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into orbits the length of each of them excepting {a} being divisible by p. Hence 1<1>1 == 1 mod p. Let f3 E 'i'\
o 15.2 Lemma (Gleason 1956). Let 21 be a finite affine plane and H a collineation group of 21. Put H (P) = H n T(P). If h is an integer greater 68
Therefore, T(P) 1= {l} for all P r/oo' By 1.1, 1.6a) and the finiteness of T, it follows that T is an elementary abelian p-group. Let I be a line with IITI maximal and let t be the number of point orbits of T in 12r. Then n 2 = ITlt. Furthermore, t + n + 1 is the number of point orbits of T in the projective closure 21 V 100 of 21. By the Theorem of Dembowski-Hughes-Parker, t + n + 1 is also the number of line orbits of T in 21 V 100 , Hence t + n is the number of line orbits of T in 21. By the maximality of IITI, we get the inequality I/TI(t + n) ;;:. n(n + 1). From n 2 = ITlt, we obtain t < n. Therefore, 21/Tln > n(n + 1) which implies 211TI> n + 1 > n. As IITI is a divisor of n, this yields IITI = n. As n is a divisor of ITI = p r, the lemma is proved. 0 15.4 Lemma. Let G be a finite permutation group acting transitively on n and assume that n = Inl is even. If 2: is a Sylow 2-subgroup of G and (J E 8(2:), then the number of elements left fixed by (J is distinct from and ..[rl+T - 1.
rn
PROOF. Let
is invariant under 2:. Therefore,
1. If \<1>1 = Ii;, then 22a divides n. Hence a = 0, i.e. n is odd: a contradiction. If 1
70
II I. Rank-3-Planes
by * the restriction map oJG onto I\{P}. lJv E N\{l} and v* E ~3(2:*), then lJ
is an elation oj IU. In particular, every involution in 8(2:) is an elation.
m
15.6 Lemma. Let be a finite projective plane of odd order, let P and Q be distinct points of I.l3 and G a group of collineations fixing P and Q and acting transitively on PQ \ {P, Q }. Let N be an elementary abelian 2-subgroup of the Sylow 2-subgroup 2: of G and denote by * the restriction map of G onto PQ\{P,Q}.Ifv E N\{l} and v* E 3(2:*), then v is an involutory homology of In particular, every involution in 3(2:) is a homology.
m.
PROOF. Let v E N\{l} and v* E 3(L). If v* = 1, then v is a perspectivity with axis I or PQ. Thus we may assume v* =1= 1. Assume that v is not a perspectivity. Then v is a Baer involution by 4.6, i.e. v fixes a subplane of order m with m 2 = n pointwise. As IV = I resp. (PQY = PQ, we see that v fixes exactly m + 1 points of I resp. PQ, i.e. v has m = fixed points in the case of 15.5 or m - I = 1 + 1 - 1 fixed points in the case of 15.6 contrary to 15.4. 0
vn -
rn
15.7 Lemma (Wagner 1965). Let ~ be a finite affine plane and G a group of collineations oJ~. Then G operates transitively on the set of points of ~, if and only if G p operates transitively on the set of affine lines through P for all Pl/ oo '
PROOF. Let PI' ... , Ps be the orbits of G which are contained in 100 and put ~i = {III n 100 E pJ, i = 1,2, ... , s. Then ~i is invariant under G for all i. Assume that G operates transitively on the set of points of ~. Then G has exactly s + 1 point orbits in ~ V 100 , By the Theorem of Dembowski-Hughes-Parker, G has exactly s + I line orbits in ~ V 100 , As {/oo} is one such orbit and as the ~i are invariant under G, we see that each ~i is an orbit of G. Let PI 100 and let I and m be affine lines through P. Then there exists an i with P E Pi' Hence I, m E ~i' As ~i is an orbit of G, there exists I' E G with IY = m. This implies (l n looF = m n 100 = I n 100 , i.e. I' E Gp . This establishes the first part of 15.7. Conversely, assume that Gp operates transitively on the set of affine lines through P for all PI 100 , Then ~i is an orbit of G for all i. Thus G has exactly s + I line orbits in ~ V 100 , Therefore, G has exactly s + I point orbits in ~ V 100 , As G splits 100 into s point orbits, G operates transitively on the set of points of ~. 0 ~
be a finite affine plane of even order and G a group of collineations of ~. If G operates transitively on the set of points of~, then G contains a non-trivial translation of 2!. In particular, if G operates primitively on the set of points of ~, then ~ is a translation plane and G contains the translation group of ~.
15.8 Lemma. Let
71
15. Line Transitive Affine Planes
PROOF. Let PI 100 , By 15.7, Gp operates transitively on the set of affine lines through P. By the dual of 15.5, Gp contains an involutory elation To If T is not a translation, then there is a line I through P with 1=1= 100 and T E G(P, I). As Gp permutes the affine lines through P transitively, there exists an integer h E N\ {I} with IG(P, x)1 = h for all affine lines x through P. Denote by E the group of all elations with centre P which are contained in G. Then lEI = 1G(P, 100 )1 + n(h - 1). As n and lEI are divisible by h, the order of G(P,/ oo ) is divisible by h. Hence G(P,/ oo ) =t= {l}. The remaining statement is now trivial, as the group of all elations which are contained in 0 G is a normal subgroup of G.
15.9 Lemma. Let ~ be a finite plane of even order and G a group of collineations of~. If G, operates transitively on the set of affine points on I for all affine lines I of~, then G(P) =1= {I} for all P I 100 , PROOF. Let II' ... , In be all the affine lines through P. Then G(P, I,.) =1= {I} for at least one i E {l,2, ... , n, oo} by ]5.5. If i = 00, then we are done. Let i =1= 00. Obviously, G operates transitively on the set of points of ~. Hence Gp operates transitively on {II' ... , In} by 15.7. Therefore, there exists h E N\ {l} with G(P, Ii) = h for all i E {I, 2, ... , n}. This yields that h divides IG(P)I = IG(P, 100 )1 as in the proof of 15.8. 0 1
I
15.10 Lemma. Let n be an odd integer and G a permutation group of order 2n which operates transitively on a set Q of length n. If Ga, b = {I} for all distinct a, b E Q, then for each pair a, b E Q with a =1= b there exists an involution a E G with aO = b. PROOF. As n is odd, every involution in G has exactly one fixed point. Therefore, G contains exactly n involutions. Let I =1= I' E Ga' Then I{ {x, x Y} Ix E Q\ {a} } I = -i( n - I). Let I =1= 8 E Gb and assume { {x, x Y} I x E Q\ {a} }
n { {x, X O} Ix Y
Then there exists x E Q\{a,b} with x hence (1'8)2 = 1. From this we infer 1'8 therefore, a = b. Thus j
U
{{X,Xy(a)}IXEQ\{a}}j=
aEQ
= xO. = 81'.
2:
E Q\ {b} } =1= 0.
This yields x yo = x and This implies a O = a and
1{{x,xy(a)}lxEQ\{a}}1
aEQ
=
1n(n - 1),
where yea) is defined by Ga = {l,y(a)}. As 2-subsets of Q, the lemma is proved.
1n(n -
1) is the number of 0
15.11 Lemma (Wagner 1964). Let Q be a set of even length and put
IQI = n + 1. Let G be a group of permutations of Q with the property:
For all a E Q, the stabilizer Ga contains a subgroup Fa of even order which operates
72
Ill. Rank-3-Planes
transilively on Q\{a}. Furthermore, Fa,b,e = {l} for all b,c b oF c. Then Ga operales primitively on Q\{a}.
E
Q\{a} with
PROOF. Fa operates as a Frobenius group on Q\ {a}. Hence Fa has a Frobenius kernel Ka , i.e. a normal subgroup Ka which operates transitively and regularly on Q\ {a}. Furthermore, Fa contains an involution y. As IKa I = nand n is odd, y tl Ka. Hence Ka
°
°
15.12 Corollary (Wagner 1964). If in addilion to the assumption of 15.11 the Frobenius kernel Ka of Fa is normal in Ga , then Ka is an elementary abelian p-group and n is a power of p.
PROOF. Ka is abelian, as Fa, b contains an involution. Since Ga operates primitively on Q\{a}, it follows that Ka is characteristically simple. Hence Ka is an elementary abelian p-group. 0 15.13 Corollary (Wagner 1965). Let I.l3 be a finite projective plane and (0,1) a non-incident point-line-pair of 1.l3. If there exists an involutory (P, OQ)-homology for all pairs of distinct points P and Q on I, then the order of \+.~ is a power of a prime.
This follows from 13.9 and 15.12. 15.14 Lemma (Ostrom & Wagner 1959). Let I.l3 be a projective plane of order n and let r be a 2-group of collineations of 1.l3. If the fixed points and fixed lines of r form a subplane of order m, then there exists an integer g with 2g n = n1 .
PROOF. We make induction on Ifi. If r = {I}, then m = nand g = O. Assume r oF { 1}. Let y be an involution in8(r). As r fixes a subplane poin twise, Y is a Baer involution, i.e. Y fixes a subplane ~ of order s with .1,2 = 17 elemcillwise. Since Y E ,'~(I'), the group r induces a 2-grollp f* of collincalions in -;:"" I:urlhermore. the Sllhni;lllf' fiXf'd piPTnpntwi,p h" l~ ;" .,
15. Line Transitive Affine Planes
73
subplane of 0. Since y* = 1, we have Ir*1 < Ifl. By induction, there exists g 1 D an integer g such that m 2 - = s. H ence m 2g = s 2 = n. 15.15 Theorem (Wagner 1965). Let W be a finite affine plane and r a group of collineations of W. Then the following statements are equivalent:
r r
operates transitively on the set of lines of W. operates transitively on the set of affine points of Wand also on the set of points on 100 , c) r acts transitively on the set of flags of W.
a) b)
PROOF. Assume a). Let P I 100 , Then r pacts transi tivel y on the set of affine lines through P. Hence r acts transitively on the set of ~oints ~f. W by 15.7. It follows immediately from the line transitivity that r IS transItIve on I . Thus b) is a consequence of a). A~sume b). By assumption, r has a point orbit of length ~~ and a point orbit of length n + I in W V 100 , As (n 2 ,n + 1) = 1, the stablhze~ ~ p of a~ affine point P is still transitive on 100 , Hence r is flag tranSItIve. ThIS establishes c). D a) is obviously a consequence of c). 15.16 Theorem (Wagner 1965). Let W be a finite affine plane. If G is a group of collineations which acts transitively on the set of lines of W, then W is a translation plane and G contains the translation group of W.
PROOF. If G contains a translation 0:/= 1, then IG(P)I = IT(P) n G I = h > 1 for a fixed hEN and all PI 100 , since G operates transitively on 100 , This yields the assertions of 15.16 by 15.2. Hence it suffices t.o prove that G contains a non-identity translation. If the order n of W IS even, then G contains such a translation by 15.15 and 15.8. Therefore, we may assume that n is odd. a) If (P,l) is a flag of Wand k = IGp"I, then I~I = n 2 (n + l)k. This follows from 15.15 and the remark that n (n + 1) is the number of flags of W. . As n(n + 1) is the number of lines of W, we obtam by a): b) IG,I = nk. As n 2 is the number of points of W, we have by a): c) IGpl = nk. Moreover, if A I '00' then IGp,AI = k. The last assertion follows from Gp,A = Gp,PA and a). Since GA permutes transitively the points of W, we have: d) IGAI = n 2k for A 1100 , As G acts transitively on the set of points of W, it follows from 15.7 that GA , : operates transitively on the set of affine lines through B fo: all B I 100 , Hence we have proved: e) Let A, B I 100 , If I and m are two affine lines through B, then IGA,B"I = IGA,B,ml· Next we prove that n is a power of a prime. Put SU = S2l. V 100 and let @3 hI" 8 sllhplane of 1-l3 with the properties:
74
III. Rank-3-Planes
(I) There exists a 2-subgroup of C the fixed points and fixed lines of which
are the points and lines of
(S.
(2) There does not exist a proper subplane of @3 with the property (1). As SU and {I} satisfy (1) and as SU is finite, there exists such a subplane From I~ = 100 we infer that foo is a line of 6. Let L be a 2-subgroup of C and assume that L is maximal with respect to the property that the fixed points and fixed lines of L are the points and lines of 6. Let ~ be the affine plane one obtains from 6 by deleting f eo and put A = G'1J' Furthermore, let 2G be the highest power of 2 dividing n + I and 2b the highest power of 2 dividing k. Put ILl = 2e • f) Each flag of ~ is left fixed by an element of A which induces in ~ an involutory homology. Let (P,/) be a flag of~. Then L c: C P .,' Hence, by a), Y divides 2b. As n is odd, a + b > b + 1. Therefore there exists a 2-group L* of C with L c: L* and IL*: LI = 2. This implies that L is normal in L* and hence L* c: A. By (2), L* induces an involutory homology in ~. Thus L* fixes a flag (Q,m) of ~. Since IGQ.ml = k and L c: L* c: GQ,m' we obtain c < b. Therefore, L is not a Sylow 2-subgroup of G p ,,' Hence, there exists a 2-group L** with L c: L** c: Gp , and IL** : LI = 2. This L** induces an involutory homology in ~ which' fixes (P, I). g) Let A, B I leo and assume that A, B belong to 6. Furthermore, let I and h be lines of ~ through B. If there exists a collineation in A inducing an involutory homology in ~ and fixing A, Band f, then there exists a collineation in A inducing an involutory homology in ~ and fixing A, B and h. Let A be the subgroup of A which fixes 6 pointwise. Then L c: A and A;;;;J A. Hence A <) AA,B". By assumption AA,B,,/A contains an involution. This shows that L is not a Sylow 2-subgroup of GA B ,. Bye), IGA,B"I = IGA,B,hl. Hence L is not a Sylow 2-subgroup ~r' GA,B,h' Therefore, there exists a 2-group L* with L c: L* c: GA , B, hand IL* : LI = 2. Then L* induces an involutory homology in 6 which fixes A, Band h. h) The order of mis a power of a prime. By 15.14, it suffices to show that ~ has prime power order. As A/A is isomorphic to a group of collineations of 6, we may assume A = {I} and we may identify A with the group of collineations induced on 6 by A. Case 1: A does not contain an involutory homology with centre a point of ~. By f), A contains an involutory homology G. Let f be the axis and A the centre of a. The A I leo' Using g) and 4.8, we see that each affine line through I n I eo is the axis of an involutory homology in A with centre A. Therefore, 6 is (A,/eo)-transitive by 13.10. We may assume that ® is not a translation plane. Then A is the centre of every involutory homology in A. By this and f), each line of ?B which does not pass through A is the axis of an involutory homology. This implies by 13.10 that 6 is the dual of a translation plane. Hence the order of ~ is a power of a prime.
15. Line Transi tive Affine Planes
75
Case 2: There is exactly one point 0 in 'i~ which is the centre of an involutory homology in A. In this case, f) and g) together with 15.13 yield that the order of ~ is a power of a prime. Case 3: There exist two distinct points of ~ which are cen tres of involutory homologies of A. Assume that every line of ~ carries a centre, and let v be the number of centres. Furthermore, denote by k, the number of centres on f. Then v(m+ l)=2.,k" where m is the order of~. As k, > 1 for all f and since there exists I with k, > 2, we have v(m + 1) > m(m + 1). Hence v> m. Therefore, by 13.9 and 15.3, m is a power of a prime. We may thus assume that there exists a line I of ~ which does not carry a centre. Put B = f n I eo and let B =1= A I 100 and suppose P is an affine centre. As there exists an involutory homology fixing A, Band AP, there is an involutory homology p fixing A, Band f by g). If I is the axis of p, then A is its centre. If f is not the axis, then B is the centre, as I does not carry an affine centre. In either case, A is the only fixed point of p in leo \{B}. Hence A B " acts transitively on foo \{B} by 15.1. As there are two centres, A contains a non-trivial translation T by 13.9. If B is not the centre of T, then the translation group of ~ has order greater than m by the transitivity of AB on 100\ {B }. Hence m is a power of a prime by 15.3. Therefore, we may assume that all translations in A have centre B. This implies that all affine centres lie on a line h through B. Thus hI>. = h. Moreover, if a is an involutory homology in A fixing f, then B is the centre of a, as h a = h. Using the transitivity of A B " on I eo \ {B}, we see by 13.9 that there exist at least m - 1 non-trivial elations with centre Band axis distinct from leo' As A(B, leo) =1= {I}, the group of all elations with centre B has order greater than m. Hence, by the dual of 15.3, m is a power of a prime. By h), we have n = pr for some prime p. Let IT be a Sylow p-subgroup o~ Then IT acts transitively on the set of p2r points of W. As p does not dIVIde n + 1, there exists A I foo with AIT = A. Let B be a point on f distinct from A. Then IBITI divides IITI and IBITI < n = pr. As p2r divide~ IITI, we have th.at pr divides IITBI. Choose B =1= A such that IITBI > IITei for all C I 100 wIth C =1= A. Let F be a fixed point of ITB with A =1= F I f eo . Then ITB = ITB.F c: IT F · As IITBI > IITFI, we have ITB = IT F . From the transitivity of IT on the set of points of m, we infer from 15.7 that ITF and hence ITB operate transitively on the set of affine lines through F. Furthermore, ITB also acts transitively on the set of affine lines through A, as we shall see now. Let L be a Sylow p-subgroup of G which contains B IT B' By d), 1GB I = n 2k. Hence L IS. a Sylow p-subgroup of G, as 2 IGI = n (n + l)k. Furthermore, ITB c: LA' This implies ITB = LA' since L and IT are conjugate. By 15.7, LA and hence ITB have the required transitivity property. Finally, let 1 =1= T E 3(IT B ). Then T fixes the points A and B. Thus T fixes at .least one line I of m. If I n 100 is not a fixed point of IT B , then there eXIsts p, E ITB such that IP. n I is an affine point which is obviously fixed by
fl.
76
III. Rank-3-Planes
Thus (Ill n I)A is a fixed line of 1'. As «(Ill n I)A) n 100 = A, we infer that there is always a fixed line I of l' such that I n III is a fixed point of TI B • By what we have seen, I/TIBI = n. Hence all the lines through I n 100 are fixed by 'T. As 0(1') = pS, we infer that 'T is an elation. This and A 1" = A aI?-d B 1" = B yield that 'T is a translation. 0 1'.
15.17 Theorem. Let mbe a finite affine plane. If G is a group of collineations which acts transitively on the set of lines of m, then G acts primitively on the set of points of m. PROOF. By 15.16, mis a translation plane and G contains the translation group T of m. Furthermore, Gp acts transitively on the set of lines through P for all points P of m. Let ?B be a set of points of 2:r such that I?BI ;> 2 and ~Y = ~ for all y E G with ~Y n ~ =1= 0. We have to show that ~ consists of all the points of m. In order to prove this, we note first that ITlEI = I~I. Pick a point P E~. As Gp operates transitively on 100 , there exists an hE N\{l} such that ITIE n T(A)I = h for all A I 100 , It follows from 15.2 that T(A) C; TQ"l and hence T = TQ"l' 0
16. Affine Planes of Rank 3 Let G be a permutation group on the set Q. We shall call G a rank-3-group on Q, if it acts transitively on Q and if Go for ex E Q splits Q into exactly 3 orbits. An affine plane mwill be called a rank-3-plane, if it admits a group of collineations acting as a rank-3-group on the set of points of m. Any such collineation group will be called a rank-3-collineation group of 2:r. Every nearfield plane is a rank-3-plane. For, if P and Q are the two "special points" on 100 , then G = TLl(P, OQ)Ll(Q, OP) permutes the points of m transitively. Furthermore, Go = Ll(P, OQ)Ll(Q, OP) has exactly four orbits, namely {O}, OP \ { O}, OQ \ { O} and the set of the remaining points. As there exists a collineation p fixing 0 and interchanging P and Q by 3.11, we see that is a rank-3-collineation group of the given nearfield plane. Every Hall-plane is a rank-3-plane: Let the Hall plane H(q) be derived from the desarguesian plane A (q2) via D. Consider the collineation group G of A (q2) which fixes D globally. As we have seen, G splits the points of 100 into two orbits. Furthermore, IG( 0,/ 00 )1 = q2 - 1. Hence the stabilizer of every line 1 in G acts doubly transitively on I. Hence G operates as a rank-3-group on the set of points of A (q2). As G is also a collineation group of H(q), we see that H(q) is a rank-3-plane. The Luneburg planes which we shall construct later on are also rank-3-planes. The next theorem was proved independently by Johnson & Kallaher 1974 and by myself 1973.
G
2. But then 49.1 and 49.4 yield p > 5 and hence q > 5. Moreover k = 11 0, as (X - 1)4 is the minimal polynomial of all elements of order p of S. Conjugating 0 by Pa- I yields ka 2s =1a2s = a-Ia-'PIs= ka-Ia-'Ps. Hence a'P = a -3 for all a E K. Using the matrices
16. Affine Planes of Rank 3
77
16.1 Theorem. Let ~{ be a finite affine plane and let G be a group of collineations of m. If G, acts doubly transitively on the set of points on I for all lines I of then is a translation plane and G contains the translation group
m,
m
afm. PROOF. Let n be the order of m. a) For UI 100 either G(U) = {I} or IG(U)I = n. . Assume G( U) =1= {l} and let I be an affine line through U. As G( U) IS normal in G! and as G( U) operates faithfully on I, we have IG( U)I = n by' the 2-transitivity of Gt on I. Case 1: n is even. Then G( U) =1= {l} for all U I 100 by 15.9. Hence we are done by a). Case 2: n is odd. Obviously, we have: b) G is point transitive. Next we prove: c) If G contains an involutory homology with affine centre, then 16.1 holds. This follows from b) and 13.9. d) If G( U) = {I} for all U I 100 , then there exists V I 100 such th.at the axis of every involutory homology in G passes through V. In partIcular, V G = V. Let a be an involutory homology in G and let 1 be the axis of a. By c), 1 =1= 100 , Then V = 1 n 100 , Let h be a second line thro~gh V. ~y b) and 15.7, there exists y E Gv with IY = h. Hence there eXIsts an Involutory homology p with axis h. Let U be the centre of a and W be the centre of p. If U = W, then 13.9 implies G ( U) =1= { 1}. Hence U =1= W. As there are exactly n affine lines through V, each point on 100 which is distinct from V is the centre of a homology with axis through V. By 15.1, G v acts transitively on 100 \{ V}. As G canno~ be transiti~e on 100 by 15.16, G = G v · Let l' be an involutory homology III G. If V IS the centre of 1', then G contains an involutory homology with affine centre by 4.9 contradicting c). Hence V is on the axis of 1'. e) If G has no fixed points, then 16.1 holds. This follows from d) and a). Therefore, we may assume that G has a fixed point V I 100 , f) If the axis of every involutory homology in G passes through V, then 16.1 holds. If I is an axis, then we are done by c). Assume that 100 is not an axis. Let U I oo and U =1= V. Furthermore, let h be an affine line through U. As h is fixed by an involutory homology in Gh by 15.6, U is the cen.t~e of an involutory homology in G. As Gil = G V . il ' the group Gil acts tranSItIvely on the set of affine lines through V. (Remember that Gh acts doubly transitively on I.) Using 13.9, we obtain I G( U)I = n. As this is true for all U =1= V, the plane 9[ is a translation plane and C contains the translation group.
I
78
III. Rank-3-Planes
By f), d) and a), there exists a point U I 100 with IG( U)I = n. If U is not a fixed point of G, then 16.1 holds. Hence we may assume V = U. Furthermore, we may assume that there exists an involutory homology in G with centre V. Let X I 100 , X =1= V. Choose an affine line h through X. Then X is fixed by an involutory homology p. We may assume that 100 is not the axis of p. Then V and X are the only fixed points of p on 100 , Hence, by 15.1, G = G v acts transitively on 100 \{ V}. Therefore, every point U I foo with U =1= V is on the axis of an involutory homology with centre V. Therefore·, we may assume by 4.8 that V is the centre of every involutory homology in G. Applying 15.6 again, we see that every line h not through V is the axis of an involutory homology in G. This yields by 13.9 that the dual of W V f 00 is a translation plane with respect to V. This implies by 1.12 that there is exactly one involutory (V, h)-homology for all lines h not through V. Again let h be a line not through V and let 2: be a Sylow 2-subgroup of Gh . Furthermore, let A be an abelian normal subgroup of 2: and put N = {o: 10: E A, 0: 2 = I}. Then N is a characteristic subgroup of A and hence a normal subgroup of 2:. Let * be the restriction of Gh onto h. Let 1 =1= a E N with a* E 3(2:*). By 15.6, a* is a homology. Therefore, h is the axis of a. Hence a* = 1, i.e. N* n 3(2:*) = {l}. This implies N* = {l}, as 2:* is a 2-group. Then N consists of (V, h)-homologies only. This yields INI < 2 by the above remark. Therefore, A is cyclic. We infer from this by Gorenstein [1968, 5.4.10] that 2: is either cyclic, or dihedral, or semi dihedral, or a generalized quaternion group. In either case, 2:* contains a cyclic subgroup Z* of index 1 or 2. Let Z be the preimage of Z*. Then \2:: Z\ < 2. Hence Z n 3(2:) =1= {I}, whence Z contains the only central involution a of 2:. (Recall that 3(2:) is cyclic.) Let l' be an involution in Z. Then 1'*2 = 1. Hence 1'* E Z* n 3(2:*). This yields 1'* = 1 by 15.6. Therefore, l' = a and Z contains exactly one involution. As n is odd, 2: has an affine fixed point P on h. Let Q and R be affine points distinct from P with Q I PV, R I hand
16. Affine Planes of Rank 3
/':J
As n is odd, (*) yields ILQI = ILRI. Since the involution a which lies in Z has no fixed points on PV other than P, we have LQ n Z = {I). This implies ILQI < 2. On the other hand a E 2: R . Thus ILRI;;;. 2. Hence ILQI = I~RI = 2. Let 1 =1= l' E 2: Q , then T has at least 2 ~ffine fixe~ point~ on pv. As PV cannot be the axis of T, we infer that T IS a Baer InvolutIOn. Hence l' fixes a point F on h which is distinct from P and I n 100 , As a ELF' we obtain the contradiction 2 = ILRI ;;.IL: F \;;. 4. This final contradiction proves 16.1. 0
16.2 Lemma (Kallaher 1969b). Let m be a Jinite aJfine plane and G a rank-3-collineation group oj W, then one oj the Jollowing holds: a) G acts transitively on the set oj lines oj mand G{ acts as a rank-3- group on 1 Jor all lines I oj m. b) G has exactly two orbits on 100 and G{ operates doub(v transitively on I Jor all lines I oj m. Let P be a point of Wand let p and q be the orbits of Gp other than {P}. Furthermore, let 1 and m be lines through P and assume that I carries points of p as well as points of q. As m meets at least one of p and q, we may assume that m carries a point R of p. Let Q be a point of p which is on I. Then there exists y E G p with QY = R. Hence IY = m. This implies that G is flag transitive. Furthermore, G{ acts as a rank-3-group on I for the particular 1 under consideration and hence for all lines, as G acts flag transitively on W. This is case a). Thus we may assume that for all lines I through P either I\{P} p or I\{P} q. Therefore, the set of lines through P is split into two orbits, one of them consisting of all the lines I with I\{P} C;:p and the other one consisting of all the lines I with 1\ {P} c;: q. Therefore, G has at most two orbits on 100 , But G cannot be transitive on 100 , as in this case G would be flag transitive by 15.15. Hence G splits 1.:1J into two orbits. As Gp , { operates transitively on 1\ {P}, we see finally that G{ acts doubly transitively on I. This is case b). 0 PROOF.
c:
c:
\2:QI = max{ l2:xlip =1= X I PV} and
12: R I = max { 12: y Ii P =1= Y I h}. We show that 2: Q is a Sylow 2-subgroup of Gp , Q, h' Let ~ be a Sylow 2-subgroup of G p , Q, h which contains 2: Q . Then there exists y E G p , h with ~Y 2:. As 1:1 Gp,Q,h' we have ~Y 2: Qy • Hence
c:
c:
c:
I2:QI ;;. I~QYI
;;.
I~YI = I~I ;;. I~QI·
This establishes 1:1 = L Q . Similarly, ~R is a Q is contained in an orbit of length n on PV. Furthermore, R lies in an orbit IG(V,PV)I = n and as Gp .{ is transitive on
(*)
IGpl = (n
Sylow 2-subgroup of Gp , R' 1 of Gp, as Gpv is 2-transi tive of length n(n - 1) of Gp , as I\{P}. Therefore, we have:
- 1)IG p ,QI = n(n - 1)IGp ,RI.
16.3 Corollary (Kallaher 1969b, Liebler 1970). Let W be a Jinite aJJine plane. IJ G is a rank-3- group oj collineations oj m, then W is a translation plane and G contains the translation group oj m. This follows from 16.2, 16.1 and 15.16.
16.4 Theorem (Kallaher 1969b). Let mbe a finite aJJine plane and let G be a rank-3-collineation group oj m. Then one oj the Jollowing holds:
a) G has no fixed point on 100 and G operates primitively on the set oj points
oJm. b) G has a fixed point on 100 and G operates imprimitively on the set oj points
oJm.
This follows from 16.3, 16.2, 15.17 and 1.3.
80
III. Rank-3-Planes
17. Rank-3-Planes with an Orbit of Length 2 on the Line at Infinity Our aim in this section is to prove: 17.1 Theorem (Kallaher & Ostrom 1971, Kallaher 1974, Liineburg 1974, 1976 a + b, ~Lineburg & Ostrom 1975). Let m be a finite affine plane of order n. If G IS a rank-3-coflineation group of msuch that G has an orbit of length 2 on I eo' then ~( is a generalized A ndre plane or n = 52 72 112 23 2 29 2 59 2 . ' , , , ,
The numbers listed in this theorem are really pla.nes. over the seven exceptional nearfields satisfYIng the hypotheses of the theorem. But it that we have done so far to prove that these Andre planes.
exceptions, as the nearfield admit collineation groups is an easy exercise after all planes are not generalized
PROOF. According to 16.3, 91 is a translation plane and G contains the translation group T of ~!. Let 0 be an affine point of 21 and put H = Go. Then G = T H. As T operates trivially on leo, the groups G and H have the same orbits on leo' Let {P, Q} be the orbit of length 2 of H on I . H has three point orbits on ~l. One is {O}. The other two consist of the points on OP and OQ other than 0, resp. the points off the lines OP and OQ. The length of the first of these is 2( n - 1) and the length of the second (n - 1)2. Furthermore, IH : H P, QI = 2. Since m is a translation plane, n = p r where p is a prime. Let 'TT be a p-primitive prime divisor of p r - 1 and let 'TT S be the highest power of 'TT dividing p r - I. If 'TT = 2, then p is odd and hence r = l. In this case mis desarguesian and hence a generalized Andre plane. Therefore, we may assume I' > 1 whence'TT > 2. As (n - 1)2 divides IHI, the highest power of 1r dividing IH[ is 'TT 2s + 1 for some non-negative integer t. Since 'TT 0:/= 2 and 2s 1 IH: H p, QI = 2, we see that 'TT + divides IHp, Q[' Put K = H p, Q and den0te by K(P, OQ) the group of all (P, OQ)-homologies contained in K. Define K(Q,OP) similarly. Let 2: be a Sylow 'TT-subgroup of K. Then 2: n K(P, OQ) is a Sylow 'TT-subgroup of K(P,OQ), as K(P,OQ) is normal in K, The order of K(P, OQ) divides n - l. Hence [2: n K(P, OQ)[ -( 'TT s . Let 2:* be the centralizer of T(Q) in 2:. Then 2:* <: K(P, OQ), since 2:* fixes 0 and T(Q) acts transitively on OQ. In particular, 2:* <: K(P, OQ) n 2:. By 3.8, K(P, OQ) n 2: <: 2:*. Hence 2:* = K(P, OQ) n 2:. . As 2:*. is the centralizer of T(Q) in 2:, the quotient group 2:/2:* is IsomorphIc to a subgroup of OL(r, p). Hence 2:/2:* is cyclic and 12:/2:*1 :;;;; TT S by 10.5. Since 7T 2s + I
~e i,nequalities 12:/2:*1 :;;;; -
7T .
'TiS
=
[2.:[
=
12: / 2: *112: * I,
and 12:*1 -(
'TT
s
imply r = 0 and 12:/2:*1
=
12:*1
81
17. Rank-3-Planes with an Orbit of Length 2 on the Line at Infinity
Similarly 12:**1 = 'TT s where 2:** = 2: n K(Q, OP). Hence 2: = 2:*2:**, as 2:* n 2:** = {I}. Furthermore, ~** ~ ~/~*. Hence ~** is cyclic. Likewise, 2:* is cyclic. As L* and L** centralize each other, 2: is abelian. Assume that L* is normal in K(P, OQ). As L* is a Sylow 7T-subgroup of K(P,OQ), it is the only Sylow 7T-subgroup of K(P,OQ). Furthermore, K(P,OQ) and K(Q,OP) are conjugate in H. Therefore, 2:** is the only Sylow 7T-subgroup of K(Q, OP). Therefore, 2: is normal in K(P,OQ)K(Q, OP) and hence in H. The group H acts transitively on leo \{ P, Q}. Therefore, 2: splits leo \ {P, Q} into orbits of equal length. As this length is a divisor of n - 1 and is, at the same time, greater than or equal to 7T s , it is equal to 7T S. Therefore, ILwl = 7T s for all W I leo with W 0:/= P, Q. Let a E 2: ween tralize T( W). Then a fixes each point on 0 W, as it fixes 0 and as T( W) acts transitively on the set of affine points on 0 W. This together with po = P and QO = Q implies a = 1. Thus 2: w operates faithfully on T( W). It follows therefore from lO.5 that 2: w operates irreducibly on T( W). Hence mis a generalized Andre plane in this case by 9.3. Our next aim is to show that 2: is almost always normal in K(P,OQ). 17.2 Lemma (Hering). Let V be an elementary abelian p-group of order pr and let 7T be a p-primitive divisor of p r - 1. Furthermore, let 2: be a non-trivial 7T-subgroup and A an abelian subgroup of OL(r, p). If 2: normalizes A, then 2:A is cyclic.
PROOF. 2: and hence 2:A act irreducibly on V by 10.5. Therefore, by Clifford's Theorem (see e.g. Huppert [1967, V.I7.3, p. 565]), V is a completely reducible A-module. Let Vi' ... , Vs be the homogeneous components of the A-module V. Then A2: and hence 2: operate transitively on {VI' ... , VJ. Thus s divides [2:[. This implies that 7T divides s provided s 0:/= l. Let I VII = pl. Then pr = piS, as V = EB~= i Vi and I Vii = pi for all i. As 7T is a p-primitive divisor of pr - 1, we see that I' is the order of p modulo 7T. Hence I' = st divides 7T - 1. Thus s is not divisible by 'TT, whence s = 1. Therefore, V = VI is the direct sum of isomorphic irreducible A-modules. We infer from 9.1 that IAL: ~AL(A)[ divides r. As A is abelian, IA2:: ~AL(A)I is a power of 7T. On the other hand I' divides 7T - 1, as we have seen above. Thus A2: ~ ~AL(A). Since 2: is cyclic by lO.5, A2: is abelian. Schur's Lemma then implies that A2: is cyclic. 0 17.3 Corollary. Let V be an elementary abelian p-group of order pr and let 7T be a p-primitive prime divisor of pr - l. Let N be a subgroup of OL(r, p) all Sylow subgroups of which are cyclic. Then N contains exactly one Sylow 7T-subgroup.
PROOF. By Passman [1968, Prop. 12.11], there exists a cyclic normal subgroup M of N such that N 1M is also cyclic. Let 2:: be a Sylow 7T-subgroup of N. Then ML is cyclic by 17.2. Furthermore, N' M, as N/M is abelian. This yields Nt <: M2:. Hence M2: is normal in N. As 2: is characteristic in ML, it is normal in N. 0
c:
82
III. Rank-3-Planes
We now return to the proof of 17.2. By 3.5, we have: (A) If K(P, OQ) is soluble, then K(P, OQ) contains a normal subgroup N all Sylow subgroups of which are cyclic and such that K(P, OQ)/ N is isomorphic to a subgroup of S4' (B) If K(P,OQ) is not soluble, then K(P,OQ) contains a normal subgroup L of index 1 or 2 which is the direct product of a group M ~ SL(2, 5) and a group N with cyclic Sylow subgroups. Furthermore, (INI, 30) = 1. If 1T =1= 3 and if K(P, OQ) is soluble in case 1T = 5, then 2:* is normal in K(P,OQ) by (A) and 17.3. Therefore we may assume 1TE{3,5} for all p-primitive prime divisors 1T of pr - 1 and furthermore that K(P,OQ) is not soluble, if 5 is a p-primitive prime divisor of p r - 1. If 1T = 3, then r = 2 as p2 == 1 mod 3. If 1T = 5, then p2 == 1 mod 5 or p4 == 1 mod 5. Thus r = 2 or 4 in this case. If r = 4, then 5 divides p2 + 1. As IK(P,OQ)I divides n - 1 = p4 - 1 and SL(2, 5) ~ K(P, OQ), we have that 120 divides p4 - 1. Consequently p =1= 2,3. This yields p2 - I == 0 mod 12 and hence p2 + 1 = 2 . Y, since every odd prime divisor of p2 + 1 is a p-primitive prime divisor of p4 - 1. Furthermore 5' = 12:*1. It follows from (B) that 12:*1 = 5. Hence t = 1, i.e. p2 + 1 = 10. As a consequence p = 3: a contradiction. Thus r = 2 in either case. 17.4 Lemma (Luneburg 1974). Let W be a finite affine plane and let I be an affine line of Wand P a point on 100 with P,f I. If G is the collineation group of Wand if I n 100 is not fixed by Gp, then either 1:::.( P, I) is cyclic or T(P)={I}.
PROOF. Assume T(P) =1= {l}. If I:::.(P,/) = {l}, there is nothing to prove. Hence we may assume I:::.(P, I) =1= {l}. Lety E Gp be such that IY is not parallel to I. Then I:::.(P,IY) = I:::.(P,/)Y =1= {l}. Therefore, by 13.9, there exists an elation € with centre P and I~ = IY. As I and IY are not parallel, we have € Ii T(P). It follows that the group E of all elations with centre P has a non-trivial partition 1T, all components of which are normal in E. Hence E is abelian by l.l and K(E,1T) is a Galoisfield by 1.6 a). As I:::.(P, I) is isomorphic to a subgroup of the multiplicative group of K(E,1T), we see that I:::.(P, I) is cyclic. 0 17.5 Lemma. Let p be a prime and let W be a translation plane of order p2. Furthermore, let 0, P, Q be three distinct points of W with P, Q I 100 and O,f 100 , Denote by G the collineation group of m. If G(P,OQ) contains a subgroup isomorphic to SL(2,5) and if Go. p operates transitively on OP\{O,P}, thenp = 11,19,29,59. PROOF. As T(P) =1= {I} and G(P,OQ) is not cyclic, 17.4 yields Gp = Gp,Q' Furthermore, by 3.5, G(P, OQ) contains a subgroup L of index 1 or 2 and L = S X N where S ~ SL(2, 5) and (I N 1,30) = 1. This yields that S is the
83
17. Rank-3-Planes with an Orbit of Length 2 on the Line at Infinity
only subgroup of L which is isomorphic to SL(2, 5). As SL(2,5) does not contain a subgroup of index 2, the group S is the only subgroup of G(P,OQ) which is isomorphic to SL(2, 5). Hence S is normal in Go,P,Q = Go,p' Since Go,p operates transitively on OP\{P,O}, it induces a group of automorphisms in T(P) which acts transitively on T(P)\ {l}. Hence Go, p induces a group Go, p in PGL(2, p) which acts transitively on ~e projective liEe o~er GF(p). Hence p + 1 divides I pI. Obviously S ~ As. Hence S = Go, p by 14.6. Therefore, p + 1 divides 60. This yields p E {2,3,5, 11, 19,23,29,59}. On the other hand 120 divides IG(P,OQ)I which is a divisor of p2 - l. Thus P > 11 and p =1= 23. 0
Ga.
Hence, by 17.5, if 1T E {3, 5} and if K(P,OQ) is not soluble, then
n = 11 2 , 19 2 , 29 2 or 59 2 . As we have seen, the numbers 11 2 , 29 2 and 59 2 are really exceptions. We shall prove in section 20 that 19 2 does not occur among the exceptions. We may now assume that 3 is the only p-primitive prime divisor of pr - 1 and that K(P, OQ) is soluble. We may further assume that 2:* is not normal in K(P, OQ). The next lemma takes care of this situation. 17.6 Lemma. Let p be a prime and let W be a translation plane of order p2. Furthermore, let 0, P, Q be three distinct points of III with P, Q I 100 and o ,f 100 , Denote by G the collineation group of Ill. If G(P, OQ) is soluble and contains more than one Sylow 3-subgroup and if furthermore Go. p operates transitively on OP\{ 0, P}, then p E {5, 7,11, 23}.
PROOF. Go,p = Go,P,Q by 17.4. Hen~e G(P,OQ) is normal in Go,p' By assumption, Go, p induces a group Go. p in PGL(2, p) which acts transitively on ..!,he set of points of the projective line over GF(p). Thus p + 1 divides IGo, pI. The group G(P,OQ) is mapped by - onto a normal subgroup M of Go, p' As G(P,OQ) contains more than one Sylow 3-subgroup and as G(P, OQ) is soluble, we infer from 14.1, using the fact that p does not divide IG(P, OQ )1, that M contains a subgroup A with A ~ A4 and 1M: A I = 2. Since A is generated by its Sylow 3-subgroups, A is char~cteristic in M and hence normal in Go, p' Therefore, by 14.4 and 14.5, IGo, p: A I < 2. Hence p + 1 divides 24. This yields p E {2, 3, 5, 7,11, 23}. As 12 divides p2 - 1, we find p E {5, 7,11, 23}. 0 By 17.6, n E {52, 11 2 , 23 2 }, if 3 is the only p-primitive divisor of p r - I and if K(P, OQ) is soluble. We may assume from now on that there does not exist a p-primitive prime divisor of p r - 1. Applying 6.2 yields p r = 26 or r = 2 and p + I = 2S for some integer s. Assume pr = 26. As H acts transitively on the set of 63 2 points off the lines OP and OQ and since IH : KI = 2, the group K also permutes this set of 63 2 points transitively. Assume that 3 does not divide IK(P,OQ)I. Then
84
III. Rank-3-Planes
the S~low 3-s~bgroup ?f K operates faithfully on T(Q). As 34 divides IKI and SInce 34 IS the hIghest power of 3 dividing IOL(6,2)1, each Sylow 3-subgroup of K is isomorphic to a Sylow 3-subgroup of OL(6, 2). Let ~ be a Sylow 3-subgroup of K. As K permutes the 63 2 points off the lines OP and OQ transitively, ~ operates regularly on the set of these points. Furthermore, ~ has an orbit @3 of length 9 on OP. Let X E @3. Then I~xl = 9. M.areover, ~x is elementary abelian, as we shall now prove. Let T E T(P) with OT = X. Then ~x = ~~(T). Thus it suffices to show that the Sylow 3-subgroup of OL(6,2)v is elementary abelian for a non-zero vector v in the vector space V of rank 6 over OF(2). Let V = VI EB V 2 EB U3 with 1Vii = 4. Let OJ (i = 1,2) be elements of order 3 in OL( Vi) and define Pi by (u l
+ U 2 + U 3t l = ufl + u 2 + u 3
resp. (UI
+ U2 + u 3t 2 = u l + U~2+
u3 •
Then
we infer that K* contains a normal subgroup 2 which is isomorphic to K(Q,OP). As K(Q,OP) is conjugate to K(P,OQ) in H, the order of K~Q, OP) is divis.i~le by 3. Thus 121 is divisible by 3. Assume 121 = 3. As K operates .t~ansltlvely on T(Q)\{ I}, the 2-irreducible subgroups of T(Q) form a partItIOn '7T of T( Q) the components of which all have the same order, namely 4. Using l.6 a) and the finiteness of T(Q), we find that K* C;; fL(3, 4). As 72 divides IKI and as IK(P, OQ)I = 121 = 3, the order of K* is divisible by 72. Hence 7 2 divides IfL(3,4)1 = 27 . 34 • 5 ·7: a contradiction. Therefore IK(P, OQ)I > 3, i.e. IK(P, OQ)I E {9,21,63}. If IK(P, OQ)I = 63, then W is either desarguesian or a nearfield plane. In the first case, W is a generalized Andre plane. Let W be a nearfield plane. As 3,9 oF- 1 mod 7, there is exactly one Sylow 7-subgroup 2 in K(P,OQ). Fur~hermore, IAut(Z)1 = 2 . 3 and all Sylow 3-subgroups of K(P, OQ) are cyclIc. Hence the centre of K(P,OQ) has an order divisible by 3. Thus ~here exis~s a normal subgroup N of order 21 in K(P, OQ). As K(P, OQ) is Isom?r~hlc. to the multiplicative group of a coordinatizing nearfield F, the multtphcatlve group of F contains a normal subgroup A of order 21. Obviou.sly, A operates irreducibly on F( +). Therefore, F* C;; fL(l, 64) by 9.1. ThiS shows that F is a generalized Andre system.
17. Rank-3-Planes with an Orbit of Length 2 on the Line at Infinity
If IK(P, OQ)I
=
85
9, then IK(P, OQ)K(Q, OP)I = 81 and A = K(P,OQ)K(Q,OP)
is normal in H. This implies that all the orbits of A in 100 \{P, Q} have the same length A. As A divides 63 and 81 and is divisible by 9, we have A = 9. Therefore, IAwl = 9 for all WI foo with W =1= P, Q. By 3.5 a), K(P,OQ) and K(Q,OP) are cyclic. Moreover A is abelian and A w is a diagonal of A = K(P, OQ)K(Q, OP). Hence Aw is likewise cyclic. From this we infer that Aw operates irreducibly on T( W). By 9.3, W is a generalized Andre plane in this case. Finally if IK(P, OQ)I = 21, then K(P,OQ) is cyclic by 3.5 c). Hence A = K(P, OQ)K(Q, OP) is abelian. Again all orbits of A which are contained in 100\ {P, Q} have the same length A. Hence A = 21 or 63. If A = 21, then it follows as above that W is a generalized Andre plane. Thus assume A = 63. If WI 100 and W =1= P, Q, then IAwl = 7. Let S be a Sylow 3-subgroup of A. Then S n A w = {I} for all such W. Hence S, acts regularly on 100 \{P, Q}. As the group K* of automorphisms induced on T(Q) by K contains a cyclic normal subgroup of order 21, namely K(Q,OP)*, we deduce, with the aid of 9.1, that K* C;; fL(l, 64). Therefore 2 IK*I divides 2.3 3 .7. On the other hand, 63 divides IKI· Moreover, 3 IKI = IK(P, OQ)IIK*I = 21IK*I. Thus 3·63 = 3 ·7 divides IK*I· Therefore, IKI = 63 2 or IKI = 2.63 2. In either case, K contains a normal 2 subgroup B of order 63 2. As K permutes the 63 points off the lines OP and OQ transitively, B does also. Moreover B acts regularly on the set of these points. In particular, Bw acts sharply transitively on OW\{ 0, W}. Let ~ be a Sylow 3-subgroup of Bw' Then ~ has order 9. Hence ~ is abelian. Moreover, S n ~ = {I}. Let SI be a Sylow 3-subgroup of K(P,OQ) and S2 a Sylow 3-subgroup of K(Q,OP). Then S = SI S 2' Furthermore ~ normalizes each Si' As ISil = 3 and I~I = 9, it follows 4that ~ centralizes S I and S2' Hence S ~ is an abelian group of order 3 . In particular, S~ is a Sylow 3-subgroup of K. This yields that (S~)* is a Sylow 3-subgroup of K*. As I( S~)* I = 33 and K* C;; fL(l, 64), we see that (S ~)* is a Sylow 3-subgroup of fL(l, 64). This is a contradiction, since (S ~)* is abelian but the Sylow 3-subgroups of fL( 1,64) are not. This 6
settles the case n = 2 . It remains to consider the case n p
=
p2 where p is a prime with
+ 1 = 25.
