kr are nonabelian simple groups. Let T be a subgroup of K. For i s 1,...,r, let Tiiis Ät g K N ᭚ t g T such that t s t1 иии tirиии t 4. Then T F T1 = иии = Tri and T s Ž.T Ł j/ ijK l K i. If H normalizes T, then H normalizes T1 = иии = Tr .

Proof. It is clear that T F T1 = иии = Tri. Next we prove T s Ž.T Ł j/ ijK l K i. Let t s t1 иии trig T and i g Ä41,...,r . Then t s y1 tŽ.Ł j/ ijt . This shows that T iF Ž.T Ł j/ ijK l K i. Conversely, suppose u s tŁ j/ ijk g K i, where k jg K jfor j / i. Since K is a direct product, this implies u s tijand so Ž.T Ł / ijK l K iF T iand the equality holds as desired.

Since K1,...,K r are nonabelian simple groups, H permutes these h h subgroups. Let h g H. Suppose Ž.K iks K for i g Ä41,...,r . Then Ž.Ti h h h h h s Ž.T Ł j/ ijK l Ž.K i s ŽŽT . Ž.Ł j/ ijK .Ž.l K i s Ž.T Ł j/ kjK l K k. h Therefore Ž.Tiks T .So H permutes T1,...,Tr . Hence H normalizes T1 = иии = Tr as desired.

LEMMA 2.2. A collineation 2-group fixes at least one point.

2 Proof. This is because n q n q 1 is odd.

THEOREM 2.3. Let G be a collineation group with e¨en order. Let ␴ be a Baer in¨olution in G. Then K [ Ä g g G N PŽ.g s P Ž␴ .4 is a tightly embed- ded subgroup of G. If K contains all in¨olutions of H [ Ä g g G N g lea¨es PŽ.␴ in¨ariant4, then H is a strongly embedded subgroup of G.

x Proof. Suppose < K l K < has even order. Let i be an involution in the x x intersection. Then PŽ.i s P Ž␴ .as i g K, and P Ž.i s P Ž␴ . as i g K . x x Hence PŽ.␴ s P Ž␴ .ŽŽ..s P ␴ . Therefore x leaves PŽ.␴ invariant. So x is in H. Since K is the kernel of the action of H on PŽ.␴ , K is x normal in H. Hence K s K. This proves that K is a tightly embedded subgroup of G. x Assume K contains all involutions of H. Suppose < H l H < has even x order. Let i be an involution in the intersection. Then i g K l K . Thus x g H from what we proved in the last paragraph. This proves that H is a strongly embedded subgroup in G as desired. 536 CHAT YIN HO

LEMMA 2.4wx Ho2, 2.4 . Suppose U is a group of multipliers of an s-group S. Then PŽ.U s CUS Ž.. If P Ž.U / 1, then FixŽ.U is a triangle or a subplane.

THEOREM 2.5. Let S be an s-group of a projecti¨e plane and let ␴ be a multiplier of S. Ifww S, ␴ x, ␴ xs 1, then w S, ␴ x s 1. Proof. We identify the points of the plane with the elements of S.By

way of contradiction, assume that wxS, ␴ / 1. Let C s CSŽ.␴ . By Lemma 2.4, we see that C is a triangle or a subplane. Suppose <

COROLLARY 2.6. Let S be an abelian s-group and ␣ a collineation normalizing S. Ifwx S, ␣, ␣ s 1, then wx S, ␣ s 1. Proof. Since S is regular, the normalizer of S in the collineation group can be expressed as the semidirect product of S with the stabilizer of a point in the collineation group, which we can identify with the multiplier group M of S. Thus ␣ s xy, where x g S and y g M. By assumption, S is abelian. This implies that wxwxS, ␣ s S, y and it follows that 1 s wS, ␣, ␣ x s wxS, y, y . By Theorem 2.5, this implies wxS, y s 1. Hence wxS, ␣ s 1as desired.