17.7 Lemma (Luneburg 1976 a + b). Let p be a prime with p == 3 mod 4 and let I2f be a translation plane of order p2, Furrhermore, let 0, p, Q be three distinct points of I2f with P, Q I ! x and 0 J I x ' If G is a group of col/ineations of W fixing 0, P and Q and permuting transitive{v the affine points off the
fines OP and OQ, then the order of a Sylow 2-subgroup of G(P,OQ) is either 25 or 25 + I, where 25 is the highest power of 2 dividing p
+ I. Moreover
Ill.
KanK-J-nalle~
if the order of a Sylow 2-subgroup of G(P,OQ) is 2s, then the Sylow
2-subgroups ofG(P,OQ) are cyclic, unless 25 + 1 divides IG(Q,OP)I. PROOF. Let 2: be a Sylow 2-subgroup of G. By our assumption, (p2 - 1)2 divides IG I. Hence ILl = 22s + 2 + b with b > O. From P2:. = P we infer that L normalizes T(P). Put L* = (I2:.(T(P». Then L/L* is a subgroup of GL(2, p). As IGL(2, p)1 = p(p2 - I)(p - I), we have IL/L*I :< 2s +2. On the other hand, 02:.* = 0 implies L* ~ G( Q, OP). Since G( Q, OP) centralizes T(P), we obtain finally L* = L n G(Q,OP). Hence 2:* is a Sylow 2-subgroup of G(Q,OP). If IL*I = 2 1 , then t:< s + 1, as IG(Q,OP)I divides p2 - 1. Furthermore 22s+2+b-1 = IL/L*I :< 25+ 2 and hence s + b 5 s t :< O. This yields s :< s + b :< t :< s + 1. Hence IL*I = 2 or 2 +). Assume IL*I = 25. Then b = 0 and IL/L*I = 2s+2. Therefore, L/L* is a Sylow 2-subgroup of GL(2, p). Hence, by 14.7, L/L* is semi dihedral. Put L** = G(P, OQ) n L. Then IL**I = 25 or 2s + I, as our assumptions are symmetric in P and Q. Assume IL**I = 25. As L** ~ L**L* /L*, we have that L** is isomorphic to a normal subgroup of index 4 in L/L*. But a semi dihedral group has only one normal subgroup of index 4 and this normal subgroup is cyclic. 0 17.8 Lemma (Liineburg 1976 a + b). Let p be a prime with p == 3 mod 4 and let 2.f be a translation plane of order p2. Furthermore let 0, P, Q be three distinct points of 2.f with P, Q I 100 and 0 I' 100 , If G is a group of collineations of 2.f fixing 0, P and Q and if G splits the set of affine points off the lines OP and OQ into two orbits of length 1- (p2 - 1)2, then the order of a Sylow 2-subgroup of G(P, OQ) is either 2s - I , 2s or 2 s+ I, where 2s is the highest power of 2 dividing p + 1. Moreover if the order of a Sylow 2-subgroup of G(P, OQ) is 2s - I , then the Sylow 2-subgroups of G(P, OQ) are cyclic unless 2s divides IG(Q, OP)I· The proof of 17.8 is similar to the proof of 17.7 and is left as an exercise to the reader. N ow we are able to finish the proof of 17.1 in the case where there are no p-primitive prime divisors of pr - 1. We were left with the case, where the order of 2.f is p2 with P + 1 = 2s > 4. Using (B) and Lemma 17.5, we see that K(P,OQ) is soluble. By 17.7 and 17.8, the Sylow 2-subgroups of K(P, OQ) have order either 2s - I or 2s or 2s + I. If they have order 2s - I , then they are cyclic, as K(P,OQ) and K(Q,OP) are conjugate in H. If s = 2, then p = 3. In this case we are done by 8.4. If s = 3, then p = 7. This is one of the exceptions. Therefore we may assume s > 4. This yields s > 5, as 24 - 1 = IS is not a prime. As K(P,OQ) is soluble, K(P,OQ) contains a normal Z-subgroup N such that K(P, OQ)/ N is isomorphic to a subgroup of S4' We show that all Sylow subgroups of odd order of N are in the centre of K(P, OQ). Let a be the largest prime dividing /N /. If a = 2, then there is nothing to prove. Let a > 2. As N is a Z-group, the Sylow a-subgroup A of N is normal in N
17. Rank-3-Planes with an Orbit of Length 2 on the Line at Infinity
87
by a theorem of Burnside (see e.g. Gorenstein [1968, Theorem 7.4.3]). As INI divides p2 - 1 = 25 (p - I), we see that IA I divides p - 1. Therefore, A operates reducibly on T(P). Let L be a Sylow 2-subgroup of K(P,OQ). Then ILl> 25 - 1 ;;;. 16. Furthermore, L normalizes A, as A is a characteristic subgroup of N. The centralizer of a subgroup of order p of T( P) in L has order 2, as p == 3 mod 4. Therefore, A leaves invariant at least 8 subgroups of order p of T(P). This is more than enough to assure that A is contained in the centre of GL(2, p) and hence in the centre of K(P, OQ). As a normal Hall subgroup A has a complement B in N. As A ~ B(N), the complement B is normal in N. Since normal Hall subgroups are characteristic subgroups, B is also normal in K(P,OQ). The assertion now follows by induction. Put LO = L n N, where, L is a Sylow 2-subgroup of K(P, OQ). Then LO is a Sylow 2-subgroup of N. Furthermore LO is normal in N as all Sylow subgroups of odd order of N are in the centre of K(P,OQ). Hence LO is normal in K(P, OQ). Moreover L is normal in LN. Assume that 2: is not normal in K(P,OQ). Then 3 divides K(P, OQ)/ N. For otherwise K(P, OQ) = LN. But L is normal in LN. Let M / N be the largest normal 2-subgroup of K(P, OQ)/ N and let L) be a Sylow 2-subgroup of M. Furthermore let C be the complement of LO in N. Then M = L) C and LI :2 LO' As C centralizes L I , we have that LI is normal in M and hence normal in K(P, OQ). If N =1= M, then ILI/Lol = 4, as K(P, OQ)/ N is isomorphic to a subgroup of S4 and as LN / N is not normal in K(P, OQ)/ N. Moreover LI/LO is elementary abelian. This implies that LI is a generalized quaternion group. Since there exists an element in K(P,OQ) which induces an automorphism of order 3 in LI (recall that LI is normal in K(P, OQ», we deduce that L) is the quaternion group of order 8, as this is the only generalized quaternion group admitting an automorphism of order 3. Hence ILII = 8 and ILl = 16. As L is not cyclic, we obtain 16 = 12:1 > 2s , i.e. s :< 4: a contradiction. This proves M = Nand LO = L). Moreover IL: Lol = 2, as L is not normal, and therefore ILol ;;;. 2s - I. As LO is the intersection of all the Sylow 2-subgroups of K(P, OQ), we have that LO is the only normal subgroup of order ILol in K (P, OQ). As LO is cyclic of order ;;;. 2S - I, it contains a cyclic subgroup L * of order 2s - I • Obviously, L* is characteristic in K(P,OQ). We infer from 2s -) > 16 that L* is the only cyclic normal subgroup of order 2s - l . Assume that L is normal in K(P,OQ). Then L contains a unique normal cyclic subgroup L * of order 2S - I: This is true by 17.8 in the case that ILl = 2s - I. If ILl;;;. 2s, then L is cyclic or a generalized quaternion group. In either case, L contains such a unique normal cyclic subgroup, as s ;;;. 5. It follows that L* is the only cyclic normal subgroup of order 2s - I of K(P, OQ). Let L* be the only normal cyclic subgroup of order 2s - 1 of K(P, OQ). As K(Q, OP) is conjugate to K(P, OQ), it also contains exactly one cyclic normal subgroup L** of order 2S - I. Therefore A = L*L** is an abelian
III. Rank-3-Planes
88
normal subgroup of H of order 22s - 2 • As H acts transitively on 100 \{ P, Q), the orbits of A contained in (XJ \ {P, Q) all have the same length A.. As A. is a divisor of p2 - 1, we have A. < 25+1. Let WI 100 with W=r'= P,Q. Then
22s -
2
= IAI = I WAllAwl < 2 + I IA w l· 5
Hence IAwl ~ 25 - 3 ~ 4. One proves as usual that ~Aw(T(W)) = {I}. Hen~e A w is isomorphic to a subgroup of the group G L(2, p). Furthermore A IS the direct product of the two cyclic groups :L* and :L**. Therefore A contains exactly three involutions. This implies that A w contains but one involution. Hence Aw is cyclic. This yields finally that Aw acts irreducibly on T( W). Therefore, 2r is a generalized Andre plane by 9.3. Thus th~ pr?of of 17.1 will be finished if we can show that a plane of order 192 satIsfymg the hypotheses of 17.1 is a generalized Andre plane. This will be proved in section 20.
18. The Planes of Type R '" P
89
29 or 59. In these cases, the number of orbits of S is 3, 7 resp. 29. Furthermore Zo permutes the orbits of S. As Zo operates regularly on V\{O} and since neither 9 nor 7 nor 29 divides 120, we see that Zo permutes the orbits of S transitively. For p = 29 or 59, the group SZo operates regularly on V\{O}. In the case p = 19, the subgroup of order 3 of Zo fixes all the orbits of S. Put W= V EB V, V(O) = {(x,O)lx E V), V(oo) = {(O,x)lx E V) and W* = W\( V(O) U V( 00 )). The group G = ZoS X ZoS operates on W by (x, yi a , (3) = (x a , y(3). Put H = S X S and A = {(r, n IrE Zo). Then AH is a subgroup of order 1202 Zol of G. Furthermore, G fixes V(O) and V( (0). 1
18.2 Lemma. A H decomposes W* in case
18. The Planes of Type R *P
p = 11 into p = 19 into P = 29 into P = 59 into
In this section we follow Liineburg 1974. Let p be a prime and let 2r be an affine plane of order p2. We shall call mof type R *p, if mpossesses the following properties: .. (j) If G is the collineation group of 2r, then Gt acts doubly transItIvely on the set of affine points on I for all affine lines I of m. (ij) There exist three distinct points P, Q and 0 of 2r with P, Q I 100 and o l' 100 such that G(P,OQ) contains a subgroup So and G( Q, OP) contains a subgroup S 00 with So;;;;:; S 00 ;;;;:; SL(2, 5). Moreover, the set of orbits of S o on I 00 is equal to the set of orbits of Soo on 100. , We shall determine all planes of type R * P in this sectIOn.
PROOF. Let (x, y) E W*. Furthermore let ex, f3 E Sand r E Zo be such that (x, yiL.i'Xa, (3) = (x, y). Then x.la = x and y.\(3 = y. If P = 29 or 59, we obtain rex = 1 = rf3, as SZo acts regularly on V\{O}. This proves 18.2 in these two cases. Let p = 19. Then x and x.l are in the same orbit of S. Therefore 3 = 1. As ex is uniquely determined by r and x and as f3 is uniquely determined by rand y, we have consequently I(AH)(x. y)1 < 3. On the other hand, if r 3 = 1, then r fixes all the orbits of S. Therefore there exist ex, f3 E S with x ~a = x and y.\(3 = y. Thus I(AH)(x. y)1 = 3. Finally, if p = 11, then S acts transitively on V\{O}. Therefore Hand hence AH operate transitively on W*. This proves 18.2. 0
18.1 Theorem. If 2r is an affine plane of type R * p, then plane and p is one of the primes 11,19,29,59.
mis a translation
PROOF. 2r is a translation plane by 16.1. Furthermore, p E ell, 19,29,59} o by 17.5. Let p E {II, 19,29, 59} and let S be a subgroup of S~(2, p) which is isomorphic to SL(2, 5). As p2 - 1 :::::: 0 mod 10, there eXIsts such an S. Denote by V the vector space of rank 2 over GF(p). Let. o.E. S and assume that 1 is an eigenvalue of o. As det(o) = 1, the multIplIcIty of 1 as an eigenvalue of 0 is 2. Hence 0 is a transvection. Therefore oP = 1. As p does not divide IS I = 120, we obtain 0 = 1. Thus S operates regularly on V\ CO}, i.e. all the orbits of S in V\{O} have the length 120. Let Z be the maximal subgroup of odd order of 3(GL(2, p)). Then IZol = 5,3 2,7,29 for p = 11,19,29,59 resp. The group SZo acts transitively on V\{O}: This is certainly true for p = 11, as }}2 - 1 = 120. Let p = 19,
one orbit of length' 1202 ,
3 orbits of length 3 . 1202, 7 orbits of length 7 . 1202, 29 orbits of length 29 . 1202.
r
Put B = Zo X ZOo Then G = BH. Furthermore IBI = IZoI2. As B centralizes AH, it permutes the orbits of AH. Since SZo acts transitively on V\{O}, the group G = BH acts transitively on W*. Therefore, we have:
18.3 Lemma. B permutes transitively the orbits of A H which are contained in W*. Put Z
= 3(GL(2, p))
and define V(n for all r E Z by
V(r)
=
{ex,x.l)lx E V}.
If Sl = {(ex, ex) I ex E S}, then SI is the stabilizer of Vel) in H. Moreover AS I operates transitively on V(l)\{(O,O)}. Therefore V(l)\{(O,O)} is completely contained in an orbit B j of AH. Moreover, B j ~ W* and
BI
= U EHeV (l)T1\{(O,O)}).
Bi+
I=
B1(1,
w').
Then we have:
Let
Zo
be
generated
by
wand
put
'7V
IlS. 1 ne t'lanes 01 1ype
18.4 Lemma. The orbits of AH which are contained in W* are in case
p = II the set B I' P = 19 the sets B 1 ,B2,B3, p=29 the sets B 1 , ••• ,B7 , p=59thesetsB 1 , ••• ,B 29 ·
Bf = B I • If i =F 1, then
= (y, x),
then we have:
AH. }P
=
{B 1 , B 2'
':II
Let ai be the number of X E ffi\{ V(O), V(oo)} with X\{(O,O)} ~ Bi . As B centralizes AS] and since B permutes the B/s transitively by 18.3, we
+
1 we then have:
18.7 Lemma. We have
18.5 Lemma. If p is the mapping defined by (x, YY p normalizes {B I , B 2' . . .
P
have a l = a 2 = ... = a. Because of Iffil = p
PROOF. This is trivial for p = 11. Let p > 11. By 1~.3, B acts transitively on the set of the orbits in question. Put C = {(l,wl)li = 0, 1, ... , IZol-l)· Then B = A C. Hence C acts transitively on the set of orbits. This proves the assertion in case p = 29 and p = 59. Therefore assume f ~ 19 and. let C be the subgroup of order 3 of C. Then Co fixes all the orbIts m questlOn. A~ C = Co U Co(l,w) U Co(l,w 2), we obtain finally that B I , B2 and B3 are 0 the orbits of AH which are contained in W*.
(j) (ij) (iij) (iv)
K •
if P = 11, a=6Ifp=19, a = 4 If P = 29, a = 2 if P = 59.
a = 10
Let V be a vector space and let a be a set of subspaces of V. We shall call a a partial spread of V if a contains more than one element and if V = X EEl Y for all X, YEa with X =F Y.
18.8 Lemma. If X E ffi\{ V(O), V(oo)} and 'TT = {XY \ yEAH}, then we have: (a) U Y E'lT (Y\ {(O, O)}) is an orbit of AH contained in W*. (b) 'TT is a partial spread of W.
. . . }.
B/ =F Bi •
PROOF. (j) is trivial and (ij) is a consequence of (j). (iij) follows from (ij) and V(lY = V(l). In order to prove (iv) assume that B/+ I = Bi + I' By 18.4, V(w i)\ {(O, O)} C; Bi + I' Therefore, (XWI, x) = (x, xw'y E B i + I. for all x E V\{O}. Because of BI = U1)EH(V(l)\{(O,O)P, there .exIst,w/ or _everI x E V\{O}, an element y E V and elements Cl, f3 2IE S wIth (x ,x) - (y , 1f3 I yf3 w). This yields y = x wh·- and therefore x = xw a- . This .proves that x W2i and X are in the same orbit of S. Hence 2i == mod 3, If P = 19, and 2i = 0, if P = 29 or 59. In either case i = O. 0
PROOF. a) follows from 18.4. (b) Let UYE'lT(Y\{(O,O)}) = B i . Consider the incidence structure (Bi' 'TT, E). This is a tactical configuration, as Bi and 'TT are orbits of AH. The parameters of (Bi,'TT, E) are v = IBil, b = I'TTI, k = IX n Bil and r. We have to show r = 1. As AS 1 C; (AHh, we have b .; ; ; 120. Furthermore, k = p2 - l. Hence vr = bk .; ; ; 120(p2 - 1). From 18.2 we infer
°
18.6 Lemma. Let V be a vector space of rank 2 over GF(q) and let W, Z, V(O), V( 00), V(n be as before. If S is a non-cyclic subgroup of GL(2, q) acting irreducibly on V, then ffi = { V(O), V( oo)} U { ~(n I ~ E Z} is the set of subspaces of rank 2 of W which are left Invarzant by SI = {(Cl,Cl)ICl E S}. PROOF. Let X E ffi. Then X is left invariant by SI' Thus assume that X is a subspace of rank 2 of W which is left invariant by S I' We may ass.ume X =F V(O), V(oo). From XSI = X and the irreducibility of S, we mfer X n V(O) = {(O, O)} = X n V( 00). Hence X is a diagonal of W = V(O) EEl V(oo), i.e. there exists aEGL(2,q) with X={(x,x")lxEV}. In particular X = {(x a, X a,,) I X E V} for all Cl E S. On the other hand X = {(xa,x"a)\x E V} for all Cl E S, as XSI = X. Hence Cla = aCl for all Cl E S. Thus a lies in the centralizer C of S. By Schur's Lemma \C\= q - 1 or IC I = q2 - 1, as Z ~ C and IZ I = q - 1. If IC I = .q2 - 1, then C ,":ould be its own centralizer, as C is cyclic. This is impossIble, because S IS not cyclic. Hence ICI = q - 1, i.e. Z = C, whence a E Z. 0
1202r .; ; ; 120( 112 - 1) 3 . 1202r .; ; ; 120( 19 2 - 1) 7· 1202r .; ; ; 120(29 2 - 1)
= = =
1202
if P = 11,
3 . 1202 7.120 2
if P = 19,
29 . 1202r .; ; ; 120(59 2 - 1) = 29 . 1202
if P = 29, if P = 59.
Thus r';;;;; 1 in all cases. As k = p2 - 1 =F 0, we find r b = 120 in all cases.
> 1.
Hence r = I and
0
18.9 Corollary. (AHh = AS I • We call 'TT as constructed in 18.8 an i-partial spread, if UYE'lT Y= Bi U {(O,O)}.
18.10 Lemma. For each i there exist
exactly exactly exactly exactly
five i-partial spreads if p = 11, three i-partial spreads if p = 19, two i-partial spreads if p = 29, one i-partial spread if p = 59.
92
III. Rank-3-Planes
PROOF. By 18.7, it suffices to show that [77 n ffi[ = 2 for all i-partial spreads 77. As B operates transitively on the set of all 77'S in question, we may assume '77= {V(l)Y[y EAH}. Since A fixes V(l) we have 77= {V(l)Y[y E H}. Let 1 =1= 0 E 3(S). Then 0 = - 1 and V(I), V( - 1) E 77 n ffi. Hence [77 n ffi [ > 2. Assume X E'77 n m. Obviously, X =1= V(O), V(oo). Thus X = V(O for some E Z. On the other hand there are a, /3 E S with V(l)(a, f3) = V(O. As
r
V(I)(a, f3)
=
V(I)(a,a)(l,a- 'f3 ) = V(I)(l,a- 'f3 \
we may assume a=l. This yields {(x,xf3)[XE V}={(x,x~)[xE V). Therefore /3 = E Z n S = 3(S) = {l, -I}. 0
r
For each i pick an i-partial spread ({Ji' Then K = {V(O), V(oo)} U ({Jl U U . " is a spread of W by 18.8. Hence K( W) is a translation plane. If YJ E Z X Z and Kl1 = {V(O), V(oo)} U ({J( U ({Ji U .. " then Kl1(W) and K( W) are isomorphic. Hence we may always assume that ({JI = { V(l)Y [y E H}. Thus we obtain up to isomorphism only one plane K( W), if p = 11. If P = 59, then there is also only one plane K (W) by 18.10. In all cases, A H is a subgroup of the collineation group of K( W). Moreover, So = {(a, 1) [a ES} and Soo={(I,a)[aES} are groups of homologies which are isomorphic to SL(2,5) and the set of orbits of So on 100 is the same as the set of orbits of S 00 on 100 , Hence K( W) is always of type R * p. ({J2
18.11 Lemma. Let N be a subgroup of H which is isomorphic to SL(2,5). If N leaves a subspace X of rank 2 of W invariant and if X =1= V(O), V( (0), then Nand SI are conjugate in H. PROOF. As H ~ SL(2, 5) X SL(2, 5), it contains only three involutions, namely the involutions 0 1 , O 2 , 0 3 defined by
(x, y)a
l
= (- x, y),
Let 0 be the involution in N. If N did not act faithfully on X, we would have that 0 were the identity on X, as 0 is contained in all non-trivial normal subgroups of N. But then 0 = 0 1 or 0 = O 2 and hence X = V(O) or X = V( (0): a contradiction. Hence N acts faithfully and therefore irreducibly on X. This yields X n V(O) = {(O, O)} = X n V( (0). Therefore there exists /3 E GL(2,p) with X = {(x,x f3 )[x E V}. Let 0 E Sand X (a, I) = X. Then, for all x E V, there exists y E V with (x a , xf3) = (y, y f3). As a consequence x = y and x a = x for all x. Therefore So n N = {(l, I)}. Similarly soon N = {(1, l)}. Hence N is a diagonal of H. So there exists y E Aut(S) with N = {(a, a Y) [a E S}. From this we infer { (x a, X af3) [x E V} = X = X (a, ar) = {( X a, X f3a r ) [x E V}
for all a E S, whence a{3 = (3a Y. This proves f3 E 9(GL(2, p)(S). By 14.6, 9C GL(2. p)(S) = ZS. Therefore we may assume /3 E S. But then (/3,1) E H. Moreover X(f3, I) = V(I) and (/3, I)-INC /3, I) C HV(I) = SI' 0
93
18. The Planes of Type R * P
18.12 Theorem. If 2l is a plane of type R * p, then 2l is isomorphic to one of the planes K( W). PROOF. By 18.1, 2l is a translation plane of order 11 2 , 19 2 ,29 2 or 59 2 • Thus we may identify the points of 2l with the elements of W. By 2.1, we may further assume that OP = V(O) and OQ = V(oo). Applying 14.6, we see that we may identify So X S 00 with H, where H is the group used in the construction of the K( W)'s. Let 0 be the spread of W defining the lines of 2l and let X E o\{ V(O), V(oo)}. Then [Hx[ = 120 by the assumption made on the orbits of So and S 00' This yields that H x is a diagonal of H = S X S. Hence Hx ~ SL(2, 5). Therefore Hx is conjugate to SI in H. Thus X H is an i-partial spread. This proves that 0 is the union of i-partial spreads and the set { V (0), V ( 00 ) } . 0 18.13 Lemma. If K( W) and K'( W) are isomorphic, then there exists an isomorphism 0 from K( W) onto K'( W) with {V(O), V( (0) t = { V(O), V( oo)}. This follows immediately from 14.8. 18.14 Lemma. If 0 is an isomorphism of K( W) onto K'( W) with {V(O), V(oo)t = {V(O), V(oo)}, then 0 E (Z X Z)H
and D
=
{U,O[t
E
Z}
and 77 = {V(1)Yly
E AH}.
18.15 Lemma. (a) ITT = DH
YJ E Z X Z. Then ffil1
V(l)c a , f3) = V(l)C a ,a)(I,u- 'f3 ) = V(l)(I,a- 'f3 ) = V(a- f3). 1
Therefore a = /3 or a immediately from (a).
= -
/3. This implies
1)
E DH proving (a). (b) follows 0
III. Rank-3-Planes
94
19. The Planes of Type F * P
95
2
18.16 Theorem. There exists up to isomorphism exactly one plane of type R * 11, namely the plane over the exceptional nearfield of order 112 the
29 .1fG is the collineation group fixing {V(O), V(oo)}, then G = DH(p) lor seven of the remaining eight planes and G = DH for the last one. PROOF. We have
multiplicative group of which is isomorphic to SL(2,5).
III
This follows from lS.12 and the remark after lS.10.
18.17 Theorem. There exist up to isomorphism exactly three planes of type R * 19. If G is the collineation group of any of these planes, then G (V(O), V(oo)} is conjugate to D H (p) in I. PROOF. By lS.lO, there are exactly 3 3 = 27 planes K(W). By lS.11, lS.12 and lS.14, the number of orbits of I on the set of the K( W)'s is precisely the number of isomorphism types in question. Let K( W) be one of these planes and Kits collineation group and let G = K{v(o), V(oo)}' By lS.14 and lS.15, we have DH ~ G ~ I. Put k = \G: DH\. Then
=
I(Z X Z)H(p)1
=
2IZI2IHII(Z X Z) n H\-I
=
!\Z\2IHI
=
!2S 2IHI.
Moreover DH ~ G ~ r. Let k = \G: 'DHI. As IDHI = !2SIHI, we have II: G\ = 2Sk- l . This is the number of planes in K(W( By IS.3, B acts transitively on {B I , ••• , B7}' Put \EI = {(r,~-I)\r E Zo): Then B = A E and A n E = {(1, I)}. Therefore E acts sharply transItl:ely on {B I , ••• , B7}' L~t KO =.{ V(O), V(oo)} U U'EE1T'. Then KO( W) IS a p~ane of type R * 29 whIch admIts E as a group of collineations. As p normalIzes E and 1T P = 1T, we see that p is also a collineation of KO( W). H~nce, by lS.15, G'1r = I'1r' whence IG\ = 7\I'1r1 = 14IDH\. Thus k = 14 in thIS ~a.se and II: GI = 2. On the other hand, if G permutes {B], ... , B 7 } transItlvely, then we infer from DH ~ G ~ I that E C G and that K( W) E KO( W)l. -
If G does not act transitively on {B I' . . . , B7}' then G / DH is a 2-group as II: DHI = 2S. Therefore we may assume BIG = Bl and also 1T ~ K. Therefore G ~ I'1r = DH(p). By lS.lO there are 26 planes K( W) with 1T ~ K. Of these 23 = S admit p as a collineation .an~ for exactly one of them G DH(p). The remaining seven are paIrWIse non-isomorphic as G = I in these cases. The corresponding orbits of I contain 28: 2 = 14 plan;s each. Hence if n is the number of orbits which contain those planes for which G = DH, then
is the number of planes in the orbit K( W)l. Now
rt.
and
\DH\
=
\D\ \H\\D n H\-I
=! IS\H\.
Thus ak = IS. Assume that G acts transitively on {BI'B 2 ,B 3 }. As DH induces the identity on this set, there exists ~ E Z with (1, nEG such that (1, n induces a 3-cycle on {B I ,B2 ,B 3 }. We may assume that is a 3-element. 3 3 Furthermore we may assume 1T ~ K. Then 1T(U ) = 1T and hence V(r ) E 1T. 3 Therefore ~3 = 1 or ~3 = -1. As is a 3-element, we have = 1. By IS.3, Zo X Zo and hence {I} X Zo act transitively on {B I' B 2 , B3}' As Zo is cyclic of order 9, we have consequently that (1, fixes each Bi : a contradiction. Thus G induces a 2-group on {B I ,B 2 ,B 3 }· Hence there exists an i with BiG = Bi . We may assume i= 1. Let CP21,CP22,CP23 be the three 2-partial spreads of W. By IS.5, CP!J.i (i = 1,2,3) are the three 3-partial spreads of W. Put
I2S
=
27 = 2Sn
+ 7 . 14 + 2 = 28n +
100,
whence n = 1.
r
r
r
n
Ki = {V(O), V(oo)} U
1T
U CP2i U CP!J.i
(i = 1,2,3).
If. G; has the meaning for K;( W) that G has for K( W), we have G; = I'1r for all i. Hence k i = 2 for all i by IS.15. Therefore the orbit of Ki(W) under I contains exactly 9 planes. If i =1= j, then Ki ( W) and K;( W) are nonisomorphic, as every isomorphism from K i ( W) onto K/ W) fixes 1T. Therefore /U7= I K;( wi/ = 3 ·9= 27. 0
18.18 Theorem. There exist up to isomorphism exactly nine planes of type R * 29, one being the nearfield plane over the exceptional nearfield of order
o
Finally, we have by the remark after 18.10 and by 18.12:
18.19 Theorem. There exists up to isomorphism exactly one plane of type R * 59, namely the plane over the exceptional nearfield of order 59 2 •
19. The Planes of Type F *P In this section we follow Liineburg 1974. Let p be a pr~me and let m: be an affine plane of order p2. We shall call m: o~ type ~ *p If m: possesses the following properties: (J) If G IS th~ colli~eation group of m:, then G{ acts doubly transitively on the set of affme pomts on I for all affine lines I of m:. (ij) There exist three distinct points P, Q and 0 of m: with P, Q I I and o I 100 such that ~(P, OQ) contains a subgroup So and G(Q, OP) co~tains a subgroup Soo wIth So;;;:::: Soo ;;;:::: SL(2, 3) and such that the orbits of S on 100 are the same as the orbits of Soo on 100 , 0
96
III. Rank-3-Planes
As a consequence of 3.2, 3.3 and 3.9, we have that So is normal in G(P, OQ) and that Soo is normal in G(Q, OP). Furthermore, by using 3.5, we see that So is the only subgroup of G(P,OQ) which is isomorphic to SL(2,3) and that the analogous statement for S 00 is also true. Applying these remarks, 16.1, 17.4 and 17.6 yields: 19.1 Theorem. If lli: is a plane of type F * p, then lli: is a translation plane and pE{5,7,11,23}. Let p E {5,7, 11,23} and let V be a vector space of rank 2 over GF(p). Let S be a subgroup of SL(2, p) which is isomorphic to SL(2, 3). (Such a subgroup exists.) As p does not divide ISL(2,3)1 = 24, the group S acts regularly on V\{O}. We let S X S operate on W = V EEl V by defining (x,y)(a,/3)=(x a,y/3). Denote by SI the stabilizer of V(l)={(x,x)lx E V} in S X S. Then Sl = {(a, a) 1a E S}. Put Z = 3(GL(2, p)) and let V(O), V( 00), V(O for E Z have the usual meaning. Finally, put 9:l = {V(O), V(oo)} U {V(Olr E Z}. By 18.16,91 is the set of all subspaces of rank 2 of W which are left invariant by S l' Denote by N the normalizer of S in GL(2, p). By 14.4 and 14.5, IN:SZI=2.
r
19.2 Lemma. The set of subgroups of S X S which are isomorphic to SL(2,3) and which fix exactly p + 1 subspaces of rank 2 of W split into two conjugacy classes. These two classes fuse under N X N into one conjugacy class. Moreover, if H is a subgroup of S X S which is isomorphic to SL(2,3) and if H fixes a subspace of rank 2 which is distinct from V(O) and V( 00), then H is in one of these classes. PROOF. Pick pEN\SZ and put S2={(a,a P)laES}. Then S2 is a subgroup of S X S which is conjugate under N X N to SI' Hence S2 fixes exactly p + 1 subspaces of rank 2. Assume that S I and S2 are conjugate under S X S. Then there exist 0,1' E S with Sl(a,T) = S2' i.e. {({30, ,BT)I ,B E S} = {(a,aP)la E S}. This yields,BT = ,Bop for all,B E S. Hence OPT- I centralizes S. Therefore OpT - I E Z, i.e. p E ZS: a contradiction. Thus we have at least two such conjugacy classes. Let X be a subspace of rank 2 of V with X =1= V(O), V( 00) and let H be a subgroup of S X S isomorphic to SL(2,3) and assume XH = X. The argument using the involutions we have made in the proof of 18.11 yields also in the present situation that H acts faithfully and hence irreducibly on X. Therefore X n V(O) = {(O,O)} = X n V(oo). It follows that there exists o EGL(2,p) with X={(x,xO)lxE V}. Likewise we obtain that H is a diagonal of sxS. Hence there exists yEAut(S) with H={(a,al')la E S}. From this we infer
19. The Planes of Type F * P
97
Therefore a I' = 0 - lao for all a E S, i.e. 0 E N. This yields that H is conjugate to SI or to S2' 0 19.3 Lemma. If
r E Z, then
V(Os x S n 91 = { V(O, V( -
n}.
PROOF. Put Soo = {I} X S. Then S X S = SlSoo' Hence V(n SXs = V(n soo . Let VC'I1) E V(n s x s. Then there exists a E S with V(7]) = Vera). Therefore 7] = ra. This yields a E S n Z. Hence 7] = or 7]=
r
-r.
0
19.4 Lemma. If lli: is a plane oj type F * p, then we may identify the points of lli: with the elements of W such that V(O), V(l) and V(oo) are lines of lli: and such that So = S X {l} and S 00 = {I} X S. PROOF. By 19.1 and 2.1, we may identify the points of lli: with the elements of W such that OP = V(O) and OQ = V( 00). By 14.4 and 14.5, we may assume So=SX{l} and Soo={l}XS, and, by 19.2, we may assume tha t S 1 = {(a, a) 1a E S} fixes a line X through 0 distinct from V (0) and V( 00). It follows from 18.6 that X = V(n with r E Z. As {l} X Z acts transitively on lR\{ V(O), V(oo)}, we may assume finally that = 1. 0
r
19.5 Theorem. There exists up to isomorphism exactly one plane oj type F * 5, namely the plane over the exceptional nearfield of order 25. PROOF. Let lli: be of type F * 5. By 19.4, lli:;;;;; K( W) with K = { V(O), V(oo)} U V(l)sxs. This establishes the uniqueness part of the theorem. The existence part is trivial. 0 19.6 Theorem. There exists up to isomorphism exactly one plane of type F * 23, namely the plane over the exceptional nearfield oj order 23 2 • PROOF. By 19.4, we may assume lli: = K( W) with V(O), V(l), V(oo) E K and So X S 00 = S X S. As 23 2 - 1 = 22 . 24, the group S X S decomposes K \ { V(O), V( oo)} into 22 orbits each of length 24. Denote by S 1 the stabilizer of V(l) in S X S and denote by S2 a subgroup of S X S which is isomorphic to SL(2,3), which fixes a subspace of rank 2 distinct from V(O) and V( 00) and which is not conjugate to S I' Such an S2 exists by 19.2. Let X E K\,{ V(O), V( oo)} and put H = (S X Sh. Then H is a diagonal of S X S, 'whence H;;;;; SL(2,3), By 19.2, H is conjugate to Sl or S2' By 18.6, S I fixes exactly 24 subspaces of rank 2, namely the subspaces belonging to 91. Therefore, by 19.3, there exist at most 11 orbits of the form X S x S with H conjugate to Sl' As the same assertion holds for S2 and since there are exactly 22 orbits of the form X SXS , we see that lli: is unique up to isomorphism. The existence is clear. 0 19.7 Lemma. Let lli: and lli:' be two planes of type F * P and identify the points of ill as well as the points of 2-(' with the elements of Waccording to 19.4. If ill
-
and 21' are isomorphic, then there exists an isomorphism a from with {V(O), V(oo)r = {V(O), V(oo)}.
~r
onto 21'
19.8 umma. Let 21 and 21' be two planes of type F * p. The points of 21 as well as the points of 2(' are identified with the elements of W as in 19.4. If a is an isomorphism from 21 onto 2(' with { V(O), V( oo)} a = { V(O), V( oo)}, then a E (N X N)
n
ID(S
X
X
S) is normal in D(S
S)M : D(S
-
.............. d
V<
...
J t, ...
·
. f
invariant under G. Furthermore,
p - I =
101 =
IG : G1TJ I ,; ;
Ie : D(S n S)M
=
P - 1,
whence G1T = D(S X S)M
This follows from 14.8.
The group D(S
•• -
X
X
S)M. Hence
S)I = 1M : M n (D(S
X
= 1M :D(M n (S
X
S))I
S))I = 1M :DS1I = IMIIDS11- = INIIDS11- I = 2IZSIIDS II- I = 2.
VU/= {(xS,x)lx E V} = {(xs,x~~-)Ix E V} = VU-
I
),
we infer from 19.3 that V(~ -I) E {V(O, V( - O}. If Va -I) = V(O, then 2 7Ts = 7T 1 , as 7T1 = 7T _I' If V(r -I) = V( - 0, then r = -1 which is impossible, as p == 3 mod 4. (Here we have used the assumption p i:= 5.) This proves b). 0 ~2 = 1 and hence
19.10 Theorem. There exist up to isomorphism two planes of type F * 7. One of them is the plane over the exceptional nearfield of order 72 . If H is the collin eat ion group of the second plane, then, up to conjugacy, H{V(o),v(oo)} =
D(S
X
S)M<(l,v)p)
where v E (N n SL(2,7))\S. PROOF. We infer from 72 - 1 = 2 . 24 that S has exactly two orbits on V\{O), since S acts regularly on V\{O}. It follows from IZI = 6 and Iznsl=2 that Z fixes each of these orbits. Using 14.4, we see N \: SL(2, 7)Z. Hence N = N n (SL(2, 7)Z) = (N n SL(2, 7))Z.
1
As IN: ZS I = 2, the order of N is 2
'1 (7 -
1)24 = 3 ·48. Therefore
3·48 = IZIIN n SL(2, 7)IIZ n N n SL(2, 7)1- 1 = 31N n SL(2, 7)1.
This yields 2
ID(S X S)MI = 2ID(S X S)I = 2 '1(P - 1)IS1 = (p - 1)242. Therefore I(N X N) : D(S X S)M I = p - 1. Let (a,T)E(NXN)"'J' As NXN=D(SXS?M({I}XN), we may assume a = 1. Since S X S = SI({l} X S) there eXIsts a E S WIth
{(x,xT)lx
E
V)
=
{(x,xa)lx
E
V).
Hence T = a E S, i.e. (N X N)"'J = D(S X S)M. Thus 17T~ XNI = p - 1. By 19.3, l7Tt n 911 = 2. Hence there are exactly 1(P - 1) such 7T~. All of them belong to 7T~ XN. Put G = (N X N)
19.9 Lemma. Assume p =1= 5. Then we have: a) G",\ = D(S X S)M
X
S)M \: G"'J' Furthermore 7T I , we see that 0: is
p E G"'J' Since N X N is normal in G and since p fixes
Thus N* = N n SL(2,7) is a group of order 48. As 7 does not divide 48, we have consequently that N* operates transitively on V\{O}. In particular, N acts transitively on V\{O}. Put BI = (UXE1TJX)\{(O,O)}. (a) If S E Z, then V(n ~ #1 U {(O, O)}: Obviously B I is invariant under S X S. Hence (x,x~) E BI for all x E V\{O} and all g E S. If 0 i:= x E V, then there exists ~ E S with x~ = x~, as Sand SZ have the same orbits on V. Thus (x,xt) = (x,x~) E B I • This proves (a). The group D(S X S)M( {I} X Z) permutes {7Tt IrE Z} transitively. Therefore D(S X S)M({l} X Z) C (N X N)B J' Moreover I(N X N): D(S X S)M({ I} X Z)I = 2. Let (1, v) E (N X N)\(D(S X S)M({ I) X Z)) and put B2 = B 1(1, p). (b) Bin B2 = 0: Let (x, y) E Bin B 2 . Then there exist u, v E V\{O} and a,{3,y,8ES with (u a,u f3 ) = (x,y) = (vY,V OP ). This yields uf3= uay-lop. Now uf3 and u ay - Jo lie in the same orbit of S in V\{O}. As N* and hence N operate transitively on V\{O} and since v E N\ZS, the elements u ay-\o and u ay-J op = uf3 are in different orbits of S. This contradiction proves (b).
100
III. Rank-3-Planes
(c) BI UB 2 = W*= W\{V(O)U V(oo)}: Let (x,y)E W*. As N acts transitively on V\{O), there exist r E Z, a E Sand i E {O, I} with Y = x~av'. Furthermore there exists f3 E S with x/3 = x~. Hence
( x, y)
=
(x, X /3av
j
)
E B I U B2 .
From B2 = B 1(1, v), (b) and (c) we infer: (d) IBil = -! I W*I = -! (72 - 1)2. Consider the incidence structure (B I ,7T~, E). This is a tactical configuration with parameters v = -! (72 - 1)2, b = IS x S II S 11- I = 24, k=7 2 -1 and r. As -!(72-1)2r=vr=bk=24(72-1), we have r=1. Thus 7T~ is a I-partial spread of W. As 7T~ = 7T _ ~, we have a total of three I-partial spreads and a total of three 2-partial spreads. The union of {V(O), V(oo)} with a 1- and a 2-partial spread yields a spread K of W. Thus we obtain 32 = 9 translation planes K( W). It follows from 19.1 that D(S X S)M is contained in the collineation group of each of these planes. Therefore these planes are all of type F * 7. We have to determine the number of isomorphism types of planes of type F* 7. In order to do this, it suffices by 19.4, 19.7 and 19.8 to determine the number of orbits of (N X N)
101
19. The Planes of Type F '" P
D(S D(S
X X
S)M«I,v)p) are the only subgroups of index 2 in K which contain S)M. Therefore L = D(S X S)M<(1, v)p). 0
As L acts transitively on {V(O), V( oo)} and KI \ { V(O), V( oo)} and since the same is true for K and K 2 , we have:
19.11 Corollary. If I2l is of type F* 7 and if G is its collineation group, then TG {V(O), V(oo)} is a rank-3-collineation group of 12l. Finally, we have to determine the planes of type F
* 11.
19.12 Theorem. There exist up to isomorphism exactly four planes of type
F * 11, one of them being the plane over the exceptional nearfield of order 112 with soluble multiplicative group. If G is the collineation group of one of the other planes, then G{V(O), V(oo)} is conjugate to D(S X S)
c
n·
I UL
cases, p is also a collineation of K( W). Hence k = 10, i.e. there are exactly two planes K( W) where 5 divides k and these two planes are isomorphic. If 5 does not divide k. then k = 1 or 2. Let ao be the number of i~omorphism types with k = 1 and a 1 the number of isomorphism types with k = 2. Then 32
= 25 =
a 020
+ all 0 + 2.
There are exactly 22 = 4 planes with 7T I S; K and p E H. For one of them, 5 divides k. All these planes are pairwise nonisomorphic. Hence a 1 ~ 3. This yields ao = 0 and a 1 = 3. 0
20. Exceptional Rank-3-Planes We are now able to finish the proof of 17.1 and to say more about the exceptions which occur.
20.1 Theorem. Let 91 be a finite affine plane of order n. If G is a non-soluble rank-3-collineation group of 91 with an orbit of length 2 on 100 , then n = 11 2, 2 2 29 , or 59 and W is the plane over the exceptional nearfield of order n, where in the case n = 112 the nearfield with non-soluble multiplicative group is meant. It follows from 14.11 that 91 is not a generalized Andre-plane. Let {P, Q} be the orbit of length 2 on 100 and let 0 be an affine point. The work of section 17 shows that G(P, OQ) is non-soluble. Hence G(P,OQ) contains a subgroup So isomorphic to SL(2,5). Likewise G(Q,OP) contains a subgroup S 00 isomorphic to SL(2,5). Furthermore n = 11 2 , 192 , 292 , or 59 2 . The group SoSoo is normal in Go. Thus the orbits of SoSoo on 100 \{P,Q} all have the same length A. As n - 1 = 120s with s = 1, 3, 7, or 29, either A = 120 or A = n - 1. As 7 and 29 do not divide 120, we have A = 120 or 2 n = 19 and A = 3 . 120. If the second possibility occurs, then So permutes the three orbits of Soo in 100 \{P, Q} transitively. If follows that So contains a normal subgroup of index 3 or 6 which is not the case. Hence A = 120. This proves that 91 is of type R * p. Theorem 20.1 follows now from 18.16, 18.17,18.18 and 18.19. 0 PROOF.
This finishes also the proof of 17.1.
20.2 Theorem. Let 91 be a finite affine plane of order n and let G be a soluble rank-3-collineation group of 91 with an orbit of length 2 on 100 , If 91 is not a generalized Andre-plane, then n = 52, 7 2, 112 or 23 2. If n = 52, 112 or 23 2 , then W is the plane over the exceptional nearfield F of order n, where F is the
exceptional nearfield with soluble F* in the case n = 112. Moreover, the two planes of type F * 7 satisfy the assumptions oj this theorem.
By 17.1,20.1, and the remarks after 17.3, we know that n = 52 ,7 2, 112 or 23 2 • If n = 72, then the planes of type F * 7 satisfy the hypotheses of the theorem. (It is easily seen by a look at their collineation groups that they are not generalized Andre-planes.) Assume that n = 52, 112 or 23 2 . Then 3 is a p-primitive divisor of n - 1 and 3 divides IG(P,OQ)I = IG(Q,OP)I where {P,Q} is the orbit of length 2 on 100 of the group G and 0 is some affine point. Moreover, if L: is a Sylow 3-subgroup of G(P, OQ), then 2: is not normal in G(P,OQ). By 3.5, the group G(P,OQ) contains a normal Z-group N such that G(P, OQ)/ N is isomorphic to a subgroup of S4' As L: is not normal in G(P,OQ), we have by 17.3 that L: is not contained in N. Furthermore, IL:l = 3, as 9 does not divide n - 1. Hence N'J:. is also a ~-group. By 17.3, the group 'J:. is normal in N'J:.. Since 'J:. is not normal In G(P,OQ), we deduce that N'J:./ N is not normal in G(P, OQ)/ N. Thus G(P, OQ)/ N contains a subgroup isomorphic to A 4 . This implies that the Sylow 2-subgroups of G(P, OQ) are generalized quaternion groups of order 8 or 16. This yields that 4 does not divide INI, since the Sylow 2-subgroups of N are cyclic and 'J:. permutes the subgroups of order 4 of a quaternion group of order 8 transitively. We infer therefore from n - 1 = 23 . 3, 5.2 3 .3, 11 . 24 . 3 that N is abelian. Let C be the maximal subgroup of odd order of N. Then C is a normal Hall subgroup of N and hence a normal Hall subgroup of G(P, OQ). Therefore C has a complement D in G(P, OQ) by the Schur-Zassenhaus Theorem. It follows that ID I = 23 . 3 or ID I = 24 . 3. It is now easily seen that D contains a subgroup So ~ SL(2, 3). As I C I = 1, 5 or 11, we have that Aut(C) is cyclic. This implies that So centralizes C. We obtain therefore from 3.6 that So is the only subgroup of CS o which is isomorphic to SL(2, 3). As SL(2,3) does not contain a subgroup of index 2, we finally see that So is the only subgroup of G(P,OQ) which is isomorphic to SL(2, 3). Likewise we obtain that G(Q,OP) contains exactly one subgroup S 00 which is isomorphic to SL(2, 3). SoS 00 is a normal subgroup of Go. Hence all the orbits of SoS 00 in 100\ {P, Q} have the same length A and A is divisible by 24. As 5 and 11 do not divide 24, we find A = 24 or n = 23 2 and A = 2 . 24. In the latter case, So permutes the orbits of S 00 in cycles of length 2. It follows that So contains a subgroup of index 2 which is not the case. Thus A = 24 in all cases and 91 is of type F* p. Theorem 20.2 now follows from 19.5, 19.6 and 19.12. PROOF.
105
21. The Suzuki Groups S(K, a)
We have 1 = 2 - 1 = 0 2 - 1 = (0 + 1)(0 - 1) = (0 - l~(o + 1). Thus 0 + 1 is an automorphism of K*. This and 0 + 2 = (1 + 0)0 YIeld that o + 2 is also an automorphism of K*. 0
CHAPTER IV
PROOF.
The Suzuki Groups and Their Geometries
IfJl =
21.2 Lemma. If K;;;;; GF(q), then
q2
+ l.
The proof is trivial. For a, b E K we denote by rea, b) the mapping defined by
+ a, y + b + a£1x,z + ab +a£1+2 + b £1 + ay + a£1+Ix + bx).
(x, y,zyrCa,b)= (x
Obviously, rea, b) is a collineation of \{S E and hence extends to a collineation of \{S which we also denote by rea, b). 21.3 Lemma. T = {r(a, b) Ia, b E K,~is a group which fixes nand U, and ~
acts transitively and regularly on n\{U}. If a =1=0, then o(r(a, b)) b =1= then 0(1'(0, b)) = 2. Moreover SeT) = {r(O,b)lb E K}. In this chapter we investigate the Suzuki groups as well as the Mobius planes and translation planes belonging to them. The investigations culminate in R. Liebler's characterization of the Luneburg planes. This chapter is a blending of my set of lecture notes 1965b and an unpublished set of lecture notes by A. Cronheim. I would like to thank him very much indeed for allowing me to incorporate his notes into this chapter.
°
PROOF.
= 4; If
We have
(x, y,zyrCa,O)= (x
+ a, y + a£1x,Z + a £1+2 + ay + a £1+1) x
and
(x, y,zyrCO,b)= (x, y
+ b,z + b £1 + bx).
Hence
2l. The Suzuki Groups S(K, a)
(x, y,zyrCa,O)'I'CO, b) = (x
Let K be a commutative field of characteristic 2 with IKI > 2. Moreover let 2 o be an automorphism of K such that x£1 = x 2 for all x E K. We denote by \{S the 3-dimensional projective space over K and we let (xo, x I' x 2 , x 3 ) be ths coordinates of the points of \{S. Let E be the plane defined by the equation Xo = and put U = (0, 1,0,0)K. We introduce coordinates x,y,z in the affine space \{SE by
°
Finally let fJ be the set of points of \{S consisting of U and all those points of ~ E whose coordinates (x, y, z) satisfy the equation
= (X + a, y + b + a ax, Z + a £1+2 + ay +a£1+I(X = ( x,y,z ) Therefore rea, 0)1'(0, b) On the other hand (
x,y,z )
'1'(0,
+ a) + b £1 + b(x + a))
'I'ca, b)
b)'I'(a, 0) = (x Y + b Z + b a + bx yr(a, 0) " = (x + a, y + b + a£1x,z + b O
104
+ I and
o
+2
0
+ 2 are automorphisms of K*.
Thus 1'(0, b)T(a, 0) = T(a, b).
+ bx
+ a(y + b) + a£1+l x )
= (x, y,zfca,b).
First of all we note: 0
.
= rea, b).
+a
21.1 Lemma.
+I
2
+ a, y + a£1X,Z + a£1+ + ay + a£1 x
)'I'CO, b)
IV. The Suzuki Groups and Their Geometries
106 Furthermore
From (0,0, oy(x, y) = (x, y, xy + x o +2 + yO) we infer that T(X, y) = T(X', y') if and only if x = x' and y = y'. Moreover T(X, y) = 1 = T(O, 0) if and only if x = and y = 0. Thus, if a =1= 0, we have T(a, 0)2 = T(a + a, a 1+0) = T(O,a l +o) =1= 1 and hence o(T(a, 0)) = 4. This yields o(T(a,b)) = 4 for all a,b E K with a =1= because of T(a,b) = T(a,O)T(O,b) and T(O, b) E S(T). Let T(a, b) E S(T). Then T(a, 0) = T(a, b)T(O, b) E S(T). Hence
°
(x, y,zr(O, b)T(O,C) = (x, y + b,z + bO + bxr(O,c)
= (x, y + b + c,z + bO + bx + CO + ex) =
(x, y,zf(O,b+C).
= T(O, b + c). This yields in particular T(O, b)2 = 1. T(O,c)T(a,b) = T(O,c)T(a,O)T(O,b) = T(a,O)T(O,C)T(O,b)
°
T(a + c,ac O) = T(a,O)T(e,O) = T(e,O)T(a,O) = T(C + a,ea
Hence T(O, b)-reO, c)
= T(a,O)T(O,b)T(O,C)
T(a,b)T(O,C).
=
Therefore T(O, c) E S(T) for all c E K. We have 2
(x, y,zf(a,O)T(b,O)= (x + a, y + aOx,z + aO+ + ay + aO
107
21. The Suzuki Groups S(K, a)
+I
x
)T(b,O)
= (x + a + b,y + aOx + bOx + bOa,z') where z' = z + ao+ 2 + ay + aO+ IX + f1i.0+2 + by + baox + bO+ IX + bO+ la. On the other hand we have
(x,y,zf(a+b,baa)= (x + a + b,y + bOa + aOx + bOx,zll) where
O ).
Therefore ac ° = ca 0 for all c by the remark made above. Putting e = 1 yields a = aO. Therefore a(e O- c) = for all c. As (J =1= 1, we infer a = 0. Thus S(T) = {T(O,b)lb E K}. Let £)* be the orbit of (0,0,0) under T. Then £)* = £)\ { U}. Moreover T acts transitively and regularly on £)*. It remains to show that U is fixed by
°
T.
°
Let I be the affine line which is defined by the equations x = = y and let (0, XI' x 2' x 3)K be the intersection of I and E. Then x 2 = X3 = and therefore U = I n E. The line I is mapped by T(a, b) onto the line defined by the equations x = a,y = b which is parallel to I. Hence U T = U. 0
21.4 Corollary. For all a, b, e, d E K, we have T(a, b)T(e, d) = T(a + ac O ).
+
°
e, b
+
d
Z"
z + (a + b)bOa + (a + bt+ 2+ b0 a + (a + b)y 2
=
For k E K* we define the collineation Yj(k) by
+ (a + bt+ IX + bOax. Now
(a + b)bOa + b0 2a + (a + bt+ 2= bOa 2 + bo+la + b 2a + (aO + bO)(a + b)2 = b O+la + bOa 2 + b 2ao + aOa 2 + aOb 2 + bOa 2 + bOb 2 = b o+1a + a o+2 + bo+ 2 and
(a + bt+ 1 + bOa
(aO + bO)(a + b) + bOa = a O+I + aOb + bOa + bO+ I + bOa = a O+I + aOb + b O+I.
=
T(a,O)T(c,O)T(O,b + d)
+ b, abO). =
T(a
= T(a + C,O)T(O,acO)T(O,b + d) =
21.5 Lemma. For all a, b E K and all k E K* we have 1](k)-IT(a, b)1](k) = T(ka,ko+lb), i.e., Z normalizes T. Also U Z = U and OZ = 0, where 0 is the affine point with coordinates (0,0,0). Moreover, Z leaves £) invariant and acts (via inner automorphisms) transitively on B( T) \ { I}.
=
Hence z' = Zll and thus T(a, O)T(b, 0) = T(a
T(a,b)T(c,d)
It follows from 21.1 that Z = {1](k)lk E K*} is a group isomorphic to K*.
T( a + c, O)T(O, acO + b + d)
= T(a + c,ac o + b + d). This proves that T is a group.
Therefore we obtain
+ c,acO)T(O,b + d)
PROOF. It follows immediately from the definition of Z that Z fixes 0 and the line I defined by the equations x = 0= y. Hence Z fixes U. A straightforward computation shows T(a, b)1](k) = 1](k)T(ka, kO+ Ib). Hence Yj(k)-IT(a,b)1](k) = T(ka,ko+lb). As Tf(k) normalizes T, the set £)*1'J(k) is an orbit of T. From 0 = OT/(k) we infer £)*T/(k) = £)*. Hence Z fixes £). Finally, it follows from T(O, l)1'J(k) = T(O, k O + I) as well as 21.1 and 21.3 tha t Z acts transi ti vel y on S( T) \ {l } . 0 Let w be the collineation of I.l5 defined by (XO,XI,X2,x3t=(XI,XO,X3' = 1 and U W = 0 and 0 W = U.
x 2 ). Obviously w 2
108
IV. The Suzuki Groups and Their Geometries
Let X = (x,y,z) E D\{O, U}. Then z = xy + x o+2 + yO. This yields O o 1 XZ + z(x + + y) = x(xy.° + x2+20 + y2) + (xy + x o+2 + yO)(x o+1 + y) = xO+yO + X20 +3 + xy2 + xo+y + xy2 + X20+3
+xo+y
+ yOxo+1 + yO+1
= yo+l. If z = 0, then this yields y = 0 and hence x = 0, as x o+2 = Z + xy + yo. Thus z =i= 0, as X =i= O. Let X = (x O,X p X2,x 3)K. We may assume Xo = Z-I. Th~n ~ = X2Z, Y = X3 Z and Z = XIZ, Therefore X = (Z-I, l,xz-I,yz-I)K. ThIs YIelds XW= (l,z-I,yz-I,xz-I)K. This is the affine point with the coordinates x' = yz -I, y' = xz -I, Z' = Z -I Therefore
X'y'
As xy
=
+ X'0 +2 + y'o = xyz -2 + yO+2z -0-2 + xoz-o
= z-o-\xyzo + yO+2 + xoz2). z + x o+2 + yO, we find
xyzo
Also
and
Hence
+ yO+2 + x Oz 2 = zO+1 + x O+2z o + yOzO + yo+2 + xOz2 = zO+I.
= Z-0-2z o+1 = Z-I = z'.
21.8 Theorem (Tits 1962). Let K be a commutative field oj characteristic 2 with IKI > 2 and assume that K admits an automorphism a such that 2 x 0 = x 2 for all x E K. If 0 is the point set defined above in the projective space of dimension 3 over K, then S( K, a) acts doubly transitively on D. Moreover, If 'Y E S(K, a)u, then there exists exactly one triple (k, a, b) E K* X K X K with 'Y = T)(k)T(a, b), and if 'Y E S(K, a)\S(K, a)u, then there exists exactly one quintuple (k, a, b, c, d) E K* X K X K X K X K with 'Y = T)(k)T(a,b)wT(c,d). We remarked already that S(K, a) acts 2-transitively on D. Put and let 11 E Gu,o, Then El1 = E and E6 = Eo by 21.7. From UW = 0 we infer EW = Eo. Hence Eo is defined by the equation z = O. Therefore (x, y, Z)l1 = (x', y', Jz) with f E K*. The line 0 U is in variant under T). Hence (0,0, Z)l1 = (0, 0, fz). Therefore x' = ax + by and y' = cx + dy. The point with coordinates (0, y, yO) is on D. Therefore the point with coordinates (by, dy, fy0) also belongs to D. Hence we have PROOF.
+ yO+2 + x Oz 2 = zO+1 + xO+2zo + yOzO + yO+2 + x Oz 2.
Therefore we obtain X'y' + X,0 +2 + Y'O have proved
in E. Then there exists a, b E K such that I consists just of U and the affine points having coordinates (a, b, z) with z E K. Thus there exists exactly one point other than U of 0 which is on /, namely the point with the coordinates (a,b,ab + a o+2 + b O). 0
G
Furthermore
xyzO
Thus we
= S(K,a)
for all y E K. As IKI > 2, there are two elements y,z E K* with Y =1= z. If (y2,yo+2,yo)=k(Z2,zo+2,zo) for some kEK, then k=(yz-I? and hence yo+z = (yz -1?Z 0+2 = y2z o. This yields yO = ZO and thus y = z: a contradiction. This proves (yZ, yO+z, yO) t£ (Z2,zo+2,ZO)K. This shows that (*) has the two linearly independent solutions (y2, yo+2, yO) and (ZZ,zo+2,zo). From (*) we deduce for all k, y E K.
21.6 Lemma. w leaves D invariant.
~e denote by S(K, a) the group of all projective collineations of ~ WhIC~ . leave D Invariant. Then TZ C; S(K, a) and wE S(K, a). As T is transItIve on D\{ U} and w switches 0 and U, we deduce that S(K a) acts doubly transitively on D.
109
21. The Suzuki Groups S(K, a)
'
As every solution of (*) also solves (**), we see that (**) has two linearly independent solutions. Thus (**) defines for all k E K* the same plane in the vector space of rank 3 over K. Therefore, given k E K, there exists an I E K with (f + dO)k O= (f + dO)l, bdk 2 = bdl and b o+2k o+2 = b o+2l. Assume b =1= O. Then k O+z = I and hence bdk 2 = bdk o+2; thus
21.7 T~eorem (Tits 1962). If P ED, then there exists exactly one plane Ep of SU wah 0 n Ep = {P}. If / is a line through P which is not contained in Ep, then / carries exactly one more point of D.
bd(kO - 1) = O. Choosing k =1= 1 we obtain d = 0 and therefore fko = jl yielding k O= I, asf=l= O. Thus k O+z = 1 = k O. From this we deduce k 2 = 1 and hence k = 1, a contradiction. This proves b = O. As a result, we have 0= (f + dO)yO and thus f = dO.
Because of the transitivity of S(K, a) on 0, we may assume that P = U. Clearly EnD = { U}. Let f be a line through U which does not lie
Thepoint (x,O,x a + 2) is on:D. Therefore (ax,cx,jx a + 2) is on D. Hence fxo+2 = acx 2 + a o+2x o+ 2 + coxo. Similar arguments as above yield c =
PROOF.
and f
=
a o + 2.
°
22. The Simplicity of the Suzuki Groups
Now (ao+1y = a o+2 = f= dO and hence d= a o+ l. Therefore,
III
Because of 21.9 we may write Seq) instead of S(K, 0) if K;;;; G F( q). The groups Seq) are the Suzuki groups named after M. Suzuki who found them in 1960. The generalizations S(K, 0) are due to 1. Tits 1962.
Let y E G u . Then there exists exactly one T(a, b) E T with oy = 01'(a,b). Hence there exists k E K* with rr(a, b)-I = YJ(k), whence y = YJ(k)T(a, b). This decomposition is unique, as ZT is the semidirect product of T with Z. Let y ti. G u . Then U i= UY. Therefore there exists exactly one T(c,d) E T with UY = 01'(c,d). This yields UY1'(c,d)-l w = U and consequently YT(C,d)-l w = YJ(k)T(a,b) for some k,a,b. Thus
21.11 Theorem. IS(q)1 = (q2 IS(q)l·
+ 1)q2(q -
1). Moreover 3 does not divide
PROOF. The first assertion follows from 21.2 and 21.8 and the second from the fact that q = 22r+ 1 == 2 mod 3. 0
Y = YJ(k)T(a,b)wT(c,d).
Assume Y = YJ(k)T(a, b)WT(C, d) = YJ(k')T(a ', bl)WT(C ' , d '). Then 01'(c,d) = UY = 01'(c',d'). Hence c = c' , d= d ' and therefore
YJ(k)T(a,b)
= YJ(kl)T(al,b ' ),
22. The Simplicity of the Suzuki Groups In this section we shall prove that the Suzuki groups are simple.
o
whence k = k', a = ai, and b = b',
Let K be a commutative field of characteristic 2 with GF(4) s: K. If o E Aut(K), then GF(4) = GF(4t, since the multiplicative group of GF(4) is just the set of all cube roots of unity of K. As IAut(GF(4»1 = 2, we deduce xol = x for all x E GF(4). Hence 0 2 i= 2, Therefore, if K;;;; GF(2n) and if K admits an automorphism 0 with xol = x 2 for all x E K, then n = 2r + 1 > 1. (Recall that we are always assuming IKI > 4.) 2 Let K;;;; GF(22r+ I) with 2r + 1 > 1. Put 0 = 2r+I. Then x 0 = X 22 ,+2 = x 2 . Hence K admits an automorphism of the required type. Finally, let a E Aut(K) be such that X ",2 = x 2 for all x. Then x'" = X 2' 221 with 1 < t < 2r + 1. This implies x 2 = X for all x. This yields 2t == 1 mod 2r + 1. Hence 2t = 2r + 1 + 1 and t = r + 1. Thus we have proved
22.1 Lemma. a) Let Y E S(K, 0). If Y has three fixed points on n, then y b) If y E ZT\ T, then y has a fixed point other than U on n.
PROOF. a) We may assume wlog that UY = U and OY = 0, since S(K, 0) is doubly transitive on n. Hence (x, y, z)Y = (kx, k 0+ k 0+2Z ). By assumption there is (x, y, z) E n\ {u, O} with (x, y, z) = (kx, k 0+ k o+2z). As (x, y, z) i= (0,0,0), at least one of the equations 1 = k, 1 = k O+ 1, 1 = k o+2 holds. It follows from 21.1 that k = 1. This proves a). b) We have y = YJ(k)T(a,b) with k i= 1. Hence
y,
( x, y, z ) Y = (kx
+ a, k 0+ Y + b + a 0kx, . . . ).
(x,y,xy
is
+ x o+2 + yO)Y = (x,y,xy + x o+2 + yO).
(Recall that one need not compute the third coordinate of the image of (x,y,xy + x o+2 + yO), since n is fixed by y.) 0 22.2 Lemma. S(K, 0) is generated by all its involutions which have a fixed point on n.
From 21.9 we obtain immediately 21.10 Theorem. Let K be an algebraic extension of GF(2) with IKI
y,
Put x=(k+1)-l a and y=(ko+l+l)-I(b+ak(k+l)-' a ). This possible by 21.1, as k i= 1. An easy computation then shows that
21.9 Theorem. a) If K;;;; GF(22r+ I) with I' > 1, then K admits exactly one automorphism 0 with x02 = x 2 for all x E K. b) If K;;;; GF(2 2r ), then K does not possess an automorphism 0 with x02 = x 2 for all x E K.
= 1.
> 4.
a) If K does not possess a subfield isomorphic to GF(4), then K admits exactly one automorphism a with x02 = x 2 for all x E K. b) If K has a subfield isomorphic to GF(4), then K does not admit an 2 automorphism 0 with x 0 = x 2 for all x E K.
PROOF. Put G = S(K, 0). Then W normalizes Go, u' Furthermore (x, y,zt = (yz -I, XZ -I, Z -I), if z i= 0. Therefore
(1,1, l)wlJ (k)w= (k,k o+ l ,k o+2)w = (ko+I-0-2,kl-0-2,k-0-2)
= (1, 1, 1)7J(k-
I ).
As wYJ(k)wYJ(k-1)-1 fixes 0, U and (1,1,1), we obtain by 22.1 that wYJ(k)w = YJ(k -I). This yields that wYJ(k) is an involution.
112
IV. The Suzuki Groups and Their Geometries
2 The mapping x ---7 x is bijective. Hence there exists I E K with 12 Consequently (l,/o+ I, 10 + 2)W1J(k) = (kl-l,ko+l/-o-l,ko+2/-0-2)
= k.
E K} If b =!= •
= (1,/ 0+1,/ 0+2).
(_;:'i:~-=:~ II;
23. The Luneburg Planes
°
/ \:)~ r;\'~~\,-1"
,-<..1,
Uf(\~~i~~~~'~~~:~~)~ ~Q "~t'- =Ii;,':' .r~ /.' .....
then (0 b bO)W = (b'-O,O,b-O). Moreover
,
,
,
(b1-0,0,b-of(O,C)= (bl-O,c,b- o + CO
\
+ Cbl-O). V0f!;~;'i~;';~!
As - 1 + a is the inverse of 1 + a and hence an automorphism
Therefore wT/(k) has a fixed point. From this we infer that T/(k) = wWT/(k) is the product of two involutions with fixed point. Let T E T and 1 =!= k E K*. Then T/(k)T has a fixed point other than U on U by 22.1. Therefore T/(k)T is conjugate to some element in Go, u whence T/(k)T is a product of two involutions with fixed points. Thus T = T/(k) -IT/(k)w is the product of four involutions with fixed points. The lemma now follows from 21.8. D 22.3 Lemma. All involutions of S(K, a) which have a fixed point on Dare conjugate. If S(K, a) is finite, then all involutions on S(K, a) are conjugate.
:[",r.•• ,. ~\~~1:\O
of~we
see that U\{ U} = {(bJ-o,c,b- o + CO
+ cb'-O)lb,c E K}. w w .(\ { U} U u 3( T) U u 3( T)w3( T) . This yields (B( T), w) There fore Y..J = b 22 2 d 22 3 0 d 3(Tg) for all g E S(K,a). Thus (3(T),w) = S(K,a) y . an .. 22.8 Corollary. If g,h E S(K,a) with ug =!= U h , then S(K,a) = (3(Tg), 3( Th». 22.9 Corollary. If g E S(K, a) and if T is an involution of S(K, a) which has a fixed point on U and if T t1. 3(Tg), then S(K,a) = (3(Tg),T).
This follows from 2l.S and the transitivity of S(K, a) on U and the fact that lui is odd if K is finite. 22.4 Lemma. S(K, a)' = S(K, a). PROOF.
Let a
E
23. The Luneburg Planes
K with a =!= aO. Then
First we want to determine the action of T(a, b) on the plane E. We have
+ b + aO~,~ + ab o 2 + a + + b O+ aT/ + a O+I~
(~,T/,~r(a,b)= (~+ a,T/
is an involution with fixed point. 22.4 now follows from 22.2 and 22.3.
D
22.5 Theorem (Iwasawa). Let G be a finite or infinite permutation group on the set @3 with the following properties:
a) Gaels primitively on @3. b) There exists P E @3 such that C p contains a normal soluble subgroup A which together with its conjugates generates C. c) G=C'.
°
+
b~).
' k (t. >') =!= (0 0) and let G be the line whose affine points have P IC <;,Yj,!\ , , ' t h l' CT(a,b) are the coordinates (A~, AT/, An. Then the affine pomts on e me points with coordinates ' O o 2 (A~ + a, AT/ + b + aOA~,A~ + ab + a + + bO + aAYj + a + IA~ +, bA~)., Thus the line G T(a, b) is parallel to the line H consisting of the pomts wIth coordina tes (A~,AT/
Then C is simple.
+ aOA~,A~ + aAT/ + ao+IA~ + bA~).
n
PROOF. Let 1 =!= N ~ C. Then N acts transitively on @3, as G is primitive. Therefore NA contains all the conjugates of A whence G = NA. From G/ N = NA/ N ~ A/(N n A) we infer that G/ N is soluble. This yields D G = N, since C = G'.
The point (~, Yj, is projectively represented by (1,~,~, T/~K . Hence G = (1,O,O,O)K + (l,~,~,T/)K. Therefore G n E = (O,t,~,T/)K. SImIlarly we obtain
22.6 Theorem (Suzuki 1962, Tits 1962). The Suzuki group S(K, a) is simple.
Consequently
In 22.5, set @3 = £, P = U and A consequence of 22.2, 22.3, 22.4, and 22.5. PROOF.
22.7 Theorem. S(K, a) = PROOF.
From B(T)
=
= B( T). Then
22.6 is a D
CH T), w).
{T(O,b)jb
E
H n E = (O,~ + aT/ + (a o +1 + b)~,~,Yj + aO~)K.
«0 ,XI ,x
2 '
X3 ) K)T(a,b)=
(0 ' x I + (a o +1 + b)X2 + ax 3 ,x 2 ,a ox 2 + X3)K.
Consider the mapping T *( a, b) defined by
T*(a,b)=(x I +(ao+l+b)x2+ax3,x2,aox2+x3)' ( XI' X 2 ' x) 3 K} we infer (O,O,O)~)(T)
=
{(O,b,bO)lb
Then the restriction of T(a, b) to E is induced by T*(a, b).
114
I V. The Suzuki Groups and Their (jeometnes
°
If I is the line defined by the equations Xo = = x 2 , then r(a,b) = I. Furthermore U ~ J and (0, 1,0y*(a,b) = (ao+ 1 + b, I,aO). Given XI and x 3 ' there exist a, b E K such that a ° = X3 and a 0+ I + b = XI' Therefore, T acts transitively on the set of points of E which are not on I. Consequently, I is the only fixed line of Tin E. We have
IW
=
{(O, XI ,0, x 3) I XI ,x 3 E K
r = {(XI' 0, X3 ,0)
Hence the coordinates of the affine points on I
W
I XI ' X3 E
K}.
satisfy the equations
y = z = 0. Moreover
23.2 Theorem. 7T = {goo} U {ga,b I a, b E K} is a spread of V = KEEl K EEl K EB K. Moreover there exists a subgroup of SL( V) isomorphic to S(K, a) which acts doubly transitively on 7T. Finally K( V, 7T) ~ K and rank K ( v, 17) (V)
= 4.
PROOF. We first consider the elements of 7T as lines of the projective 3-space SU over K. Then S(K, a) permutes 7T doubly transitively. It follows from 23.1 that S(K, a) operates transitively on the set of points of SU which do not belong to O. Hence every point of SU is on a line belonging to 7T. Consequently V = UXE17X, Also
goo + go, 0 = {(O, s, 0, t) Is, t
(x,O,Or(a,b)= (x + a,b + aOx,ab + a o+2 + b O+ ao+lx + bx). Therefore
IWT(a,b) = {(x 0' abx 0 +a o+2x 0 +box 0 +ao+lx 3 +bx 3' X3 + axe , bx o + a °x 3) I Xo ' X3 E
K}.
Put goo = {(O,s,O,t)ls,t E K) and
ga, b = {( s, [ ab + a 0 + 2 + b 0 ] s + [ a 0 + I + b ] t, t + as, bs
+ a °t) Is, t E
K }.
We want to prove that 7T = { goo} U { ga, b Ia, b E K} is a spread. In order to do this we first prove
23.1 Lemma. S(K,a)u operates transitively on the set of points SUE which do not belong to O. PROOF. Let (x, y, z) be a point of SUE which is not in O. Then z+xy+x o+2 +y°=l=O. By 21.1 there exists kEK* with k o+2 =z+ xy + x o+2 + yO. Furthermore there exist a,b E K with k + a = x and b + a Ok = y. From this it follows (1, 0, O)l](k)T(a,b)= (k + a,b + aOk,ab + a o+2 + b O+ ao+lk + bk)
E
K} + {( s, 0, t, 0) Is, t
E
K} = V.
We deduce from this and the double transitivity of S(K, a) on 7T that V = X EB Y for all X, Y E 7T with X =1= Y. Thus 7T is a spread of V. The determinant of w is 1. Hence w E SL( V). As SL( V) is a normal subgroup of GL( V), we deduce from 22.3 that every involution of S(K, a) having a fixed point is induced by an involution in SL( V). Let G be the group generated by all involutions of SL( V) which induce an involution in S(K, a) having a fixed point in O. Let G be the group induced by G on ~. By 22.2 we have G = S(K, a). Moreover, by the general theory of projective spaces, G ~ G/3(G) and 3(G) = G n 3(SL(V)). As the centre of SL(V) consists of all the mappings v~ kv of V onto itself with k4 = 1, we deduce 3(SL(V)) = {l}. Hence G ~ G = S(K,a). Certainly K( V, 7T) contains a subfield Ko isomorphic to K such that V is of rank 4 over Ko. Therefore rank K ( v, 17) (V) = 2 or 4. If it is 2, then 7T( V) is desarguesian. It then follows that G is isomorphic to a subgroup of GL(2,K(V,7T»). Therefore the stabilizer of goo in GL(2,K(V,7T)) contains elements of order 4, a contradiction. Hence rank K ( v, 17) (V) = 4. 0 The planes 7T( V) constructed in this section are called the Luneburg
planes.
= (x,y,ab + a o+2 + b O+ ao+lk + bk). Now we have
ab + a o+2 + b O+ ao+lk + bk = a(aOk + y) + a o+2 + a 2k o + yO + a O+Ik + a Ok 2 + yk
24. The Subgroups of the Suzuki Groups
= ay + a o+2 + a 2k o + yO + a Ok 2 + yk = xy + (k + X)0+2+ (k + x)2kO
We start with a lemma that will be useful in the sequel.
24.1 Lemma. Let G be a finite group and H a subgroup of G. If (£c(h) C H for all h E H having prime order, then H is a Hall subgroup of G.
+yo+(k+x)Ok2 = xy + (k + xt x2 + k o+2 + k ox 2 + yO = xy + x 0+2 + yO + k a + 2 =
Hence (1,0, O)l](k)T(a, b) = (x, y, x).
z.
0
PROOF. Let S be a Sylow subgroup of H. We have to show that S is also a Sylow subgroup of G. Assume by way of contradiction that T is a Sylow subgroup of G with SeT and S =1= T. Then there is a subgroup S* of T which properly contains S with S normal in S*. Since S is normal in S*,
116
IV. The Suzuki Groups and Their Geometries
~e have S n 3(S*) *- {I}. Hence there exists an element s of prime order In S n 3(S*). Consequently S* C Q:G(s) C H, a contradiction. 0 x?
2
Let q
= 2 ,\+ 1 and
=
for all x E GF(q).
x
2
G
= Seq). Put r = 2'\+ 1. Then r2 = 2q and hence
The element x of the group X is called real, if there exists y E X with x Y = X-I
24.2 Theorem (Suzuki 1962). Let S be a Sylow 2-subgroup of G, then we have: a) exp(S) = 4. b) 3(S) = {a/a E S, a 2 = I} -and /3(S)/ = q. c) 91 G(S) is a Frobenius group of order q2(q - 1). The Frobenius kernel is S and each Frobenius complement is cyclic of order q - 1. d) If 1 *- s E S, then Q:G(s) C S. e) The element s E S is real, if and only iJ s E 3(S). f) S does not contain a quaternion group. g) If Z is a Frobenius complement of 91 G(S), then Z acts transitively on 3(S)\{l}. PROOF. a), b), c), and g) have been proved earlier. In order to prove d), e), and f), we may aSSume S = T. d) Let 1 *- T(a, b) E T and y E Q:G( T(a, b». As U is the only fixed point of T(a, b) on §J, we have UY = U. Since G u is a Frobenius group and y E G u , we deduce yET. Thus Q:G( T(a, b» C T proving d). e) Let T(a, b) E T be real and assume T(a, b)2 *- 1. Using Lemma 21.5, we see that we may assume a = 1, since T(a, b)2 *- 1 implies a O. As T(l,b) is real, there exists y E G with T(l,b)y = YT(l,b)-I. Obviously, y2 centralizes T(l, b). Hence y2 E T by d) and thus yET. Therefore y = T( C, d) for some c, d E K. We thus have
'*
T ( 1, b )T ( c, d) = T ( c, d )T ( 1, b) - 1 = T ( c, d}r ( 1, b
+ 1).
From 21.4 we then obtain b + d+ CO = b + d+ 1 + c. Thus CO = 1 + c. o2 2 This implies c = c = 1 + CO = 1 + 1 + c = c. Therefore c = 0 or c = l. This yields CO = c and hence 1 + c = c, a contradiction. This contradiction shows that T(a,b)2 = 1 and hence T(a,b) E 3(T). Conversely, if T E 3( T), then T = T - 1 and T is real. f) is implied bye), as the elements of order 4 in the quaternion group of order 8 are real. This finishes the proof of 24.2. 0 For the lelement g of the group X we put Q:.t(g) = {xlxEX, x g c{g,g-}}.
W. e call ~he element g of the group X strongly real, if g is the product of two InVOlutlOns of X.
24. The Subgroups of the Suzuki Groups
117
24.3 Lemma. Let 1 =1= g E G be strongly real. Then g is an involution or g is conjugate to TW where T is some element in 3(T)\{l}. If T,T' E 3(T)\{l} are such that TW and T'W are conjugate, then T = T'. Hence there are exactly q - 1 conjugacy classes of strongly real elements which are not involutions. If g is not an involution, then IC£G (g)1 is odd. PROOF. There exist involutions hI and h2 such that g = h lh 2. Let Sj be the Sylow 2-subgroup containing hi' If S I = S2' then g is an involution. Assume S I *- S2' Let S3 be the Sylow 2-subgroup containing wand let R be the fixed point of S3' Then G u n GR is a Frobenius complement of GR which by 24.2 operates transitively on 3(S3)\{1}. As G is 2-transitive on §J, there exists 1] E G with S( = T and Si = S3' Furthermor;, there exists o E Gu n GR with hio = w. It follows from UO = U that T = T. Hence T = h(8 E 3(T)\{l}. Thus g1J 8 = TW is of the required form. Let a = TW and b = T'W. Assume that o(a) is even. Then a centralizes a 2-element which is different from 1. Hence a is a 2-element by 24.2 d). Thus 0(4) = 2 by 24.2 e). This yields that T centralizes w. Therefore T fixes U and R whence T = 1, a contradiction. Hence o(a) is odd. Similarly, we see that o(b) is odd. As a centralizes no 2-element other than 1, we have that IQ:G(a)1 is odd. Therefore w f!. Q:G(a), but wE Q:(j(a). Thus <w) is a Sylow 2-subgroup of Q:(j(a), since 1C£(j(a): C£G(a)1 < 2. Similarly, <w) is a Sylow 2-subgroup of Q:(j(b). Let g E G and assume a g = b. Then there exists h E Q:G(b) with <w)gh = <w) whence gh E C£G(w). Furthermore a gh = b h = b. This yields T gh = T'. From this we infer Ugh = U, as U is the only fixed point of T and T'. Therefore gh E Gu n Q:G(w) C Gu n S3 = {I} whence T = T'. 0
24.4 Lemma. Let g E G be strongly real and assume g2 *- 1. Then A = Q:G(g) is abelian. If 1 *- g' E A, then g' is strongly real and Q:G(g') = A. Moreover A is a Hall subgroup of G and Q:G(A)/ A is cyclic of order 2 or 4. PROOF. IA I is odd by 24.3. As g is strongly real, there exist involutions p and T with g = pT. Therefore p-Igp = g-l whence pc 91 G(A). We infer that I?nG(A)/ A I is even. Let a, b c A, let 13 be an involution in ?nG(A) and assume that a -lf3af3 = b -lf3bf3. Then ab -113 = f3ab -I. Hence ab - I E ?n G( 13) n A = {I} by 24.2 d). Thus a = b. As A is finite, we obtain A = {a - If3af3l a E A}. Now f3(a- If3af3)f3 = f3a- If3a = (a- 1f3af3)-I. Thus af3 = a-I for all a EA. This implies that A is abelian. Moreover (af3f = 1 whence af3 is an involution for all a EA. As a = (f3a -1)13, we have that a is strongly real forallaEA. Let l=l=g'EA. Then A ~C£G(g'). As g' is strongly real and o(g') is odd, Q:G(g') is abelian. Hence Q:G(g') C Q:G(g) = A. Therefore Q:G(gl) = A. From this and 24.1 we infer that A is a Hall subgroup of G.
- •. - --- .....
It remains to show that 9C G (A)/ A is cyclic. As A is a normal abelian Hall subgroup of 9C G(A), there exists a complement C of A in iRG(A) by the Schur-Zassenhaus theorem. Moreover, since A is its own centralizer, 9C G (A)/ A is isomorphic to a group of automorphisms of A. Since af3 = a - I for all involutions [3 in 9C G(A) and all a E A, the group 9C G (A)/ A contains exactly one involution. We infer from C;;;;;; 9C G (A)/ A that C contains exactly one involution [3. Therefore C C Q:G( [3) whence C is a 2-group by 24.2 d). This implies that C is either cyclic or a generalized quaternion group. Now 24.2 a) and f) yield that C is cyclic of order 2 or 4.
o
24.5 Corollary. If g, g' E G\{l} are strongly real and then Q:G(g) n Q:G(g') = {I}.
if
PROOF. This is certainly true, if g2 = l. We may thus assume that g2 =1= 1. Then Q: G (g) has odd order by 24.2 d). Let h E G such that gh = g - I. Then h 2 E Q:G(g) and h =1= l. Thus h is a 2-element by 24.2 d) whence h 2 = 1, as IQ:G(g)1 is odd. Therefore g = hg-Ih is the product of the involutions hg-I and h. 0
24.7 Theorem (Suzuki 1962). If G ;;;;;; Seq), then all elements of odd order of G are real. If 1T is the set of centralizers of the real elements of G which are different from 1, then 1T is a normal partition of G. The partition 1T consists of four classes of conjugate subgroups of G: the Sylow 2-subgroups of G, one class of cyclic groups of order q - 1, one class of cyclic groups of order q + r + 1, and one class of cyclic groups of order q - r + 1, where r2 = 2q. PROOF. The normalizer of Go, u is a dihedral group of order 2(q - 1). Therefore all the elements of Go,u are strongly real. Let ~O'~I' . , . ,~s be all the conjugacy classes of centralizers of strongly real elements of odd order and let the labelling be such that Go, u E ~o. Let A 0' A], ... ,As be a transversal of ~O'~]' . . • , ~s' Put ai = IAil and bi = I9C G (AJ : AJ Then a o = q - 1, b o = 2 and bi = 2 or 4 by 24.4.
= 1 for i =1=
k. In particular ai divides q2 + 1 for i
I
~
......... ....., ...... t-'u
I
s
(ij)
q - 1=
2: bi-I(ai -
1).
i=O
Let d be the number of non-real elements of odd order. Then we have s
g
(iij)
= IGI = I + (q2 + 1)( q2 - I) +
2: (a
i -
1)gbi-Iai- I + d.
i=O
By 21.10 we know that 3 does not divide ai • Hence ai > 5. Therefore 1 - l)ai = 1 - ai-I> I - t = ~. Let u be the number of i's with bi = 2 and put v = s + 1 - u. Then
(al
1 = g-l + g-l(q2 + 1)(q2 - 1) +
s
2: (ai -
4
I
4
1
> '5 u 2' + '5 v"4
2
=
l)bi- l ai- 1 + dg-]
1
'5 u + '5 v.
+ v , ;;; 4. As bo = 2, we have u > 1. Thus the only solutions of 2u + v , ;;; 4 are Cu, v) = (2,0), (1, 0), (1,1), (1, 2). Assume u = 2 and v = O. Then s = l. By 0), we have 2(q - 1) = q - 2 + a l - 1 whence a l = q + l. But this contradicts the fact that a I divides q2 + 1. Hence u = 1. Assume v = O. Then 2(q - 1) = q - 2 by (j) implying q = O. Assume v = 1. Then 4(q - 1) = 2(q - 2) + a l - 1 which yields a l = 2q + l. By (j), we have q2 + 1 = k(2q + 1) for some integer k. We deduce k == 1 mod q. As q > 2, we obtain k > 1 and hence k > q + l. Thus q2 + 1 > (q + 1)(2q + 1), a contradiction. Therefore u = 1 and v = 2. This yields 4(q - 1) = 2(q - 2) + a 1 - 1 + a 2 - 1 and hence Hence 2u
a 1 +a 2 =2(q+I).
(iv) Now
(a l
-
1)g(4a 1)-I+ (a 2 - l)g(4a2)-I= g(4a l a2)-1[2a]a 2 - a 1- a2]
= g(4a 1a2)-1[2a]a 2 - 2(q + I)J by (iv). Using this and (iij) we obtain
i
g(q - 2) g(2a 1a2 - 2(q + 1») g = 1 + (q2 + 1)( q2 - 1) + 2( q _ 1) + 4a I a + d. 2 Since g = (q2 + l)q2(q - 1), we get (q2 + 1)q2( q - 1)
=
1 + (q2 + 1)( q2 - 1) +
+ (q2 + 1)q\ q - 1)
±(q2 + 1)q2( q - 2)
a a2
> 1.
This follows, using 24.4, from a theorem of Wielandt (see e.g. Huppert [1967, III.5.8, p. 285]). We infer from 24.5 that for X, Y E ~i either X = Y or X n Y = {l}. Hence Ux ESt; X consists of bi- I( ai - 1) classes of conjugate elements, each
17
elements. Therefore it follows from 24.3 that
i=O
24.6 Lemma. If g EGis real, then g is strongly real.
(ai' ak )
..... -t" .... ""'4 ...... - .........................
class consisting of I G la i -
Q:G(g)=1= Q:G(g'),
PROOF. Assume Q:G(g) n Q:G(g') =1= {l}. If g is involutorial, then Q:G(g) is a Sylow 2-subgroup of G. We infer from Q:G(g) n Q:G(g') =1= {I} that Q:G(g') is also a Sylow 2-subgroup of G. Hence Q:G( g) = Q:G( g') in that case. If g is not an involution, then g' is not an involution. Hence IQ:G( g)1 and IQ:G(g')1 are odd. Pick h E Q:G(g) n Q:G(g') with h =1= l. Then 24.4 implies Q:G( g) = Q:G(h) = Q:G( g'). 0
(j)
o~
= 1 + (q2 + 1)( q2 - 1) +
I
-
q- 1
2a 1a2
+d
~ (q2 + 1)q2( q -
2)
+ l( 2+ 1) 2( -1)- (q2+ l)q2(q2-1) +d. 2 q
q q
2a a
I 2
',1
120
IV. The Suzuki Groups and Their Geometries
A straightforward computation shows that
- t q\ q2 Put d
= cala2q2(q -
1) = -
t (q2 + 1)q\ q2 -
l)a i 1a 2- I
+ d.
1). Then we obtain
(v) Let x be non-real of odd order. Then Q:G(x) does not contain a real element othe~ than ,1. As the A/s are nilpotent Hall subgroups of G, it fo~lows by Wielandt s theorem (Huppert loco cit.) that IQ:G(x)1 is relatively p~l~e to 2(q - l)a 1a2 . As 2,q - I,a l ,a 2 are relatively prime, q2(q - l)a a dIVIdes IG: Q:G(x)l. Hence d is divisible by q2(q - I)a a whence c is la~ i~t~ger. Since a 1a2 divides q2 + 1 and (q + I,q2 + 1) = {, ~e see that q + 1 dIVIdes C.
Assu~e c > O. Then (q2 + l)(q + 1) ~ a 1a 2(q + 1) + 2(q + l)a~ai and hence q 2 + 1 > a 1a2(1 + 2a 1a2)· We deduce from (iv) that a 1a 2 ~ q + 1. Hence q + 1 > (q + 1)(2q + 3) > q2 + 1, a contradiction. Thus c = 0 and therefore d = O. This proves that 1T is a normal partition. We have a l + a2 = 2(q + 1) by (iv) and a 1a2 = q2 + 1 by (v). Thus a l = 2(~ + 1) - .a 2 and hence q2 + 1 = 2(q + I)a 2 Solving this quadrat~c equatIOn yields a2 = q + r + 1 or a 2 = q - r + 1 where r2 = 2q. In the first case a l = q - r + 1 and in the second a l = q + r + 1. Thus we may assume a 1 = q + r + 1 and a 2 = q - r + 1. The group PGL(4, q) contains a cyclic subgroup Z of order (q + I)(q2 + 1) (see e.g. Huppert [1967, 11.7.3 c), p. 188]). As q is even and IPGL(4, q)1 = q6(q2 + l)(q - 1)3(q2 + q + 1)(q + 1)2, the subgroup U of Z of order q2 + 1. is a Hall subgroup of PGL(4, q). By Wielandt's theorem, Al and A2 are conjugate to subgroups of U. Hence A 1 and A2 are cyclic. 0
ai.
Case 1: All Sylow subgroups of Ware cyclic. By a theorem of Burnside, W contains a cyclic normal subgroup V distinct from {1}. As V is cyclic, there exists X E 1T with V c:: X. Let yEW, then XY E 1T, since 1T is a normal partition. On the other hand {I} -=/= V = V Y c:: X n XY. Hence X = XY, i.e., W ~ \)(S(q) (X). Case 2: The group W contains a normal 2-subgroup LO distinct from {I}. Then the same argument works by taking for X a Sylow 2-subgroup of Seq) containing Lo. By 24.9, it remains to consider Case 3: The group W contains more than one Sylow 2-subgroup. Let Lo be a Sylow 2-subgroup of W. As 1T is a normal partition, we may assume Lo C T. If LO is cyclic, then by 24.9 and Burnside's theorem, LO is normal. Hence Lo is not cyclic. Also Lo is not a quaternion group by 24.2 f). Thus LO contains more than one involution. As Lo is not normal in W, there exists an involution in W\Lo. Using that Seq) acts doubly transitively on n, we may as well assume that wE W. Put 1= Lo n B(T) and III = qo' Then qo > 4. Furthermore put N = 9C w (Lo). Obviously N = Wu' Put I W: NI = k + 1. Then k + 1 is the number of Sylow 2-subgroups of W.
(a) In order to prove this let p and a be involutions in W with PL o = aL o. Then pa E Lo. If p -=/= a, then pa -=/= 1 and U is the only fixed point of (pa). As (pa) is normalized by p and a, we have UP = U = u a • Hence p, a E Lo. Therefore, if x E W\Lo then XLo contains at most one involution. The number of involutions in W is (k + 1)(qo - 1) and qo - 1 is the number of involutions in Lo. The number of left cosets of LO being distinct from Lo is I W: Lol - 1. Hence
(k + 1)(qo - 1) ~ qo - 1 + IW: Lol- 1 = qo - 2
24.8 Corollary (Suzuki 1962). If 1 -=/= Y E Seq), then Q:S(q)(Y) is nilpotent. PROOF. If Y is a 2-element, then ~S(q)(Y) is a 2-group by 24.2 d). If Y is not a 2-element, then y has odd order and is strongly real by 24.7 and 24.6. Hence ~S(q)(Y) is abelian in that case by 24.4. 0 24.9. Corollary (Suzuki 1962). The Sylow subgroups of odd order of Seq) are cyclic. PROOF. Let L be a Sylow subgroup of odd order of Seq). Pick E B(L)\{1}. Then L c:: ~S(q)(n. Moreover ~S(q)(n E 1T. Hence Q:S( len, and therefore L, is cyclic by 24.7. q
t
24.10 The~rem (Suzuki 1962). Let 1T be the partition of Seq) described in 24.7. If W IS a subgroup of Seq), then there exists X E 1T with We\)( (X) " Isomorp h'Ie to S(qo) with qo ~ q. - Seq) or W IS
PROOF. We may assume W -=/= {I}.
121
24. The Subgroups of the Suzuki Groups
=
+
I W: NIIN : Lol
qo - 2 + (k + 1)1 N : Lol·
This yields qo - 1 ~ (qo - 2)(k + 1)-1 + IN: Lol· Let L J be a Sylow 2-subgroup of W which is distinct from Lo. Then I{LgI~ E B(Ll)}1 > qo' Hence k + 1 > qo and thus (qo - 2)(k + 1)-1 ~ 1 and qo - 1 ~ IN: Lol· As N= W u , there exists a cyclic subgroup A with N=LoA and Lo n A = {I}. Also, the order of A divides q - 1. This implies that A operates regularly on I\{ I}. Hence IN: Lol = IA I ~ qo - 1. Therefore IN:Lol = qo-I. From q - 1=0 mod IAI and IAI = qo - 1 we deduce:
q = qo for some integer s.
(b) Next we prove:
(c)
If x E W\N, then XLo contains exactly one involution.
ILL
We have I W: 2:01 = (k + l)(qo - 1) and IN: 2:01 = qo - l. Also, all involutions of N belong to LO' Therefore, the (k+ l)(qo-l)-(qo-l) = k(% - I) involutions outside of N lie in the union of the k(qo - 1) cosets of 2:0 which are not contained in N. As each of these co sets contains at most one involution, (c) is proved.
(d) The points U, U WT with 7" E 1 are pairwise distinct. If U = Uw-rw, then 7" since Q = U W= Uw-r = QT. Assume U WT = U WT jWT 2. Then QT' = QTjW, where 7"' = 7"7"2' Now
7"(~,ka+lg), where ex is again the automorphism of GF(q) satisfying
x
a
= x 2 for
all x EGF(q). From this and 21.1 we infer
1= {7"(O,k a+ lg)lk EGF(qo)}. Now UWT(O,b) = (O,b,b a) and hence, if b i= 0, Uw-r(O, b)WT(O, c) = (0, b, b atT(O, c) = [( b a, 1, b, O)GF( q) reo, c)
= [(1, b -a, b I-a, O)GF( q) reo, c) =
= 1,
= (bl-a,c,b- a + c a + cbl-a). This implies in particular that the y-coordinates of the points in U w\ { U} are in gGF(qo)' Putting b = c = g we obtain
QTjW = (O,O,Or(O,aj)w= (O,a l ,aft= [(l,af ,0,al)GF(q)r = (af, 1,a l ,0,)GF(q) and
UWT(O, g)w-r(O, g)w
QT' = (O,O,Or(O.a)= (O,a,a O ) = (l,a O ,O,a)GF(q).