COROLLARY 2.7. Any two normal abelian s-groups in a collineation group are equal. FINITE PROJECTIVE PLANES 537

Proof. Suppose K and L are two normal abelian s-groups in a collineation group. Then wxK, L F K l L. Since L is abelian, this implies wwK, L x, L x s 1. By 2.6, this implies wK, L x s 1. Hence L F K. Since both L and K have the same group order, K s L as desired. LEMMA 2.8. Suppose a collineation group acts transiti¨ely on the points of a projecti¨e plane. Then a nontri¨ial element of an abelian normal subgroup of this collineation group does not ha¨e any fixed point. Proof. Let ⌫ be such a normal subgroup. As a normal subgroup of a transitive group, all point orbits of ⌫ in the plane have a common cardinality. Being a collineation group, ⌫ has a faithful point orbitŽ see, e.g.,wx HP, Corollary 1, p. 255. . Since ⌫ is abelian, this faithful orbit is a regular orbit, and so it has cardinality <<⌫ . Hence each point orbit of ⌫ has cardinality <<⌫ , and so each point orbit is regular. This proves Lemma 2.8.

3. PROOF OF THEOREM 2

In this section, let ⌸ be a projective plane of order n, and let S be an abelian s-group of ⌸. Suppose G is a solvable collineation group of ⌸ containing S. We use induction on <

G s SGXX, S l G s 1, where G Xis the stabilizer of a point X of ⌸.By way of contradiction, we assume that S is not normal. The rest of the proof is divided into three steps.

STEP 1. S contains all minimal normal subgroups of G. Proof. Let N be a minimal normal subgroup of G. By way of contra- diction, assume N is not contained in S. Since G is solvable, N is an elementary abelian p-subgroup for some prime p. By Lemma 2.8, <

From G s SN and S l N s 1, we obtain <

STEP 2. S contains all normal p-subgroups of G for any prime p.

Proof. Let p be a prime and let P be a normal p-subgroup of G with minimal order subject to P Fu S. Since P is normal in SP, we may assume G s SP by induction on <image of a subset F of G under the natural homomor- phism from G to GrE. Then G s SP and S l P s 1. Since S l GX s 1 and E F S,so E l GXXs 1. Hence G ( GX. We now prove Q F ZGŽ.. Suppose Q does not centralize P. Then CQP Ž.is a nontrivial proper subgroupŽ as these two groups are p-groups. of P normalized by S Ž.as S is abelian . Let R be the preimage of this subgroup. Since PЈ F E F R and R - P, R is a nontrivial proper normal subgroup of P.As R is S-invariant, R is S-invariant. Hence R is normal in SP s G. The minimality of <

We now prove that PQ is abelian. Suppose S l GX/ 1. Let W be the preimage of this group. Then E - W. From W s W l EGXXs EWŽ.l G , we see that W l GX / 1. However, this contradicts the fact that W F S and S l GXXs 1. This contradiction proves S l G s 1. Therefore <

However, this contradicts 1 / GX F QP. This contradiction establishes Step 2.

STEP 3. Final contradiction. Proof. By Step 2, S contains FGŽ., the Fitting subgroup of G. Since S is abelian, this implies S F CFGGGŽŽ...AsG is solvable, CFG ŽŽ..F FG Ž. Žsee, e.g., Theorem 1.3, p. 218 of wxG .ŽŽ..Ž.. Hence S F CFGG F FG F S. Therefore S s FGŽ., which is a normal subgroup of G. This final contra- diction proves Theorem 2.

4. SOME CONSEQUENCES OF THEOREM 2

COROLLARY 4.1. Let G be a collineation group. If G contains a normal abelian s-group, then this subgroup is the only abelian s-group contained in G. Proof. Let S be an abelian normal s-group of G. Suppose T is any abelian s-group of G. Then ST is a solvable subgroup of G. By Theorem 2, T is normal in ST.SoS and T are two abelian normal s-groups of ST. By 2.7, S s T as desired. Remark. PGŽ.2, 4 admits a cyclic Singer group and a noncyclic Singer group. Also 4.1 does not claim that S contains all semiregular collineation subgroups of G.