°
This implies a l = and hence 1 = 0, a contradiction. Thus the orbit of U under W contains at least 1 + qo + (qo - l)qo = q6 + I points. Hence k + 1 > q6 + 1, i.e., k > q5· Each left coset of N contains exactly qo - I left co sets of LO' Therefore, any left coset of N, including N itself, contains exactly qo - 1 involutions of W. If P is an involution in W\N, then UP i= U. Put P = Up. Then, of course, pP = U. Let P = PI> ... ,PI' where t = qo - 1, be all the involutions in pN. Then Pi = PVi for some Vi E N = Wu' Hence PPi = PPI'i = U. Thus UPi = P for all i. Consequently PIPi E Wu,p for i = 2, ... , t. Hence IW u , pi > qo - 1 and therefore IWu,pl = qo - 1. Since there are k left cosets of N other than N, we obtain in this way at least k complements of 2:0 in N. As N is a Frobenius group, we therefore obtain 12:01 + k(qo - 2) < INI = (qo - 1)I L ol· Thus k < 12:01· Therefore q5 < k < 12:01· As III = qo we have in particular 2:0\1 i= 0. l Pick 0 E 2:0\1. Then = 7"(a, b) with a i= 0. Also 7"(a,b? = 7"(O,a +a) 2 2 where ex is the automorphism of GF(q) with xa = x for all x E GF(q). If P E 2:0\1 is such that p2 = 0 2 and if P = 7"(c, d), then c l +a = al+a and hence c = a by 21.1. This implies po E 1. Hence there are at most qo such P's with p2 = 0 2. Therefore ILol < q5 whence q5 < k < 12:01 < q6. This proves (d). As an immediate consequence we have:
a
(e) The group A = Wo , u is cyclic of order qo - 1. If YJ(k) E A, then k qo Hence:
I
=
(gl-a, g, g-a + ga + g2-a)w
= [(1, g-a + ga + g2-a, gl-a,
=
(I,A, gA, gl-aA)GF(q)
= (gA, gl-aA,A), where A = (g-a + ga + g2-a)-I. (It is easily seen that g-a + ga + g2-a i= 0.) By the remark made above gl-aA E gGF(qo). Hence g-aA E GF(qo) whence ga(g-a + ga + g2-a) E GF(qo)' This yields 1 + g2 + g2a E GF(qo) and therefore (g + ga)2 = g2 + g2a E GF(qo). This implies g + ga EGF(qo)' Consequently ga + g2 = (g + gat E GF(qo) and hence g + g2 EGF(q~). From this we infer [GF(qo)(g): GF(qo)] < 2. As q = 22,\+ I, we obtam GF(qo)(g~ = GF(qo), i.e., g E GF(qo). Therefore 1 = {7"(0, b) Ib EGF(qo)}' From thIS, 22.7 and the fact that wE W, we finally see that W~
S(qo)'
0
24.11 Corollary (Suzuki 1962). If qo is a divisor of q and if qo - 1 divides q - 1, then Seq) contains a subgroup So isomorphic to S(qo)' Moreover, if SI c;: Seq) and SI ~ S(qo), then SI and So are conjugate in Seq).
It should be mentioned that this character-free proof of 24.10 is due to A. Cronheim.
= 1.
25. Mobius Planes An incidence structure IJJ( following hold:
Next we prove: 7"(0, b) E I,if and only if bE GF(qo). Let 7"(0, g) be a fixed involution in 1. By 21.5 we get 7"(0,
g)GF(q)f
= (g-a + ga + g2-a, 1, g, gl-a)GF(q)
(f) (g)
(b I-a, 0, b -a) T(O, c)
g)l1(k)
=
= (s:rs, G£, I)
is called a Mobius plane, if the
(1) If A, B, C are three distinct points in s:rs, then there exists exactly one circle kin G£ with A,B, elk.
124
IV. The Suzuki Groups and Their Geometries
(2) Let A, E E SU and k E G£. If A I k and Elk then there exists exactly one k' E G£ with A,E I k' and k n k' = {A}. (3) Each k E G£ is incident with at least one A E SU and there exist four points in SU which are not concircular. For P ESU define SUp, G£p, and Ip by SUp = SU\{P}, G£p = {klk EG£, PI k} and Ip = I n (SUp X G£p). Furthermore put weep) = (SUp,G£p,I p)'
25.1 Lemma. If points P of we.
we is
f) g) h) i)
t
The automorphism a of the Mobius plane if a =t= 1 and if a fixes a circle pointwise.
we is called a reflection
25.3 Lemma. If a is a reflection of a Mobius plane, then a 2
We shall say that the circle k avoids, touches, respectively intersects, the circle k', if Ik n k'i = 0, 1, respectively 2. A set ?B of circles is called a pencil with carrier P if ?B is a parallel class in weep).
a) b) c) d) e)
uniquely and as there are ~ n(n + 1) point-pairs on k, there are exactly n 2 (n + 1) circles intersecting k. Finally, i) follows from b), g), and h). This finishes the proof of the theorem. 0
we
of
we
a Mobius plane, then weep) is an affine plane for all
PROOF. Call the circles of G£p lines of weep). Let Q and R be two distinct points of weep). By (1), there exists exactly one line of weep) joining Q and R. Condition (2) gives us Euclid's axiom of parallelism for weep) and (3) guarantees non-degeneracy. 0
25.2 Theorem. Let integer n > 2 with:
125
25. Mobius Planes
= l.
PROOF. Let k be the circle fixed pointwise by a and pick P on k. Then a induces a perspectivity with affine axis in weep). Hence there exists a pencil ?B(P, a) with carrier P which is fixed elementwise by a. Let Q be on k and distinct from P. Then we obtain similarly a pencil ?B( Q, a) with carrier Q which is fixed elementwise by a. Let R be a point off k. Then there exist two circles I and m with I E ?B(P,a) and m E ?B(Q, a) and R I I, m. As R I k, we have PI m and hence I =t= m. It follows from la = I and m a = m that {R, R a} C;; I n m. Furthermore R =t= R a, as a =t= 1. Thus 2 a2 {R,R a } = I n m and {R a,R } = la n m a = I n m. Hence R = Ra 2 whence a = 1. 0
be a finite Mobius plane. Then there exists a positive
The number of points of we is n 2 + 1. The number of circles of we is n(n 2 + 1). Each circle carries exactly n + 1 points. Each point is on exactly n(n + 1) circles. Given distinct points A and E, then there exist exactly n + 1 circles k with A,E I k. Every pencil contains exactly n circles. There exist exactly n 2 - 1 circles touching a given circle. There exist exactly 1n 2(n + 1) circles intersecting a given circle. There exist exactly 1n(n - l)(n - 2~ circles avoiding a given circle. The integer n is called the order of
we.
PROOF. Pick a point P of we. Then weep) is a finite affine plane. Let n be its order. Then weep) has n 2 points. Therefore a) is true. This implies that weep) is an affine plane of order n for all points P of we. Thus c), d), e), and f) are true. As we is obviously a tactical configuration, we have (n 2 + l)n(n + 1) = ben + 1), where b is the number of circles. Hence b = n(n 2 + 1) proving b). By c), each circle belongs to n + 1 pencils. This and f) imply g). Given a point-pair on k, then bye) there exist exactly n circles intersecting k in the given point-pair. As three points determine a circle
25.4 Lemma. If k is a circle of a Mobius plane, then there exists at most one reflection fixing k pointwise. PROOF. Let a and 7' be reflections fixing k pointwise. Then a 2 = 7'2 = (a7')2 = 1 by 25.3. Hence a7' = 7'a. If PI k then ?B(P, a) = ?B(P,7'), as a and '1' centralize each other. Likewise ?B(Q,a) = IB(Q, 7') for P =t= Q I k. Repeating the argument of the proof of 25.3, we then see that {R,Ra} = I n m = {R,R"r}, whence a = 7. 0
25.5 Lemma. Let we be a finite Mobius plane and let a be an involutory automorphism of we. Denote by 0: the set of fixed points of o. Then one of the following holds: (1) (2) (3)
0: = 0. \0:\ = 1. \0:\ = 2.
(4) There exists a circle k with 0:
= {P \ PI
k}.
PROOF. Let '0 =1= 0 and P EO:. Then a induces an involutory collineation in weep). If a is a perspectivity in weep), then (2), (3) or (4) holds. Assume that a is not a perspectivity. Then a induces a Baer involution in the projective closure of weep). As 100 is fixed by a, we infer that '0\{P} is the set of points of an affine Baer subplane U of "JR(P). In particular 1 where m 2 is the order of 9)(. Let S flO:. Then there exists exactly one line of U which passes through S; i.e., there exists exactly one
10:1 = m 2 +
D. MoblUS
126
Ik n 01 = m + 1. As P was conclude that m + 1 is a divisor of m2 + I, a contradiction. circle k through Sand P with
A non-empty point set 0 of a projective 3-space satisfies the following conditions:
arbitrary, we
0
s,n is called an ovoid if it
(A) No three points of 0 are collinear. (B) If P E 0, then there exists a plane E of s,n with 0 n E = {P}. . (C) If P EO and if E is a plane of s,n with E n 0 = {P}, then all lInes I through P which are not contained in E carry a point of 0 distinct from P.
E
If 0 is an ovoid, then given P EO there is exactly one plane E with = {P}. This plane is called the tangent plane at 0 in P.
n0
25.6 Theorem. Let 0 be an ovoid in the projective 3-space s,n. Define 9)((..0) as follows: 1° The points of 9)((0) are the pOints of U. 2° The circles of 9)((0) are the planes of s,n which carry at least two points of O. 3° Incidence in 9)((0) is equivalent to incidence in s,n.
Then 9)((0) is a Mobius plane. PROOF. a) Three distinct points of 9)((0) are on exactly one circle of 9)((0), as no three points of ..0 are collinear. b) Let P, Q be points and k a circle of 9)((0) with PI k and Q l' k. Let T be the tangent plane at 0 in P. Then T n k is a tangent line of 0 which lies in k. Let I be the plane determined by T n k and Q. As P, Q I I and P =1= Q, we have that I is a circle of 9)((0). As Q l' k, we see that k n 1= T n k. Hence k n I n 0 = {P} whence the circle I touches the circle k in the point P. Let I' be a second circle through Q touching k in P. Then k n l' is a tangent line of O. It follows from (C) that k n I' = k n T whence I' = I. c) The points of 0 are not complanar by (B) and (C). Hence there exist four non-concyclic points in 9)((0). 0 A Mobius plane 9)( is called egglike, if there exists an ovoid 0 in a projective 3-space such that 9)( and 9)((0) are isomorphic.
25.7 Theorem. If 9)( is an egglike Mobius plane, then 9)((P) is desarguesian for all points P of 9)(. The proof is obvious.
Planes
127
IV. The Suzuki Groups and Their Geometries
25.8 Theorem. If U is an ovoid in PG(3, q), then lui = q2 + 1. If E is a plane of PG(3, q), then E is either a tangent plane oj [) or E n [) consists of the q + I points oj an oval oj E. PROOF. Let P E O. There are q2 + q + 1 lines through P. As exactly q + 1 of them are tangent lines of 0, it follows from (A) and (C) that 101 = q2 + 1. Using 25.6 and 25.2, we see that exactly q(q2+ 1) planes of PG(3,q) intersect 0 in the q + 1 points of an oval. Moreover there are exactly q2 + 1 tangent planes at O. As q(q2 + 1) + q2 + 1 = (q + 1)(q2 + 1) is the total number of planes in PG(3, q), the theorem is proved. 0
25.9 Theorem (Qvist 1952). Let 0 be an oval in a Jinite projective plane s,n of even order. Then there exists a point K in s,n not on 0 such that the lines through K are just the tangent lines of o. PROOF. Let P E o. Then P is on exactly n + 1 lines, where n is the order of = n + 1 that there is exactly one tangent through P at o. Hence the number of tangents of 0 is n + l. Let P and Q be two distinct points on 0 and let X I PQ with X =1= P, Q. As 10\{P,Q}1 = n - 1 is odd, there exists at least one tangent of 0 through X. Hence every point of a secant is on a tangent of o. As there are exactly n + 1 tangents and as every secant carries n + 1 points, we deduce that every point X of \U which is on a secant is on exactly one tangent. Hence if K is the intersection of two tangents, each line through K is either a tangent or does not meet 0 at all. Since every point of 0 is joined by a line to K, all the lines through K are tangents. 0
s,n. We infer from 101
K is called the knot of o.
25.10 Theorem (Segre 1959). Let U be an ovoid in PG(3, q) where q is a power of 2. Define the mapping cp from the set of planes of PG(3, q) into the set of points as Jollows: IJ E is the tangent plane of 0 at P, then E'P = P. If E is not a tangent plane, then E'P is the knot of E n 0 in E. Then cp is bijective and '!T = (cp - I, cp) is a symplectic polarity of PG(3, q). Moreover, each collineation of PG(3, q) which fixes 0 centralizes '!T. PROOF. Let E be a plane which is not a tangent plane of 0 and put P = E'P. Let I" ... ,lq+1 be all the lines of E which pass through P. Then II' ... ,Iq+ I are tangents of 0 by 25.9 and 25.8. Hence Ii is contained in exactly one tangent plane Ei of O. Thus there are at least (q + 1)(q - 1) + 1 = q2 planes through P which are not tangent planes. As there are exactly q2 + q + 1 planes containing P, there are exactly q + 1
128
IV. The Suzuki Groups and Their Geometries
tangent planes and hence exactly q + 1 tangents through P. This shows that cp is injective and therefore bijective. Let P be a point and E a plane of PG(3, q) and assume PIE. If P = E'P, then P'P-' = E and hence E'P I pcp-I. If P E'P, then PE'P is a tangent line and hence E'P I PfP-'. This proves that 'TT is a symplectic 0 polarity. The remaining statement of 25.10 is now trivial.
*
26. The Mobius Planes Belonging to the Suzuki Groups Let q = 22r+ 1 ;;;:. 8 and assume that 9)( is a Mobius plane of order q admitting a group G of automorphisms isomorphic to Seq). 26.1 Lemma. Let IT be a Sylow 2-subgroup of G. Then IT has a fixed point and acts sharply transitively on the set of the remaining points.
PROOF. As q2 + 1 is odd, IT has a fixed point P. Let a E IT be an involution. If a has more than one fixed point, then a is a reflection by 25.5, because q2 - 1 is odd. Let k( a) be the circle fixed pointwise by a. As a E S(IT), we have that k(a) is fixed by IT. Also PI k(a), since a has no fixed points not on k(a). Let l' be another involution in II. Since all involutions of G are conjugate by 22.3, l' also is a reflection. If Ik(a) n k('T)1 ;;;:. 2, then T fixes more than two points of k(a), since q - 1 is odd. (Recall that k(a) is fixed by IT.) Therefore Ik(a) n k('T)1 ;;;:. 3 and thus k(a) = k(T). Hence a = T by 25.4, a contradiction. Therefore we have k( a) n k( 1') = {P}. As IT contains exactly q - 1 involutions, there exist exactly q - I circles through P which touch each other mutually and each one of which is fixed pointwise by an involution in IT. Let p be the pencil with carrier P to which all these circles belong. Then p is fixed by II. Moreover Ipi = q by 25.2 f). Hence there exists a circle k E P with kIT = k and k k(a) for all involutions a E II. Let Q I k and Q P. Then III Q I ;;;:. q, as IIII = q2. Therefore ITQ n S(II) {I}, a contradiction. Therefore, if a E S(IT)\ {1}, then P is the only fixed point of a. Thus ITQ = {I} for all Q P whence the orbit of Q under IT has length q2. 0
*
*
*
*
26. The Mobius Planes Belonging to the Suzuki Groups
shows that II and L have distinct fixed points. 26.2 now follows from 26.1, 21.8, and 22.1. 26.3 Lemma. G acts transitively on the set of circles of 9)(. The stabilizer of a circle has order q(q - 1).
PROOF. Let P be a point of 9)( and put H = Gp • Then H = 91 G (TI) for some Sylow 2-subgroup TI of G. The number of pencils with carrier P is q + 1. Hence there is a pencil p with carrier P which is fixed by II. If k E p, then ITIkl = q by 26.1. Therefore IHkl = qt where t divides q - 1. Assume Gk Hk . Then Gk acts transitively on k, as Hk acts transitively on k\{P}. Thus q + 1 divides IGkl and therefore IGI = (q2 + l)q2(q - 1), a contradiction. This proves Gk = H k • Let ~ = {k O Ia E G}. Then we have
*
Therefore I~I
; ;:.
q(q2
+
1), as
t
each element of G \ { I} has at most two fixed points. PROOF. Let IT and L be distinct Sylow 2-subgroups of G and assume that the point P is fixed by II and L. As II n L = {I}, there are exactly 2(q - 1) involutions in II U 2:. It then follows from 26.1 that TI U 2: contains translations of WC(P) with distinct centres. Therefore B(ll) and 3(L) centralize each other in contradiction to 24.2 d). This contradiction
o
<; q - 1.
26.4 Theorem (Hughes 1962, Luneburg 1964). If q = 22r+ I ;;;:. 8, then there exists up to isomorphisms exactly one Mobius plane of order q admitting a group of automorphisms isomorphic to Seq).
PROOF. The existence follows from section 21 and 25.6. To prove uniqueness let 9)( be a Mobius plane of order q and let G be a group of automorphisms of 1))( which is isomorphic to Seq). Pick a circle k of we. Then Gk has a fi;s.ed point P on k. Also IGkl = q(q - 1). Hence Gk acts sharply 2-transitively on k\ {P} by 26.2. Let II be the Sylow 2-subgroup of G whose fixed point is P. Then IIk is the Frobenius kernel of the Frobenius group Gk . Hence ITk is elementary abelian. Therefore TI k, = S(II). Put IIk = Z and GK = A and let Po ~ P, PI' ... ,Pq be all the pomts on k. Furthermore put Gi = Gp,. Fmally let ~ = {G i I i = 0, 1, ... ,q} and 11={ylyEG, GJE~}. If i>O, then, obviously, ~= {Go} U {G/ IrE Z}. Define the mappings cp and 1J; by if and only if if and only if
26.2 Lemma. G acts doubly transitively on the set of points of 9)(. Moreover,
129
po = Q, kT = I.
We infer from 26.2 that cp is a bijection of the set of points of 9)( onto the set of right cosets of Go and 26.3 implies that 1J; is a bijection of the set of circles onto the set of right co sets of A. Let Goa be incident with A l' if and only if po I k T • Obviously, Goa is incident with A'T, if and only if a'T- 1 E 11. Let 9)(* be a second Mobius plane of order q admitting a group G* of automorphisms isomorphic to Seq). Let Gti, A*, ~*, ~* and Z* have the corresponding meaning to Go, 11, ~, 11 and Z. If Gi E ~*\{ Gti}, then there
uu
1 V.
1
ne
;'ULUKI
vruul-'~
allU
1lleu
"'''v,,,,,,,
'''~
exists an isomorphism p from G onto G* with Gt = Gti and Gf = Gi. This implies ZP = Z*. This again implies ~P = ~* and hence t:J.P = t:J.*. As A = Z(Co n C1), we have AP = A*. Therefore 9]( and 9](* are isomorphic.
o
21. S(q) as a Collineation Group of PG(3, q)
jjl
27.2 Corollary. If \U is a finite desarguesian projective plane, then the collineation group of SU does not contain a subgroup which is isomorphic to a Suzuki group. PROOF. This follows at once from 27.1 using the simplicity of the Suzuki 0 groups and the fact that PfL(3, q)/PSL(3, q) is always soluble.
27. S( q) as a Collineation Group of PG(3, q)
27.3 Theorem (Liineburg 1965a). Let q = 2 2r + 1 ;> 8 and let I.l3 be the
27.1 Lemma. Let I.lS be a desarguesian projective plane. Then I.lS does not admit a collineation group G isomorphic to S(K, a) such that all involutions of
I.l3 contains subgroups isomorphic to Seq). Moreover all these groups are
3-dimensional projective space over GF(q). Then the full collineation group of
G are central collineations.
PROOF. Assume by way of contradiction that G is a collineation group of I.l3 isomorphic to S(K, a) for some K and a and that all involutions of G are central collineations. Let ~ be a 2-subgroup of G which is mapped by an isomorphism of G onto S(K, a) onto the group T. Let 1 =1= a E 3(~). Then a is an involution. Thus a is a (C, l)-perspectivity for some C and I. Let L be the coordinatizing field of 1.l3. Case 1: The characteristic of L is distinct from 2. Then C II. As t:J.( C, I) is isomorphic to the multiplicative group of L, we have that a is the only involutory (C, I)-perspectivity of 1.l3. Let a I be a second involution in 3(~). Then a l is a (D, m)-homology. As aa l = ala, we have Cal = C and 101 = I. It follows therefore from a =1= a I that D I I and C I m. Let a 3 be a third involution in 3(L). Then similar arguments show that a 3 is an (l n m, DC)-homology. Hence 13(~)1 < 4, a contradiction. . . Case 2: The characteristic of L is 2. Then C I I. Let a l be an InVOlutIOn in 3(~) distinct from a. Let D be the centre and m the axis of a l . If m =1= I, then C = I n m, as Cal = C. Likewise D = I n m. This shows that all involutions in 3(L) either have the same centre or the same axis. We may assume wlog that they all have the same axis. Let y E G and Ll =~i' =1= ~. Then all the invohltions in 3(~1) have axis fY. If fY =1= I, then I n fY is a fixed point of <3(~), 3(L l » = G. If fY = I, then I is a fixed line of G. We may therefore assume wlog that G has a fixed line I. Let E be the group of all elations with axis I. Then EnG is normal in G. As G is a simple, non-abelian group by 22.6, we obtain as a result EnG = {l}. Now E acts transitively on the points of 1.l3{ by 1.14. We may therefore assume that G fixes a point P not on I. As G is generated by its involutions with fixed points, we conclude G ~ SL(2, L). Again let 1 =1= p E 3(~). Then the centre C of p is on I. Furthermore, ~ fixes C, whence ~ C; SL(2, L)c. But this contradicts the fact that SL(2, L)c does not contain an element of order 4. This proves 27.1. 0 27.1 is wrong without the assumption made about the involutions of G, as one can easily construct commutative fields of any characteristic whose automorphism groups contain subgroups isomorphic to Seq).
contained in the projective group of \U and are conjugate under this group. Let G be a subgroup of the collineation group of I.l3 which is isomorphic to Seq). Then G splits the set of points of SU into two orbits. One of these orbits is an ovoid [). The set of planes is also split by G into two orbits: the set of tangent planes of [) and the set of the remaining planes. The set of secants of [) as well as the set of exterior lines of [) are line orbits of G. The set of tangent lines of [) is split into two orbits, one of them being a spread of SUo
PROOF. Let G be a collineation group of SU isomorphic to Seq). ( 1) G has no fixed point. Assume that G has a fixed point P. The lines and planes through P together with the inclusion relation as incidence form a desarguesian plane. Hence by 27.2, the group G fixes all the lines through P individually. Thus G consists only of perspectivities with centre P. As the group of all perspectivities with centre P has order q3(q - 1), we arrive at the contradiction that (q2 + l)q2(q - 1) divides q\q - 1). (2) G is contained in the projective group of 1.l3. This follows from the fact that G is simple and that PfL(4, q)/PGL(4, q) is cyclic. (3) If II is a Sylow 2-subgroup of G, then II has exactly one fixed point. The set of fixed points of the Sylow 2-subgroups of G is an ovoid of 1.l3. The centre of II cannot fix a plane pointwise by (1), as two planes always have a point in common and since G is generated by the centres of two distinct Sylow 2-subgroups. Let P and Q be distinct points fixed by II. As q - 1 is odd, II has a third fixed point on PQ. It follows from (2) and Baer [1952, p. 66-69] that II fixes PQ pointwise. There are q + 1 planes containing PQ. Hence there exists a plane E with PQ C; E = En. As PQ is fixed pointwise by II, the group II induces a group II* of elations with axis PQ in E. As E is desarguesian, all elations of E are involutions. Furthermore, each involution of II is contained in a cyclic group of order 4. Therefore 3(II) induces the identity on E, a contradiction. This proves that II has at most one fixed point. As the number of points of SU, being (q2 + 1)(q + 1), is odd, II has exactly one fixed point. Let £) be the set of fixed points of the Sylow 2-subgroups of G. Then by (1) and the remark just made, I[)I = q2 + 1. It follows by 24.10 that G acts
132
IV. The Suzuki Groups and Their Geometries
in its natural representation on O. We consider the following incidence structure ~: (a) The points of ~ are the points of O. (b) The blocks of ~ are the lines of I,l3 which meet 0 in more than one point. (c) Incidence is inherited by incidence in 1,l3.
':,1,'
\'
1
i
As G is twofold transitive on 0, we have that ~ is a block design with v = q2 + 1 points, b blocks, k points per block and r blocks through each point. Moreover, any two points are joint by exactly ,\ = 1 block. It follows that r(k - 1) = '\(v - 1) = q2. Assume k > 3 and let P, Q, R be three distinct collinear points of O. As G acts in its natural representation on 0, we have IGp , QI = q - 1 and IGp , Q, R I = 1. Therefore all the points on the line PQ belong to O. Thus k = q + 1 whence r = q. Now (q2 + 1)q = vr = bk = b(q + 1), a contradiction. Hence no three points of 0 are collinear. The dual arguments show that each Sylow 2-subgroup IT of G fixes exactly one plane E. As the number of points on E is odd, the fixed point P of IT is on E. If IE n 01 > 2, then 0 C; E, since IT acts transitively on O\{P}. Hence EG = E in contradiction to the dual of (1). Hence E n 0 = {P}. This proves that 0 is an ovoid. (4) G has exactly two point- and two plane-orbits. One of the point orbits is 0 and one of the plane orbits consists of the tangent planes of O. G acts transitively on the set of tangent planes of 0, as the above arguments show. Furthermore, by (3), 25.6, 25.8, and 26.3, the group G acts transitively on the set of the remaining planes. Moreover G operates transitively on O. Let ;( and Y be two points not on 0 and let 7T be the polarity defined in 25.10. Then X'7I', Y'7I' are planes which are not tangent to O. Hence there exists y E G with X'7I'Y = Y'7I'. Using 25.10 again, we obtain XY = XY'7I'2 = X'7I'Y'7l' = y'7I'2 = Y. This proves (4). (5) The secants as well as the exterior lines of 0 each form a line orbit of G. The secants form a line orbit, since G acts doubly transitively on O. From this and 25.10, it follows that the exterior lines also form an orbit. Now let 0* be the ovoid defined in section 21. Then 9)1(0) and 9)1(0*) are isomorphic by 26.4. Let cp be an isomorphism of 9)1(0) onto 9)1(0*). We have to show that we can extend cp to a collineation of 1,l3. Let 7T be the polarity defined by 25.10 with respect to 0, let 7T* be the polarity defined with respect to 0*, and define the collineation K as follows: If X is a point of 0, then X K = X'P. If X is not on 0, then XK = X'7I'
28. S(q) as a Collineation Group of a Plane of Order q2
133
automorphisms of GF(q) induce collineations in I,l3 leaving 0* invariant, as is easily seen from the definition of 0*. The remaining statement that G splits the set of tangent lines into two 0 orbits one of them being a spread now follows from 23.2 and 23.1.
28. S(q) as a Collineation Group of a Plane
of Order q2 We shall assume throughout this section that I,l3 is a projective plane of order q2 where q = 220:+ 1 > 8 and that G is a group of collineations of I,l3 which is isomorphic to S( q).
28.1 Lemma (Dembowski 1966). If G has a point orbit p of length q2 then either p is an oval of I,l3 or all the points of p are collinear.
+ 1,
PROOF. It follows from 24.10 that G acts in its natural representation on .p. Consider the following incidence structure ~: The points of ~ are the points of p and the blocks of ~ are the lines of I,l3 carrying more than one point of.p. Then ~ is a block design with parameters v = q2 + l,b,k,r, and ,\ = 1. Hence r(k - 1) = '\(v - 1) = q2. Therefore r = 2S with s < 2(2a + 1). If s = 0, then r = 1 and all the points in p are collinear. Thus assume s :> l. As G operates in its natural action on p, we have k = 2 + l(q - 1) with t :> O. Therefore
1 + t (q - 1) = q2r - 1 = 22( 20: +
I) - s.
This implies that 2 0:+ 1 - I divides 1 which yields that 2a + 1 divides 2(2a + 1) - s and hence s. Therefore s = 2a + 1 or s = 2(2a + 1), . 2 2 l.e., r = q or r = q . If r = q, then k = q + 1 and thus (q + l)q = vr = bk = b(q + 1), a contradiction. Thus r = q2 and k = 2. 0 2
2 2 (20:+ I) -s -
28.2 Lemma (Dembowski 1966). G has a fixed element, if and only involutions in G are elations.
if all
PROOF. Assume that G has a fixed element. We may then assume that G fixes a line f. If G fixes f pointwise, then all involutions of G are elations. Therefore we may assume that G operates non-trivially on f. In this case, G acts faithfully onf, as G is simple. Using 24.10, we see that G operates in its natural action on the set of points incident with f. Hence all involutions of G have exactly one fixed point on f. Therefore they cannot be Baer involutions. Conversely, assume that all involutions of G are elations. Let IT be a Sylow 2-subgroup of G and 1 =!= a E 3(11). Then a is an involution and hence an elation. Let C be the centre and a the axis of a. Then C and a are both fixed by IT. Pick T E s(n) \ { 1, a}. Let D be the centre and b the axis of T. If b *- a, then D = C, as aT = Ta. In other words: If there exist two
134
IV. The Suzuki Groups and Their Geometries
involutions in n with distinct axes, then all involutions in IT have the same centre. Thus all involutions of IT either have the same centre or the same axis. We may assume that they all have the same axis, a say. Let 2: be a Sylow 2-subgroup of G distinct from IT. Then all involutions in 2: also have the same axis, b say. If a = b, then a is fixed by G, as G is generated by 3(2:) U 3(IT). If a =1= b, then a n b is a fixed point of G for the same reason. 0 28.3 Lemma (Dembowski 1966). A II involutions of G are elations.
Assume that G contains a Baer involution. Then all involutions of G are Baer involutions by 22.3. If a is an involution, we denote by @(a) the Baer subplane of I.l3 fixed pointwise by a. Let IT be a Sylow 2-subgroup of G and assume @(a) = @(r) for all a, r E 3(IT) \ {l }. Then IT acts regularly outside of @( a). Hence q2 divides q4 + q2 + 1 - q2 - q - 1 = q(q3 - 1), a contradiction. Thus there exist involutions a, r E IT with @(a) =1= @(r). As ar = Ta, we then have that r induces an involutory collineation r* in @(a). Since q = 22a + [ is not a square, r* is an elation. Let C be the centre and I the axis of r*. As all the fixed lines of IT are fixed by a and r, they all belong to @( a) and hence pass through C. Similarly, all the fixed points of IT belong to @(a) and are on I. Therefore, if IT has more than one fixed point, then I is fixed by the normalizer of IT. If IT has just one fixed point, then this fixed point is fixed by the normalizer of IT. As this normalizer is a maximal subgroup of G, we see that IT has a fixed element such that the orbit of under G has length 1 or q2 + 1. The first case cannot occur by 28.2. Hence G has an orbit of length q2 + l. We may assume that it is a point orbit. As G has no fixed elements, it follows from 28.1 that it is an oval. But this again contradicts 28.2, as the knot of the oval is fixed by G. This final contradiction proves the lemma. 0 PROOF.
It follows from 28.3 and 28.2 that G has a fixed element. We may assume that G has a fixed point P.
28.4 Lemma. G acts in its natural representation on the set of lines through P. PROOF. As IGI = (q2 + l)q2(q - 1) does not divide q4(q2 - 1), the group G cannot consist only of perspectivities with centre P. Therefore G acts faithfully on the set of lines through P since it is simple. 28.4 then follows D from 24.10.
28.5 Lemma. P is not the centre of an involution in G. PROOF. 28.4 yields that every involution fixes exactly one line through P. Therefore P cannot be the centre of such an involution. D
135
28. S(q) as a Collineation Group of a Plane of Order q2
28.6 Lemma. Let I and m be two distinct lines through P. Then G" m has exactly two fixed points on I. The remaining points are split by G" minto q + 1 orbits of length q - 1. PROOF. G" m is cyclic of order q - 1. Let 8 E G" m and let Q and R be points on I with P =1= Q =1= R =1= P and QO = Q and RO = R. Furthermore, let H be the group generated by 8. By 28.4, there exists a E G with la = m and m a = I. We have Gt m = Gl,m and therefore H a = H, as Gl,m is cyclic. Thus Q, R, Qa, and R a are fixed points of H no three of which are collinear. This yields that H fixes the three lines I, m, and P(QR a n QaR). Therefore H = {I} by 28.4. This shows that any non-identity element of Gl , m has at most two fixed points on I. Let y be an element of prime order in Gl,m' As (q - l,q2) = 1, we have that y fixes a point Q on I other than P. As this is the only fixed point of y on 1\ {P} by the remark made above and as G" m is abelian, we see that Q is a fixed point of G" m' The remaining statement is now trivial. 0
28.7 Lemma. Let IT be a Sylow 2-subgroup of G. If P I I at least q + I fixed points on I.
= lIT,
then IT has
PROOF. P is a fixed point of IT on I. Let a be an involution in IT. Then a is an elation with axis I. Let Q be the centre of a. Then P =1= Q by 28.6. As a E 3(IT), we also have Q IT = Q. As q2 - 1 is odd, IT has a third fixed point R on I. Let m be a line through P which is distinct from I. Then Gl , m normalizes IT. It follows from 28.6 that at least one of the points Q and R lies in an orbit of length q - 1 of Gl , m' Thus IT has at least q + 1 fixed 0 points on I.
28.8 Lemma. Let IT be a Sylow 2-subgroup of G and assume that there exist involutions in IT with distinct centres. Then G has, besides {P}, three point orbits of length q2 + 1, (q2 + 1)(q - 1) and (q2 + l)q(q - 1) respectively. The point orbit 0 of length q2 + 1 is an oval whose knot is P. The tangent lines and the secants of 0 each form a line orbit of G. The exterior lines of 0 are split into two orbits of length -}(q - r + l)q2(q - 1) and t(q + r + 1) q2(q _ 1) respectively, where r2 = 2q. PROOF. Let I be the line through P fixed by IT. By assumption, there exist distinct points Q and R on I which are centres of involutions. Futhermore, Q, R =1= P by 28.5. We then deduce from 28.6 that no two involutions of IT have the same centre. Thus there exist exactly q - 1 points on I which are centres of involutions in IT. Let m be a line distinct from I and assume ImITI <; q2. Then m n I is a centre. As there are q2 + 1 - (q - 1) = q2 - q + 2 points on I which are not centres, there exist q2(q2 - q + 2) lines distinct from I and meeting I in a point which is not a centre. The set of all these lines is split by IT into q2 - q + 2 orbits of length q2.
I,
136
IV. The Suzuki Groups and Their Geometries
28. S(q) as a Collineation Group of a Plane of Order
137
q2
Jji If m n 1 is the centre of an involution, then IIm is cyclic of order 2 or 4, as II is of exponent 4 and does not contain a quaternion group by 24.2 and as ITm contains exactly one involution. Therefore ImIlI = ! q2 or ImIlI = :t q2. Assume that ImIlI = q2 for all lines m such that m n 1 is a centre. As there are q - 1 centres on I, there are 2( q - 1) line orbits of II whose length is q2. As {I} is a line orbit of IT, we thus have that II possesses 1 + 2(q - 1) + q2 - q + 2 = q2 + q + 1 line orbits. Now we count the point orbits of II. Since II operates regularly on the set of points off I, there are q2 point orbits of length q2. Moreover, II has at least q + 1 point orbits of length l. As there are still q(q - I) points left, the number of point orbits of IT is greater than q2 + q + 1 which contradicts the Dembowski-Hughes-Parker theorem. Hence there exists a line m with ImIlI = :i- q2. Assume that IhIlI = :i- q2 for all lines h with IIh =!= {I}. Then II has exactly 1 + 4(q - 1) + q2 - q + 2 = q2 + 3q - 1 line orbits. As II has q2 point orbits of length q2 and at least q + 1 point orbits of length 1, there exist q2 + 3q - 1 - q2 - q - 1 = 2(q - 1) further point orbits all of which are contained in I. Let j be a line through P distinct from I and let Q be a fixed point of GI • j on I other than P. If QIl = Q, then IQGI = q2 + 1. Therefore, by 28.1, either Q Gis a line or Q Gis an oval. If Q Gis a line, then all the involutions in II have the same centre 1 n QG, a contradiction. Thus QG is an oval. Let R be a centre on 1 and s a secant through R. Then ISIlI = q2, a contradiction. Thus Q is not fixed by IT. As GI . normalizes IT, it fixes Q II II ,J as a set. Therefore IQ I ~ q and hence IQ III = q, as the centre of II fixes all the points on f. Now II has at least q + 1 fixed points on 1 and P is the only common fixed point of II and GI , j; hence II has at least 2(q - I) + 1 = 2q - 1 fixed points on I, since GI , j permutes the fixed points of II in orbits of length q - 1. Of these, q + 1 have been taken into account already. Thus among the remaining 2(q - 1) point orbits, there are 2q - 1 - (q + 1) = q - 2 of length 1. There remain 2q - 2 - q + 2 = q orbits, one of which has length q as we have seen. The remaining q - 1 orbits are permuted transitively by GI ,), as is easily seen using 28.6 and the fact that the only fixed point of GI ,) on I\{P} lies in the orbit of length q already counted. Thus the remaining orbits all have length 2a • Consequently
t
t
t
2q - 1 + q + 2a ( q - 1) = q2 + 1 which yields (2 a + l)(q - 1) = (q - Ii, a contradiction. Hence there also is a line n with InIlI = t q2. As GI normalizes IT and acts transitively on the set of centres, there are q - 1 line orbits of length t q2 and 2(q - 1) line orbits of length ..t q2. Thus II has a total of I + q2 - q + 2 + q - 1 + 2( q - I) = q2 + 2q line orbits.
IT splits the points off 1 into q2 orbits of length q2. Using th.e Dembowski-Hughes-Parker theorem, we see that the points on I are splIt by II into 2q orbits, q + 1 of th~m having length 1. Let i\, -'1' . ~ the lengths of the remaining orbIts. Then 1 <; Aj <; q and L,1= I A; - q(q . 1). Hence A; = q for all i. This implies tha: GI ,) per~utes these ~ - 1 orbIts transitively. Hence GI splits the set of pomts on 1 mto four orbIts. Two of these have length 1. The other two have len~th q - ~ and q(q -,1) respectively. This yields that G splits the set of pomts of mmto four orbIts, one being {P}. The others have length q2 + 1, (q2 + 1)( q - 1) and (q2 + l)q(q - 1) respectively. . ' By the Dembowski-Hughes-Parker theorem, G splIts the set of Ime~ of into four orbits. Let 0 be the point orbit of length q2 + 1. By 28.1, 0 IS a line or an oval. As the points of 0 are not centres, 0 must be an oval ,,:ho.se knot obviously is P. Again it follows from 24.10 that G operates m Its natural action on o. Hence the secants of 0 form a line orbit. Therefore the set of exterior lines of 0 is split into two orbits (tl and (t2' Put l(til = Ai' Then AI + A2 = q4 + q2 + 1 - q2 - 1 - ! (q2 + 1)q2 = ! q2( q2 - 1).
0q-1
m
Let C be a centre on the line I and let {Q} = 0 n I. Then II acts transitively on o\{ Q}. Therefore, the line orbit of length! q2 of II all of whose lines pass through C consists of secants. Th~refore the o~her, two line orbits of II consisting of lines through C contam only extenor lm~s. Therefore 4 divides IGel for some exterior line e. Moreover IGel ~ 4m WIth m odd. We may assume e E @I' Then AI is even and therefo~e A.2 IS. Thus 4 divides IGIl for all f E @1 U (t2' Let Gi b,e the stabilizer of a lme m (t i' ,T?en mi dIVIdes IG 1 = 4m' where m· is odd. Moreover, It follows from 24.102 that q _ir i + 1 or q + r + 1, where r2 = 2p. Now Ai = (q + 1) q2( q- 1) / 4m. and thus I
,
{- (q' + I)q'(q -
1)[ ~l + ~21 = ~ q'(q' -
I) 2
This implies (q2 + 1)(mj + m 2) = 2(q + l)mjm2' From (9 . + 1,2(q + 1)) = 1 we infer that q2 + 1 divides mjm2 and 2(q + 1) dIVIdes m J + m 2· Furthermore m l + m 2 <; 2( q + r + 1) = 2( q + 1) + 2r < 2( q + 1) + 2q, whence m l + m 2 = 2(q {q- r+ l,q+ r+ I}.
+ 1) and m l m 2 = q2 +
1. This implies {m l ,m 2} ~
28.9 Lemma. Let II be a Sylow 2-subgoup of G and assume that all involutions of IT have the same centre. T~en G fixes a line l,not through .P. The points of I form an orbit of length 1 + 1: T~e set of pozn~s of 1{5 which
are not on I and distinct from P IS spilt mto two orbllS of length (q2 + 1)(q - 1) and (q2 + 1)q(q - 1) respectively. The set of lines oJm which
138
IV. The Suzuki Groups and Their Geometries
do not pass through P and are distinct from I is split into two orbits of length
(q2 + l)(q - 1) and (q2 + 1)q(q - 1) respectively.
PROOF. We already proved a) and b). c) is the dual of b) and d) IS a consequence of a), b) and c). D
PROOF. Let 2: and T be two distinct Sylow 2-subgroups of G and let hand m be the two lines through P fixed by 2: and T respectively. Let Q be the centre of the involutions in 2: and R the centre of the involutions in T. Then Q I hand RIm. From 28.5 we get P =1= Q, R. From this and 2: =1= T we obtain P.f 1= QR. As I is fixed by 3(2:) and 3(T), it is fixed by
<3(2:),3(T»
=
G.
139
29. Geometric Partitions
By now we have proved:
28.11 Theorem (Luneburg 1965a). Let 1{3 be a projective plane of order q2, where q = 220: + I :> 8 and let G be a group of collineations of 1{3 which is isomorphic to Seq). Then one of the following holds:
Let X be a point with P =1= X.f I and XTI = X. It follows from 28.6 that Gx = n. Hence IXGI = (q2 + 1)(q - 1). As there is always such an X by 28.7, G has a point orbit of length (q2 + 1)(q - 1). As I is a fixed line of G, dual arguments show that G also has a line orbit of length (q2 + 1)(q - 1). Let P be a point orbit of length (q2 + 1)( q - I) and let m and n be lines such that m n P =1= 0 =1= n n P and P.f m, n. In order to prove the existence of y E G with mY = n, we may assume that m and n have a point X of P in common. Put n = Gx . Then n acts transitively on I\{PX n I}. Hence there exists yEn with mY = n. Therefore, g = {m I P.f m, m n P =1= 0} is a line orbit of G. Now (p, g) is a tactical configuration with parameters v = (q2 + l)(q - 1), b, k, and I' = q2. Moreover the points X and Y of pare not joined by a line of g, if and only if P, X, and Yare collineaJ.'. Hence
a) G fixes a non-incident point line pair (P, I) of 1{3 and G acts doubly transitively on the set of lines through P as well as on the set of points on I. Moreover, G has two further non-trivial point orbits PI and P2 and two further line orbits gl and g2 such that IpII = IgII = (q2 + l)(q - I) and Ip21 = Ig21 = (q2 + I)q(q - I). b) G fixes an ovalo and its knot P. It operates doubly transitively on the set of lines through P and on o. It also acts transitively on the set of secants of o and splits the set of exterior lines into two orbits of length ±(q - I' + l)q2(q - I) and ±(q + I' + l)q2(q - I) respectively, where 1'2 = 2q. Furthermore, G has two point orbits of length (q2 + I)(q - I) and (q2 + I)q(q - 1) respectively. c) The dual to b).
q2( k - 1) = (q2 + 1)( q - 1) - (q - 1) = q2( q - 1),
Remark. All three possibilities really occur. We have seen that a) occurs in the Luneburg planes and it follows from results of Pollatsek 1971 that c) also occurs in these planes. Hence b) occurs in the dual planes.
whence k =
q.
Therefore
q2(q2
+ I)(q - I) =
VI'
= bk = bq.
Thus b = q(q2 + I)(q - I). Therefore, we have found four line orbits of length 1, q2 + 1, (q2 + I)(q - 1) and (q2 + l)q(q - 1) respectively. Since this accounts for all the lines, G has exactly four line orbits and hence also four point orbits. As three of them have length 1, q2 + 1, and (q2 + I)(q - I) respectively, the last one must have length (q2 + I)q(q - 1).
o 28.10 Corollary. Same assumptions as in 28.9. Let PI,P2,gl,g2 be the point and line orbits with Ipll = IgII = (q2 + 1)(q - 1) and Ip 21 = Ig21 = (q2 + 1) q(q - I). Then we have: a) (PI' gl) is a tactical configuration with parameters v = b = (q2 + 1)(q - 1) and I' = k = O. b) (PI' g2) is a tactical configuration with parameters v = (q2 + 1)( q - 1), b = (q2 + I)q(q - I), I' = q2 and k = q. c) (P2' 9 1)2 is a tactical configuration with parameters v = (q2 + I)q(q - 1), b = (q + 1)(q - 1), r = q and k = q2. d) (P2' g2) is a tactical configuration with parameters v = b = (q2 + 1)q( q I) and I' = k = q(q - 1).
29. Geometric Partitions Let 7T be a partition of the additively written abelian group A. The subgroup U of A is called 7T-admissible, if for all X E 7T we have that X n U =1= {O} implies X <.: A non-trivial partition 7T of A is called geometric, if X + Y is 7T-admissible for all X, Y E 7T.
u.
29.1 Lemma (Baer 1963). Let 7T be a geometric partition of the abelian group A. Then 7T is a spread, if and only if there exist X, Y E 7T such that A
=
X
+
Y.
PROOF. If 7T is a spread, then, of course, A
=
X
+
Y for all X, Y E
7T
with
X =1= Y. (This also shows that every spread is a geometric partition.) In order to prove the converse, let X, Y E 7T be such that A = X + Y. Furthermore, let Z E 7T\{X}. As the lattice of subgroups of A is modular, we have X
+ Z = (X + Z) n
A
= (X + Z) n (X +
Y)
=
X
+ ((X + Z) n
Y).
140
IV. The Suzuki Groups and Their Geometries
Since X is a proper subgroup of X + Z, it follows that (X + Z) n Y =1= {O}. Therefore Y ~ X + Z, as 1T is geometric. This proves X + Z d X + Y = A, l.e., A = X + Z. Let U and W be distinct components of 1T. We may assume that U =1= X. Then applying the above argument twice yields A = X + U = W + U. 0 If 1T is a geometric partition which is not a spread, then we call 1T a spatial partition. As we did in section 1, denote by K(A,1T) the kernel of the partition 1T.
141
29. Geometric Partitions
4) If a, b, c are three non-collinear points of @, then there exists exactly one plane P with a, b, c E P. As a, b, c are not collinear, we have a - c =1= 0 =1= b - c. Hence there exist X, Y E 1T with a - c E X and b - c E Y. If X = Y, then a, b, c E X + c, a contradiction. Thus X + YEa and a, b, c are on the plane X + Y + c. To prove uniqueness, let SEa be such that a, b, c E S + c. Then a - c, b - c E S whence X + Y C; S by 3). It follows from 29.1 that X + Y = S. This proves 4). 5) Each plane of @ contains three non-collinear points.
29.2 Theorem (Baer 1963). Let 1T be a spatial partition of the abelian group A. Then K(A, 1T) is a skewfield. Moreover, the 1T-admissible subgroups of A are exactly the subspaces of the K(A, 1T)-vector space A. In particular, 1T consists of all the subspaces of rank 1.
PROOF. Let 0 =1= K E K(A, 1T). Then K is injective by 1.6 a). To prove that K is surjective, it suffices to show that X K = X for all X E 1T. Let X E 1T and pick Y E 1T\{X}. Furthermore, let 1To = {Z I Z E 1T, Z ~ X + Y}. Then 'lTo is a geometric partition of X + Y, as 1T is geometric. We infer from 29.1 that 1To is in fact a spread of X + Y. It follows therefore from 1.6 b) that the restriction of K to X + Y is surjective. Hence X K = X. Therefore, K is an 1 1 automorphism of A. Let Z E 1T. Then ZK- = ZKK- = Z. Thus K- 1 E K(A, 1T) whence K(A, 1T) is a skewfield. Denote by a the set of all X + Y with X, Y E'lT and X =1= Y. We consider the geometry @ whose points are the elements of A, whose lines are the co sets X + a with X E 1T and a E A, and whose planes are the co sets S + a with SEa and a EA. We show now that @ is an affine space: 1) Let a and b be distinct points of @. Then there is exactly one line of @
containing a and b.
This is trivial. 6) There exist four points not contained in a plane.
As 1T is spatial, there exist X, Y, Z E 1T such that Z ct X + Y. Hence Z n (X + Y) = {O}. Pick x EX, Y E Y, Z E Z with x, y,z =1= O. Assume that 0, x, y, z are contained in the plane P. Then PEa, as E P. It then
°
follows from 5) that X, Y, Z C; P. We infer from 29.1 that X + Y = P d Z, a contradiction. Two lines X + a and Y + b of @ are said to be parallel, if they are contained in a plane and if either X + a = Y + b or (X + a) n (Y + b) = 0. the proof of 1.4 shows that X + a and Y + b are parallel, if and only if X = Y. 7) Let Land M be parallel lines and P and Q be planes of @ such that L C; P and M ~ Q. If P and Q have a point in common, then P n Q contains a line.
There are a, b E A and X E 1T such that L = X + a and M = X + b by the remark made above. Moreover there exist S, TEa with P = S + a and Q = T + b. It follows that X C; S and X C; T. Let c be a point in P n Q. Then P = S + c and Q = T + c. Therefore X + c C; P n Q. This proves
Let X E 1T and c E A be such that a, b EX + c. Then a - b EX. As a - b =1= 0, the component X is uniquely determined. Moreover X + c = X + b. Hence there is at most one line joining a and b. Conversely, there exists X E 1T such that a - b EX. This yields a, b EX + b. This proves 1).
7).
2) Each line of @ contains at least two points.
9) If L is a line of @, then there exists a line parallel to and distinct from
This follows from the fact that
IX I > 2 for
all X E 1T.
3) Let a and b be two distinct points of @ and let P be a plane such that a, b E P. Then the line joining a and b is contained in P.
There are X, Y, Z E 1T such that a, b E P = X + Y + b and a E Z + b. Therefore 0 =1= a - b E Z n (X + Y) whence Z ~ X + Y, as 1T is geometric. This yields Z + b ~ X + Y + b.
8) Let L, M, N be lines of @ and assume that M and N are parallel to L. Then M and N are parallel to each other. This is trivial. L.
This is trivial, too. From 1) to 9) we conclude that @ is an affine space which is not an affine plane (see e.g. Lenz [1967, p. 136]). Therefore the group D. of all homologies with centre 0 acts transitively on X\{O} for all X E 1T (see e.g. Lenz [1967, p. 141]). The mapping T defined by x'T(a) = x + a is an isomorphism of A onto
142
IV, The Suzuki Groups and Their Geometries
143
30. Rank-3-Groups
the translation group T of ®, If 8 E b., then oo-l r (a)8
Hence 0 -IT(a)o
T(a Therefore sitively on rank 1 of subgroups
= T(al5).
As Ll 7T (x)y and LlY"'(x ) both belong to O(xY), it suffices to show that their intersection is not empty. Pick z E /11T(X)Y. Then zy-I E /11T(X). There -I 8-1 exists 8 E G with z = x 15. Hence x 8y E b. 1T (x) whence x y E b. by 30.1. ly 15 -18 This implies x y8 - E /1Y. Now /1Y E n(xY). Therefore z = x = (xY)Y Y E b. Y7T(X ) . 0 PROOF.
= or(a)8 = a 8 = or(a~).
This yields
+ b/) = 8 -IT(a + b)8 = 8 -17(a)88 -17(b)8 = 7(al5 + b 15 ). (a + b)15 = a 15 + bl5. X \ {O} for all X E the K(A, 7T)-vector of A are exactly the
This proves b. C; K(A,7T). As b. acts tran7T, we see that each X E 7T is a subspace of space A. This yields that the 7T-admissible subspaces of the K(A, 7T)-vector space A. 0
Y
30.3 Lemma. If G is finite, then 1111
= 1/17T(a) I for all /1 E n(a).
PROOF. Let 7T = 7T(a) and let /1 = {XI' ... ,xn }, There exist YI' ... , Yn E G such that Yi- I maps a onto Xi' Denote by Yi the image of a under Yi' Then Yi E /17T for all i. If Yi = Yj' then Yi Yj - \ E Ga' This yields Xj = Xi 2 whence i = j. This proves 1111 < 1117T1. From this and 30.1 we infer 1b.7T 1 < 1b.1T 1= 1b.1.
o
30. Rank-3-Groups In this section we follow D. G. Higman 1964. Let G be a permutation group which acts transitively on a set st. If a En, then n(a) denotes the set of orbits of Ga' For b. E n(a) we put b. 1T (a) = {aYly E G, a E b. Y }. 30.1 Lemma. 7T(a) is a permutation of n(a) with 7T(a)2
=
1.
PROOF. Let 7T = 7T(a) and let b. E n(a). As G is transitive, there exists 1 1 with a 15 E/1. Therefore a Eb. 15 - and hence a 15 - E/1 1T . This proves /1 i= 0. Next we prove that b. 1T is an orbit of Ga' Let x, Y E b. 1T . Then there exist l 1 y,8 E G with x = a Y andy = a 15 and a y- ,a 15 - E b.. Hence there is 11 E Ga wi th a Y- I = a 0- \ This implies y - 111 - 18 EGa' Furthermore
o1TE G
x y-Lry- 115
aT)- '15 = a 15
=
= y.
This proves that there exists /1' E n( a) with b. 1T S;; 11'. Let z E /1'. By the definition of b. 1T and the remark that b. 1T is not empty, there exists y E G with a Y E /1 1T and a E /1Y. Furthermore there is 0 E Ga I I with z = a Yo , and a(yo)-I = a o - Iy - = a y - E /1. Thus z E b.1T whence 111T = b.' E n(a). 1T2 I Pick x E /1 . Then there exists y E G with x = a Y and a y - E /1 1T . This 1 1 y 15 15 yields a 8 E G with a - = a and a - E b.. We infer that E Ga and
oy
x This proves that /1
1T2
= a Y = a 15 - Il5y
n /1 i= 0
E
and hence /1
b. 15y 7T2
= b..
= 11,
as /1
~.
30.2 Lemma. It is b. 1T (x)y y E G.
7T2
and Ll are orbits of
0 =
b. Y1T(X
Y )
for all x E
Q,
all b. E O(x), and all
30.4 Lemma. Let G be finite. Then 7T(a) fixes an orbit ofGa other than {a}, if and only if IG I is even. PROOF. Let 7T = 7T(a), let {a} i= b. E n(a) and assume /1 = /1 7T . Then there I exists Y E G with a Y E /1 7T = /1 and a y - E /1. Therefore there ex~sts 0 E Ga ya with a = a y - ' • This yields yo fi. Ga , but (YO)2 EGa' as a(Yo) = aO = a. Therefore the order of yo is even, whence I G I is even. Assume conversely that IG I is even. Then there exists Y E G with Y i= 1 = y2. Pick bEn wi,th bY i= b. As G, acts t~~nsitively on st, ~~ rna;: assume b=a. There eXIsts /1 En(a) WIth a Y E/1. Hence a Y =a E b. n 111T, whence b. = b. 1T . Finally /1 i= {a}, since a i= a Y. 0
From now on, we shall assume that G is finite and that In(a)1 = 3 for one and therefore all a Est, i.e., we shall assume that G is a rank-3-group. Let n(a) = {{a},/1,f} and pick Y E G. Then n(aY) = {{a Y},/1 Y,fY}. Assume that there exists 0 E G with /18 = f Y. Then fY is an orbit of Gb and of G where b = a Y and c = a 15 . Moreover 1b.1 = 1b.81= jfYI = IfI, whence Istl =c 1 + 21/11. The group G cannot act imprimitively, since in that case either {a} U 11 or {a} U f would be a set of imprimitivity implying that 1 + 1/11 divides I + 21/11 which is impossible. Hence Gb and Gc are maximal subgroups of G. If Gb i= Gc , then G = Gb , Gc )' As fY is an orbit of Gb and of G , this would imply G being intransitive. Thus Gb = Gc , whence b = c and c E Ga' Moreover, f y8 - I = /1 and this is a contradiction. This shows that there are mappings /1 and f from st into UaEnn(a) such that n(a) = {{a}, b.(a), f(a)} and /1(a)Y = b.(a Y) as well as f(a)Y = f(aY) for all a E n and all Y E G. Put Istl = n, ILl(a)1 = k and If(a)1 = I. Then n = k + 1 + 1.
<
yo -\
30.5 Lemma. a) There exist two non-negative integers A and i-L with the property: If a,b Est, then Ib.(a) n b.(b)1 = A If b E b.(a), and Ib.(a) n b.(b)1 = i-L If b E f(a).
144
IV. The Suzuki Groups and Their Geometries
b) There exist two non-negative integers '\1 and fJ-I with the property: If a,b En, then \f(a) n f(b)\ ='\1 if b E f(a), and \f(a) n f(b)1 = fJ-I if bE
~(a).
Therefore 1 = If(a)1 = 1 + \f(a) n ~(b)\ + '\1' This implies k - 1= i-L - 1 '\1' i.e., Al = 1 - k + i-L - 1. Assume b E t::.(a). Then a E t::.(b) by 30.7. Therefore
l
a) Let a, b, aI, b En and assume b E ~(a) and bl E ~(al). There l exists 'Y E G with a Y = a . Hence bY E ~(a)Y = ~(aY) = ~(al). Therefore there is an YJ EGa' with b'(T1 = bl. This implies PROOF.
\~(a)
n ~(b)\ =
\~(a)'(T1n ~(b)'(T11
= I~( a
l)
=
1~(a'(T1)
n ~(bl)l·
30.6 Lemma. Let ?T(a) be defined as at the beginning of this section. Then we
have: a) If \G I is odd, then ~(ayr(a) = f(a) for all a En. . b) If \G \ is even, then ~(ayr(a) = ~(a) and f(ay(a) = f(a) for all a En.
o
This follows immediately from 30.4.
a) If \G \ is odd, then b E ~(a) If and only If a E feb). b) If IGI is even, then b E ~(a) if and only if a E ~(b).
a) Let b E ~(a). There exists 'Y E G with a y - = b. Hence a Y E ~(ayr(a) = f(a) by 30.6 a). Therefore a E f(a)Y-' = f(a y - ' ) = feb). If a E feb), then bY = a E feb). Therefore by-I E f(byr(b) = ~(b) and hence b E t::.(b)Y = ~(bY) = t::.(a). b) is proved similarly. I
30.8 Lemma. If IGI is even, then '\1 1.
,\ +
= 1-
k + fJ- - 1 and fJ-I
= 1- k +
Let a, bEn and assume b E f(a). By 30.7, we have a E feb). Now {a} U t::.(a) U f(a) and hence
PROOF.
n=
~( b) = ~(b)
n n = [ t::.( b) n {a} ]
= [t::.(b)
n t::.(a)]
U
[t::.(b)
U [~( b)
n t::.( a )]
U [ ~( b)
n f ( a) ]
n f(a)].
Hence k = 1t::.(b)1 = fJ- + I~(b) n f(a)l. On the other hand
f(a) = f(a)
nn
=
[f(a)
n {b}]
U [f(a)
n t::.(b)]
11.
=
{b} U [f(a) n t::.(b)] U [f(a) n f(b)].
= {a}
n {a} ]
U [D.(b)
U [ t::.( b)
n t::.( a)]
U [ t::.( b)
n f ( a) ]
n D.(a)] U [D.(b) n f(a)].
30.9 Lemma. If IGI is odd, then 1 = k, n
= 2k +
1 and A = i-L = Al = i-Ll'
As t::.(ayr(a) = f(a), we have k = I by 30.3. Let a,b En and bE t::.(a). Then a E feb) by 30.7 a). Therefore fJ- = \t::.(b) n D.(a)\ = \t::.(a) n t::.(b) \ = A by 30.5. Similarly Al = i-LI' Moreover D.(b) = [D.(b) n t::.(a)] U [t::.(b) n f(a)], as a fI- t::.(b), and hence k = A + ID.(b) n f(a)\. Similarly rea) = [rea) n D.(b)] U [rea) n feb)], whence 1= Ir(a) n t::.(b) \ + AI' As k = I, we get A = '\1' 0 PROOF.
30.10 Lemma. The following statements are equivalent:
(j) G is imprimitive and k ..;;; I. (ij) Ga =1= Gr(a) . (iij) There exist a, bEn with a =1= band f(a) = feb).
30.7 Lemma. Let a, bEn.
PROOF.
t::.( b) = [t::.( b)
Thus k = 1 + ,\ + It::.(b) n f(a)l. Similarly 1= If(a) n t::.(b) I + i-Ll' whence 0 k - 1 = 1 + A - i-LI'
n ~(b'(T1)1
This proves the existence of ,\. The remaining statements are proved similarly. 0
PROOF.
145
30. Rank-3-Groups
U [f(a)
n f(b)]
PROOF. (j) implies (ij): Let be a set of imprimitivity with a E . As \{a} =1= 0, n\{a}, we have either «1> = {a} U t::.(a) or = {a} U f(a). From = {a} U f(a), we deduce that 1+ 1 divides 1 + k + I and hence k. But then 1+ 1 ..;;; k ..;;; I. Hence = {a} U t::.(a). Now Ga c;: G¢> = Gr(a)' Moreover, G
(ij) implies (iij): We infer from Grea) = G{a}ul1(a) and G a =1= Gr(a) that there exists a 0 E Gr(a) with b = as =1= a. Obviously f(a) == feb). (iij) implies (j): As G acts transitively on n, the set f(a) cannot be invariant under G. Hence Ga , Gb ) =1= G. Therefore Ga is not maximal. This yields that G acts imprimitively. Let be a set of imprimitivity with a E «1>. As above, either = {a} U t::.(a) or = {a} U f(a). Assume that = {a} U f(a). Then Grea) = Ga' There exists y E G with a Y = b. Hence Y={b}Uf(b)={b}Uf(a). This yields Y= whence a=b, a contradiction. Therefore = {a} U t::.(a). This yields that k + 1 divides k + I + 1 and hence I. Thus k < I. 0
<
30.11 Corollary. If G is imprimitive and k..;;; I, then k + 1 divides I. Furthermore G has exactly one system of imprimitivity, namely the system consisting of all the sets {a} U D.(a). Finally Ga = G~(a) for all a E 5'2, and
D.(a)
= t::.(b)
implies a
= b.
146
IV. The Suzuki Groups and Their Geometries
30.12 Corollary. IJ G is imprimitive, then G acts doubly transitively on its system oj imprimitivity. 30.13 Corollary. Every rank-3-group oj odd order acts primitively. We consider now the incidence structure (st, st, I) where a I b, if and only if a E ~(b). Obviously, G induces a group of automorphisms on (st, st, I). As G acts transitively on st, we see that (st, st, I) is a tactical configuration. Let c E st. We count the pairs (a, b) Est X st with a I b, c and b -=1= c. As (st, st, I) is a tactical configuration with k points per block, there also are k blocks through a given point. Let a E ~(c). Then there are k - 1 blocks b with a E ~(b) and b -=1= c. Hence there are k(k - 1) such pairs (a, b). On the other hand if b -=1= c, then there are A points a with a E ~(b) n ~(c) if b E ~(c) and fL such a's, if b E fCc). Hence the number of (a, b)'s is also Ak + fLl. Thus Ak + fLl = k(k - 1). This yields
30.17 .Corollary. If
IGI > 3,
then the following statements are equivalent:
(a') G is imprimitive and I ,;;;; k.
(b ' ) i\ = k - I - 1. (c') fL = k. PROOF. (a') implies (b'): As G is imprimitive, it has even order. Furthermore fLl = 0 by 30.15. By 30.8, we then have 0 = 1- k + A + 1. Thus
A=k-I-1. (b') implies (c'): By 30.14, we have
fLl = k (k - A-I) = k (k - k + I
+ 1 - 1) =
kl
whence fL = k. (c') implies (a'): If fL = i\, then kl = k(k - k - 1) and thus 1= - 1, a contradiction. Thus fL -=1= A. Therefore IG I is even by 30.15. Hence i\1 = 1- k + fL - 1 = 1- 1 by 30.8. Thus G is imprimitive with I ,;;;; k by 30.16. 30.18 Corollary. If I G I > 3, then G is primitive, if and only if fL
30.14 Lemma. fLl
=
From 30.14 and 30.9 we get: 30.15 Corollary. IJ
-=1=
0, k.
k(k - A-I).
IGI
is odd, then A = fL
31. A Characterization of the Luneburg Planes = ~(k -
1).
30.16 Corollary. The following statements are equivalent: (a) I G I = 3 or G is imprimitive and k <; I. (b) fL = o. (c) A = k - l. PROOF. (a) implies (b): Let G be imprimitive and k <; I. Pick a, bEst. Then {a} U ~(a) and {b} U ~(b) are sets of imprimitivity by 30.1l. If b E rea), then b tl {a} U ~(a) whence [{a} U ~(a)] n [{b} U ~(b)] = 0. Therefore ~(a) n ~(b) = 0, if b E f(a). This proves fL = O. If IG I = 3, then 3 = n = k + 1+ 1. Hence k = 1. Thus fL = 0 by 30.15. (b) implies (c): This follows immediately from 30.14. (c) implies (a): If A = 0, then k = 1, as A = k - 1 by assumption. Therefore ~(a) = {b} for some b -=1= a. In this case Ga = Ga. b = Gb' If 1= 1, t~en I G I = 3. If I ;> 2, then k = 1 < 2 <; I. Moreover a and b are the only ~lxed elements of Ga = Gb . Hence 9(c( Ga. b) -=1= Ga' Therefore G is Imprimitive. We may thus assume that A > O. Then it follows from A = k - 1 and 30: 15 that I G I is even. Let b E ~(a). Then I~(a) n ~(b)1 = A = k - l. As \ G \ IS even, a E ~(b) by 30.7 b). Since a ft:- ~(a), we therefore have that {a} U (~(a) n ~(b)) = ~(b). Therefore {b} U ~(b) c;:;; {a} U ~(a), whence {a} U ~(a) = {b} U ~(b). This implies rea) = reb). Therefore, by 30.10, the group G is imprimitive and k <; t. 0
The following theorem is due to R. Liebler 1972.
31.1 Theorem. Let q = 22a + I ;> 8 and let mbe a translation plane of order q2. IJ m admits a collineation group G isomorphic to Seq), then m is a Luneburg plane.
m
PROOF. Let T be the translation group of and put H = TG. Then T n G = {l}, since G is simple. As T acts transitively on the set of points, we have Ho ;;;; G where 0 is some point of m. Thus we may assume that Ho = G. Hence G fixes a non-incident point-line pair in the projective closure I,l3 of m, namely (O,loc,). By 28.11, G therefore has a point orbit PI with IP!I = (q2 + 1)(q - 1) and a point orbit P2 of length (q2 + l)q(q - 1). In partIcular, H acts as a rank-3-group on the affine plane m. Moreover, H has two line ~rbit~ gl and g2 with Igil = (q2 + l)(q - 1) and Ig21 = (q2 + 1) q(q - 1). If lIS a lme through 0, then HI acts as a rank-3-group on the set st of points on I by 16.2 a). The lengths of the orbits of HI,0 are 1, q - 1 and q(q - 1) respectively. Let ILl,(O)1 = q - 1 and If,(O)1 = q(q - 1); also let A and fL be the intersection numbers defined in 30.5. Then 0<; A <; q - 2. By 30.14, we have fLq(q - 1) = (q - l)(q - 2 - A) and hence fLq = q - 2 - i\ whence A + 2 == 0 mod q. Furthermore 2 <; A + 2 ,;;;; q and therefore i\ + 2 = q. This yields fL = O. It follows from 30.16 that H acts . .. . ~mpr~m~t~v~ly on st. Thus by 30.11, /={O}U~,(O) is a set of lmpnmltlvlty of HI'
'
148
IV. The Suzuki Groups and Their Geometries
Put T",={TITET, /=~'}' Then T", is a subgroup of order q contained in T(l n lex,). As G acts transitively on 100' we infer from Igil = (q2 + 1)(q - 1) that each point of 100 is on exactly q - 1 lines of g I and on q (q - 1) lines of g2' It follows that Hp acts as a rank-3-group on the set of affine lines through P for all P I 100' As above, we see that Hp acts imprimitively and that '¥ 0, p = {OP} U {h Ih E gl' PI h} is a set of imprimitivity. Put Lp = {T IT E T, '¥~,p = '¥ o,p}. Then T(P) C; Lp and hence ILpl = q3. Let I and m be two distinct lines through 0 and put P = I n 100 and T"m = Lp n T(m n 100)' Then IT',ml = q. Let 7T be the set of all T", and T"m' We shall show that 7T is a partition of T. Let Ip . . . , Iq _ I be the lines of g I parallel to I and denote by XI' ... ,Xq _ 1 the points of tJ I which are on m. Then
Tm,m = {l} U {TIT ET and OT E {XI"'" X q _ I }} and T"
= {1}
m
U { TIT E T and 0
E {II n
T
m, ... , Iq _ I
n
m} }.
By 28.10 a) we obtain T m, m nT" m = {I}. Moreover, for y E G we have y -IT, mY = T,r mr for all I and m distinct or not. As G has two point orbits other'than {O}, we therefore obtain that T = U X E'7T X. Furthermore, if 1=1= m then y -IT" mY = T" m' if and only if Y E 3(II)G"m
where II is the Sylow 2-subgroup of Gm . This is seen as follows: We have Y - IT" mY = T" m if and only if r T" m = {l} U { T Y IT E T and 0 E {II n m, ... , Iq _ I n m} } T
and hence if and only if {II n m, ... ,
Iq_1
n m
r = {II n m, ... ,
Iq_1
n m}.
From this and {II n m, ... , Iq _ I n m} C; tJ 2 we obtain the above assertion. Hence there are (q2 + l)q groups of type T"m with 1=1= m. As there are q2 + 1 groups of type T"" we obtain 17T1 = (q2 + l)q + q2 + 1 = (q2 + 1) (q + 1). Hence ITI < 1 + (q2 + 1)(q + 1)(q - 1) = q4 = ITI. Thus 7T is indeed a partition of T. N ext we prove that 7T is geometric. If 1=1= m, then T, mT m m = T(m n IC(J Hence T, mT m m is 7T-admissible. ' , Again let '1 =I=:n. Then IG{I,m}1 = 2(q - 1). Hence there exist q - 1 elations in G which switch I and m. The axes of these elations are mutually distinct. Let a be one such elation and denote by a its axis. Then T"
a =
{
I}
U
= {1} U
{T IT
{T IT
E
T and OT E {II n a, ... ,
E T and 0
T
E
lq_1
{11° n a, ... , 1;_1
n a} } n
a} } =
This implies T"a C; T"m T m ,,' From this we infer
T"mTm"
.
=
T"m U Tm,1 U
Ua T"a
where the union is taken over all those a's which are axes of elations in G (t, m)' Hence T I, m T m, 1 is 7T-admissible. Next we remark: T/,mTm" = Tu,vTv,u if and only if {I,m} = {u,v}. Assume first {u,v} n {I,m} = 0. As Tu,v C; T"m T m", we have Tu,v = T"v by what we have just proved. Let Y E G with u Y = I and v Y = v. Then Y is in the normalizer of Tu, v as well as in the normalizer of T" v' Therefore Y E 3(II)Gu, v n B(II)GI , v = B(II) where II is the Sylow 2-subgroup of Gv' Hence v Y = v, u Y = I and fY = u. Moreover, as T m" C;Tu,vTv,u' we have Tm ,= Tu I' There also exists an involution 0 E G with U O = v and 10 = I. He'nce v oy = u Y = I, lOY = fY = u, u oy = v Y = v. Thus 3 divides the order of or and hence of G contradicting 21.lO. Hence {u, v} n {l, m} =1= 0. We may assume wlog that v = m. Then T"m = Tu,m' Assume u =1= I. From T m, u C; T I , mT m,l we deduce that there exists an involution Y E G with u Y = u and mY = I, and from T m,l C; T m, uT u, m we deduce that there also exists an involution 0 with 10 = I and m 0 = u. Hence u y8 = U 0 = m, m y8 = 10 = I, fyo = m 0 = u. We thus reach again the contradiction that 3 divides the order of G. Hence {u, v} = {m, I}. We have thus shown that there exist t q2(q2 + 1) subgroups of the form T, mT m , all of which are 7T-admissible. Furthermore, no subgroup T, mT m , co'ntai~s a group of the form Ta,a' ' , Let I and m be two distinct lines through O. Put S = T, mT(l n 100)' If x is a line through 0 distinct from I then S n T(x n 100) = 1', x' Hence S is 7T-admissible. Furthermore G, acts transitively on the set of ~ll T" x as well as on the set of all T, x T x ,. Therefore the incidence structure whose points are the T" x and wh~se blocks are the T" x T x,, is a tactical configuration with the parameters v = b = q2 and r = k = q. Let Y" ... , Yq be those T" y T y " which contain T" x' Then
Is \V, (y,\T"x)l- q3 - q(q2 - q) - q',
Moreover S\Ul=I(Yi \T"J is a union of components of 7T. Finally (T " x T 1,1)
n
Yi
= T " xC T ',' n
Yi ) = T " x
.
As T", <;;;; S, we thus obtain
(T" x T/,I) n [ s Hence T"
T m, a
149
31. A Characterization of the Liineburg Planes
T",
\ ,V, (Y, \ T"x) ]- T"x T/,I'
is also 7T-admissible. then 1= m, as there are q groups T" y contained in T"x TI,/' Using once again the fact that T,.xTf.I contains exactly q x
If T l , x Tl,l
= Tm, y Tm, m'
150
IV. The Suzuki Groups and Their Geometries
subgroups of the form T /, y we see that q(q2 + I) is the number of groups
of the form
T" x T",.
Consider the group R m, I = T m, mT(l n I rYJ). Let 2: be a Sylow 2-subgroup of GI • As 3(2:) fixes all the lines through I n I rYJ and as 3(2:) centralizes T(lnl rYJ ), we have Tma,ma~Rm,1 for all aE3(2:). Now Tma,maTma,1 = T ma maT m I is 7T-admissible for all a E 3(2:). This implies that Rm, 1\ T(l n IrYJ) is the union of the sets T ma, maT m, 1\ T m, I where a ranges over 3(2:). Hence Rm,1 is 7T-admissible. From RI,mRm,1 d T(l n IrYJ)T(m n IrYJ) = T and
q6
=
IRI,mIIRm,/1
=
ITIIRI,m
n Rm,/1 =
q4lRI,m
CHAPTER V
Planes Admitting Many Shears
n Rm,/1
we conclude that RI m n Rm I = TI IT m m' Hence TI IT m m is also 7Tadmissible. "" , , Let I, m and x be three distinct lines through O. Then T x x n (T/, ,Tm, m) = {l}: For assume Tx,x n (T"/Tm,m) =1= {I}. Then there ~xist lines II ,m l with II parallel to I and m l parallel to m such that II n m l E PI' By 28.10, we have therefore II,m l E g2 and III n PII = q = Iml n PI I· Hence there exist involutions p and a in G with IP = I, m P= x and m = m, 1(J=x. This implies that 3 divides IGI, a contradiction. Thus Txx n (T/,/Tm,m) = {l}. Therefore T/,/Tm,m = Tu,uTv,v, if and only if {u,~} = {I,m}. N ow there are (J
(q2 + I)q(q + 1) pairs X, Y E 7T with XY E
K,
where
K
=
{T(P) I PI IrYJ },
+ 1) pairs X, Y E 7T with XY = T/,m Tm,I' tq2(q2 + I)q(q + 1) pairs X, Y E 7T with XY = T/,ITm,m' q(q2 + l)q(q + 1) pairs X, Y E 7T with XY = T"m T/,I' Hence there are q(q + 1)(q2 + I)(q2 + q + 1) pairs X, Y E 7T with X 1q2(q2 + l)q(q
=1= Y
such that XY is 7T-admissible. The total number of pairs X, Y E 7T with X =1= Y is also q(q + 1)(q2 + 1)(q2 + q + I) as 17T1 = (q2 + I)(q + I). Hence 7T is geometric. It follows from 29.2 that T is a vector space of rank 4 over GF(q). Hence GF(q) is the kernel of W. Furthermore, it follows from 1.10 and the simplicity of G that G acts as a group of linear transformations on the GF(q)-vector space T. Therefore W is a Liineburg plane by 27.3. 0
Next we collect results about unitary groups, we prove a characterization of A5 which I extracted from J. Assion's Diplomarbeit, and we give a characterization of Galoisfields of odd characteristic due to A. A. Albert. All this is done in order to prove Hering's & Ostrom's theorem on collineation groups of translation planes generated by shears. I would like to thank J. Assion for allowing me to make use of his unpublished Diplomarbeit.
32. Unitary Polarities of Finite Desarguesian Projective Planes and Their Centralizers We start with a remarkable theorem of R. Baer 1946 b.
32.1 Theorem. Let 7T be a polarity of a finite projective plane. Then 7T has at least one absolute point.
sn
PROOF (Ryser). Let PI' ... , P v be the points of the plane and lable the lines of by Ii = Pi" . Furthermore, define the incidence matrix A = (a i .) of ~ by aU = I, if and only if Pi I ~, and aU = 0 otherwise. Then we have au = 1, if and only if Pi I lJ7r which is equivalent to Pj I Pi7r , as 7T 2 = 1. Therefore au = I if and only if aji = 1. Hence A is symmetric. If n is the order of the projective plane SU, then A 2 = AA I = nE + J where E is the (v X v)-identity matrix and J is the matrix all of whose entries are l. It is easily seen that (n + Ii is a simple eigenvalue and that n is an eigenvalue of multiplicity n 2 + n of A 2. 151
sn
152
V. Planes Admitting Many Shears
The row sums of A are all equal to n + 1. Hence n + 1 is an eigenvalue of A. Let e l , • .• ,ev - I be the remaining eigenvalues of A. As A is symmetric, A is diagonalizable. Hence e? are eigenvalues of A 2. Therefore e? = n whence ei E for all i. Let a be the mUltiplicity of as an eigenvalue of A and let b be the multiplicity of -/;;. Then trace(A) = n + 1 + (a - b)~-;;. Assume trace(A) = O. Then a - b =1= 0, as n + 1 > 3. Hence is rational. This yields Ell whence n + 1 is divisible by which is possible only for n = 1. Thus trace(A) =1= O. Now trace(A) = 2:~= I aii and aii = 1 if and only if Pi I Pt, i.e., if and only if Pi is absolute. As trace(A) =1= 0, we have that the number of absolute points is not nought. 0
{rn, - rn}
rn
rn
rn
rn
32.2 Corollary (Baer 1946 b). If'TT is a polarity of a finite projective plane of order n and if n is not a square, then 'TT has exactly n + 1 absolute points.
rn
PROOF. As n is not a square, is not rational. Therefore trace(A) = n + 1 + (a - b)rn yields a = b, i.e., trace(A) = n + 1. 0
32.3 Lemma. Let 'TT be a polarity of a projective plane. If P is an absolute point of 'TT, then P is on precisely one absolute line of 'TT. PROOF. As P is absolute, PI p 7T = I. Then 111" = p1I"2 = P I I, whence I is an absolute line passing through P. Let P I m and m7T I m. Assume that 1=1= m. Then Q = m1l" =1= 17T = P. Furthermore, P = I n m and hence 1= p 7T = 11I"m 7T = PQ. On the other hand Q = m7T I m. Therefore m = PQ = I, a contradiction. 0
32. Unitary Polarities of Finite Desarguesian Projective Planes and Their Centralizers
non-degenerate symmetric semibilinear form (a, j) induces a polarity 'TT on I,l3 defined by X 7T = {y lyE V, f(X, y) = O} for all subspaces X of V. We shall say that 'TT is represented by the pair ((X, j).
32.4 Lemma. Let the polarity 'TT be represented by (a, f) and let I = uK EB vK be a line such that V = 1 EB 17T. Then there exist vectors u VI E I such that I = u K EB VI K, VI K = L n (vKt and (u + v' x O:)K = 1 n [(u + vx)Kr for all x E K. l
,
(1) f(x + y, z) = f(x, z) + fey, z) for all x, y, z E V, (2) f(x, yk) = f(x, y)k for all x, y E V and all k E K, (3) f(x, y) = f(y,xt for all x, y E V, 7T =
such that X {y lyE V, f(X, y) = O} for all subspaces X of V. From {O} = V 7T we infer (4) f( V, y) = 0 implies y = O.
Let (X be an involutory antiautomorphism of K and let f be a mapping from V X V in K such that the pair ((X, I) sa tisfies (1) through (4). Then we say that (Ct, J) is a non-degenerate symmetric semibilinear form. Each
l
l
PROOF. For a point P on 1 we define po by po = 1 n absolute, p 7T =1= I. Thus po is a point. Furthermore
P7T.
As 1 is not
since P C; I and 1 n 17T = {O}. This shows that a is an involutory permutation of the set of points which are on I. The fixed points of a are the absolute points of 'TT which lie on I. uK, vK, and (u + v)K are three distinct points on I. Hence (uKY, (vK)O, and «u + v)KY are three distinct points on I. Therefore there exist vectors U",V" E I with (uKY = u" K, (vKY = v" K, and «u + v)KY = (u" + v")K. Put t = f(v, u") and assume t = O. Then u" K C; (vKt n I = v" K, a contradiction. Thus t =1= O. Set u = u" t - I and 0" = v" t -I. Then we have (uK)O = u' K, (vKY = Vi K and «u + v)KyJ = (u + v')K. Moreover feu, u' ) = f(v, Vi) = feu + v, u + VI) = 0 and f(v, u /) = 1. Hence / 0= feu + v,u + VI) = f(u,u /) + f(u,v /) + f(v,u /) + f(v,v /) = f(u,v ) + 1, l
l
l
l
i.e., feu, VI) = - 1. Therefore
feu + vX,u + VIXO:) l
Let V be a vector space of rank 3 over the skewfield K and let I,l3 be the projective plane whose points are the subspaces of rank 1 of V and whose lines are the subspaces of rank 2. If 'TT is a polarity of 1,l3, then it follows from 32.3 that not all points of I,l3 are absolute points of 'TT. Hence, by Baer [1952, Chapt. IV. In particular Prop. 1, p. 110], there exists an involutory antiautomorphism (X of K and a mapping f: V X V ---7 K with the properties:
153
=
f(u,u
l )
+ f(u,v/)xO: + xO:f(v,u') + xO:f(v,v/)xO:
=
o. o
32.5 Lemma. Let 'TT be a polarity represented by (Ct, I). If I is a line all of whose points are absolute, then (X = 1. PROOF. I is not absolute by the dual of 32.3. Hence V = I EB 17T. Let I = uK EB vK and choose u', VI according to 32.4. Since (uKt = uK, (vKt = vK and «u + v)Kt = (u + v)K, there exist A, Il, v E K* with (u + V)A = u + VI, u = ull, and VI = vv. As u and v are linearly independent, we obtain A = Il = v. Moreover, for each x E K there exists p E K* with (u + vx)p = u + Vi xO:, as a = 1. This implies up + vxp = UA + VAXo:, whence p = A and xO: = A- I XA for all x E K. Therefore Ct is an automorphism as well. From this we deduce that K is commutative. Hence xO: = A-IXA = x. 0 l
l
l
The polarity 'TT is called unitary, if a =!= l. If K is finite and a =!= 1, then K = GF(q2). Conversely, if K = GF(q2), then K admits an (anti-) automorphism a with a 2 = 1 =1= a, namely the one defined by xo: = x q . It is
154
V, Planes Admitting Many Shears
then easily seen that the desarguesian plane over GF(q2) admits a unitary polarity. Therefore, a finite desarguesian plane admits a unitary polarity, if and only if its order is a square. Let 7T be a unitary polarity of the desarguesian plane over GF(q2). Then 7T has absolute points by 32.1. Let P be an absolute point and let I be a line through P other than P7T. Then V = I EB 17T by 32.3. Let 7T be represented by (a, f). As a =1= 1, there exists a point Q on I with Q =1= ~a. Since P = pa, the points P, Q, Qa are mutually distinct. Hence there eXIst vectors u ,and v with Q = uK, Q a = vK, and P = (u + v)K. Pick u' and v' accordmg to 32.4. Then u' K = (uKt = vK, v' K = (vKt = uK, and (u ' + v')K = «u + v)Kt = (u + v)K. Hence u' = VA, v' = Ull, and u' + v' = (u + v)v with
A, Il, v E K*. It follows t~at A = Il = ,v. , , *' The point (u + vx)K IS absolute, If and only If there eXIsts p E K WIth (u + vx)p = u' + v' X a = VA + uAx a; hence if and only if xp = A and p = Ax a. This implies 1 = x a+ I, as A =1= O. Conversely, if 1 = x a+ 1, then a there exists p E K with xp = A whence p = x a+lp = XaA = AX . As , f . 1+a q xl+a = xl+ , there are exactly q + 1 elements x E K satIs ymg x =.1 Therefore I carries exactly q + 1 absolute points of 7T. Let U be the incidence structure consisting of the absolute points of 7T and those lines of SU which carry at least two absolute points. Each line belonging to U carries precisely q + 1 points of U, as we have just seen. Therefore U is a block design with parameters v, b, k = q + 1, r, A = 1. By 32.3 we see that r = q2. Hence v-I = A(V - 1) = r(k - 1) = q3 whence v = q3 + 1. As vr = bk, we finally get b = q2(q2 - q + 1). The design U is called the unital belonging to 7T. Thus we have proved:
32.6 Theorem. If U is the unital belonging to the unitary polarity 7T, then U is a block design with parameters v = q3 + 1, b = q2(q2 - q + 1), r = q2, k = q + 1, and A = 1. 32.7 Theorem. If SU is a desarguesian projective plane of order q2, then all unitary polarities of SU are conjugate. PROOF. Let 7T be a unitary polarity of SUo By 32.3 there exists a point P=bIK which is not absolute. Thereforef(bl,bl)=I=O. Fromf(b,l,blt = f(b l , b l ) we infer f(b l , b l ) E GF(q). 2 As each ele~ent of GF(q~ IS t.he norm of an element in K = GF(q), there eXIsts A E GF(q) wIth AI + a = f(bl,bl)-I. Therefore f(bIA,bIA) = AI+af(bl,b l) = 1. We may therefore assume that f(b l , bl) = 1. Let U be the unital belonging to 'TT. Then (b I KY is a line of U. This follows from 32.6, as q3 + 1 + q2(q2 - q + 1) = q4 + q2 + 1. Therefore (b K)7T carries q + I absolute points. Hence there exists b2 E(b,K)7T with j( b2 , b 2 ) =1= O. As the above argument shows, we even find such a b 2 with f(b 2,b 2) = 1. Obviously f(b l ,b 2) = 0 = f(b 2,b l). There are exactly q + 1 elements x E K with x I + a = - 1. Let x be one such element. Then f(blx + b2,b lx + b2) = xl+a + 1 = O. Hence the line
32. Unitary Polarities of Finite Desarguesian Projective Planes and Their Centralizers
155
b l K + b2 K is not absolute by the dual of 32.3. It follows that (bIKY n (b 2KY is not absolute. Hence there exists b) E (blKt n (b 2Kt withf(b 3,b3) = 1. Now {b l ,b 2,b 3} is a basis of V and we have
fC~1 b;x; ,
t
bJ';) -
;~l x;"y; .
c::
Let aK be a non-absolute point of 7T. Then f(a, a) =1= O. Hence we may assume f(a, a) = 1. Let a be a homology with centre aK and 7Ta = a7T. Then (aKt is the axis of a. Therefore a is induced by the mapping (which we also call a) defined by x a = x + aAf(a,x) where A is a suitable element in K. As a is linear and as 7T is centralized by a, there exists Il E K with Ilf(x, y) = f(x a, ya) for all x, y E V. This yields
Ilf(x, y)
=
f(x, y) + Af(x,a)f(a, y) + Aaf(a,xt f(a, y)
+ AI +af( a, xt f( a, y). There exist x, y E V such that f(x, y) =1= 0 = Ilf(x, y) whence Il = 1. Thus we obtain
=
f(x, a). Therefore f(x, y)
0= (A + Aa + AI+a)f(x,a)f(a, y) for all x, y E V. Putting x = y = a yields 0 = A + Aa + Al +a. This is equivalent to (A + 1)I+a = 1. Conversely, if (A + 1)I+a = 1, then x a = x + aAf(a,x) defines a homology with centre aK and 7Ta = a7T. As there exist exactly q + 1 such A.'s, the group 6.(aK) of all homologies with centre aK which centralize 7T is cyclic of order q + 1. (The cyclicity of 6.(aK) follows from 3.1.) We note this as
32.8 Lemma. If P is a non-absolute point Of'TT, then D.(P) is cyclic of order q + 1. If the characteristic of K is 2, then f( x, y) = - f(y, x t. Assume that the characteristic of K is not 2. Then there exists t E K with t a = - t =1= O. Put g = tf. Then 'TT is also represented by the form (a, g) (which is, of course, not symmetric). Moreover
g(x,y) = tf(x,y) = -tj(y,xt= -(tf(y,x)t= _g(y,x)a. Hence in either case there exists a form
(a,
g) representing
7T
with
g(x, y) = - g(y,xt. Denote by PGU(3, q2) the centralizer of 7T in PG(3, q2) and let T be a non-trivial elation in PGU(3, q2) with centre aK. Then (aK)7T is the axis of T whence aK ~ (aK)7T. Thus aK is absolute. This implies g(a, a) = O. As (aKt is the axis of T, we have x'T = x + aAg(a,x) with some A E K. Moreover, there exists Il E K with Ilg(X, y) = g(x'T, y'T) for all x, y E V. A trivial computation shows that
Ilg(x,y)
=
g(x,y) + (A - ;\a)g(x,a)g(a,y)
156
V. Planes Admitting Many Shears
for all x, y E V. This implies fL = 1 and A = Aa, whence A E GF(q). Conversely, if A E GF(q), then A = Aa and the mapping defined by x T = X + aAg(a, x) is an elation with centre aK belonging to PGU(3, q2). Thus we have proved the first part of
32.9 Lemma. a) If P is an absolute point of 1T and If E(P) denotes the group of all elations with centre P in PGU(3, q2), then IE(P)I = q. b) If P and Q are two distinct absolute points of 1T, then <E(P), E(Q» ~ SL(2,q). PROOF. Let P = aK and Q = bK. As P + Q is not absolute, we may choose a and b such that g(a, b) = 1. Then g(b, a) = - 1. Let P'" n Q'" = cK. Then {a,b,c} is a basis of V. Pick a E E(P). Then XO = x + aAg(a,x) for some A E GF(q). Therefore a is represented with respect to {a,b,c} by the matrix
100
A
I
0
001 If 7' E E(Q), then there exists fL E GF(q) with x T Hence 7' is represented by the matrix
[~ Therefore
(E(P), E(Q»
~\[ ~ ~
I
II!I)
II'II Ii
Ii
I) l,i
'
0 1 0
SL(2,q).
fL I 0
= x - bfLg(b,x).
H
n[~
fL 1 0
~j
A, p. E
32. Unitary Polarities of Finite Desarguesian Projective Planes and Their Centralizers
157
(g) If P is a point of U and TI the Sylow p-subgroup of G containing E(P), then E(P) = 3(TI). (h) Let II be a Sylow p-subgroup of G. If P = 2, then B(II) = {T IT Ell, 7'2 = I}. If P > 3, then 7'P = 1 for all 7' E TI. (i) All elations of G are contained in H and conjugate under H. (j) If q =1= 2, then H is simple.
PROOF. (a) Let P be a point of U. Then E(P) is a p-group, where p is the characteristic of GF(q2). As P is the only fixed point of E(P) in U, the group G acts transitively on the set of points of U by Gleason's lemma 15.1. If I is a line of U, then 32.9 b) implies that G, acts doubly transitively on I n U. Hence we need only show that Gp acts transitively on the set of lines of U passing through P for some point P of U. Let II' ... , Iq2 be all the lines of U which pass through the point P of U. . Put qj = Ij "'. Then L1( Qi) has order q + 1 by 32.8. From P I Ii we mfer Qi I P"'. The group L1(Qj) consists of homologies with axis Ii' Therefore L1( Qi) acts regularly on {II' ... , Iq2} \ {O. Applying Gleason's lemma once again using a prime divisor of q + 1, we obtain that Gp acts transitively on
{iI' ... ,1q2}'
(b) If y E G p , Q fixes a third point of P + Q, then y induces the identity on the line P + Q. Hence Gp , Q/ L1«P + Q)"') is isomorphic to a group which acts regularly on the set of q - 1 points of P + Q other than P and Q belonging to U. Hence IGp , QI = (q + l)s, where s is a divisor of q - 1. Let P = blK, Q = b2K and P'" n Q'" = b3K. Then {b l ,b 2,b 3} is a basis of V. Define p by bf = blA q, b~ = b2A -), bj = b 3 where A E K*. Then
f(bf ,bf) = f(bIAq,bIA q) = 0 = f(b l ,bl)' f(bf ,'bD = f(b IA q,b 2A -I) = f(b l ,b 2)Aq 2_ 1 = f(b l ,b 2),
GF(q») 0
32.9 b) is actually all we shall need later on. But for the sake of completeness we also prove:
f(bf ,bn = f(b IAQ,b 3)
=
0
= f(b l ,b 3),
,b 2A -I) = 0 = f(b 2 ,b 2),
f(b~ ,bD
= f(b 2A-
f( b~ , bn
= f( b 2A -I, b3 ) = 0 = f( b2 , b3 ),
I
f( bf ' bn = f( b3 , b3)· 32.10 Theorem. Let 1T be a unitary polarity of the projective plane over GF(q2) and U the unital belonging to 1T. If G = PGU(3, q2) and H = PSU(3, q2) = G n PSL(3, q2), then we have: G acts doubly transitively on the set of points of U. If P and Q are distinct points of U, then GP, Q is cyclic of order q2 -:- 1. IGI = (q3 + l)q\q2 - 1). IHI = (3,q + 1)-I(q3 + 1)q3(q2 - 1). If P is a point of U, then Gp contains a normal subgroup of order q3 which acts transitively on the set of points of U distinct from P. (f) If TI and TIl are two distinct Sylow p-subgroups of G, where p is the characteristic ofGF(q2), then TI n TIl = {l}.
(a) (b) (c) (d) (e)
This implies pEG. Hence Gp , Q contains a cyclic subgroup of order q2 - I, whence IG p , QI = q2 - 1. (c) is an immediate consequence of (a) and (b). (d) From IPGL(3,q2):PSL(3,q2)1 =(3,q2_1) we infer IG:HI E {l,3}. Therefore H acts also doubly transitively on U. This implies IG: HI = IGp,Q: Hp,QI· Furthermore <E(P), E(Q» ~ SL(2,q) by 32.9 b). Hence t(q - 1) divides Hp,Q as, obviously, all elations of G are contained in H. If 3 does not divide q+ 1, then ~(q2-1) divides IHp,Qi,·whence iHp,Qi = q2 - 1. Assume that 3 divides q + 1 and let P = aK with j(a, a) a E t1(P), then XO = x + aAj(a,x) with (A + l)q+ 1= 1. As aO = a[1
= 1. If + Af(a,
158
V. Planes Admitting Many Shears
we see that ,\ + 1 is an eigenvalue of a, Moreover X O = x for all x E P"', Hence 1 is an eigenvalue of multiplicity 2, Therefore det( a) = A + 1. We infer that a belongs to H if and only if A + 1 is a third power in GF(q2). Therefore H =1= G as 3 divides q + 1. Hence IG: HI = 3, (e) Let P and Q be two distinct absolute points and let R = p 7T n Q7T, There exist b l E P and b2 E Q such that f(b l , b2) = 1. Pick b 3 E R such that f(b 3 ,b 3 ) = 1. Then f(bl'b l ) = f(b 2,b2) = f(b l ,b 3 ) = f(b 2,b 3 ) = 0. Let A, 11 E K be such that 11 q + 11 = - AI +q and define T(A, 11) by
a)]
= a(l + ,\),
b;(J..,
Putting
p.) T
= bl
,
= T(A,
f(b~ ,bD
b;(J..,
p.)