LEMMA 4.2. Let G be a collineation group. Suppose that G contains more than one abelian s-group. Then any abelian s-group of G is not normal. Also G is not sol¨able and a Sylow 2-subgroup of G is not cyclic. Proof. Let S be an abelian s-group of G. By 4.1, S is not normal in G. So G is not solvable by Theorem 2. Therefore 2 divides <

LEMMA 4.3. Let S be an abelian s-group, and let Q be a collineation group normalized by S. Then S centralizes Q if one of the following holds: Ž.1 Q is abelian. Ž.2 <

Singer group S s S7 = S7Ј. Then Q s Sf7 ²:is a nilpotent subgroup of order 73, not abelian, normalized but not centralized by S. Note that 27 is not a square. In PGŽ2, q 2 ., let f be the involutory multiplier of the cyclic Singer group S. Then wS, ²:ffx²:s Q is normalized by S but not centralized by S. Note that Q is solvable but not nilpotent. These examples show that Lemma 4.3Ž. 3 is in some sense best possible. FINITE PROJECTIVE PLANES 541

5. PROOF OF THEOREM 1

In this section, let ⌸ be a projective plane of order n, admitting a collineation group G, which contains more than one abelian s-group. We show that ⌸ is Desarguesian. By way of contradiction we may assume that ⌸ is non-Desarguesian.

STEP 1. Each in¨olution in G is a Baer in¨olution. Proof. Suppose an involution is not a Baer involution, then it is a perspectivityŽ see, e.g.,wx HP. . Since G acts transitively on the points of ⌸, ⌸ is Desarguesian and G contains the little projective group of ⌸, in this case by a result of Wagnerwx HP, p. 260 . This shows that we may assume that each involution in G is a Baer involution.

2 STEP 2. We ha¨ens m for some positi¨e integer m. Let S be an abelian s-subgroup of G. Then the Sylow 2-subgroup of the multiplier group MŽ. S of S is a direct factor of MŽ. S , and is a nontri¨ial cyclic group. ŽNote that M Ž. S may not be in G..

2 Proof. By 4.2 and Step 1, G contains a Baer involution. So n s m . Since S is abelian, Hall’s multiplier theoremwx HP shows that m3 induces a multiplier of order 2. The second conclusion in Step 2 is a result inw Ho2, Theorem 3x . STEP 3. Let ␴ be the unique in¨olution in MŽ. S . Then ␴ normalizes at least two conjugates of S in the full collineation group L of ⌸.

L gg Proof. For ␶ g ␴ , let WŽ.␶ s ÄS N g g L and S is normalized by ␶ 4. L We now compute the cardinality k of the set ÄŽ.␶ , U N ␶ g ␴ and U g WŽ.␶ 4 in two ways. Let e s

We now introduce some notation. Let ⍀ s FixŽ.␴ , B s CS Ž.␴ . Sup- pose X is a collineation group containing more than one abelian s-group, and S F X. Let UXŽ.s Ä x g Xx< leaves ⍀ invariant4 . Then B F UXŽ.. Let B11Ž.respectively, U be the restriction of B ŽŽ..respectively, UX on ⍀. Bywx Ho1 , B1 is an abelian s-group of ⍀. Let WXŽ.be the kernel of the action of UXŽ.on ⍀.

STEP 4. If B11 is not normal in U , then ⌸ is Desarguesian.

Proof. Suppose B11is not normal in U . By induction on the planar order, we see that ⍀ is a Desarguesian plane, say PGŽ.2, q , where q is a power of a prime. Also U12contains a subgroup U isomorphic to PSLŽ.3, q acting naturally on ⍀ and m s q.

Case 1: q s even. In this case U1 contains an elementary abelian 2 2-subgroup R1 of order q consisting of elations of ⍀ with a common axis ␣ l ⍀, where ␣ is a line of ⌸. Let R be a Sylow 2-subgroup of the preimage of R11in UXŽ.. Thus Rr ŽR l WX Ž..( R , and R fixes each point of ␣ l ⍀. Let y g R. Suppose y fixes one point in ␣ outside of ⍀. 2 22 Since y is in WXŽ., y is planar. As y fixes more than m q 1 points on 2 the line ␣, y s 1. If y / 1, then y is a Baer involution. However, y fixes more than m q 1 points on ␣, which is impossible. This proves that y s 1. 2 Therefore R acts fixed point freely on the q y q points of ␣ outside ⍀. 2 2 Hence <