=
bill
= f(b l ,bl)'
+ b 2 - b 3 A q)
=
f( b] ,b 2 ),
feb; ,b;) = feb] ,blA + b3 ) = f(b l ,b3 ), f( b; ,bD = f( bill
+ f( b2 ,b2) + AI +q = f( b 2 ,b 2), f( b; ,b;) = f( bill + b2 - b3 A q, blA + b 3 ) = A - A =
°=
f( b2 , b3 ),
feb; ,b;) = f(b 1A + b3 ,blA + b 3 ) = f(b 3 ,b 3 )·
Therefore T E Gp • The mapping 11-----') 11 q + 11 is a homomorphism of the additive group of GF(q2) onto the additive group of GF(q). Hence for each A E GF(q2) there exist precisely q elements 11 such 11 q + 11 = - Aq+ I. Therefore there exist exactly q3 elements T(A, 11)' Now b~(J..,
p.)'T(J..', p.')
= b 1,
b;(J.., p.)'T(J..', p.') = (b]A + b3 r(J..', p.') = b 1(A + A') + b 3 , b;(J.., p.)'T(J..', p.') = (bill + b - b A qr(J..', p.') 2
3
= b l( 11 + 11/ - A'A q) + b 2
bP-''T(J.., p.)p = b I
I'
bP-''T(J.., p.)p = b I rIIK 1+q 2 br''T(J.., p.)p = b1AKq
-
+ b 2 - b 3AqK ,
+ b3 •
Furthermore =
-(AK)I+q= -(AKq)l+q,
as Kq(l+q) = KI+q. Hence p-1T(A, Il)p = T(AKq, flK1+q). Hence II is normal in Gp , since Gp = llGp,Q' (f) is a consequence of (e). (g) Obviously E(P) = {T(O, fl) I fl E GF(q2), fl q + fl = O}. Let T(A, 11) E S(ll). Then A'A q = A'qA for all A' E GF(q2). Hence with A' = 1 we get A = Aq. If A =1= 0, then A' = A'q for all A', a contradiction. Hence A = 0. Conversely, if A = then T(O, fl)T(A', 11/) = T(A/, 11 + 11/) = T(A', 1l')T(O, 11). (h) Induction shows that rCA, flY = T(iA, ill - ~ i(i - I)A I +q) for all i EN. Thus if P = 2, we have 1 = T(A, fl)2 = T(O, AI +q) if and only if A = 0. If P > 2, then 2 divides p - 1, whence T(A, IlY = T(pA, Pfl - ~p(p - 1) A I +q) = 1.
°
+ b2 - b 3A q, bill + b 2 - b 3A q)
= 11 q + 11
159
~s we have seen under b), the group Gp , Q consists of all the mappings P defmed by bi = b l Kq, bi = b2 K- I, bj = b3 , where K E K*. This yields
(Il K1 +q)q+ Il K1 +q = (Il q + Il)K 1+q
+ b2 - b3A q,
11) we then obtain
f( b; , bD = f( b I ,b I 11
32. Unitary Polarities of Finite Desarguesian Projective Planes and Their Centralizers
b3 (A
+ A')q.
(i) The groups E(X) are contained in H, as we remarked earlier. Moreover they are all conjugate, as H acts transitively on the set of points of U. Therefore, it suffices to show that all non-trivial elations of E(P) are . conjugate under H p. Now we know already (see the end of the proof of (e» that p -IT(O, Il)p = T(O, flK 1+ q). Furthermore, the mapping K-----') K1+ q maps the group of all third powers of GF(q2)* onto GF(q)*. Since the set of all fl such that 11 q + 11 = is a subspace of the GF(q)-vector space GF(q2), the result follows. (j) As q > 2, there exists K E GF(q2) with K 3(q+ I) - 1 =1= 0. Consider now the mapping p with K replaced by K3. Then p E H. Furthermore
°
T(O, fl)-Ip-IT(O, fl)P = T(0,(K 3 (q+l) - 1)11) E H;.
Furthermore (11 + 11/ - A'A q) q + 11 + 11/ - A/A q = 11 q + 11 + Il'q + 11/ - A/A q - A'qA
= -(Aq+1 + A'q+1 + A'Aq + A'qA) = - (A q + A'q)(A + A') = -(A+A/)q+l. Therefore T(A, Il)T(A" 11/) = T(A + A" 11 + 11/ - A'A q). This proves that the T(A, Il)'s form a group II of order q3. Moreover the orbit of Q under II has length q3.
Hence E(P) = S(ll) C H;. Moreover T(A, 11)-1 = T(-A, -AI+q - fl) and hence T(A, 1l)-1 p - l T(A, Il)p = T«K 3 q - I)A,z) with a suitable z. Since there exists K such that K3q - 1 =1= 0, since A is arbitrary, and since T(A, Il)T(O, 11/) = T(A, 11 + 11/), we find that II C H;. On the other hand Hp/ll is cyclic. Therefore II = H;. Let y be the mapping defined by bi = b2 , bI = b l , and bI = - b3 • Then y is in H. Furthermore y-IHp,QY = Hp,Q' As H acts doubly transitively and y r£ H p , we therefore have H = Hp U HpyHp = Hp U llHp, Qyll.
160
V. Planes Admitting Many Shears
Let 1"(A, }.L) =F 1. Then
br(A,
~)Y
= bI(A,
~)y
= (b l [.L + b2 -
b 3 A q)y = b 2 [.L
+ b l + b3 A q fl P.
PROOF. We may assume x = (123). Then X-I = (132). The remaining elements of order 3 are (124), (142), (134), (143), (234), (243). Furthermore
= (14)(23) (123)(142) = (234) (123)(134) = (124) (123)(143) = (12)(34) (123)(234) = (13)(24) (123)(243) = (143)
(123)(124)
Theref ore y1" (A, [.L)-Y E II H P, Q YII. Since we have to in terpret this projectively, we have (*) There exist A', [.L',A", [.L",a E GF(q2) and p E Hp,Q such that v yr(t.., ~)y = V r(t..', ~')pyr(A", ~")a for all v E V. From (*) we get py
= 1"(A', [.L')-I1"(A, [.L)1"(A, [.L)-IY1"(A, [.L)y1"(A", [.L")-I.
Therefore py E H' for all p that occur in (*), as II ~ H' and as y = y-I. N ext we show that e,ach p E H P, Q occurs in (*). In order to do this we compute (*) for v = b l and v = b3 • Recall that there is a K E K* such that bf = blK q, b~ = b2K- 1 and bj = b3 •
bt(A, b;(A',
~)y
~')pyr(A", ~"\x
bt(A,
~)y
b{(A', ~')pyr(t..", ~")ex
= b l + b2 [.L + b3Aq, = (b [.L" + b - b A" q)K qa, i 3 2
(132)(124)
= (134)
(132)(142)
= (13)(24)
(132)(134)
= (14)(23) = (243)
(132)(143)
= (142) (132)(243) = (12)(34). (132)(234)
0
B) If A4 = <x, y) with o(x) = o(y) = o(xy) = 3, then o(xyx) = 2. PROOF. If o(xyx) =1= 2, then o(xyx) = 3. But then, by A), we obtain the contradiction 2 = O(y-Ix-IX) = O(y-I) = 3. 0 1 y C) If As = (x, y) with o(x) = o(y) = 3, then <x, yX - ) ~ A4 and 1 yXy - = X -lxY. Also o(xy - yX) = 2. 1
We may assume x = (123) and y = (345). Then yXy = (345)( 123)(354) = (145)<354) = (134) proving (x, yXy- 1 ) ~ A 4 • Moreover x -IX Y = (132)(123)(345) = (132)(124) = (134). Finally (123)(354)(145) = (12)(34). 0
PROOF.
= - b2A + b3 , =
161
33. A Characterization of As
[bl([.L"KqA' - A") + b2KqA' - b 3 (1 + A"qKqA')] ex.
Thus we get in particular [.L = Kqex, Aq = -A"qKqa, -A = KqA' ex, 1 = - a A"qKqA' ex. The last two equations yield a = M"q - 1. From the first and second equation we obtain Aq = - A"q}.L. Hence a = - AI +q}.L -1 - 1. Now A and [.L are linked by [.L + [.Lq = -AI+q. Therefore ex = (}.L + [.Lq)}.L-I - 1 = }.L q-I. Using the first equation once again, we find ex = }.L q-2Kqa whence 1 = [.Lq-2 Kq. As [.L can be any element in GF(q2)* and as (q2 - 1,q - 2) = (q2 _ 4 + 3,q - 2) = (3,q - 2) = (3,q + 1), we see that py E H' for all p E Hp,Q' This yields y = ly E H' whence Hp,Q ~ H'. Thus Hp ~ H'. As this is true for all P, we have H = H'. Since Hp is soluble, the simplicity of H now follows from Iwasawa's theorem (22.5).
D) Let As PROOF.
= (x, y) with o(x) = o(y) = 3. Then o(xy) = 5 and o(yyX) = 2.
We may assume x
= (123)
andy
= (345).
yyX = (14)(35).
Then xy = (12453) and 0
= (x, y) with o(x) = o(y) = 3. !f z E = 2 and o(zy) = 5, then z E <xY, x y ). Let x = (123) and y = (345). Moreover
E) Let As
<x, y) with o(z) = 3,
o(xz)
PROOF. put z = (abc). Then \{1,2,3}n{a,b,e}\=2 and \{3,4,5}n {a,b,e}\= 1. If 3E{a,b,e}, say a = 3, then {b,e} n {3,4,5} = 0 whence {b,e} = {I,2}. Thus z E (x) and hence o(xz) = 1 or 3. Thus 3 fl {a,b,e}. We may therefore assume that {a,b} = {I,2} and e E {4,5}. Hence z = (124), (125), (214) or (215). Using 1 o(xz) = 2 one gets z = (124) or (125). Now x y = (124) and x y - = (125). 0
= <x, y) and o(x) = o(y) = o(z) = 3. If (xz)2 = 1 = (zy)2, l then z=yxy or z=yX- y- • In the first case (z,y)~A4' and in the second (z, x) ~ A 4 •
F) Let z E As
33. A Characterization of As
Let x = (123), Y = (345), and z = (abc). = 2 = 1{3,4,5} n {a,b,e}\. Hence 3 E {a,b,e}, {b,e}1 = 1 = 1{4,5} n {b,e}l, whence
PROOF.
Before we state the theorem characterizing As, we shall make a few observations about A4 and As·
A) Let (x, y) = A4 and o(x) o(x-y) = 2.
=
o(y)
=
3. Then o(xy)
= 3,
if and only if
z
Then \{I,2,3} n {a,b,e}\ say a = 3. Then \{I,2} n
E {(314), (315), (324), (325), (341), (351), (342), (352)}.
162
V. Planes Admitting Many Shears
Computing xz for all these cases and using (XZ)2 = 1 yields z E {(314), = 1 yields z E {(315),(342)}. On the other handyxy = (315) andyx-y-' = (342). 0
(315), (342), (352)}. Finally (zy)2 G) If As
= (x, y) and o(x) = o(y) = 3, then (xY, yX) ~ A4 ~ (x Y, yX-I).
PROOF. We may assume x = (123) and y = (345). Then x Y yX = (145). Finally yX- = (245). This proves G).
= (124)
1
and 0
H) If A4 = (x, y) with o(x) = o(y) = o(xy) = 3, then (xy -I, y -IX) is the elementary abelian 2-group of order 4 of A 4· PROOF. Let x = (123) and y = (142). Then xy -I = (14)(23) and y -IX = (13) (24). 0
33.1 Theorem (Assion). Let G be a finite group and let D be a conjugacy class of elements of order 3 of G. Assume furthermore: (1) G = (D). (2) If a,b E D and b tt. (a), then (a,b) ~ A4 or As' (3) There exist a,b E D with (a,b) ~ As·
Then G
~
As.
PROOF. As all elements of order 3 in As are conjugate in As, it follows from (3) that a E D implies a -I E D. From this and (2) we infer that for all a,b E D the set (a,b) n D consists of all elements of order 3 in (a, b). We shall make continuous use of these remarks without mentioning them. (a) Let x,y,z E D be such that (x-~i = (x-Izi = (y-Izi = 1 and 4 z tt. (x, y). Then (x, y,z) is a Frobenius group of order 2 .3 with Frobenius kernel (xy-l,y-IX)(xZ-I,Z-I X ) and Frobenius complement (x). I Put u = x-yz. As (X- IZ)2 = (y-Izi = (x-yi = 1, we have zx= xz-I, zy-I = yz-I, and x-~ = y-I X. Therefore
u 2 = x-yzx-yz = x-yxz-~z = x-~xz-yz-Iz-I = X-~xz-Izy-Iz-I
= y-Ixxy-Iz-I
=
y-Ix-y-Iz-I
=
(zyxy)-I.
This implies u 2 =1= 1, as z tt. (x, y). Assume o(u) = 5. Then u 3 = u- 2 = zyxy. Furthermore y-I = x-yx-I, as (x-Iyi = 1. Therefore y-Ix-y-I = x-yx- 3yx- 1 = x-y-IX-I. This and B) imply y - IX I = xyx, whence u 2 = xyxz - I. Thus
Y-
3
zyxy=u =xyxz
-I
-I
-I
X yz=x(z)
x-y-I
y
-I
163
33. A Characterization of A 5
a contradiction to B). Hence o(u) =1= 5. As (y - IZ )2 = I, it follows from A) that yZED. This together with 0(u)=I=5 implies (x-l,yz)~A4' Therefore o(u) = 3. From this and A) we obtain o(xyz) = 2. Hence 1 = xyzy-Ix-y-Ix-I z , as o(xy) = 3. But x-y-Ix-I = xyx and zy-I = yz-I. Thus 1 = xy2z-lxyxz and hence xy-I = yx- I = (xyxy. Now (xy-I)2 = xy-Ixy-I = xx-yy-I = 1. Similarly (xz- I? = 1 and (yz -1)2 = 1. Therefore, replacing x, y, z by x-I, Y -I, Z -I in the above argument, we obtain y-I x = (X-y-Ix-Iy-I, i.e., y-I x = (xyxy-I. From this and H), we deduce that M=(xy-I,y-I X ) is a normal subgroup of (x, y,z) = H. As (X- Iz)2 = (X-y)2 = I and z-yz-y = z-I zy - = 1, we may interchange the roles of y and z in the above argument. Therefore N = (xZ-I,Z-I X) is a normal subgroup of H. Obviously H = MN(x), whence IHI = 22 +$.3 with 1 ~ s ~ 2, as z tt. (x, y). The group (x, y) contains 4 Sylow 3-subgroups of H. As (z) (x, y), the group H contains more than 4 Sylow 3-subgroups. Since the number of Sylow 3-subgroups is congruent to 1 modulo 3, the number of Sylow 3-subgroups of H is 16. Hence IHI = 24 .3 and H is a Frobenius group with Frobenius kernel MN and Frobenius complement (x). (b) Let a,bED and assume (a,b)~As' If eED is such that (a,e) ~ (b,e) ~ A 4, then e E (a,b). Replacing a by a - I if necessary and b by b - I, we may assume by A) that (aei = 1 = (bef Assume e tt. (a,b). Then
y
ct
2
(ab - Ie) = ab -Ieab - Ie
= ab - la - laeaee - Ib - Ie = ab - la - Ie - Ib - Ie
= ab -I a -Ie -Ib -Ie -Ib -lbe 2 = ab -la -lbe 2 =1= 1. It follows from A) that b-Ie E D. Hence (a,b-Ie) ~ A4 or As. Assume (a,b-Ie) ~ A 4 . Then (ab- Ie)3 = 1, since (ab- Ie)2 =1= 1. Therefore we have ab-Ieab-Ie = e-Iba- I. From (ae)2 = 1 = (be)2 we obtain ea = a - Ie - 1 and be = e - Ib - I. Therefore e - Iba - I = ab - la - Ie - Ib - Ie = ab- Ia- Ibe 2 = ab-Ia-Ibe-I, whence (ba-Iy = (b-Iy-' b . But o(ba- I) = 5 by D) and 0«b-Iy-1b) ~ 3. This contradiction proves (a,b-Ie) ~ As· As (a,b-Ie) ~ As, we obtain from D) that o(b-Ie(b-Iey) = 2. Since (b-Ie)a = a-Ib-1ea = a-1b-Ia-1e- 1, we therefore have 1 = b-Ie(b-Ie)ab-Ie(b-Ie)a = b-Iea-Ib-Ia-Ie-Ib-Iea-Ib-Ia-Ie-I =
b-Iea-Ib -I a -lbe 2a- Ib -Ia-Ie-I
( _I)ba c b -I = b - Ica a -I e -I
z.
This yields
(z-If- y - ' = x-Izyxyz-y = z-Ixyxzy-y = (xyx)Z,
=
b -Iea-Ib-Ia-Ibaeb -I a -Ie-I
b -I ca ( _I)bac b -I a -Ibbe - I -I
=
I ba c (a - I) bcb b - Ic (a -)
=
b-Ie(a-I)ba(a-I)be-'(b-Ic)-I.
=
=
I ba (a -) I be - I c 2b b - Ie (a -)
164
V. Planes Admitting Many Shears
Hence abc-I = (a-I)ba. Moreover (b-lay = e-lb-lae = bee-la-I = ba- I. As (aba,b-Ia) is a subgroup of (a,b) generated by an element of order 5 (by D)) and an element of order 3, we have (aba,b-1a) = (a,b). Thus (a,b) is a normal subgroup of (a,b,e). We deduce that (a,b) n (a,b-Ie) is normal in (a,b-Ie) ~ As. This proves (a,b) = (a,b-Ie) and hence e E (a, b). This contradiction establishes (b). (c) Let a,b,e E D be such that (a,b) ~ (a,e) ~ As and (b,e) ~ A 4 • Then e E (a,b). We may assume by A) that (bei = 1. Assume furthermore e fl (a,b). Then b-Ie E D by A). Therefore (b-Ie,a) ~ A4 or As, as b-Ie fi. (a), since otherwise e E
(b- Ia- Ia VI )2 = b-la-lbab-lb-Ia-Ibab-I
= b-Ibabbab- I
. b-\bb a)2b- 1 = b 3 = 1. A S a - I a C = e ac - I an d a C = a b - I ,we th us h ave ( b - I eac - 1)2 = 1, i.e., ((b- Iye a)2 = 1. Moreover (b-Iy = (e- lb- I)2be 2 = be- I whence (eb- I, ea) ~ A4 and ((eb-I)-Ieai = 1. Hence H = (e-I,eb-I,e a) is a Frobenius group of order 24 '3 by (a). Moreover b E H whence (b, ea) ~ A 4. As I (ea,a C) ~ A4 by G), we have e a E (aC,b) by (b). As a C= a b- , we have 1 (aC,b) = (ab-I,b) = (a,b). Hence e E (aC,b)a- = (a,b). This proves I that
34. A Characterization of Galois Fields of Odd Characteristic
165
I Therefore a-laC = e ac - E (a,eb- I), i.e., a C E (a,eb- I). Moreover (aa c)2 = 1 and a ceb - I = e -lac -Ib - I = (aby has order 5. Therefore, by E), a C E {aCb-l,abC-I}. If a C= acb-I, then bE (a C) ~
34. A Characterization of Galois Fields of Odd Characteristic All algebras to be considered in this section are algebras with lover fields of characteristic =t= 2. Algebras are not assumed to be associative unless otherwise stated. Moreover, only those homomorphisms are considered which map 1 onto 1. Let K be a field and let 1 be a K-algebra. We shall call 1 a lordan algebra if a 0 b = boa and ((a 0 a) 0 b) 0 a = (a 0 a) 0 (b 0 a) for all a, bEl. Examples of Jordan algebras are easily obtained as follows: Let A be an associative algebra. Then we define a new operation 0 on A by aob = ~(ab + ba) for all a,b EA. It is easily checked that A(+, 0) is a Jordan algebra. This Jordan algebra is usually denoted by A +.
166
V. Planes Admitting Many Shears
The Jordan algebra J is called special, if there exists an associative algebra A and a monomorphism of J into A +. Let J be a Jordan algebra over K and let A be an associative K-algebra. If a is a homomorphism of J into A +, then the mapping a is called an associative specialization of 1 into A. Let 1 be a Jordan algebra and let s be an associative specialization of 1 into V. The pair (V, s) is called a special universal envelope of 1, if for all pairs (A, a) where a is an associative specialization of J into A there exists a unique homomorphism 71 of V into A with a = S7J.
34.1 Theorem. Let J be a Jordan algebra and let (V,s) be a special universal envelope of J. Then we have: a) If (V', S') is a special universal envelope of J, then there exists exactly one isomorphism t from V onto v' with s' = st. b) V is generated by 1 s . c) There exists exactly one involutory antiautomorphism a of V with x sa = X S for all x E J. d) If rankK(J) = n < 00, then rank K ( V) < 2n - l • PROOF. a) As (V,s) and (V',S') are special universal envelopes of 1, there exists exactly one homomorphism t of V into V' and exactly one homomorphism t' of V' into V with s' = st and s = Sit'. Hence s = stt' and s' = S'!'t. This yields tt' = id u and t't = id u ' by the universality of U repectively V'. Hence t is an isomorphism. This proves a). b) Put V = <1S). Then s is an associative specialization of 1 into V. Hence there exists a homomorphism 71 of U into V with s = S71. As V t;;;;; U, the homomorphism 71 is in fact an endomorphism of U. Hence, by universality, 71 = 1, i.e., U = V. c) Let f} be the opposite structure of U. Then JL = 1 is an antiisomorphism from V onto V. This yields that JL is an isomorphism of V + onto f}+. The mapping (3 = SJL is a homomorphism from J into f}+. Hence there exists a homomorphism y of V into f} with SJL = (3 = sy. Put a = Yil- I . Then a is an antiendomorphism of V with s = sa. This yields s = sa = sa 2 and hence a 2 = 1 by universality. This proves the existence part of c). The uniqueness part follows from b). d) Let {l, b l , . . . , bn - I } be a K-basis of J and put
V
= { 2.: xiki I Xi
=
Yi~ ... Y/ rj 'Yij E 1, k i E K }.
Then V is a sub algebra of V and hence V = V by b). As every Yij is a K-linear combination of 1 and the b/s, it suffices to show that every product of the form bi( l)bl(2) . . . bi(l) is a K-linear combination of F and the elements b:c I) • . . b:(r) with 1 < a(l) < a(2) < ... < a(r) < n - 1. Let N be the number of pairs (k, I) with 1 < k < I < t and i(k) > i(l). Assume that t is minimal with respect to the property that bi( I) . . • bi(t) is not a K-linear combination of the required form. Moreover choose the
34. A Characterization of Galois Fields of Odd Characteristic
16/
product for the given t in such a way that N is minimal. Then N > 1. Hence there exists k E {I, 2, ... , t} with i(k) > i(k + 1). Assume i(k) = i(k + 1). Then bi(k) = bi(k+ I)' Hence (bi(k) 0 bi(k+ I)Y = bi(k)bi(k+ I)' Now bi(k) 0 bi(k+ I) = ~;~ci bjkj , where bo = 1. This yields n-\
bi(I)' .. bi(l) =
2.:
(bi(I)' .. bi(k-l)b/b i(k+2) ... bi(t))kj .
)=0
Using the minimality of t, we obtain that each summand of the right hand side and hence the left hand side of the above equation is a K-linear combination of the required form. This contradiction proves i(k) > i(k + 1). Therefore
and hence
As bi(k)
0
bi( I)
bi(k+ I) is a K-linear combination of I and the b/s we have •..
bi(l) = - bi(l) ... bi(k-I)bi(k+ l)b i(k)b i(k+2) ... bi(l) + L,
where L is a linear combination of the required form by the minimality of t. From the minimality of N we duduce that
bi( I) . . • bi(k-I)bi(k+ l)b i(k)b i(k+2) ... bi(l) is also of the required form thus reaching again a contradiction.
0
34.2 Theorem. Let 1 be a 10rdan algebra. Then 1 possesses a special universal envelope. PROOF. Let T be the tensor algebra of 1 considered as a vector space over K and let I be the ideal of T generated by the elements a 0 b - tea ® b + b ® a) with a,b E 1 and IJ - I where IJ is the unit element of 1 and 1 is the unit element of T. Put U = T / I and let s be the mapping defined by as = a + I. We show that (V,s) is a special universal envelope of 1. First of all we note that s is linear. Moreover
(a 0 b f
=
a0 b+ I
=
t (a 0
b + b 0 a) + I
and
1+ I
l~ =
by the definition of I. Hence s is an associative specialization of 1 into V. Let a be an associative specialization of 1 into A. In particular, a is a K-linear mapping of 1 into A. Therefore there exists a homomorphism 8 of T into A with x 8 = xO" for all x E 1. For a,b E 1 we then have
(a o b - t(a 0 b + b ® a))8= (ao b)8- t(a 8b 8 + b 8a 8) = (a o bt- !(aO"bO" + bO"aO")
=
0,
168
V. Planes Admitting Many Shears
as a is an associative specialization of J. Hence I ~ Kern(8). Therefore there exists a homomorphism 1] of U into A with (a + 1)'rI = aO. This yields
By (B)
- (x 13. ... x·'" x·'I x·'2 )* E J r (K) ( X'I. ... x.)* 1"
a = S1].
FinallY,1] is unique, as T is generated by I and J. This proves 34.2.
D
If the algebra A admits an involutory antiautomorphism a, then we shall say that A is an algebra with involution a. For an associative algebra A with involution a we put H(A,a) = {xix a E A, x = x}. Obviously, H(A,a) is a subalgebra of A +. Let Fr(K) be the free associative K-algebra in the free generators Xl' . . . , X,. Then F,(K) admits an involution p with (Xii'" xi)P = Xik ... Xii' This involution p is called the reversal operator of Fr(K). Let Jr(K) be the Jordan subalgebra of Fr(K)+ generated by Xl' . . . , x r . Then J,(K) ~ H(F,(K), p). P. M. Cohn proved that Jr(K) =i= H(Fr(K), p) if r > 4. He also proved:
34.3 Theorem. If r <; 3, then Jr(K)
= H(F,(K), p).
=
WI
Denote by W the set of words in the X J' ••• , x r • Then W is a for Fr(K). From this we infer that Wo = {t(w + wP)lw E W} is a for H(Fr(K),p). Hence it suffices to show Wo C Jr(K). For we put w* = t (w + w P). If the length of w is less than or equal to w* E Jr(K). Let w = Xii' . . xin with n > 3 and 1 <; m < n. Put Xii' . . xim and w2 = X im + • • • x in ' Then WI w2 = w. Furthermore 1
4wi
0
w;
=
t(wl + wD(w2 + wn + (W2 + wn(w l + wD)
= t (w + WIW~ + wfw2 + wfw~ + w2w J + w2wf + w~wJ + wP) = w* + (wJwn* + (wfW2)* + (w 2w J)*. By the induction hypothesis, 4wi
(A) For m
0
w; E Jr(K), i.e.,
+ (wlwn* + (wfw 2)* + (w 2w J)* E J,(K). = I we obtain WI = wf and hence (wJw~)* + (wfw 2)* = t(WlW~ + W2Wl + WlW2 + WiWl) w*
= t (Wl(W2 + = 2Wl ° W;' As this is in Jr(K), we have induction we get
( B)
(x II. ... x.)* In
(Xii' . .
w~)
Xi)*
Ik
+ (w2 + wnwl)
+ (Xi2 ... XinXi)*
E
J,(K). By
I
. .. x·In x·II ... x.)* E J (K). Ik r
This yields for n odd 2w* E Jr(K) and hence w* E Jr(K). Thus we may assume that n is even. Take m = 2. Then (A) yields
+ (x
II
= (x. x· x· ... Xi )*. Therefore 2«x . ... x.)* + (X. x· x· ... Xin )*) E J,(K).
Trivially (x In. ... x·I)x·II x·12)* II
x.12x·In ... x·I) )*
+ (x x· x· ... X )* 12
II
I)
12
II
I)
12
In
n
II
I)
Thus
( XII. ... x.)* In
== sgn(l 2)(x. x· x· ... Xi)* '12
II
I)
n
mod Jr(K).
As n is even, sgn(1, 2, ... ,n) = -1. Hence by (b)
== sgn(l, 2, ... , n)(xi2 ... Xi"Xi)* mod Jr(K). ( x 'I. ... x.)* 'n As (1,2) and (1,2, ... , n) generate Sn' we obtain ( x II. ... x.)* In
== sgn('lT)(x.
1,,( I)
x·z,,(2) ... Xi,,(n) )* mod Jr(K)
for all 'IT E Sn' . Since n > 2 and n is even, n > 4> r. Thus there eXIst a,b E {l, 2, ... , n} with a =i= band xia = x ib ' Let 'IT = (a, b). Then, by (C), == - (x II. ... Xin )* mod J,( K). ( x II. ... x.)* In Therefore
(Xii' •.
D
xJ* E Jr(K).
34.4 Theorem (Jacobson). Let J be a Jordan algebra generated by at m~st 3 elements and let (U, h) be a special universal envelope of J. Then ~ IS an epimorphism of J onto H ( U, a), where a is the involution of J havmg the property that ha = h. PROOF. As ha = h, we have Jh C H( U, a). Let Yl' ... ,Yr generate J. If Fr(K) is the free K-al~ebra with the free g~nerators ~J' • '0' ,:",X'h then ther~ exists a homomorphIsm a from F (K) mto U WIth Xi - Yi for all l. Obviously a is also a homomor;hism of F,( K) + into U + and the , h hUh restriction of a to J,(K) maps Jr(K) onto J . As J generates , t : oa homomorphism a is surjective. Moreover, if X E J,(K) then x = X = xpa. Hence aa = pa, as Jr(K) generates F,(K). Pick b E H( U, ex). Then there exists a E Fr(K) with a O = b. Therefore
b
+ (- l)k-I(X. +
(x II. ... x In )*
and
(C)
PROOF. K-basis K-basis w E W, 2, then
169
34. A Characterization of Galois Fields of Odd Characteristic
By 34.3, J,(K)
= t(b + b a )
= t(a O
+ a Oa ) = tea + aP)o. h
= H(F,(K), p). Thus H( U, a) ~ J,(Kt = J .
o
If J is a Jordan algebra, if a E J and if n E No, then we define ao inductively by aO o = 1 and ao(n+ 1) = a on 0 a.
In
34.5 Lemma. If J is a special Jordan algebra, then J is power associative.
n
170
V. Planes Admitting Many Shears
34. A Characterization of Galois Fields of Odd Characteristic
171
PROOF. We may assume J ~ A + for an appropriate associative algebra A. Then aO o = 1 = aD. Assume ao n = an. Then
inverse. Jordan division algebras may have zero-divisors, however. Every Jordan division algebra 1 is special by 34.2, as {O} and J are the only ideals of J.
Hence a on =
34.8 Theorem (Albert). If J is a finite Jordan division algebra (of characteristic not 2), then J is a Galois field.
for all n. From this we infer
an
o
PROOF (McCrimmon). As J is special, J is power associative by 34.5. Hence the sub algebra generated by the element a is assocIatIVe. Moreover, by 34.7, a - I is a polynomial in a, as J is finite. Hence
34.6 Lemma (McCrimmon). Let A be an associative ring with involution a.
a-IE. Let a,b,c E J and put J o =
Let A be generated by H(A,a) and assume that H(A,a)\{O} consists only of units. If {O} and A are the only a-invariant ideals of A, then A is a skewfield or A = D EEl D a where D is a skewfield.
PROOF. Let 2r be the set of all elements of A which do not have a right inverse and define 2, similarly. Then ZrA S 2r and A 2, C; Z,. Let z E 2r and put H = H(A,a). Then zza E Zr n H = {O}. Hence zza = O. This yields z E Z" i.e., Zr S 2,. Similarly Z, C; Zr' i.e., Zr = Z, = Z. If Z = {O}, then A is a skewfield. Assume 2 =1= {O}. As za = Z and A 2 C; Z as well as ZA s Z, there are x, y E Z with x + y ft. z. Otherwise, Z = A and 1 E Z, a contradiction. Therefore, x + y is a unit. Put e = xu- I I andf= yu- , where u = x + y. Then e,f E 2 and 1 = e + f. Furthermore eO:e E Z n H = {O}. This yields e a = eO(e + j) = ej. Similarly fa = fOe. Hence f = eO:f = ea. This proves 1 = e + ea. Moreover eHe a, eaHe C; Z n H = {O}. Hence H = IHI = (e
This implies A
+
ea)H(e
+ ea) s eHe + eaHe a C; eAe + eOAe a. + eaAeO:. Therefore A = eAe + eO:Ae a. Pick
34.7 Lemma. Let A be an associative algebra over a field of characteristic not 2. If a, b E A are such that 1 = ~ (ab + ba) and a = ~ (a 2b + ba 2 ), then a is a unit in A and a-I = b.
PROOF. We infer from 1 = ~ (ab + ba) that ~ (a 2b + aba) = a = ~ (aba + 2 2 2 ba ). Hence a b = ba • Therefore a = ~ (a 2b + ba 2 ) = a 2b = ba 2 • This yields 2 ab = ba b = ba. Thus I = ab = ba. 0 Let J be a Jordan algebra and let x E J. The element y E J is called a Jordan inverse of x, if x ° Y = 1 and (x 0 x) ° y = x. The Jordan algebra J is called a Jordan division algebra, if every non-zero element of 1 has a Jordan
0
associative.
34.9 Corollary (Glauberman). Let V be a finite vector space over GF(pS) where p is an odd prime. If B C; EndGF(ps) (V) satisfies:
a) 1 E B, b) If 0 =1= x E B, then x E GL(V) and c) B + B C; B, then BB
s
Band B
~
X-I
E B;
GF(pr) for some r.
PROOF. We first show that xyx E B for all x, y E B. For this we may assume x, y =1= 0 and x =1= y - I. Put z = Y - I - x and w = x - I + z -I. Then xwz = x(x- I + Z-I)Z = z + x = y-I. Hence w = x-y-IZ-I. This yields x - w- I = X - zyx = x - (y-I - x)yx = xyx. As x - w- I E B by b) and c), we have xyx E B for all x, y E B. Next we show xy + yx E B for all x, y E B. This follows from a), c) and the remark just made, since xy + yx = (l + x)y(l + x) - Y - xyx. As B( +) is an elementary abelian p-group, this yields that B is a sub algebra of EndGF(p) (V) +. Moreover, B is a Jordan division algebra, as x o X-I = 1 and (x o x)o X-I = x. Hence B(+, 0) is a Galois field by 34.8. In particular (B \ {O} )( 0) is cyclic. Let a be a primi tive elemen t of (B\{O})( 0). Then aO i = a i . From this we infer that 0 = '. 0
172
V. Planes Admitting Many Shears
35. Groups Generated by Shears In this section, we shall follow Hering 1972 b and Ostrom 1970 a, 1974. Let V be a vector space of rank 2r > 0 over a skewfield D and let v be a ?arti~l spread of V. Furthermore, let G be a subgroup of OL( V) leaving mvanant v. For 01=- X C; G we define Vx by Vx = {v Iv E V, V O= v for all a E X}. If X ~ {a} then we ~hall writ; ~o inste~d.of V{o}' We shall call a EGa v-shear, If Vo E v and If (x - 1) IS the mmlmal polynomial of a. Moreover, Va is called the axis of a. It is convenient to also call the identity a v-shear, every W E v being an axis of it. 35.1 Lemma. If 1 1=- a E G and a
if a
is a v-shear, then the following is true:
1
a) V - = Yo' b) o(a) = p, where p is the characteristic of D. c) If W E v and WO = W, then W = Yo' PROOF.
a) As v(a-l)2
2r = rk( V) = rk( va-I) 1 )
= rk(V a -
PROOF. By 3S.1 b), we need only show that S is an abelian group. Let a,T E S\{I}. Then W C; VOT . Moreover, v o- 1 = Vo = W= VT = V T- 1 by 3S.1. Therefore, v(a-l)(T-l) = V(T-l)2 = {o} = V(T-I)(a- I). This yields (a - 1)(T - I) = 0 = (T - 1)(a - 1). Hence aT = a + T - 1 = Ta. Moreover, aT - 1 = a-I + T - 1, whence vaT-I C vo- I + V T- 1 = W C; VaT'
From this we obtain (aT _1)2 = O. Hence, if W = VOT then aT E S. Assume VaT 1=- W. As Ivl > 2, there exists V E v with V1=- W. Then V = V ttl Wand hence V01' n V1=- {O}. This yields vaT = V. Therefore, V OT - 1 C; V. On the other hand, V o1'-1 C; V OT - 1 C; W. Hence Vo1'-l C; V n W = {O}. This proves aT = 1. Thus SS C; S. Finally, if a E S then obviously a -I E S. 0 35.3 Lemma. Let 'IT be a spread of V. If a is a collineation of 'IT( V), then the following statements are equivalent:
we have v o- 1 C; Kern(a - 1). Moreover
= {O},
173
35. Groups Generated by Shears
a) a is a shear of 'IT( V) fixing O. b) a is a 'IT-shear.
+ rk(Kern( a-I))
Assume a). Then a E fL( V, K), where K is the kernel of 'IT( V) by 1.10. As a fixes a component X of 'IT pointwise, a is linear. This yields that a is also D-linear, as D C; K. Moreover, (X + vt = X + v, since a is a shear with axis X. Therefore, vo- I c;: X. Furthermore, X a - 1 = {O}, whence (a - 1)2 = 0. As Vo = X E 1T, it follows that a is a 'IT-shear. Hence a) 0 implies b). The converse is trivial. PROOF.
+ rk(Va) = rk(VO- 1) + r.
Hence rk( va-I) = r = rk( Va) proving v o- 1 = Kern(a - 1) = VO' b) If P > 0, then a P - 1 = (a - lY = (a - ly-2(a - 1)2 = O. Hence a P = l. As a1=- 1, we have o(a) = p. Assume p = 0 and let o(a) = n > O. Then o(a) > 2. Therefore
From now on we shall assume that V is finite and that Ivi ): 2. Let p be the characteristic of D. Moreover, denote by ® the set of all nontrivial v-shears in G and assume ® =1= 0. For a E ® we put Therefore, (x - 1)2 divides 1 +
(- It + 227: l (- lY(7)x n -
i
and hence
n-i
(x - l)n_ 1- (-1/- i~i (-I)i(7)x n- i = xn
+ (-1/-1- (_I)n
S ( a) = {I} U {T IT E ®, V1' = Va } and L= {S(a)laE®}.
= xn - 1. i
As xn - 1 = (x - 1)[227:ci(x - 1) + n], we obtain the contradiction n == 0 mod (x - 1). This proves b). c) Let p be the restriction of a to W. Then (p - 1)2 = O. Hence Wp-I ~ Kern\p - 1). As rk(Wp-i) + rk(Kern(p - 1» = rk(W) = r > 0, we therefore obtam Kern(p - 1) {O}. From this we infer W n Vo 1=- {O}, as Kern(p - 1) C; Kern( a - I) = Va' Therefore, W = VO' as v is a partial spread. 0
*
35.2 Lemma. ~et W E ~ and put S = {1} U {a I a E G, Vo = W, (a - 1)2 = Then S IS an. abehan group provided Ivi > 2. If the characteristic p of D IS not 0, then S IS an elementary abelian p-group.
?}.
1. . [I
'.,.:,r
ij
'!
Finally, L
= (®).
By 3S.2, the Sea) are elementary abelian p-groups.
35.4 Lemma. a) If S E L, then 9(L(S) n Sx = {I} for all x E L\9(L(S), b) The elements of L are conjugate in L. PROOF. a) If ILl = 1, then there is nothing to prove. Hence let ILl> 1 and let Rand S be distinct elements of ~. If a ERn 9(L(S), then a fixes VR and Vs by 3S.1 c). Using 3S.1 c) once more, we find a = 1. b) follows from 3S.1 c) and IS.1. 0
Let R, S be two distinct elements of L: and let 1 =1= a E Rand T E S. Then V = VR ttl Vs and V; is a diagonal of VR ffi Vs' By 2.1, we may
174
V. Planes Admitting Many Shears
assume that V = U EB U and V( 00)
(0, xt
=
VR ' V(O) = Vs and V(l)
=
V;' Then
= (0, x) and (x,Ot = (x, 0) + (x, 0),,-1. As a-I is an isomorphism
175
35. Groups Generated by Shears
Furthermore
Or
from Vs onto VR , we see that (x, = (x, y) for some y. Moreover (x, 0)" E V(l). Hence (x, 0)" = (x, x). Therefore (x, y)" = (x, x + y) for all x, y E U. Furthermore V( 00 Y E v. Thus there exists ex E GL( U) with V(ooy = {(x(\x)lx E U}. As 7' - 1 maps V R onto Vs ' we have (O,xY = (XO,x) and hence (x,yY = (x + yO,y) for all x,y E U. For a, b, c, d E End K ( U), we identify the mapping A: (x, y) ~ (x a + ye, X b + Y d) with the matrix (~ ~). The mapping A ~ ~) is an isomorphism of all these mappings onto the ring of all (2 X 2)-matrices over End K ( U). We shall keep these notations fixed.
e
35.5 Lemma. Let R, S be distinct elements of ~ such that (R, S) ;;;:; SL(2, q), where q is a power of p. Then there exists a field F S; End K ( U) such that IFI = q and
(R,S) =
{(~ ~)la,b,c,d E
F,ad- cb =
I).
An easy computation then shows
(b)
1 - t'(~ -I)t(~)
=0
for all
(c)
(~ -r r C~) -t~) I
=
We infer from (b) and t'(l) = 1 that t(l)
= 1= -
=
1)(0 - 1) = 1.
t';~) )
with t'(~) E End K ( U) for all ~ E GF(q). Similarly (~ ~y = 0(0 with teO E End K ( U) for all ~ E GF(q). As i is an isomorphism, we obtain that t and t' are additive. Moreover ((1) = 1, as (6 IY = o. Let 0 =/= ~ E GF(q). Then
?)
U n(~
I )
t( - 1). Moreover
Hence
PROOF. Let R* be a Sylow p-group of (R, S) containing R. As all subgroups of order p in SL(2, q) are conjugate, R* consists only of v-shears. This yields R* = R, i.e., Rand S are Sylow p-subgroups of (R, S). This implies that there exists an isomorphism i of SL(2, q) onto (R, S) mapping {(6 f) I~ E GF(q)} onto Rand {(j i) I~ E GF(q)} onto Sand (6 i) onto o. Let pER. Then V(oy E v. From this we infer the existence of t E End K ( U) with (x, yy = (x, X I + y) for all x, y E U. Hence we have
(~ ~ r (~
-
C~ ~)(~ -r')=(~ ~~,) - t
+ (7' -
GF(q)*.
Hence by (a) and (b)
Moreover
Finally, there exists 7' E S with (0 - 1)(7' - 1)
~ E
_~-I)U n=(~ ~~').
(O~) - I )
= ( t
0)
(O~)
t (~) -
1
From this we obtain t(~1]) = t(~)t(1]) for all ~,1] E GF(q). Hence t is an isomorphism from GF(q) onto F= {t(~)I~ E GF(q)}. This proves the first assertion. Finally, if 7' = (: ?), then (0 - 1)(7' - 1) + (7' - 1)(0 - 1) = 1, as is easily seen. 0 35.6 Lemma. Assume I~I > 1. If there exists n EN such that (X, Y) ;;;:; SL(2, p n) for any two distinct X, Y E ~, then L ;;;:; SL(2, p n).
c:
a PROOF. Let R, S, 0, U and F be as above. Pick ep E 6. Then ep = 1 :d) with a,b,c,d E EndK(U). Moreover, bE F, as we shall prove now. If ep E S, then b = O. Hence we may assume ep ti S. Then S( ep) =/= S. Applying 35.5 to Seep) and S, we see that there exists pES with (p - l)(ep - 1) + (ep - l)(p - 1) = 1. As pES, we have p = (~ ~) with x E F. Thus
(~ ~) = (~ ~)(~
0)
~)(~
0
+( b) = (bX d dx o xa whence 1 = bx = xb and hence b = X-I E F. Let Q E ~\{R}. We have to show Q S; (R,S). By 35.5, there exists cp E Q with (0 - l)(ep - 1) + (ep - 1)(0 - 1) = 1. Let ep = e~a l:d)' Then bE F, as we have seen. Moreover
Therefore
(a)
-
t'(~
- I)
)
•
Therefore, c
= 1 and a = -
I - t'(~ - I)t(~) .
0=
(~
d. As ep is a v-shear, (ep - 1)2
b)2 = -a
(a
2
+b
0
ab - ba). b
+ a2
=
O. Thus
176
V. Planes Admitting Many Shears
35.8 Lemma. Let v be a partial spread of V = U EB U with V(O), V(1), V(oo),V(a)Ev, where aEGL(U). If the mapping p defined by I (x, y)p = (x a- , ya) leaves v invariant, then V(a i) E v for all i EN.
This yields b = - a 2 • Hence
q;=(l~a If x E F, then ~ =
(6 f) E
L. Now
i
By the remark made above - (a - X)2 E F for all x E F. In particular 2 a , (a - If E F. Hence 2a = a 2 - (a - 1)2 + 1 E F. Consequently a E F if p =1= 2. Mor~over, (1 + a)(1 - a) + a 2 = 1. Therefore, q; E
°
I:' :!
(6 ~ )( -i
V(a<) and V(al) are in v. Let n n. Then
PROOF.
-(a-x)2). l-a+x
i' ii,
177
35. Groups Generated by Shears
~)(6 ~)=(-~
6)
and hence all elements of the form
(-~ ~)( -; ~)(~ -6)=(6 ~ ). All these elements belong to R. Hence
<
V(a n - 2)P= {(xa-l,xan-2a)lx E U}
=
> 1 and
assume V(a
{(x,xan)lx E U}
i
=
)
E
v for all
V(a n ).
0
35.9 Lemma. Assume p > 2 and let a and l' be two v-shears with aT =1= Ta. Then either
(I)
(O,T) C;;
{(~ ~)Ia,b,e,d E A, ad - be =
I}.
Denote by /-t the set of all components of v which are axes of non-trivial v-shears belonging to
°
(6
178
V. Planes Admitting Many Shears
K';;;(, GF(9) and V(1), V( - I) E fl. Moreover, Case 2:
V(OO)(T) V ( 1) (T)
=
H';;;(, SL(2, 5).
{V(oo), {(xa,x)lx
= {
E
As a 2
U}, {(x 2a ,x)lx
= (6 E
-:),
we have
U}}.
V (1 ), { ( X I + a, x) I X E U}, { ( X I + 2a, x) I X E U} },
V(-l)(T)= {V(-l),{(x-l+a,x)lxE U},{(x- I +2a ,x)lxE U}}. This implies M d {~ + a'l] I~, 'I] E F}. Let m be the minimal polynomial of a over GF(3). Then meA) = O. Hence degree(m) > 2. Therefore, 1 and a are linearly independent over F. Hence 9 = I + a'l] I~,'I] E F} I " IM I. As SL(2,5) contains exactly 10 Sylow 3-subgroups, IMI" 9. Hence M = {~+ 'l]a I~, 'I] E F}. This proves (4). ~) Let G be the subgroup of GL( V) which leaves invariant 11. Then (! I) E G for all a E M. This follows immediately from (4). (6) E G. As 1 E M, we have p = ?) E G by (5). Furthermore a - I = (b - :). Hence = (: ?) E G. (7) -a- I EM. Put p = Then V(oyP E 11 by (6). On the other hand V(oyP = {(-xa-l,x)lx E U). Hence -a- I EM. (8) f) E G for all a E M. This follows from (6), (5), and C 7 b)(-! ci) = (b f).
a
(7 -6)
(7 -6)
c:
7)(6 -:)(:
(? -6).
(6
?)(? -
(9)
n(?