Case 2: q s odd. In this case U21contains a subgroup R isomorphic to SLŽ.2, q such that it leaves invariant a line ␣ and elements of prime order dividing q are elations of ⍀ with axis different from ␣ l ⍀. Let R be the preimage of R1 in UXŽ.. Then R leaves invariant the line ␣. Let y be an element of prime order p of R. Suppose y fixes one point Y on ␣ outside p ⍀. Then y , being in WXŽ., is planar. It fixes Y and the points on ␣ l ⍀. p This implies that y s 1. Suppose p s 2. Then y is a Baer involution fixing Y and the points ␣ l ⍀.So y fixes at least m q 1 points on ␣, which is impossible. Therefore p / 2. 2 Assume p divides q. Let the set of the q y q points on ␣ outside ⍀ be 2 ␤. Since p does not divide q y q y 1, y must fix at least one point in ␤ different from Y. Now y induces an elation in U1 with its axis ayŽ.l ⍀ ŽŽ..where ayis a line different from ␣ l ⍀.ŽŽ.Ž.. Thus P y l ⍀ s ayl ⍀. So y fixes the points of its axis in ⍀ and at least two points in ␤. This implies that y is planar. Since U2 ( PSLŽ.3, q , all root subgroups Ž of order q. of U2 are conjugated. In fact, all subgroups of order p generated by 2 elations of U11are conjugated in U . There is a subgroup T 1of order q of U2 , whose nontrivial elements are elations with the common axis ayŽ.l ⍀ FINITE PROJECTIVE PLANES 543

2 and T1 is covered by conjugates of ²:y in UX Ž .. Since there are q 2 elements of T1 and only q y q points ⌫ on ayŽ.outside ⍀, some element z / 1ofT1 must fix at least one point on ⌫. However, we know y is planarŽ²: as z is conjugated to ²:.y , and fixes all q q 1 points of ayŽ.l ⍀. So this planar element z must be 1. This contradiction proves that the order of the stabilizer H of a point of the action of R on ␤ is prime to q and 2. Thus the image H1 of H in RrWXŽ.also has order prime to q and 2. Since RrWXŽ.( SL Ž2, q ., this implies that H1 is a cyclic subgroup of order dividing Ž.q q 1 r2, if q ' 3 Ž. mod 4 , and Žq y 1.Ž.Žr2, if q ' 1 mod 4 see, e.g.,wx Hu or wx Su. . As <

By Step 3, ␴ normalizes S and at least one conjugate S1 Ž./ S of S.In proving ⌸ to be Desarguesian, we may assume G is generated by S and

S1, since this subgroup contains more than one abelian s-group. Note that 2 G is normalized by ␴ . We use induction on <

Suppose S is normal in F*Ž.GS. Since S G CSGŽ., this will imply that F*Ž.GSF MS, where M s G l MSŽ.. Bywx Ho2, Theorem 3 , MSŽ.is solvable. Hence SM is solvable as S is abelian. So F*Ž.G is also solvable. However, this contradicts EGŽ./ 1. Thus F*Ž.GScontains more than one abelian s-group. By induction, we may assume that G s F*Ž.GSs EGSŽ., and S acts transitively on the set of components of G. Note that F*Ž.GSis normalized by ␴ . Now ZGŽ.acts semiregularly on the fixed points of any involution of G. Since an involution is a Baer involution by Step 1, this implies that < ZGŽ.<<

Proof. Note that EGŽ.rZEG ŽŽ..( EGŽ.s E1 = иии = Er,where

E1,...,Er are the components of EGŽ.. Then S acts onÄ4 1, . . . , r induced by the conjugation of S. Observe G s EGSŽ.. Let H be the stabilizer of a point of the plane in G. Then G s HS and H l S s 1. Thus G s HS, H l S s 1, and H ( H. We divide the proof into two cases: Case 1: EGŽ.l S s 1. In this case <<<<<vector space V over a field of p-elements. Since the dimension of V is 1 q r G 1 q 2 s 3, HH the dimension of x is at least 2. Therefore x l²:x1,...,xr / 1. Hence there is z g ²:x1,..., xr , such that z / 1, and z centralizes x. Since x g S _ ZGŽ.Žas x / 1 . , Step 5 implies z g NSGŽ.. Thus wxS, z F S.So wxS, z F S. However, z g EGŽ.. Therefore wS, z xF S l EGŽ.s 1. This implies wxS, z F ZGŽ.. Hence wxS, z, z s 1. So z centralizes S by 2.6. Thus z g S. But this implies z g S l EGŽ.s 1, a contradiction. This contradic- tion proves that x does not normalize R1. xxpy 1 Let i be an involution in E1. Then i ,...,i belong to distinct components. So they commute with each other. Thus x centralizes the xxpy 1 involution di s i иии i . Since E111is simple and < ZEŽ.< is odd, E contains an elementary abelian group F of order 4. The above argument shows that x centralizes dj for all involutions j g F. Hence x centralizes a 4-group in EGŽ.. Since ZEGŽŽ..has odd order, x centralizes a 4-group W in EGŽ..

Let C s CxGŽ.. Then C contains W and S. By Step 5, C F NSGŽ.. However, this contradicts the fact that the Sylow 2-subgroup of MSŽ.is cyclic by Step 2. This contradiction proves that Case 1 cannot occur if r G 2. We now turn to Case 2.

Case 2: S l EGŽ./ 1. Let D s S l EGŽ., and let Di be the projec- tion of D to Ei for i g Ä41,...,r . Since S is abelian, S normalizes D.By 2.1, S normalizes R s D1 = иии = Dr. Since S is abelian, R is abelian. Let R be the complete preimage of R.As RrZGŽ.s R is abelian, so R is nilpotent. Since S normalizes R, S normalizes R. This implies, by Step 2 and 4.3, that S centralizes R.So R F S and D1F S. The commutativity of S now implies that it fixes 1 g Ä41,...,r .Sor s 1 by the transitive action of S onÄ4 1, . . . , r . Therefore r s 1 in both Case 1 and Case 2. So EGŽ.rZEG ŽŽ..is simple as desired.

STEP 7. We may assume ZŽ. G does not contain any subgroup of order n q m q 1.

Proof. Let Z s ZGŽ.. Then Z F S. Suppose Z contains a subgroup of order n q m q 1. Then the stabilizer of a point in G, which contains an involution, fixes pointwise a subplane of at least n q m q 1 points. This implies <

HZrZ. Hence E is isomorphic to Lq23Ž., Sz Ž. q ,orUq Ž., where q is a power of 2 by the Bender᎐Suzuki theorem. Note that E is generated by two Sylow 2-subgroups. Thus EGŽ.is generated by two Sylow 2-subgroups and its center. As EGŽ.is perfect, this implies that EG Ž.is generated by two Sylow 2-subgroups. If two Sylow 2-subgroups of EGŽ.have a common fixed pointŽ.Ž. recall Z is central with odd order , then EG will have a fixed point. This implies that S will act on the fixed-point set of EGŽ., which is a proper subset of points of the plane.Ž This set cannot be the point set of the plane as EGŽ./ 1. . This contradicts that S acts transitively on the points of the plane. Hence different Sylow 2-subgroups of EGŽ.have disjoint fixed point sets. Since S permutes the Sylow 2-subgroups of EGŽ., S acts on the union of the fixed points of Sylow 2-subgroups of EGŽ.. The transitivity of S on the points of the plane implies that the set of points of the plane is theŽ. disjoint union of the fixed points of the Sylow 2-sub- 2 groups of EGŽ.. Hence n q n q 1 s Žnumber of Sylow 2-subgroups in EGŽ ..Ž nq m q 1 . . The number of Sylow 2-subgroups of EG Ž . equals the i number of Sylow 2-subgroups of E. This number is q q 1, where i s 1if 2 E ( Lq23Ž.; i s 2if E ( Sz Ž. q ; and i s 3if E ( Uq Ž.. Therefore n q i ii n q 1 s Žq q 1.Žn q m q 1. . So q q 1 s n y m q 1. Hence q s n y m s mmŽ.y 1 . However, q is power of 2. The last equation implies that m s 2 and so n s 4. As any plane of order 4 is Desarguesian, we may assume <

Recall ⍀ s FixŽ.␴ . Let U [ UG Ž.s Ä g g G N g leaves invariant the Baer subplane ⍀4, W [ WGŽ.s Ä g g Gg< pointwise fixes ⍀4, and Y s YZŽ. G rZG Ž.. Set V s BW. Since ␴ normalizes a Sylow 2-subgroup of G and CGŽ.␴ F U, <

THEOREM I. A multiplier of an abelian Singer group fixes pointwise a Baer subplane if and only if it is an in¨olution.