This follows from (5), (6), (7), (8), and (6 ~)(_~-I ?)(b -II) = (~ ~-I). (10) M = A ';;;(, K. By 35.8 and (9), V(a- i ) E 11 for all i EN. On the other hand V(a- i ) = {(x,xa-')Ix E U} = {(xa',x)lx E U}. Hence a i EM for all / As M( +) is a group, A = GF(p)[a] c: M. Let m be the minimal polynomial of a over F. Then meA) = O. Hence degree(m) > [K: F]. This ~ields IAI > IKI· On the other hand, IMI = IKI. Therefore, M = A. Thus M IS a subalgebra of End F ( V). Hence, by (2), M is a field isomorphic to K. aNbow we can finish the proof of 35.9. By (1), H c: 1= {C d) I a, b, e, d E A, ad - be = I}. By .Q 0), I = SL(2, A). If Jj ~ SL(2, K), then H = I by (10). Assume p = 3 and H ~ SL(2, 5). Let ~ E H and assume that ~ induces the identity on W. Obviously ~ = (g ~) with a EA. As A = M, we infer from (2) and 1 E M, that a = 1. Hence Jj = H. 0 35.10 Theorem (Hering 1972b, Ostrom 1970a, 1974). Let V be a vector
space of rank 2r over GF(p), p being a prime, let v be a partial spread of V
179
35. Groups Generated by Shears
with Ivl > 2, and let G C; GL(V) be a group leaving invariant v. Moreover, let ® be the set of all non-trivial v-shears in G and assume @3 =1= 0, and denote by 11 the set of all components of v which are axes oj shears in @3. If L = <@3), then one of the following holds: a) 1111 = 1 and L is an elementary abelian p-group. b) L ~ SL(2, P S) for some s . Moreover 1111 = p S + 1 and L operates doubly
transitively on 11. = 3 and L ~ SL(2,5). Furthermore, 1111 = 10 and L operates transitively on 11. d) P = 2 and L ';;;(, S(2 2s + I) for some s. In this case 1111 = 22(2s+ I) + 1 and L operates doubly transitively on 11. e) p = 2 and 4 does not divide ILl. Here 1111 is odd and L acts transitively on 11.
c) p
PROOF. If 1111 = 1, then L is an elementary abelian P-group by 35.2. Therefore we may assume 1111 > 1. Case 1: p = 2. In this case, 35.4 determines the structure of L almost completely. In fact: By a theorem of Hering [1972a, Theorem 1], L is isomorphic to one of the groups SL(2, q), Seq), PSU(3, q2), or SU(3, q2), where q is a power of 2, or L = ()(L)S( a) where ()(L) is the largest normal subgroup of odd order of Land S( a) is the group of all shears with axis Va and a is any element of @3. Moreover, Sea) is a Frobenius complement in the latter case. Hence IS(a)1 = 2, as Sea) is elementary abelian. This is case e). If L ~ SL(2, q) or Seq), we obviously have case b) or d). Assume L ~ PSU(3,q2) or SU(3,q2), then
180
V. Planes Admitting Many Shears
Ther; are actually examples of non-des argues ian translation planes of order 3 where the group generated by all shears with axes through 0 is SL(2,5) (Prohaska 1977). Moreover, in the Hall plane H(q) of even order, the. group generat~d by. all those shears is dihedral of order 2(q + 1), as is easIly seen. Also, If m- IS a proper nearfield plane of even order, then the group generated by all shears of m- with axes through 0 is of type e).
CHAPTER VI
Flag Transitive Planes
In this chapter we give Huppert's description of all finite soluble 2-transitive permutation groups and Foulser's description of all soluble flag transitive collineation groups of finite affine planes. Using these and some characterizations of finite des argues ian projective planes involving the groups SL(2, q) and PSL(2, q), we are able to prove the theorem of Schulz and Czerwinski on finite translation planes admitting a collineation group acting 2-transitively on 100 ,
36. The Uniqueness of the Desarguesian Plane of Order 8 I
II IIII! I
III
According to Hall, Swift and Walker 1956, there exists up to isomorphism only one plane of order 8. As this result was found by the aid of a computer, we shall give here a proof for the weaker result that there exists up to isomorphism only one translation plane of order 8. In order to do this, we have to show by 2.6 that there exists up to conjugacy exactly one L C; GL(3, 2) with the properties: (1) 1 E L. (2) If A, BEL and if A =1= B, then A - B E GL(3, 2). (3) ILl = 7. Now IGL(3,2)1 = 23 . 3 . 7. Hence, if A E L, then A 7 = 1. Moreover, if A operates irreducibly on the vector space of rank 3 over GF(2). This yields that the minimal polynomial JlA of A is irreducible of
A
=1= 1, then
181
182
VI. Flag Transitive Planes
= X 3 + X 2 + 1 or /LA = X 3 + X + 1. From this we infer that there are two conjugacy classes of elements of order 7 in 0L(3, 2), degree 3. Thus
/LA
namely
183
37. Soluble Flag Transitive Collineation Groups
trace( C) + I, and trace(A) = trace( C) + 1. Therefore, trace(A) = trace(B), a contradiction. Thus L is a Sylow 7-subgroup of OL(3,2) and m is unique up to isomorphism. 0
Si'o= {A I 0 (A) = 7 and trace( A) = O} and
Si'l
= {A 10(A) =
7 and trace(A) =
37. Soluble Flag Transitive Collineation Groups
1}.
There are exactly 8 Sylow 7-subgroups in GL(3,2) and OL(3,2) acts doubly transitively on the set of its Sylow 7-subgroups.
~ ~ ~]. Then A
Let A =
is the companion matrix of x'
Hence A is an element of order 7. Furthermore, A 3 =
+ x + 1.
[6 1 ~]. 1
1
Therefore, A E Si'o and A 3 E Si'l'
Let B = [
~
r ~ l·
Then B =
r
I
and C =
r
lAB = [ :
~
1
n
In this section we shall determine all soluble flag transitive collineation groups of finite affine planes. At the same time, we shall determine all soluble doubly transitive permutation groups. All the results in this section are due to Huppert 1957 and Foulser 1964. 37.1 Lemma (Huppert). Let V be an elementary abelian p-group of order pr and let G be a subgroup of OL(r, p). If G contains a normal abelian subgroup A which acts irreducibly on V, then L = OF(p)[A] is a Galois field of order p' and G c;: fL(V,L) ~ fL(1, p'). This follows immediately from 9.1.
Moreover,
C' =
[i H c' 0 l
1
C' =
[~
1 0 1
=
[l
n
1
1 0 6
ll'
C =
[~
C' 0 0
1
= [~
il-
H
Computing A - C i for i = 1, ... ,6 yields that only A - C 3 is an element of OL(3,2). Also, among the elementsA 3 - C i , only A 3 - C is in OL(3,2). Thus we have proved 36.1 Lemma. If III and II2 are distinct Sylow 7-subgroups of GL(3, 2) and if 1 =1= A E III' then there exists exactly one B E I12 with B =1= 1 and A - B E OL(3, 2), Moreover, trace(A) = trace(B) + 1. Now we can establish: 36.2 Theorem. If
mis a
translation plane of order 8, then
mis desarguesian.
Let L be a subset of OL(3,2) with the properties (1), (2) and (3). Let A,B E L\{l} and assume that A and B belong to distinct Sylow 7-su bgroups. Moreover, let C E L \ {A, B, 1}. By 36.1, C belongs to a third Sylow 7-subgroup. Moreover, trace(A) = trace(B) + 1, trace(B) =
37.2 Lemma. Let V be an elementary abelian p-group of order p' and assume that 'TT is a p-primitive divisor of p r - 1. If L is a non-trivial 'TT-subgroup and N any subgroup oj OL(r, p) which is normalized by L, then V is the direct sum of isomorphic irreducible N-submodules. PROOF. L, and hence LN, operate irreducibly on V by 10.5. Therefore, by Clifford's theorem (see e.g. Huppert [1967, V.I7.3, p. 565]), V is a completely reducible N-module. Let VI"'" Vs be the homogeneous components of the N-module V. Then LN, and hence L, act transitively on {VI' ... , Vs }' Hence s is a power of 'TT, as s divides ILl. Let I VII = pt. Then IV;I = pI for all i. Therefore st = r, as V = EB~= I Vi' Since 'TT is a p-primitive divisor of p' - 1, we see that r is the order of p modulo 'TT. This yields that r = st divides 'TT - 1. Hence s = 1. 0
37.3 Lemma. Let V be an elementary abelian p-group of order pr and let 'TT be a p-primitive divisor of p' - 1. Moreover, let G be a soluble subgroup of OL(r, p) with 'TT II G I. Assume that all abelian normal subgroups of G operate reducibly on v. If A is a maximal abelian normal subgroup of G, then A = B( G) and IA I divides p - 1. Furthermore, if N is a minimal non-abelian normal subgroup of G, then N has the following properties:
PROOF.
a) N / c;: B( N) c;: A and I N /I = 2. b) N / S(N) is an elementary abelian 2-group of order 2 2m and 4.
1.8(N)1
divides
184
VI. Flag Transitive Planes
Furthermore, we have: c) r = 2m and'TT = 2m + 1 = 12::1, where 2:: is a Sylow 'TT-subgroup of G. d) G /lfG(N /S(N» is isomorphic to a subgroup of the symplectic group Sp(2m, 2). e) IfG(N /S(N»/ A is isomorphic to a subgroup of the direct product of 2m copies of SeN).
PROOF. Let 2:: E Sy1?T( G). Then 2:: ~ IfG(A) by 17.2. Moreover, 2:: n A = {I} by 10.5, as A acts reducibly on V. Therefore A is properly contained in IfG(A). Since A is normal in G, we have that IfG(A) is normal in G. Thus IfG(A) is not abelian, as A is a maximal abelian normal subgroup. Hence IfG(A) contains a minimal non-abelian normal subgroup N. As Gis soluble N is, whence N' 1=- N. Therefore N' is abelian by the minimality of N. From N' C; (;£G(A) we infer that AN' is an abelian normal subgroup of G. This yields N' ~ A by the maximality of A. Therefore N' C; SeN), as N is centralized by A. Hence N is nilpotent of class 2. It follows that N contains a non-abelian Sylow q-group. As this group is characteristic in N, we see that N is a q-group by minimality. From 17.2 and the reducibility of A, we get that IA I divides pI - 1 for some t < r. Therefore, q divides pI - 1, as N' 1=- {l}. Hence q 1=- 'TT. Let U be a normal subgroup of G which is properly contained in N. Then U is abelian and hence U ~ A, i.e., U ~ AnN C SeN). As SeN) is one such U, we obtain AnN = S(N). Moreover, N /S(N) is characteristically simple and hence an elementary abelian q-group. N' and S( N) are cyclic since N' ~ S( N) C; A and A is cyclic by 17.2. Let fl, v E N. Then [f.L, v] = fl-IV -1f.LV commutes with fl and v. Therefore, by Huppert [1967, II. 1.2, p. 253], [f.L,v]q = [flq,v] = 1 and (flvY = flnvn[v, fl]Ci) for all n EN. The first assertion yields that N' is an elementary abelian q-group. Therefore N' is cyclic of order q. If q = 2, then (flvt = fl4V4, as IN'I = 2. Therefore U = { p.1 fl E N, fl4 = I} is a characteristic subgroup of N. Moreover, exp( U) = 4, as N' 1=- {l}. If U 1=- N, then U ~ A and U is cyclic of order 4. Also, U is the only subgroup of order 4 of N. Hence N is cyclic by Huppert [1967, III. 8.4, p. 311]. This contradiction proves U = Nand exp(N) = 4. Let N'=(z). If fl,vEN, then [p.,v]=z!(Ji.,v) with f(fl,V)EGF(q). Obviously, f( fl, fl) = 0 and f( fl, v) = - f(v, f.L). Moreover if f.L' f.L - I, v' V-I E S(N), then f( f.L, v) = f( f.L', v'). Hence g defined by g( f.LS(N), vS(N» =f( f.L, v) is a mapping from N / SeN) into GF(q). As N' ~ SeN), we have [flv,P] = [fl,P][v,p] by Huppert [1967, III.1.2.c), p. 253]. Hence f(f.LV,p) = f( f.L, p) + f(v, p). Therefore g is a skew symmetric bilinear form on the GF(q)-vector space N / S(N). Moreover, g is non-degenerate, as f(~, v) = 0 for all ~ E N implies v E S(N). Therefore IN /3(N)1 = q2m. Let k be the degree of a faithful irreducible representation of N over C. As N' centralizes N, each commutator is represented by a scalar matrix by Schur's lemma. Let X be the character of the representation and let
185
37. Soluble Flag Transitive Collineation Groups
f.L E N\3(N). Then there exists v E N with [v, f.L] = cE 1=- E, where E is the (k X k)-identity matrix. Thus
As c 1=- 1, we obtain X( f.L -I) INI
=
2: MEN
= O.
IX(f.L)1
2
Therefore, by the orthogonality relation,
=
2:
2
IX(f.L)1 = k2IS(N)I.
ME3(N)
Hence k = qm. Therefore every absolutely irreducible representation of N has degree qm (see e.g. Huppert [1967, V.12.9, p. 530]). By 37.2, V is the direct sum of isomorphic irreducible N-submodules VI' ... , VI' If N does not operate faithfully on Vi for some i, then SeN) does not operate faithfully on Vi' since every non-trivial normal subgroup of N intersects SeN) non-trivially. As the are all isomorphic N-modules, SeN) does not operate faithfully on V, a contradiction. Let Ki be the centralizer of N in EndGF(p)( VJ Then SeN) C; Kt, i.e., each element of SeN) acts as a scalar multiplication on the Ki-vector space ~. There exists a finite extension Li of Ki such that Wi = ~. 0 KLi splits into a direct sum of absolutely irreducible N-submodules. Si~ce each element of SeN) acts as a scalar multiplication on Vi' every such element acts as the same scalar multiplication on Wj' Therefore, N operates faithfully on each N-submodule of Wi' By the remark made above, qm divides rkLJWJ = rkKJVJ = rkGF(p)(Vi)[Ki : GF(p)r l • From this we infer that qm divides r. 2:: centralizes SeN) by 17.2. As the centralizer of 2:: is cyclic, 2:: does not centralize N, nor does any non-trivial subgroup of 2::. Hence 2:: is isomorphic to a subgroup of Aut(N / S(N» by Gorenstein [1968, 5.3.2, p. 178]. Put W = N /S(N). As q 1=- 'TT, we have W = WI EB ... EB Wn with irreducible 2::-modules Wi by Maschke's theorem. As 2:: is cyclic, 2:: acts faithfully on at least one Wi' since 2:: acts faithfully on W. We may assume i = 1. Put IWII = qb. Then 12::1 divides qb - 1. Moreover, r divides 'TT - 1. As qm divides r, it divides 'TT - 1. Therefore, assuming b < m, we obtain the contradiction 12::1 < qb - 1 < qm < 'TT - 1. Thus m < b < 2m. Define hand g by 'TT = hqm + 1 and qb = g'TT + 1. Then h, g > 1. Therefore, qb = g(hqm + 1) + 1, whence g =: - 1 mod qm. Put g = kqm - 1. Then k > 1, as g > 1. Thus qb > (qm - l)(qm + 1) + 1 = q2m. Hence b = 2m and W = WI' Moreover, 'TT = qm + 1. As 'TT is a prime, q = 2. Furthermore, 12::1 divides 2b - 1 = 22m - 1 = (2m - 1)'TT, whence 12::1 = 'TT. Finally, 2m < r < 'TT - 1 = 2m and hence r = 2m. As r = 2m , we have that 2:: acts absolutely irreducibly on V. Therefore, by Huppert [1967, V.11.1O, p. 525], the centralizer of N in EndGF(p)(V) is the centre of EndoF(p)(V). Therefore, A C B(G) and hence A = B(G) by the maximality of A. In particular, IA I divides p - l. Therefore, the assertions of the theorem up to c) are proved.
Vi
VI. Flag Transitive Planes
186
Let yEG and /-t,vEN. Then [/-tY,V Y] = [/-t,V)y = [/-t,v], as N'c:;8(N) ~ A = 3( G). Therefore, G leaves invariant f and hence g. This proves d). S(N) is cyclic. Let S(N) = and let vIS(N), ... , v2m S(N) be a b~sis of N / 3(N). If ~ E rJc(N / S(N», then ~ -IVi~ = vi~ai. The mappmg ~ ~ Gat, ... , ra 2m ) is a homomorphism of G into the direc.t product o~ 2m copies of 8(N) the kernel of which is the centralizer of N In G. As V IS an absolutely irreducible N-Module, A = r.£c(N), as we have remarked above. This proves e). 0
187
37. Soluble Flag Transitive Collineation Groups
If m = 2, then p2 + 1 divides 2 . 3 . 15. As p2 == 0 mod 3 or p2 == 1 mod 3, we obtain that 3 does not divide p2 + l. Hence p2 + 1 = 10, i.e. p2 = 9. Let m > 3. Choose v such that 3v < p < 3V + 1. Then v > l. Moreover, 2(22 - 1)(24 - 1)(2 6 - 1) = 2 . 3 . 15 . 63 < 3 8 . Furthermore, 23 < 32 yields 2i < 3i - I for all i > 3. Therefore, 1 m1 3v2m - < p2 - + 1 < 2
m
II (22i i=I
37.5 Corollary. L: acts irreducibly and regularly on N/8(N). 37.6 Lemma. Let Q be a finite set and let G be a soluble permutation group
acting primitively on Q. Then G contains exactly one minimal normal subgroup. If V is this minimal normal subgroup, then V is an elementary abelian p-group and I VI = IQI· The proof is obvious. 37.7 Theorem (Huppert 1957, Foulser 1964). Let G be either a soluble doubly transitive permutation group on a finite set Q or a soluble jla? transitive collineation group of a finite affine plane 2t of order n. Then G lS the semidirect product of an elementary abelian p-group V of order IQI 2 respectively n 2 and a group H, where H c:; fL(l, I VI) unless I VI E {32, 52, 7 , 112,23 2,3 4}. In the latter cases, H c:; fL(l, I VI) or H is one of 16 exceptional
groups. PROOF. It follows from 37.6 and 15.16 that G = VH where V is a normal elementary abelian p-group and V n H = {l}. By 37.1, we may assume that all normal abelian subgroups of H operate reducibly on V. Let I VI = pf. Then f = 2r, if G is a flag transitive group. Assume that there exists a p-primitive divisor 'TT of pf - 1. Then 'TT divides IHI, as either pf - 1 or pr + 1 divides IHI. Hence, applying 37:3, we obtainf= 2m,.'TT =:.2m + 1, 18(N)1 = 2j with) = 1 or 2. Moreover, 2) divides p - 1, as 2) dIVIdes IA I and IA I divides p - l. Also, IHI = IH /r.£H(N /8(N))IIr.£H(N /8(N))/ A IIA I·
m
Therefore IH I, and hence p2 - + 1, divide 2m2+2jm(p - 1)II7= 1(22i - 1), since ISp(2m,2)1 = 2 m2 II7=I(2 2i -1) by Huppert [1967, II.9.13, p. 220]. If m = 1, then H permutes the subspaces of rank 1 of V transitiv~ly. As all these subspaces are fixed by A, we obtain that p + 1 divides 21 +2) ·3. If j = 2, then 4 does not divide p + 1. Hence p + 1 = 6, i.e., p = 5. (Observe 3 that'TT = 3 divides p + 1.) If) = 1 then p + 1 divides 2 . 3, whence p = 5, 11 or 23. m m If m ~ 2, then p2 - + 1 is not divisible by 4. Hence p2 - + 1 divides 2II7= I (22i - 1). 1
1
1
< 3 8 II 32i - I. i=4
Hence v2 m - I < 3 + 5 + 2:7=4 (2i - 1) = m - l. From this inequality we obtain m < 5. As 'TT = 2m + 1 is a prime, m = 4. In particular, 'TT = 17. By 37.3, 'TT is the only R-primitive divisor of pf - 1. Moreover, 'TT2 does not divide IHI. Hence p23 + 1 = 2 . 17, a contradiction. Hence we are left with the case where there is no p-primitive prime divisor of pf - 1. In this case, either pf = 2 6 or f = 2 and p + 1 = 2S by 6.2. If f = 2 and p + 1 = 2s , then H is a subgroup of GL(2, p) which induces a subgroup U of PGL(2, p) that acts transitively on the projective line over GF(p). It follows from 14.1 that U is isomorphic to a subgroup of S4' Thus P + 1 divides 24, whence p = 3 or 7. It remains to consider the case pf = 26. Assume first that G is a flag transitive collineation group on a plane 2t of order 8. As 2t is desarguesian by 36.2, we have H ~ fL(2, 8). If 4 divides IH I then SL(2,8) is contained in H, as an inspection of Dickson's subgroup list will tell us. Hence 4 does not divide IH I, since H is soluble. Moreover, If L(2, 8)1 = 23 . 33 .7 2 . Therefore IHI divides 2.3 3 .72 . Let H be the group induced by H in PfL(2, 8). Then IHI divides 2.3 3 .72 . Moreover, 2,3,9,27,6,18,54 are all t= 1 mod 7. Thus H contains only one Sylow 7-subgroup L:. If 2: =1= {I}, then IIf : L:I < 2. On the other hand, 9 divides IHI, as H acts transitively on 100 , Thus L: = {I}. This yields that there is only one Sylow 3-subgroup in H. Therefore, if IT is a Sylow 3-subgroup of H, then A = (H n 3GL(2, 8» IT is a normal subgroup of H of order 7i . 3 2 +j with i = 0 or 1 and) = 0 or 1. If ) = 0 then A is abelian. Moreover, A acts irreducibly on V. Thus H c:; fL(l, 64). If) = 1, then B = A n GL(2, 8) is an abelian group of order 7 i . 3 2. Again, B acts irreducibly on V, whence H c:; fL(l,64). We may assume from now on that H acts transitively on V\{l}. Let A be a maximal abelian normal subgroup H. We may assume that A acts reducibly on V. Let X be an irreducible A -submodule of V and put ® = {XY lyE H}. As H acts transitively on V\ {l}, we obtain that 6 is a partition of V. Moreover, A is contained in the multiplicative group of K( V, 6). Hence A is cyclic. Moreover, rk(X) divides rk( V) = 6. Therefore IA I = 3 or IA I = 7. If IA I = 7, then 6( V) is the desarguesian plane of order 8 and A is not maximal, as we have seen above. Thus IA I = 3. As IAut(A)1 = 2, we see that IH: C£H(A)I < 2. It follows that r.£H(A) acts also transitively on V\{l}. Hence A =1= r.£H(A) and r.£H(A) is non-abelian. Let N be a minimal non-abelian normal subgroup of H which is contained in (iH(A). It follows that N' c:; AnN c:; 8(N), whence N is nilpotent of class 2
37.4 Corollary. N acts absolutely irreducibly on V.
m
1)
188
VI. Flag Transitive Planes
2. The mi.nimality of N then implies that N is a 3-group. Moreover, N I SeN) IS an elementary abelian 3-group of order 32 or 33 , as ~H(A) \: GL(3,4) and IGL(3,4)1 = 26 .3 4 .7. 5. The arguments used in the proof of 37.3 show that IN IS(N)I = 32. Let .L be a Sylow 7-subgroup of ~H(A). Then ILl = 7. Moreover L centralIzes N IS(N) and SeN), as IGL(2,3)1 = 3 ·24. Hence L centralizes N by Gorenstein [1968, 5.3.2, p. 178]. Moreover, IALI = 21. Therefore, AL operates irreducibly on V. By Schur's lemma, ~ H(A L) is cyclic. On the other hand, N \: ~H(AL), a contradiction. This proves that 32' 52 , 72, 1 , 2 4 23 ,3 are the only exceptional degrees. We shall construct now all the exceptional groups. If the degree is p2, then H \: GL(2, p). Moreover, INI = 8 unless p = 5; for in all other cases 4 does not divide p - 1. If P = 5, then INI = 8 or 16. Let p = 3. Then 4 divides IHI. As H is not contained in rL(1, 9), we get that H = SL(2,3) or H = GL(2, 3). This accounts for two exceptions. . Let.p > 5.. If P divides IHI, then H contains SL(2, p) which is ImpossIble, SInce H is soluble. As S(GL(2, p)) is cyclic, it contains just one involution. Hence SeN) n SL(2, p) =1= {I}. Therefore, v 2 E SL(2, p) for all v E N. From this we deduce IN I(N n SL(2, p))1 <; 2, as GL(2, p)/SL(2, p) is cyclic. Moreover N n SL(2, p) is normal in H. If N n SL(2, p) =1= N, then N n SL(2, p) \: SeN) by 37.2 which is impossible, as N is not abelian. Hence N n SL(2, p) = N, i.e., N \: SL(2, p). Let;: = 5. Then 3 divides IH n SL(2,5)1. Hence IH n SL(2,5)1 = 23 • 3. In p~r.t1cular, H n SL(2, 5) ~ SL(2, 3). This yields that H n SL(2,5) acts tran.slt1vel~ on V\{I}. Furthermore, 'R GL (2,5)(H n SL(2,5))/(H n SL(2, 5)) IS cyclIc of order 4 by 14.5. Hence we have three possibilities for H; namely: H I(H n SL(2, 5)) is cyclic of order 1,2, or 4. As all subgroups of SL(2,5) which are isomorphic to SL(2,3) are conjugate within SL(2,5) by 14.5, we have exactly three exceptions in this case. Let p=7. As IGL(2, 7)1 =2 5 '3 2 '7, we have IHI=2 3 + i .Y with i,j E {O, 1,2}. Let A be a maximal abelian normal subgroup of H. Then A acts reducibly on V. As H acts transitively on the set of subspaces of rank 1 of V, we see that A fixes all the subspaces of V. This yields that A is the centre of H. Let N be a minimal non-abelian normal subgroup of M. Then N' \: A. This yields that N is nilpotent of class 2. In particular, N is either a 2-group or a 3-~roup. Since N is non-abelian and 33 does not divide IH I, we have that N IS a 2-group. Using the ideas developed in the proof of 37.3, we get IS(N)I = 2, as 4 does not divide 6, and hence N' = SeN). Also N ~ SL(2,7), whence N I SeN) is elementary abelian of order 4. In particular, N is the quaternion group of order 8. Let C be a cyclic subgroup of order 4 of N and assume that C is normal in H. As 4 does not divide 6, we have that C operates irreducibly on V. Hence H fL(l, 72) by 37.1, a contradiction. Thus C is not normal in H.
e
J.l:'ri~.'· : :1
I. · . I
I
.I i)
c:
37. Soluble Flag Transitive Collineation Groups
189
Since N contains exactly three cyclic subgroups of order 4, we infer that H I~H(N) is not a 2-group. Therefore 3 divides IH I~H(N)I· N acts absolutely irreducibly on V. Hence Cf-H(N) consists only of scalar matrices. Therefore ~ H(N) fixes all subspaces of rank I of V. As H acts transitively on the set of all these subspaces, we have that 8 divides IH I~H(N)I. This proves H I~H(N) = Aut(N). Therefore IHI = 24. Y with jE{I,2}. If IHI = 24 . 32, then H = 9C GL (2, 7)(N). We infer from 14.4 that there is 4 just one exception of this type. Thus we may assume IHI = 2 .3 from now on. As the number of subspaces of rank I of V is 8, each element a of order 3 of H fixes at least two subspaces of rank 1. Let 1 =1= v E SeN), then
190
VI. Flag Transitive Planes
operates irreducibly on V, we see that N is the direct product of Z-irreducible subgroups Xl ,X2 , . . . . As Z is cyclic of order 4, we obtain IXil = 9 and hence V = XI EB X 2 • Assume that {X I ,X2 } is left invariant by N. Then XI is fixed by a subgroup U of N with lUI = 24. If U centralizes Z, then the restriction of U to X I is cyclic by Schur's lemma. Hence Ulx l = Z Ix l , as N is of exponent 4. Therefore U contains a subgroup Wof order 4 fixing XI pointwise. Moreover, W must operate faithfully and hence irreducibly on X 2 . Therefore W is cyclic of order 4, whence 3(N) \: W, a contradiction. This proves that U does not centralize Z. Put C = Q:N(Z), Then C permutes XI and X 2 , as IC I = 24 and C =1= U. (Recall that IAut(Z)1 = 2.) Therefore XI and X 2 are isomorphic Z-modules. If {XI' X 2 } is not left invariant by N, then there is an irreducible Z-submodule X with X n XI = X n X 2 = {O}. We infer from this that XI and X 2 are also in this case isomorphic Z-modules. It now follows from 9.1 that N is isomorphic to a subgroup of fL(2,9) and that IN: N n GL(2,9)1
37. Soluble Flag Transitive Collineation Groups
If BD = - DB, then it is easily checked that six of these cosets consist of elements of order 4 and nine of elements of order 2. But this is impossible by 37.5. Hence BD = DB. Assume that CD = - C. Then CD is an involution with A CD = A - I and BCD = -B. This contradiction proves CD = C. Using all these informations, an easy computation then shows
l~
0 0 - 1 0 0 1 0 0 Hence we may assume that
D=
=2. The group GL(2, 9)/SL(2, 9) is cyclic of order 8. Therefore (N n GL(2, 9))/(N n SL(2, 9)) ~ (N n GL(2,9))SL(2,9)/SL(2,9) is cyclic of order I or 2, as the square of every element in N is contained in 3(N). Since Z \: 3(GL(2,9)), we see that IN n SL(2,9)1 = 8. In particular, N n SL(2,9) is the quaternion group of order 8. It follows from 14.4 that N n SL(2,9) is uniquely determined up to conjugacy within GL(2, 9). Hence N n GL(2,9) is uniquely determined, as N n GL(2,9) = ZeN n SL(2,9)). Let 0
A=
- 1
0 0
1 0 0 0
0 0 0 - 1 0 0 C= I 0
0 0 , 1
B=
0 0 0 0 I
1 0
0
1
0 I
0 - I 0 0 0
1
1 0 - 1
0
0 1
o' - 1
0 - I 0 0
Then it is easily checked that (B, C) is a quaternion group which is contained in the centralizer of A. Therefore we may assume Z = (A) and N n SL(2,9) = (B, C). N is generated by its involutions, as N is a minimal non-abelian normal subgroup of H. Hence there exists an involution DEN with AD = A -I. As all cyclic subgroups of order 4 of N contain 3(N) = N ' , they are normal in N. Hence CD E (C) and BD E (B). The cosets of N 13(N) which are distinct from 3(N) are:
{A, - A}, {B, - B }, {C, - C}, {D, - D }, {AB, - AB }, {AC, - AC}, {AD, -AD}, {BC, - BC}, {BD, - BD}, {CD, - CD}, {ABC, -ABC},
\ABD. -ABD). (ACD. -ACD). \BCD. - BeD), {ABeD, -ABeD}.
191
0 0 0 -1
l~
E {D, -D,AD, -AD}.
0 - 1 0 0
0 0 1 0
~J.
- 1
As IGL~4,3)1 is not divisible by 52, all subgroups of order 5 in 9C GL (4,3)(N) are conjugate. Therefore we may assume that -1 - 1 0 0
F=
1 -1
1 -1
I
0
-1
0
IS III H, as F normalizes N, i.e., H always contains the group Ho = (A,B,C,D,F). Moreover, Ho acts transitively on the set of non-zero vectors of V. As (F) acts irreducibly on N I 3(N), the group 3(N) is a maximal abelian normal subgroup of H. Therefore by 37.3 e), IQ:H(N13(N))1 < 25. As N C; (£H(N 13(N)), we obtain N = Q:H(N 13(N)). Therefore HI N is isomorphic to a subgroup of Sp(4, 2). Moreover Sp(4, 2) ~ S6 (see e.g., Huppert [1967, 11.9.22, p. 227]). This yields that HI N contains exactly one Sylow 5-subgroup since HI N is soluble. Thus Ho is normal in HI N. Moreover, HI N is isomorphic to a subgroup of the group of all mappings x ~ ax + b with a, b E GP(5) and a =1= 0, as this group is isomorphic to the normalizer of a Sylow 5-subgroup in S6' Hence HI Ho is cyclic of order 1, 2 or 4.
Let G=
l
6
- 1
GAG
o -I
o 1 o o o 1 - 1 = ABC,
o
o - 1
o o GBG
]. Then we have the relations:
-I
= B,
GCG
-I
G4
= CDB,
GDG - I = DB, GFG - I = DCBF2, = - B. Hence H2 = (A,B,C,D,F,G), HI = (A,B,C,D,F,G 2) and Ho are groups which occur as stabilizers of a point in an exceptional doubly transitive
192
VI. Flag Transitive Planes
group of degree 34 . (Recall that all exceptional groups of degree 34 are doubly transitive.) Let H be any such stabilizer. Then Ho H, as we have seen above. Let IH: Hoi = 2 and assume H =!= HI' Then
c:
37.8 Corollary. The three exceptional doubly transitive groups of degree 34 operate as doubly transitive collineation groups on the nearfield plane of order
32 . This follows from what we have seen above and 8.3. 37.9 Corollary. Let G be a soluble flag transitive collineation group of the finite affine plane m and let q be the order of m. If q =!= 9, then G also acts flag transitively on the desarguesian plane of order q. Moreover, the representation of G on the point set of m is permutation isomorphic to the representation of G on the point set of the desarguesian plane. This follows from 37.7 and the remark that translation planes of prime order are desarguesian by theorem 1.13. 0 PROOF.
37.10 Lemma. Let p be a prime and let mbe an affine plane of order pr. Assume that u is a p-primitive divisor of pr - 1. If G is a soluble flag transitive collineation group of m whose order is divisible by u, then m is desarguesian. PROOF. m is a translation plane by 15.16 and G contains the translation group T of m. If 0 is a point of m, then G = TG o and T n Go = {l}. As ITI = p2r, we have that u divides IGol. Moreover, Go fL(l, p2r) by 37.7 or r = 1 and p is one of a finite number of primes or pr = 32. In the second case, mis desarguesian by l.13. In the third case p r - 1 = 8 so that there is no p-primitive divisor of pr - 1. Hence we are left with the case that Go C: fL(l,p2r). Moreover, we may assume r > 2 by 1.13. Now
c:
IGo I = IGo : Go n G L( 1, P2r) II Go n G L( 1, P2r) I and I Go: Go n GL(l, p2r)1 is a divisor of 2r. As r > 1, we have u > 2. Thus u does not divide 2r, as u does not divide r. This implies that u divides I Go n GL(l, p2r)l. Therefore Go n GL(1, p2r) contains a subgroup U of order u. As Go n GL(1, p2r) is a cyclic normal subgroup of Go' the group U is normal in Go. Moreover U fixes a point on 100 , as u is a divisor of p r - 1 which is greater than 2. We therefore infer from the transitivity of Go on 100 that U fixes all the points on 100 , Thus U is a group of (O,loo)-homologies. This yields that u divides the order of the multi-
193
38. Some Characterizations of Finite Desarguesian Planes
plicative group of the kernel K of m. As u is a p-primitive prime divisor of p r - 1, we finally see that K ~ GF(p r). 0 37.11 Theorem (Foulser). Let m be a finite affine plane. If W admits a soluble collineation group acting doubly transitively on the set of points of m, then mis desarguesian or mis the nearfield plane of order 9. PROOF. Let G be a soluble doubly transitive collineation group of m. Then mis a translation plane by 15.16 and G contains the translation group of m. If p' is the order of m, then p2r(p2r - 1) divides IGI. If m is non-
desarguesian, then there does not exist a p-primitive prime divisor of 6 5 _ 1. Hence p' = 2 or r = 2 and p + 1 = 2 by 6.2. Assume pr = 26. Then Go fL(l,212) by 37.7. As IGolG o n GL(1,2 12)1 divides 12, we see that 21 divides IG o n GL(l,2 12)1. Therefore Go contains a normal subgroup U of order 21. As above, U consists entirely of (O,loo)-homologies whence it follows that mis desarguesian. Hence r = 2 and p + 1 = 2s . Then
pr
c:
Therefore 25 + 2 divides I Gol. If p > 3, then Go fL(1, p4) by 37.7. Thus IGol Go n GL(I, p4)1 divides 4. This implies that Go n GL(l, p4) and hence Go contain a cyclic normal subgroup V of order 25. Since 4 does not divide p2 + 1, there exists a subgroup U of V of order 2s - 1 fixing a point on 100 , As V is cyclic, U is a characteristic subgroup of V and hence a normal subgroup of Go. Therefore U fixes all the points of 100 , i.e., U consists of (O,loo)-homologies only. From p>3 we obtain IUI>4. Therefore I UI does not divide p - 1. This yields that the kernel of m is isomorphic to GF(p2). Hence p = 3 and mis the nearfield plane of order 9 by 8.4. This proves 37.11. 0
c:
There are plenty of finite non-desarguesian affine planes admitting soluble flag transitive collineation groups. See e.g. M. L. N. Rao 1973.
38. Some Characterizations of Finite Desarguesian Planes Let q be a power of a prime p and let IJ3 be a proj ective plane of order q. Furthermore, let !:l be a collineation group of IJ3 which is isomorphic to PSL(2, q).
38.1 Lemma. If !:l does not fix a point or a line, then ~ has a point or a line orbit of length q + 1.
1~4
VI.
t'lag
1 ransllJ vt: rt
PROOF. Let 2: be a Sylow p-subgroup of L1. As q2 + q + I == I mod p, we see that 2: has a fixed point X. Put N = 9CtI(2:) and assume XN = X. If qJ E 11 \N, then XfJJ =1= X, for otherwise X is fixed by
o
38.2 Lemma. If Ll fixes a point A, then the lines through A form an orbit of Ll. PROOF. Assume that L1 fixes all the lines through A. Then L1 consists only of perspectivities with centre A. But this implies that 1L11 = q(q2 - l)d where d = (2, q - 1) - I divides q2( q - 1) which is impossible. Thus there exists a line I through A with 1/t.1 > 2. Therefore, 2 < Ill: Ll,1 < q + 1. It follows from 14.1 that either ILl: Ll,1 = q + 1 or q = 2, 3, 5, 7, 9 or 11 and 1L1 : L1,1 = 2, 3, 5, 7, 6, or 10 respectively. We have to show that these seven exceptions cannot occur. If q = 2 or 3 and Ill: Ll,1 = q, then Ll[ is normal in Ll. From this we infer that D., fixes q and hence all lines through A. Therefore, q + 1 = ILl,1 divides q2(q - 1), a contradiction. Let q = 5, 7 or 11. As q is not a square, the involutions in Ll are homologies. Moreover, D. fixes a line I through A. As Ll is simple, Ll
38. Some Characterizations of Finite Desarguesian Planes
195
operates faithfully on I as well as on the set of the remaining lines through
A. Therefore, if a is an involution in L1, the axis of a passes through A and is distinct from I and the centre of a is on 1 and is distinct from A. Let m be a line through A other than I. Then L1m ~ A 4 , S4' respectively As for q = 5, 7, respectively 11. Since S4 and As are generated by their involutions, L1m is a group of homologies, if q = 7 or 11. But 24 does not divide 7 - 1 and 60 does not divide 11 - 1. If q = 5, then Ll m contains an elementary abelian 2-group A of order 4. As A is abelian and since all elements of A have axis m, they must have the same centre. This implies that A is cyclic as s,n is desarguesian, a contradiction. Finally, we have to rule out the exceptional case q = 9. Here L1 fixes four lines through A. Moreover, L1 fixes at least four points on each of these lines. Therefore Ll fixes a subplane 0 of s,n elementwise. As L1 =1= {I} and 0(0) > 3, we see that 0 is a Baer subplane of s,n. From this we infer that L1 acts regularly on s,n\0, whence ILlI = 360 divides 34 + 32 + 1 - 32 - 3 - 1 = 78, a contradiction. Thus we have proved 38.2. 0
38.3 Lemma. Let 0 be a point orbit of length q + I of Ll. Then p = 2 and 0 consists of the points of a line.
0
is an oval or
PROOF. Assume that 0 consists of the points of a line. If a E L1 is an involution, then a has at most 2 fixed points on I. Therefore, a is a perspectivity. If p > 2, then a has exactly 2 fixed points P and Q on l; one of them, say P, is the centre. The axis of a passes through Q. There exists o E L1 with pS = Q and QS = P. It follows from 4.8 that ao -lao is an involutory homology with axis l. This contradicts the fact that Ll operates faithfully on o. Hence p = 2. Let P, Q, REo be non collinear. If p = 2, then Ll acts threefold transitively on o. Therefore, 0 is an oval in this case. Thus we may assume p > 2. Then ILlp,QI = 1(q - 1). Assume that there exists S E (PQ\{P, Q}) no. Then it follows that I(PQ\{P,Q}) n 01 = 1(q - 1) or q - 1; note that the latter case is impossible, since R 1 PQ. Let 2: be a Sylow p-subgroup of Llp . Then 2: is transitive on 0\ {P}. This implies that 2: PQ is transitive on (PQ\{P}) no. Thusp divides 1(q + 1), a contradiction. This proves PQ n 0 = {P, Q}. As Ll acts doubly transitively on 0, we finally obtain that 0 is an oval. 0
38.4 Lemma. If a ELlis an involution, then a is a perspectivity. PROOF. Dualizing if necessary, we may assume by 38.1 and 38.2 that Ll has a point orbit 0 of length q + 1. If 0 consists of the points of a line, then p = 2 by 38.3 and a has exactly one fixed point on o. Hence a is a perspectivity. We may thus assume that 0 is an oval. If q is even, then a has exactly one fixed point in o. Moreover, the knot K of 0, i.e., the point K of s,n which is on all the tangents of 0, is also fixed by a. If a is not a perspectivity, then a is a Baer involution and hence fixes {cj + 1 lines
196
;,
i'
I
VI. Flag Transitive Planes
through K and therefore {cj + 1 points on 0, a contradiction. Thus we may also assume that q is odd. If q == 3 mod 4, then q is not a square, whence a is a homology. If q == 1 mod 4, then a has exactly 2 fixed points on o. Let P and Q be these fixed points and denote by tx the tangent of 0 which carries X E o. Then X ~ tx n tp is a bijection of o\{ P} onto tp \{ P}. Moreover, (tx n tpt = tx • n tp. Therefore, (tx n tpt = tx n tp if and only if xa = X. Thus a has exactly two fixed points on tp , namely P and tQ n tp. This proves that a is not a Baer involution. Hence a is a homology. 0
38.5 Lemma. Assume p = 2. If I:::. fixes a point P, then I:::. has exactly one point orbit 0 of length q + 1. If no two of the involutions of I:::. have the same centre, then 0 is an oval and P is its knot. If two distinct involutions have the same centre, then 0 consists of the points of a line. In either case, the points not in 0 and other than P form an orbit of 1:::.. PROOF. By 38.2, the lines through P form an orbit of 1:::.. Hence I:::. acts doubly transitively on the set of lines through P. Let I and m be two distinct lines through P. Then I:::.{, m is cyclic or order q - 1. Let 8 E 1:::." m be an element of order s where s is a prime. Then 8 fixes a point Q on I other than P. Assume that 8 fixes a third point R on I and let a be an element of I:::. switching I and m. Then I:::.',m = I:::."m' Therefore, as I:::."m is cyclic, (o)IJ = (0). This shows that fixes the quadrangle Q,R, Q6,R6. Thus fixes a subplane of 1.l3. As p t5 = P, we see that 8 fixes at least three lines through P, a contradiction. This shows that 8 fixes P and Q, but no other point on I. From this we infer that I:::.{ m fixes P and Q and operates regularly on /\{P,QI. Putting 0= {Q7JI7JEI:::.}, either lol=q+l or 10 1 = q(q + 1). Let L be a Sylow 2-subgroup fixing I. Then ~ acts transitively on the set of the remaining lines through P. As the number of these lines is q = I~I, it follows that P is not the centre of an involution in 1:::.. Therefore, I is fixed pointwise by ~. This yields 101 = q + 1. Obviously, 0 is unique. Moreover, the points not in and other than P form an orbit of 1:::.. As all involutions of I:::. are conjugate, either all or none of them have their centres in o. Since the number of involutions in I:::. is q2 - 1, there exist distinct involutions in I:::. having the same centre if and only if all involutions have their centres in o. If all involutions have their centres in 0 then, obviously, consists of the points of a line. On the other hand, if 0 consists of the points of a line x, then xll = x, whence x carries all the centres. Now 38.3 yields the desired result. 0
°
°
°
°
38.6 Lemma. Let p = 2 and assume that I:::. has a point orbit 0 which is an oval. Then !:l splits the set of points of I.l3 into three orbits: The set I.l3 1 = {K}, where K is the knot of 0, the set 1.l3 2 = 0, and the set 1.l3 3 of the remaining points. Moreover, Do splits the sets of all lines into three orbits: The set 53 1 of all the tangents of 0, the set 53 2 of all the secants, and the set 53 J of the remaining lines.
197
38. Some Characterizations of Finite Desarguesian Planes
PROOF. 1.l3 1 and 1.l3 2 are of course orbits. Applying 38.5 yields that 1.l3 3 is also an orbit. 53 is an orbit, as 1.l3 2 is. Moreover, 53 2 is an orbit, because I:::. acts doubly tra~sitively on = 1.l3 2 • Since !:l has as many point orbits. as line orbits by the Dembowski-Hughes-Parker theorem, 53 3 is also an orbIt. 0
°
38.7 Lemma. Let p = 2 and assume that I:::. has a point orbit which is an oval. Then I.l3 is desarguesian and 0 is a conic section. PROOF. Using the notation of 38.6, we have 11.l3 11 = 1, I1.l3 2I = q + 1, 11.l3 3 1= q2 - 1, 153 1 1 = q + 1, 1532 1= ! q(q + 1), and 153 3 1 = ! q(q - 1). We represent I.l3 within I:::. as follows: a) b) c) d)
To To To To
K we assign the set 11 = {91Il(~) I~ E SyliDo)}. X E 1.l3 2 we assign the group !:lx· . X E 1.l3 3 we assign the involution ax whose centre IS X. each line I of we assign the group Do,.
m
m
The mapping defined in b) is a bijection from 2 to 11. The one defined in c) is a bijection from onto' the set of all involutions of 1:::.. If we de~ote the restriction to 53 i of the mapping defined in d) by ai' then a I 1S a bijection from 53 1 onto 11, and a 2 is a bijection from. $2;2 o~.to t.he set of all dihedral subgroups of order 2(q - 1) of Do, and a3 1S a b1JectlOn from $2;3 onto the set of all dihedral subgroups of order 2(q + 1) of 1:::.. Incidence is described as follows: K I I, if and only if !:l{ En. If X E 2 , then X I I if and only if !:lx = !:l{ or IDo x n !:l,1 = q - 1. If X E 3 , then X I I if and only if ax E Do,. In order to show that is desarguesian, we have only to prove that the desarguesian plane of order q admits a collineation group !:l;;;; PSL(2, q) which fixes an oval c. Let:n be the desarguesian plane over F=GF(q). Put c= {F(O,I,O), F(f, f2, 1) If E F}. It is easily checked that c is an oval. Let !:l be the group genera ted by the rna trices
m3
m
m
m
and
[~
with
a,b E F,
a =1= 0.
It is a trivial exercise to see that Do ;;;; PSL(2, q) and that c is an orbit of 1:::.. Hence ;;;;:n. Finally, c = {F(x, y, z) Ix 2 + yz = O}. Thus c and hence 0 are conic sections. 0
m
38.8 Lemma. Let q be odd. Then !:l has a point orbit 0 of length q + 1 and a line orbit t of length q + 1. The point orbit is an oval and the line orbit t consists of the tangents of 0. If q == 1 mod 4, then the point P is the centre of an involution in Do If and only If P is an exterior point of o. If q == 3 mod 4, then the point P is the centre of an involution of I:::. if and only If P is an interior point of 0. In particular, each point of mis the centre of at most one involution in !:l.