STEP 8. <

Proof. Suppose <

ZGŽ.F B, this implies that i normalizes B. This contradicts < NBU Ž.< 2 s 1. Therefore there is a prime p dividing both <

STEP 9. All in¨olutions of U belong to W.

Proof. Deny this. Let i be an involution of U outside W. For a subset Y of ²:²:U, ␴ s U ␴ , let Yˆˆ[ YWrW. Recall B is a normal subgroup of U ˆ by Step 4. So Bˆˆis an abelian Singer group of ⍀.Ž The elements of B can be identified with the points of ⍀.. Since ˆˆi / 1, i can be viewed as an involutory multiplier of Bˆˆ. Thus CiBˆŽ.is a Baer subplane of groupal order m q ''m q 1, and wxBˆˆ, i is an arc subgroup of order m y m q 1in- verted by ˆi Žsee, e.g.,wx Ho1, Theorem B. . Since B l W s 1, there is a subgroup J F B such that Jˆˆˆs wxB, i . Note that <

From the fact that ˆˆi inverts J, there is a conjugate i1 of i in ²:iJWsuch

Let UE s U l EGŽ.s Ä g g EGŽ.< g leaves ⍀ invariant4 , and WE s W l EGŽ.. Since U is a strongly embedded subgroup in G by Step 9 and Theorem 2.3, U contains a Sylow 2-subgroup of G. On the other hand, all Sylow 2-subgroups of G are in EGŽ.as

STEP 10. All in¨olutions of UEEE belong to W and U is a strongly embedded subgroup of EŽ. G .

Proof. By Theorem 2.3, it suffices to show that all involutions of UE belong to WE. As the Sylow 2-subgroups of G are in EGŽ., the set of involutions of UE equals the set of involutions of U, which is the set of involutions of W by Step 9, which is the set of involutions of WE. This establishes Step 10.

STEP 11. UE is a strongly embedded subgroup in EŽ. G s EGZGŽ.Ž.rZG Ž.. All in¨olutions of UEE belong to W . x Proof. Suppose <

STEP 12. Let T12/ T be two Sylow two-subgroups of EŽ.Ž G . So they are Sylow 2-subgroups of G. For any 2-group T, ⍀1Ž.T denotes the subgroup generated by the in¨olutions of T..ŽŽ..ŽŽ..Then Fix ⍀11T and Fix ⍀12T are Baer subplanes with disjoint points sets.

Proof. By Step 10, FixŽŽ..⍀1 T is a Baer subplane for a Sylow 2-sub- group T of EGŽ.. By Step 11, EGŽ.is isomorphic to one of the following groups: Lq23Ž., Sz Ž. q , Uq Ž.for some q, which is a power of 2. Assume, by way of contradiction, that FixŽŽ..⍀11T and Fix ŽŽ..⍀12T have a point Y in common. FINITE PROJECTIVE PLANES 549

We will show that EGŽ.fixes Y. Suppose this is true. Then the fixed point set of EGŽ.is a proper subset of the set of points of the plane. However, this fixed point set is left invariant by S. This contradicts the transitivity of the action of S on the points. We now prove that EGŽ. fixes Y.

Suppose EGŽ.( Lq21Ž.or Sz Ž. q . Then EG Ž .is generated by ⍀ Ž.T1and ⍀12Ž.T . Thus, in this case, EGŽ.fixes Y as asserted. Suppose EGŽ.( Uq3Ž.. In this case, ⍀1Ž.T s ZT Ž.for a Sylow 2-subgroup T of EGŽ.and ²ZT Ž12 ., ZT Ž .:( SL Ž2, q . Žsee, e.g.,wx DGB, p. 107. . By Lemma 2.2 and the transitivity of the action of S on the points, GY contains a Sylow 2-subgroup of G. This Sylow 2-subgroup of G is inside

EGŽ.as we noted before

Step 7, as <ÄT N T is a Sylow 2-subgroup of EŽ. G 4<

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