°
198
VI. Flag Transitive Planes
PROOF. By 38.1 and 38.2, or the dual of 38.2, we know that ~ has an orbit of length q + I. Dualizing if necessary we may assume that ~ has a point orbit 0 of length q + l. As q is odd, 0 is an oval by 38.3. Since each point of 0 is on exactly one tangent, the set t of tangents is a line orbit of length q + 1. As t is an oval in the dual of SU, the first assertion of the lemma is proved. Let a and T be involutions in 11 having the same centre. Then aT fixes 0 pointwise. Consequently aT = 1, i.e., a = T, proving the last assertion of the lemma. Assume q == 1 mod 4. Then an involution a E 11 fixes two points, P and Q say, on o. Moreover, tp n tQ = R is also fixed by a. As a is a homology, one of these points is the centre of a. Obviously, P and Q are not centres of involutions in ~. Hence R is the centre of a. Therefore, centres of involutions are always exterior points in this case. As ! q(q + 1) is the number of exterior points as well as the number of involutions in ~, the third assertion is proved. Let q == 3 mod 4. In this case the involutions of 11 act fixed-point-free on o. Therefore, their centres are interior points. As the number of interior points is the same as the number of involutions, namely !(q - l)q, the fourth assertion is also proved. 0
38.9 Lemma. Assume q == 1 mod 4 and let 0 be the oval fixed by ~. Then ~ acts transitively on the set of flags (P, I), where P is an interior point and I is an exterior line of o. PROOF. Let a and T be distinct involutions of ~ and assume aT = Ta. Let I and m be the axes of a respectively T. By the dual of 38.8, we have 1=1= m. Hence, by 4.8, aT is an involution with centre I n m. This yields by 38.8 that I n m is an exterior point of o. Let P be an interior point of o. The number of secants through P is ! (q + 1). Therefore, ! (q + 1) is also the number of exterior lines through P. Each of the secants is the axis of exactly one involution. As P is not a centre, the above argument shows that the only secant through P fixed by an involution of ~p is the axis of that involution. Therefore, by 15.1, the group I1p acts transitively on the set of secants through P. Thus ! (q + 1) divides Il1pl. As Il1pl is even and ! (q + 1) is odd, q + 1 divides Il1pl. It follows from 14.1 that I1p is a dihedral group of order q + 1. As 111: I1pl = f q(q - 1) is the number of interior points, 11 permutes the interior points transitively. By the dual argument, 11 acts transitively on the set of exterior lines. If I is such a line, then 11, is a dihedral group of order q + 1. The number t(q + 1) of exterior lines through P is odd, as q == 1 mod 4. Therefore, each involution in I1p fixes at least one exterior line through P. Assume that there are distinct involutions in I1p n 11" where I is an exterior
199
38. Some Characterizations of Finite Desarguesian Planes
ll,1
ll,.
line through P. Then Illp n > 3. It then follows from 14.2 that IIp = Let Q be an interior point on I which is distinct from P. Then, using 14.2 again, Il1p n ~QI < 2. If Il1p n I1QI = 2, then there exists an involution T fixing P and Q. As P and Q are interior points, PQ = I is the axis of T and hence I is a secant. This contradiction proves I1p n I1Q = {l}. From this and I1p = 11, we infer that I1p acts regularly on the set of interior points on I which are distinct from P. Thus q + 1 divides ! (q + 1) - 1, a contradiction. This proves Il1p n 11,1 < 2 for all exterior lines I through P. Hence Il1p n 11,1 = 2 for all such lines, as each involution in I1p fixes at least one exterior line through P. Invoking 15.1, we see that I1p acts transitively on the set of exterior lines through P. 0
38.10 Lemma. Assume q == 3 mod 4 and let 0 be the oval fixed by 11. Then 11 acts transitively on the set of flags (P, I) where P is an exterior point and I is a secant. PROOF. Let P be an exterior point of o. Then there are exactly two tangents through P. Let Q and R be their points of contact. Then I1p = ~(Q, R) , whence I1p is a dihedral group of order q - 1. Let s be a secant through P. Then I~p,sl < 2. Thus q - 1 = l~pl < !(q - 1)ll1 p,sl and hence Il1p,sl = 2. This shows that ~p acts transitively on the set of secants through P. Moreover, I~: I1pl = ! q(q + 1) is the number of exterior points. Therefore, ~ acts transitively on the set of exterior points. 0
p
In the following theorem we consider only the case p be dealt with later on.
> 3.
The case
= 2 will
38.11 Theorem (Luneburg 1964, Yaqub 1966). Let p
> 3 be a prime and let
q be a power of p. If SU is a projective plane of order q admitting a collineation group ~ ~ PSL(2, q), then SU is desarguesian. If r is a second collineation group of SU with r ~ PSL(2, q), then 11 and r are conjugate in the collineation group of SU. PROOF. As we know, 11 fixes an ovalo. Case 1: q == 1 mod 4. We represent SU within
~
in the following way:
a) If P is an exterior point of 0, then we assign to P the involution a p whose centre is P. b) If P E 0, then we assign to P its stabilizer ~p, c) If P is interior, then we assign to P the group ~p as well as the coset II(P) = II7], where II is the stabilizer of a fixed interior point Po and 7] is such that PfJ = P. The mapping under a) is a bijection of the set of exterior points onto the set of involutions of ~. The mapping under b) is a bijection of 0 onto the set of all normalizers of Sylow p-subgroups. The first mapping under c) is a
200
VI. Flag Transitive Planes
bijection of the set of interior points onto the set of all dihedral subgroups of order q + 1, whereas the second one is a bijection onto the set of right co sets of II.
a/) If I is a secant, then we assign to I the involution a, whose axis is I.
b /) If I is a tangent, then we assign to I the group 11,. c/) If 1 is an exterior line, then we assign to 1 the group 11, as well as the coset A(l) = A YJ, where A is the stabilizer of an exterior line 10 through Po and YJ is such that I'(} = I.
What we have said about the mappings under a), b), and c) carries over mutatis mutandis to the mappings under a/), b /), and c/). Incidence is described as follows: Let P be an exterior point and I a line. If I is exterior or tangent, then PI 1 if and only if ap E 11,. If 1 is a secant, then P I I if and only if ap i= a, and apa, = a,ap. Let P Eo and 1 a line. If 1 is tangent, then P I 1 if and only if I1p = 11,. If 1 is a secant, then P I I if and only if a, E I1p. Let P be interior. If 1 is exterior, then PI 1 if and only if II(P) n A(l) i= 0 by 38.9. If 1 is secant, then P I 1 if and only if a, E D.. p . Comparison with the desarguesian plane of order q now yields the desired results. Case 2: q == 3 mod 4. We represent ~ within 11 in the following way: a) If P is an interior point of 0, then we assign to P the involution ap whose centre is P. b) If P E 0, then we assign to P its stabilizer I1p. c) If P is an exterior point, then we assign to P the group 6. p as well as the coset II(P) = IIYJ, where II is the stabilizer of a fixed exterior point Po and 1] is such that P(} = P. The mapping under a) is a bijection of the set of interior points onto the set of involutions of 11. The mapping under b) is a bijection of 0 onto the set of all normalizers of Sylow p-subgroups. The first mapping under c) is a bijection of the set of exterior points onto the set of all dihedral subgroups of order q - 1, whereas the second one is a bijection onto the set of right cosets of II. a/) If 1 is an exterior line of 0, then we assign to I the involution a, whose axis is I. b /) If 1 is a tangent, then we assign to 1 the group 11,. c /) If 1 is a secant, then we assign to 1 the group 11, as well as the coset A(l) = A1], where A is the stabilizer of a fixed secant 10 passing through Po and 1] such that I'(} = I. What we have said about the mappings under a), b), and c) carries over mutatis mutandis to the mappings under a /), b /), and c /). Incidence is described as follows: Let P be an interior point and I a line. If 1 is a secant or a tangent, then P I I if and only if apE 6.,. If 1 is exterior,
38. Some Characterizations of Finite Desarguesian Planes
201
then P I I if and only if a p =t= a, and apa, = a,ap. Let P E 0 and let I be a line. If I is tangent, then P I I if and only if I1p = 11,. If I is a secant, then P I I if and only if Il1p n 11,1 = t (q - 1). Let P be exterior. If I is a secant, then P I I if and only if TI(P) n A(l) =t= 0 by 38.10. If I is a tangent, then P I I if and only if Il1p n 11/1 = t (q - 1). If I is an exterior line, then P I I if and only if a, E I1p. Comparison with the desarguesian plane of order q now yields that ~ is desarguesian. This proves 38.11. 0 38.12 Theorem (Liineburg 1964, Yaqub 1966). Let p be a prime and let q be
a power of p. Assume furthermore that ~ is a projective plane of order q. If ~ admits a collineation group 11 ~ SL(2, q), then ~ is desarguesian. If p = 2, then the full collineation group of ~ contains exactly three conjugacy classes of groups isomorphic to SL(2, q). If P ;;. 3, the collineation group of ~ contains just one conjugacy class of groups isomorphic to SL(2, q). PROOF. Assume first that p = 2. Then SL(2, q) ~ PSL(2, q). Dualizing if necessary, we may assume by 38.1 and 38.2 that 11 has a point orbit of length q + 1. If 0 is an oval, then the plane ~ is desarguesian by 38.7. If 0 is not an oval, then consists of the points of a line I by 38.3. By the dual of 38.5, we see that D. has a line orbit $3 of length q + 1. If $3 is the dual of an oval, then ~ is desarguesian by the dual of 38.7. Therefore we may assume that $3 is not the dual of an oval. But then the dual of 38.3 implies that there exists a point P with P I x for all x E $3. Obviously, P II. By 38.5, the points of I are the centres of the involutions in 11 and the lines through P are their axes. Moreover, the points not on I and distinct from P form an orbit of 11. Assume now that p > 3. We shall show in this case that 11 fixes a non-incident point line pair (P, I), that the p-elements of 11 are elations with centres on I, and that D. acts transitively on the set of points not on I and distinct from P. As P is odd, 18(D.)1 = 2. Let 1 =t= a E 8(11). If a is not a homology, then a fixes a Baer subplane 0 of ~. Let s be the order of O. Then 52 = q. As a is in the centre of 11, the group 11 induces a group 11* of collineations of
°
°
202
VI. Flag Transitive Planes
q=9. Hence Itl*l=t9(9 2 -I)=5·9·8 divides the order of the
collineation group of U. As s = 3, the plane G is desarguesian, whence 5 ·9 . 8 divides IPGL(3,3)1 = 27 . 16 . 13, again a contradiction. Thus a is a homology. If P is the centre and I the axis of a, then pc,. = P and I!J. = I. Moreover P I I. tl cannot consist of (P,l)-homologies, as otherwise Itll = q(q2 - 1) would divide q - 1. Hence tl has a non trivial orbit t on I. If Q E t, then q(q2 _ 1) = Itll = ItlltlQI, whence q(q2 - 1) > ItlQI > q(q - 1). Assume q -=1= 5,7,9,11. Then it follows from 14.1 that ItlQI = q(q - 1), i.e., Itl = q + 1. This yields that tl acts doubly transitively on I. Assume q = 5,7,9 or 11 and Itl-=l= q + 1. Then it follows from 14.1 that q = 5,7,11 and It I = q or q = 9 and It I = 6. In either case tl fixes a point Q on I and hence the line PQ. As there are only q - 1 points other than P and Q on PQ, we infer that tl fixes at least one more point on PQ which is impossible, as a is a (P, I)-homology. Hence tl acts in its natural 2-transitive representation on I. Let Q and R be two distinct points on I. Then Itl Q,R I = q - 1 and tl Q,R/3(tl) is cyclic of order t(q - 1). Hence tlQ,R is abelian. As the Sylow 2-subgroups of SL(2, q) are generalized quaternion groups, it follows that the Sylow 2-subgroups of tl Q, R are cyclic. Hence tl Q, R itself is cyclic. Let S be a point on PQ other than P and Q and let 8 E tlQ, R be such P that Sll = S. . There . exists p E tl with QP = Rand RP = Q. As tl Q,R = tl Q,R and tlQ,R IS cyclIc, we have (8)P = (8). This yields that SP is fixed by 8. Therefore 8 fixes a quadrangle and hence a subplane. This implies that 8 fixes more than two points on I whence 8 E 3(tl). This and So = Simply 8 = 1. Therefore, tlQ, R acts transitively on the set of points on PQ which are distinct from P and Q. Hence tl acts transitively on the set of points not on I and other than P. Let IT be a Sylow p-subgroup of tl Q . Then IT fixes a point S on PQ which is distinct from P and Q. As tl Q , R normalizes IT and acts transitively on PQ \ {P, Q}, we see that IT fixes PQ pointwise. Therefore IT consists entirely of (Q,PQ)-elations. Let p again be arbitrary, i.e., 2 or distinct from 2, and denote by \U* the incidence structure we obtain by removing P and I and all the elements incident with P or I from \U. Then tl acts flag transitively on \U*: As we have seen above, tl acts transitively on the set of points of \U*. If X is a point of \U*, then tlx is a Sylow p-subgroup of tl. Since tlx acts transitively on the set of points on I which are distinct from I n PX, it follows that tlx acts transitively on the set of lines of \U* which pass through X. Put IT = tlx for a fixed point X of \U* and A = tl m for a fixed line m of \U* which passes through X. If Y is a point of \U*, then we assign to Y the coset IT 1] where X17 = Y. Similarly we assign to the line n the coset A8 where m 8 = n. Then YIn if and only if IT 1] n AS -=1= 0. Comparison with the desarguesian plane
39. Translation Planes Whose Collineation Group Acts Doubly Transitively on
'00
203
38.13 Corollary. Let p be an odd prime and let SU be a projective plane of order q = pro If ~ is a collineation group of \U isomorphic to SL(2, q), then tl fixes a non-incident point line pair (P, I). Moreover, if IT is a Sylow p-subgroup of tl, then II consists entirely of (Q, PQ)-elations, where Q is a suitable point on I.
39. Translation Planes Whose Collineation Group Acts Doubly Transitively on I co We start with the following theorem.
39.1 Theorem (Luneburg 1964). Let G = PSL(2, q) and K = PfL(2, q) and let H be a subgroup of K containing G. If H admits a faithful representation as a doubly transitive permutation group of degree n + 1, then either n = q and H is any group between G and K or one of the following is true: (1) q = 4, n = 5 and H = PSL(2,4) or H = PfL(2, 4). (2) q = 5, n = 4 and H = PSL(2, 5) or H = PGL(2, 5). (3) q = 7, n = 6 and H = PSL(2,7). (4) q = 8, n = 27 and H = PfL(2, 8). (5) q = 9, n = 5 and H = PSL(2,9) or H = PSL(2,9)
PROOF. First we show that these six exceptions really occur and that there are no other possibilities for H if q and n are as listed. (1). It is easily seen and well known that PSL(2, 4) ~ PSL(2, 5) ~ A 5' Hence PSL(2,4) has a representation as a doubly transitive group of degree 6 coming from the isomorphism PSL(2, 4) ~ PSL(2, 5). As every such representation is equivalent to the action of PSL(2,4) via inner automorphisms on the set of Sylow 5-subgroups of PSL(2, 4), we see that H can be any group between PSL(2,4) and PfL(2,4). We infer from IPfL(2,4) : PSL(2, 4)1 = 2 that we have for H just the two possibilities listed. (2). We infer from PSL(2, 5) ~ PSL(2,4) that PSL(2,5) has a doubly transitive representation of degree 5. Every such representation is equivalent to the action of PSL(2, 5) via inner automorphisms on the set of Sylow 2-subgroups, whence H = PSL(2,5) or H = PGL(2,5), as PGL(2, 5) = PfL(2, 5). (3). The groups PSL(2,7) and PGL(3,2) are isomorphic (see e.g. Huppert [1967, II.6.14 (4), p. 183]). From this we infer that PSL(2,7) has a doubly transitive representation of degree 7. Every such representation is equivalent to the action of PSL(2,7) via inner automorphisms on a conjugacy class of A 4's. As there are two such classes which fuse under PGL(2,7) into one, we see that H must be distinct from PGL(2,7). There are no further possibilities for H, since PGL(2,7) = PfL(2, 7).
204
VI. Flag Transitive Planes
(4). Let G = PSL(2,8) and let S be a Sylow 3-subgroup of G. Then 1S 1 = 9, as 1G 1 = 8(8 2 - I). It follows from 14.1 that S is cyclic. Moreover, its normalizer is a dihedral group of order 18. Hence there are 4·7 = 28 Sylow 3-subgroups in G. Therefore G and hence K = PfL(2,8) have a transitive representation of degree 28. Let T be a Sylow 3-subgroup of G distinct from S. Then S n T = {1} by 14.3. Thus S n 9C G (T) = {I}. Therefore, S splits the set :t of Sylow 3-subgroups of G which are distinct from S into orbits :t 1 , :t2 , and :t3 each of length 9. Let S* be a Sylow 3-subgroup of K which contains S. Then IS*I = 27. Moreover S* either leaves invariant all the :t/s or acts transitively on :t. The polynomial f = x 3 + X + 1 is irreducible over GF(2), but reducible over GF(8). Hence there exists a E GF(8)\GF(2) such that j(a) = 0. Moreover, a generates G F(8) *. Therefore
Ai =
(a ~ ~ i
i
i ),
= 0, 1, ... , 6
are 7 distinct matrices. Each of these matrices has order 3. Also
A.-1 = ( 1 I a-i
~i).
This proves that the A/s lie in 7 distinct Sylow 3-subgroups of G. Put 4
C = ( a a ). a4 a2 3 Then C =A o and hence o(C)=9. We may therefore assume that S =
~ ),
(~3 ~4
2 B6= ( a 4 a
),
6
a6 ) ' a
2 B2 = ( a 6 a
6
3
Ej = ( a
1
a ) a2 '
B = ( as
a ) a4 '
Es= ( a 4 a
~2
Bs =
4
aJ 6
E7= ( a 3 a
3
3
),
(~
4
a ) a6 '
:), 6
a 1
).
Hence A 1 and A6 are conjugate under S, whereas there is no element in S which transforms A 1 into Ai or Ai- 1 for i = 2, ... , 5. Hence the Sylow 3-subgroups containing A 1 and A6 belong to :tl say, whereas the Sylow 3-subgroups containing A 2 , . . . ,A s are in :t 2 U:t 3. Now, if A is the mapping defined by (x, y)A = (x 2, y2), then A -IA IA = A 2. Hence S* is transitive on :t. This proves that K acts doubly transitively on :t u {S}. Moreover, H = K is the only possibility for H, as 28·27 divides IHI. (5). The groups PSL(2,9) and A6 are isomorphic (see e.g. Huppert [1967, 11.6.14 (6), p. 183]). Hence PSL(2,9) admits a doubly transitive representation of degree 6. Each such representation is equivalent to the action of PSL(2,9) via inner automorphisms on a conjugacy class of As's.
39. Translation Planes Whose Collineation Group Acts Doubly Transitively on
'00
205
As PSL(2,9) contains exactly 12 subgroups isomorphic to A s which are all conjugate under PGL(2,9), we see that H cannot be equal to PGL(2, 9). Let a E PfL(2,9) be defined by x a = x 3 . Then 0 fixes both conjugacy classes of As by the Frattini argument. Hence H = PSL(2,9)<0) also admits a doubly transitive representation of degree 6. (6). G = PSL(2, 11) has a maximal subgroup M isomorphic to As. As IPSL(2, 11)1 = t 11(112 - 1) = 11 '60, we see that PSL(2,11) has a representation of degree 11. As 11 is a prime and as G is not soluble, G acts doubly transitively in this representation by a theorem of Burnside (see e.g. Huppert [1967, V.21.3, p. 609]). As PSL(2, 11) has 22 subgroups isomorphic to As which are all conjugate under PGL(2, 11), we see that there is only one possibility for H. N ow we prove that these are the only exceptions. Let G = PSL(2, 9) ~ H ~ K = PfL(2,9) and assume that H admits a faithful representation as a doubly transitive permutation group of degree n + 1. If q ~ 3, then G has an abelian characteristic subgroup N of order q + 1. As N is also normal in H, we have that N acts transitively and hence regularly, whence n + 1 = q + 1, i.e., n = q. We may therefore assume q > 3. Then G is simple. Since G is normal in K, it is also normal in H. This yields that G is a minimal normal subgroup of H. Therefore, by Burnside [1955, § 154] the group G acts primitively in the given representation of H. In order to determine n, we therefore have to compute the indices n + 1 of maximal subgroups of G and to check whether or not n(n + 1) divides IHI. As IHI is a divisor of IKI, we have
n(n + 1) divides ;q(q2 - 1),
(j)
where q = pS with p a prime. Using 14.1, we see that the only candidates for maximal subgroups of G are:
1. Normalizers of Sylow p-subgroups. 2. Dihedral groups of order q + 1 or 2(q + 1) according to whether q is odd or even. 3. Dihedral groups of order q - 1 or 2(q - 1) according to whether q is odd or even. 4. Groups isomorphic to PSL(2, pr) with sr - I > 3. 5. Groups isomorphic to PGL(2, pr) with p > 2 and sr- 1 = 2k > 4. 6. Groups isomorphic to A 4 , if q == ± 3 mod 8. 7. Groups isomorphic to S4' if q == ± 1 mod 8. 8. Groups isomorphic to As, if q == ± 1 mod 10. Case 1. Here we get n = q. Case 2. Let D be a dihedral group of order q + 1 respectively 2( q + 1). Then n + 1 = IG: DI = tq(q - I) in either case. Therefore t(pS - l)pSI = (p2s - pS - 2) divides sp S(p2S - 1) by (j). This yields that ~ (p2s pS _ 2) divides 2s(pS + 1), as (tpS(pS - 1), tpS(pS - 1) - 1) = 1. Hence
t
206
V1. FlagJ ranS1l1ve t'lanes 39. Translation Planes Whose Collineation Group Acts Doubly Transitively on
p2S _ pS _ 2
< 4sps + 4s.
This implies p2s - (4s
+ l)pS < 4s + 2, whence
From this we obtain the inequality
P
s _ 4s + 1 2
pS(p2s - 1) <,; 4slDil2 + 21 Dil < 8sIDY. Assume pS(p2s - 1) > 8slDlf. Then
.
< 4s + 1 +
pS+I(p2(s+1) _ 1) > pS+I(p2s+2 - p2) > 8p 3s lDil 2 > 8(s + I)IDiI2.
4s + 1 2
which yields pS < 8s + 2. Assume pI> 8t + 2. Then pl+1 > 8tp + 2p and hence pl+ I > 8(t + 1) + 2. If P > 11, then p > 10 and. hence pS > .8s + 2 for all s. Thus pS < 8s + 2 imphes p = 2, 3, 5, or 7. ThIS ~oget?er wIt.h the above argument yields pS = 2,4, 8, 16, 32, 3, 9, 5, or 7. Usmg (j), we fmally obtain q = 4 or 8. But then n = 5 respectively 27; i.e., we have case (1) or case (4). . Case 3. In this case we have n + 1 = ~ q(q + 1). It follows from (J) that ~ q(q + 1) - 1 divides sq(q2 - 1) and hence 2s(q - 1). From this we deduce p2s + (1 - 4s)pS <,; 2 - 4s < 0 whence pS < 4s - 1. Assume p > 3 and pS > 4s - 1. Then pS+ I > p(4s - 1) > 2~4s - 1) > 4(s + 1)4- 2. Therefore pS+ I > 4(s + 1) - 1. Thus pS < 4s - 1 YIelds p = 2. Now 2 = 16 > 4· 4 - 1. Assume 2s > 4s - 1. Then s > 4 and 2s+ I > 2(4s - 1) = 4 . 2s ~ 2 = 4(s + 1) + 4(s - 1) - 2 > 4(s + 1) - 1. Therefore pS <,; 4s - 1 implIes pS = 2, 4, or 8. Using (j) we finally obtain pS = 2 and n = 2 = q. . Cases 4 and 5. Here we have n + 1 = a- s-r(p2s - 1)(p2r - 1)-1 WIth a = 1 or 2. By (j) there exists kEN with
p
a- ps-r(p2s _ I)(p2r _ I)-I[ a- ps-r(p2S - I)(p2r - 1)-1 - l]k
Therefore
pS-r(p2S _
I)(p2r - 1)-1 - I]k = sapr(p2r - 1).
As s - r > 1, we see that p does not divide n. Thus a- ps-r(p2s - 1)· (p2r _ 1)-1 - 1 divides as(p2r - 1). Therefore
ps-r[ a- l (p2S _ I)(p2r _ 1)-1 - 1] Furthermore s > s - r
> 2r
< as(p2r - 1).
=
< as(p2r - 1)(p2r -
obtain n = 6, i.e., case (3), and q = 9 yields n = 14 which is impossible, as 14 does not divide 2 . 9 . 80. i = 3: Using the above inequality and q == ± I mod 10, we obtain q = 9, 11, 19, or 29. By (j), the numbers 19 and 29 cannot occur. If q = 9 then n = 5, i.e., we have case (5); and if q = 11 then n = 10, i.e., we obtain case
(6).
D
The next theorem has been known to several people for quite some time, but it has never appeared in print as far as I know. So I do not know to whom I shall ascribe it. The only thing I can trace is that Piper proved in 1963 that a plane satisfying the assumptions of the next theorem is a translation plane.
a) b) c)
mis desarguesian, G = TG p , and Gp ;;;;: SL(2, q). mis desarguesian of order 9 and G = TG p with Gp ;;;;: SL(2, 5). mis desarguesian of order q = 2r and G = TG p , where Gp is dihedral
order 2(q + 1). d) mis a Luneburg plane of order q = 22(2r+ I) and G = TG p with Gp S(22r+ I).
whence
a- l (p2s _ I)(p2r _ 1)-1_ 1 < as(p2r - I)pr-s
i=l: Thenp S(p2S-I)<8·12 2s. If p> 11, thenp(p2-1»8·12 2 • Hence we are left with p = 3, 5, or 7. It is easily seen that q = 3, 9, 5 or 7. As q == ± 3 mod 8, we have q = 5, since we have assumed q > 3. But q = 5 yields n = 4. This is case (2). i = 2: Then pS(p2S - 1) < 8 . 242s. In this case we obtain pS = 3, 9, 5, 7, 11 or 13. As q == ± 1 mod 8, we are left with q = 7 or 9. For q = 7 we
39.2 Theorem. Let mbe a finite affine plane of order q, let P be an affine pOint of mand let G be a group of collineations of mgenerated by shears. If every affine line of mis the axis of a non-trivial shear in G then one of the following is true:
= SpS(p2S - 1). [a-
207
From this and (j) we infer that IG: Dil - 1 divides 2slDJ Hence
(4s + 1)2 4s; I ) < 4s + 2 + 4 2
( Ps _
'00
1)
of ;;;;:
-]
as
and hence p2s - 1 < a 2s(p2r - 1) < 4s(pS - 1), since a < 2 and 2r < s. Thus pS + 1 < 4s which yields the contradiction pS = 2. Case 6, 7, and 8. Let Di (i = 1,2,3) be the stabilizer of a point and let the D.I be numbered in such a way that DI = A 4 , D2 = S4 and D3 = As· Then
m
PROOF. First we show that is a translation plane and that G contains the translation group T of m. Let X be a point on 100 and let II' ... , lq be the affine lines of m carrying X. Furthermore, denote by Si the group of all shears with axis Ii in G and by E the group of all elations in G whose centre is X. By 1.1 and 1.6, E is an elementary abelian p-group. Moreover ISil> 1 and q
lEI
=
IG(X)I
+
2: (ISil- 1) = i=l
q
IG(X)I- q +
2: ISJ i=!
208
VI. Flag Transitive Planes
(Recall t~at G(X) = G n T(X).) As IG(X)I ~ 1 and lEI == ISil == q == that IG(X)I > 1 for all X. As E is a p-group for all X on 100 which induces a group of permutations on 100 possessing only one fixed point on 100 , we obtain by Gleason's lemma (15.1) that G acts transitively on 100 , Therefore there exists an integer h ~ 2 with IG(X)I = h for all X I 100 , Thus mis a translation plane and G contains the translation group of mby 15.2. This yields G = TG p and T n Gp = {I}. Let YJ be a shear in G. Then there exists 'T E T such that the axis of 'T - IYJ'T = E passes through P. Hence YJ = 'TE'T -I = 'T( E'T -IE -I)€ is the product of a translation and a shear the axis of which passes through P. Let y E G. As G is generated by shears, we have by the remark just made that y . = 'T I€I'T 2€2 ... 'TnEn' where the 'T/s are translations and the €'S I are shears wIth axes through P. Using induction , we obtain y = 'TE I E2 '" En where 'T is a translation. This yields in particular that Gp is generated by shears whose axes pass through P. Therefore 35.10 applies to Gp ; we have:
o mod p, It follows
A) Gp ~ SL(2, q). B) q = 9 and G p ~ SL(2, 5). C) q = Y and Gp = C(E) where C is a normal subgroup of odd order of Gp and E is an elation with axis through P. D) q = 22 (2r+ I) and Gp ~ S(22r+ I). In case A), the plane is desarguesian by 38.12. In case B), the plane mis des argues ian by 8.4 and 8.3. In case D), the plane is a Liineburg plane by 31.1. I t remains to consider the case C). In this case Gp is soluble by the Feit-Thompson theorem. As Gp acts transitively on 100 , the group G acts flag transitively on m. Hence G p <;;;; rL(1, q2) by 37.7. Moreover, by 37.9 the ?roup Gp acts o~ the set of points of the desarguesian plane ~ of order q In the same way It acts on the set of points of the plane m. As q is even, all involutions in rL(I, q2) are conjugate. This shows that the shears of m which are contained in Gp are also shears of ~. This yields that the lines of mwhich pass through P are also lines of ~. Therefore mis desarguesian. 0 This yields finally that Gp is dihedral of order 2(q + I). 39.3 Theorem (Schulz 1971, Czerwinski 1972). Let mbe a finite translation plane admitting a collineation group G acting doubly transitively on 100 , Assume furthermore that G does not contain a Baer involution if the order of mis odd. Then mis either desarguesian or a Luneburg plane. PROOF. As mis a translation plane, we may assume G = Gp for some affine point P of m. Let q be the order of mand assume that q is even. If I is a line through P, then Gp ,/ acts transitively on the set of points on 100 which are distinct from I n 100 , Therefore, it follows from 15.5 that G p I contains an elation a =1= 1. As po = P and /0 = /, it follows that / is the' axis of a. This yields that IJt is either desarguesian or a Liineburg plane by 39.2.
39. Translation Planes Whose Collineation Group Acts Doubly Transitively on 100
209
Assume that q is odd. If G contains an involutory homology whose axis passes through P, then by 3.17 and the double transitivity of G we see that G contains non-trivial shears, whence 2{ is desarguesian by 39.2. We may thus assume that G does not contain an involutory homology with affine axis. It then follows by our assumption on the involutions in G that all involutions in G are (P, I oo)-homologies. Hence, IG I being even, G contains exactly one involution. This implies that the Sylow 2-subgroups of G are either cyclic or generalized quaternion groups. Let G be the group induced on 100 by G and assume that the Sylow 2-subgroups of G are cyclic. Then G contains a normal subgroup N of odd order such that G/ N is a 2-group (see e.g. Gorenstein [1968, 7.6.1, p. 257]). As the degree of G is q + 1, the group N having odd order cann~t act transitively on 100 , Therefore N = {1} by the double transitivity of G. But then G is a 2-group, a contradiction. Therefore the Sylow 2-subgroups of Q. are generalized quaternion groups and hence the Sylow 2-subgroups of G are dihedral, since the only involution 'in G fixes 100 pointwise. Suppose first that G is soluble. Then G contains an elementary abelian normal subgroup E. As G acts doubly transitively on 100 , we have that IE I = q + 1. Since q is odd, E is an elemen~ry abelian 2-group. This yields lEI = 4, since the Sylow 2-subgroups of G are dihedra~ Therefore q = 3 and m is desarguesian. We may thus assume that G is not soluble. Moreover, G does not contain a non-trivial normal subgroup of odd orde~ since q + 1 divides the order of any no~trivial normal subgroup of G.-:.. Therefore, by Gorenstein & Walter 1965, G is either isomorphic to A7 or G is isomorphic to a group H with PSL(2, q*) <;;;; H <;;;; prL(2, q*) for some odd prime power q*. Assume first G ~ A 7 , then q(q + 1) divides t7! = 23 . 32 .5.7. As q is odd and not a prime (otherwise mwould be desarguesian), q = 32 and mis the nearfield plane of order 9 by 8.4. But 7 does not divide the order of the collineation group of this plane by 8.3. Thus G is isomorphic to a group H with PSL(2, q*) C; H <;;;; prL(2, q*) with q* odd. Using this and the fact that q is odd, then 39.1 tells us that either q* = q or q* = 9 and q = 5. In the latter case, mis desarguesian of order 5. But the group SL(2,9) does not operate on that plane. Hence q* = q. This shows that G contains a group S which induces a group S* on 100 isomorphic to PSL(2, q) in its natural doubly transitive representation. Let S be minimal among all such subgroups of G and let K be the kernel of the homomorphism S ----;) S*. The group PSL(2, q) is simple, as q > 3. Hence KS' = S. Moreover, (S')* = S*. Therefore S' = S by the minimality of S. Since K consists only of (P,loo)-homologies, K is cyclic. Thus Aut(K) is abelian. We deduce from this and S' = S that K C; 3(S). Moreover 2 divides IKI, as the Sylow 2-subgroups of G are quaternion. Hence by Schur [1907, Satz III and Satz IX], we have either S;;; SL(2, q) or q = 9 and jKj = 6. But the latter case cannot occur, since IKI divides q - l. Hence S ~ SL(2, q) and lli IS desarguesian by 38.12. 0
210
VI. Flag Transitive Planes
40. A Theorem of Burmester and Hughes 40.1 Theorem Let mbe a finite translation plane of order q and let G be a group of collineations of mfixing two distinct points P and Q on 100 and an affine point O. If q is even, assume that G does not contain a Baer involution, and if q is odd, assume that the rank of the translation group over its kernel is twice an odd number. Then G is soluble. Using standard notations, we may assume that V(O) = OP and V(oo) = OQ. If K is the kernel of m:, then G consists of K-semilinear mappings. If 0: E G, denote by comp(o:) the companion automorphism of 0:. Then comp is a homomorphism of G into Aut(K). If G I is the kernel of comp, then G/ G I is cyclic, as Aut(K) is cyclic. Let 0: E G I . Then there exist 0: 1,0: 2 E GL(X) with (x, yt = (x ai, ya 2 ), where X is such that V(O) = {(x,O)lx E X} and V(oo) = {(O,x)lx E X}. The mapping 0:--) (det(0:1),det(0:2» is a homomorphism of G I into K* X K*. Hence G I /G 2 is abelian, where G2 is the kernel of the mapping 0: --) (det( 0: I)' det( 0: 2», Therefore, G is soluble if G2 is. N ext we show that G does not contain a Baer involution. By way of contradiction, assume that 0: is a Baer involution in G. Then the order of m: is odd by assumption. Moreover 0: centralizes a subspace U of V(O) with 2rk( U) = rk( V(O». From this we infer that 4 divides the K-rank of the translation group of m, a contradiction. Let 0: be an involution in G 2 • Then 0: is a central collineation by the remark just made. As 0: fixes 0, P and Q, we see that 0: is a homology. Therefore the order of m: is odd. Let 0: 1 and 0: 2 be as above. Then at least one of the O:j maps every x E X onto - x. For this O:j we then have 1 = det(O:j) = (- lyk(X). This shows that rk(X) is even, as the characteristic is odd. Therefore 4 divides the K-rank of the translation group, a contradiction. Thus G2 is odd, whence G2 is soluble by the FeitThompson theorem. 0 PROOF.
I
1
40.2 Corollary. Let m: be a finite translation plane of non-square order. If G is a group of collineations of m: fixing two points on 100 , then G is soluble. PROOF. Let 0 be an affine point of m: and put H = TG. Then H = THo and Ho is soluble by 40.1. Therefore H is soluble which implies that G is soluble. 0
41. Bol Planes An affine plane m: is called a Bol plane, if there exist two distinct points P and Q on 100 such that each line of m: which does not pass through either of the points P and Q is the axis of an involutory perspectivity which
211
41. Bol Planes
interchanges P and Q. By 4.9, a Bol plane is always a translation plane. Thus, by the remarks after 3.10, the plane m: may be represented by a set 2: of fixed point free linear mappings of a vector space X such that p"L -Ip = "L for all p E "L. This implies, as we have seen earlier, that "L- 1 = "L and that pi E "L for all p E"L and all i E 71. Let K be the centralizer of "L in the endomorphism ring of X and assume that X is finite. Then K = GF(q). Denote by d the rank of X over K. Assume furthermore that X may be identified with the additive group of GF(qd) such that "L C fL(1,qd). If a E "L, then there exist A(a) E {a, 1, ... , d - I} and a E GF(qd)* with x" = xt(a la . 41.1 Lemma. If p,a E "L, then A(pap) = 2A(p)
+ A(a).
PROOF. B~ the remarks made above, pap E "L. Thus A(pap) is defined. Let x P = xt(P a and X O = xt(a lb and x pop = Xt(POPlC . Then
whence A(pap) = 2A(p) congruence modulo d.)
+ A(a).
(Note that equality in this context means 0
41.2 Lemma. If a E"L and kEN, then A(a k ) = kA(a). l PROOF. Let x" = xt(a a and assume that A(a k ) = kA(a). Then
o 41.3 Lemma. A is a mapping from 2: onto the set {O, I, ... , d - I}. PROOF. As K = GF(q) is the centralizer of "L, we have that l f = «x --) XqA(a ) I a E "L) is the Galois group of GF(qd): GF(q). Therefore it suffices to prove that ~ = {(x--)xt(a»la E "L} is a subgroup of f. This is certainly true, if d = 1. Assume d > 1. Then there exists p E 2: with 0< A(p) < d. Let p be such that A(p) is minimal. Let a E "L and put f=(A(p),A(a». Then either f=(2A(p),A(a» or f=(A(p),2A(a». Let f= (2A(p),A(a». Then there exist integers rand s withf= 2rA(p) + sA(a). From this we obtain by 41.2 and 41.1 that f=A(praSpr). Similarly f= A(aSpra S), if f= (A(p),2A(a». Thus (X--)Xt(fl) E.0.. By the minimality of A(p), we have f> A(p) > (A(p),A(a» = f. Thus f= A(p), whence A(p) divides A(a). This together with 41.2 yields .0. = {(x --) XqkA(Pl) IkE N}. 0
For 0.,;;; i";;; d - 1 we put 2: j
=
{aIA(a) = i}.
41.4 Lemma. If p E "L, then p"Ljp = 2: j+ 2,\(P) This follows immediatly from 41.1.
.
( .• /1["''',,-...<-\;,:,
212
VI. Flag Transitive Planes
41.5 Lemma. 2:0 is a cyclic group. PROOF. Let a E 2: 0 , Then X O = xw i , where w is a generator of GF(qd)*. Let 11= a be such that i is minimal. Pick T E 2:0 and let x'T = xwJ. Put f = (i, i). Then either f = (2i, j) or f = (i,2). Assume f = (2i, j). Then there exist s,t E 7l such that f= 2is + it. Put P = aSTta s • Then A(p) = 2sA(a) + tA(T) = O. Thus p E 2: 0 , Moreover, x P = xw f . Therefore f > i. On the other hand f= (i,)) ..;; i. Hence f= i and i divides). This proves 2:0 ~
.
ven order, then
This follows from the fact that 2:0 is a subgroup of GL(I,qd) and that all subgroups of a cyclic group are characteristic subgroups of that group.
For more information on Bol planes see a paper by Kallaher and Ostrom which will appear in J. Algebra.
" " -_ "::"'2k' P2k ""::"'0 -_ P k "..::...op k -_ "::"'0+2kA(p)
o
41.8 Lemma. If d is odd, then 2: is a group. PROOF. Since d is odd, there exists for each i E {O, I, ... , d - l} a k E {O, 1, ... , d - I} such that i == 2k mod d. By 41.3, there exists a a E 2: with ;\(a) = k. Therefore,
2: i
= 2: 2A (0) = a2: oa = 2: oa 2 ~ 2: 0
o
',\~,, ' "
I'
41.10 Theorem (Kallaher 1972). If mis a finite Eol pia ~h mis a nearfield plane.
213
41.6 Lemma. 2:0 is normalized by 2:.
PROOF. By 41.6, 41.4 and 41.2, we have
Ii
\·Jt\~\~:k~
PROOF. Let A be the group generated by all the shears interchanging V(O) and V( 00). As each V( a) is the axis of exactly one shear in A it follows from 35.10 and the Feit-Thompson Theorem that A is soluble. This implies that
41.7 Lemma. If p E 2: 1 , then 2: 0
1
41 Bol Planes
\y)
m
4l.9 Theorem (Kallaher & Ostrom 1971). Let be a finite Eol plane. If the rank of the translation group of mover its kernel is twice an odd number, then mis a nearfield plane. PROOF. We may assume that the points of mare the elements of a vector space V = X EEl X where X is a vector space of rank d over the kernel K = GF(q) of m. We may even further assume that V(a) is the axis of an involutory perspectivity interchanging V(O) and V( 00) for all a E 2:. This 1 perspectivity is represented by the mapping (x, y) ---7 (y 0- , x 0), as we know. Combining this collineation with the collineation (x, y) ---7 (y, x) yields that (x,y)---7(X O,yO-I) is a collineation of m. Moreover, by 40.1 the group of all collineations fixing V(O) and V( 00) is soluble. Therefore, <2:) is soluble. As
21S
42. Ovals in Finite Desarguesian Planes
CHAPTER VII
Translation Planes of Order q2 Admitting SL(2,q) as a Collineation Group
Q fl Q. Let r be the number of secants through Q and s be the number of tangents. Then 2r + s = k. If k = n + 2, then there are no tangents of a, whence s = and n = 2(r - l). This proves b). 0
°
We have also proved: 42.2 Lemma. Let a be a k-arc and let Q be a point not in a. If s is the number of tangents of a through Q, then s == k mod 2. Next we prove: 42.3 Lemma. Let ~ be a projective plane of order q > 3, q being odd. If a is a q-arc of~, then there exists a point Q of ~ with Q fi:. a such that Q is on at least 5 tangents of a.
This chapter gives the complete description of all translation planes of order q2, having GF(q) contained in their kernels, and admitting SL(2,q) as a collineation group. Before we can give this description, we have to prove some results on ovals in finite desarguesian planes of odd order, among them Segre's famous result that any oval in such a plane is a conic. Moreover, we have to investigate twisted cubics in projective 3-space and similar configurations in projective 3-spaces of characteristic 2.
42. Ovals in Finite Desarguesian Planes Most of the results in this section are due to B. Segre. See Segre 1961 for bibliographical details. A set a of k points in a finite projective plane is called a k-arc, if no three points of a are collinear. Hence if I is a line of the plane I,l3 and if a is a k-arc, then II n al < 2. We shall call I a secant, a tangent or an exterior line of a, if II n a I = 2, 1 or respectively.
°
42.1 Lemma. For the number k of points of a k-arc in a projective plane of order n we have:
a) k b) k
< n + 2, if n is < n + 1, if n is
even. odd.
Let a be a k-arc and pick PEa. Then there are k - 1 secants through P, whence k - 1 < n + 1. Hence k < n + 2 in all cases. Pick PROOF.
214
PROOF. Assume that there is no such point and denote by sQ the number of tangents through a point Q of ~. Then sQ = 2 for all Q E: a, as a is a q-arc, and sQ E {1,3} for all other points by 42.2. Let t be a tangent and put P = ant. Then L(SQ - 1) = 2(q - 1) where the summation is over all points Q on t other than P. Let r be the number of those points Q for which sQ = 3. Then 2r = 2(q - 1). Hence r = q - 1. Therefore there is exactly one point X on t such that Sx = 1. Let ~ be the incidence structure whose points are the points Q of ~ with sQ = 3 and whose lines are the tangents of a. Then ~ is a tactical configuration with v = q2 + q + 1 - q - 2q = (q - 1)2, b = 2q, k = q - 1 and r = 3. Therefore 3(q - 1)2 = 2q(q - 1). This yields 3(q - 1) = 2q, i.e. q = 3, a contradiction. Therefore there exists Q with sQ > 3. Thus sQ ;> 5 by 42.2, as q is odd. 0 42.4 Lemma (B. Segre). Let I,l3 be the desarguesian plane over GF(q) = K and let a be a k-arc containing the points PI = (l,O,O)K, P 2 = (O,I,O)K, P3 = (0,0,1 )K. Then the tangent lines ti) through Pi have equations:
with
II1... \ IIJ=\ki) =
-1, where h = q - k + 2.
° °
PROOF. Pick P = (a\,a 2,a 3)K E a\{P I ,P2,P3 ). Then ai =I=- for all i, as no three points of a are collinear. Put b l = a 2a 3- I , b 2 = a 3a\-1 and b 3 = a 1a2- 1• Then the lines PPi have equations x 2 = b l x 3 , X3 = b2x I and XI = b 3 x 2 respectively. As no line I with equation x 2 = bX3 and b =I=- contains P 2 or P 3 , we have 1= tl) for some j or 1= PP 1 for some P E a\{P\ ,P2 ,P3 }. Hence h
IT k II j= \ I)
P Eo P=j=Pi
b \ ( P)
= - 1,
216
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
as the product over all non-zero elements of K is - 1. Similarly h
IT
ITkij )= I
bi(P)=-l
whence a3b l (a l + a3 ) = a3 b2(a 2 + a3 ). As a is a k-arc, we have a 3 =1= 0. Hence bl(a l + a 3) = bia2 + a 3). Similarly bia} + a 2) = bla l + a 3) and bla 2 + a3) = b l (a 2 + a)). As s is a tangent through Q and Q =1= Ql,Q2,Q3' we see that bi =1= for all i. Hence there exists "A E K* with "A(b l ,b2,b 3) = (a2 + a3,a3 + al,a l + a 2)· Thus
°
i=2,3.
for
PEa P=I=Pu
Moreover bl(P)biP)blP) = 1. Thus 3
- I
=
3
h
IT IT kij IT i=1 )=1
PEa P=I=Pu
bi( P)
=
h
IT IT kij .
o
i=1 )=1
42.5 Lemma. Let ~ be the desarguesian plane over GF(q), q being odd and assume that a is a q-arc of ~. Let t I ,t2, ... , tr be tangents of a passing through the point P. Put Qi = ti n a and denote by Si the second tangent of a through Qi' Then there is a conic c of ~ containing all the Qi and such that Si is a tangent of c for all i. PROOF. We may assume r ;> 3. We may further assume that QI = (1,O,O)K, Q2 = (O,I,O)K, Q3 = (O,O,I)K, where K = GF(q), and that Sl and S2 have the equations x 2 = - X3 and X3 = - XI respectively. Then the conic c determined by the equation X IX 2 + x 2X 3 + X 3X I = passes through QI' Q2,Q3 and has tangents SI,S2' The equations of t 1 , [2' t3 and S3 are x2=k))X3' x 3 =k 21 x l , x l =k 3I X2, XI = k 32 x 2· Hence by 42.3 we have kllk21k31k32 = -1. As t l , t2,t3 are confluent
°
o 0= det - k21 1
° - k31
Hence k32 = - 1. From this we infer that S3 is also a tangent of c. The triangle QI,Q2,Q3 and the triangle whose sides are SI,S2,s3 are perspective, their centre of perspectivity being the point (1,1, I)K as is easily seen. As QI' Q2' Q 3 can be replaced by any three of the points QI' . . . , Qr' it follows that the triangle Qi' Qj' Qk and the triangle whose sides are Si ,s) ,Sk are perspective and hence axial, since ~ is desarguesian. Pick Q E {Q4' ... , Qr}' If Q = Qi' put S = Si' Let Q = (a l ,a2,a3)K and let s have the equation bix i + b2x 2 + b3X3 = 0. Then the triangles Q, QI, Q2 and S,SI,S2 are axial. This means that the points s n QI Q2' Sl n QQ 2 and S2 n QQl are collinear. Now Ql Q2 has equation x3 = 0. Hence S n QI Q2 = (b 2, - b l ,O)K. The line QQ2 has equation a3x I - a l x 3 = 0. Hence Sl n QQ2 = (ai' - a 3 ,a 3)K. Finally QQI has equation a3x 2 - a 2x 3 = and hence s2 n QQl = (- a3,a2,a3)K. Thus
°
217
42. Ovals in Finite Desarguesian Planes
0= (a 2 + a3)a l + (a 3 + a l )a 2 + (a l + a 2)a 3 = 2(a l a2 + a2a3 + a3a l ) whence Q E c, as q is odd. Moreover, if (x I ,X 2 ,x 3)K is on s, then 0= (a 2 + a3)x l + (a 3 + a l )x 2 + (a l + a 2)x 3. Hence s is a tangent of c. This proves 42.5. 0 42.6 Theorem (B. Segre). Let 0 be an oval in the desarguesian plane GF(q), q being odd. Then 0 is a conic.
~
over
PROOF. This is certainly true if q = 3. Thus we may assume q ;> 5. Pick P E o. Then all the secants of 0 through P are tangents of the q-arc 0\ {P }. Hence there is a conic c with Cd o\{P} such that all the tangents of 0 which have their points of contact in o\{P} are also tangents of c by 42.5. Pick Q E 0\ {P}. The same argument yields a conic b with b d 0\ { Q} such that the tangents of b through points of o\{ Q} are also tangents of o. Hence c and b have at least q - 1 ;> 4 points and at least as many tangents 0 in common. This is more than enough to make sure that c = b. A collineation of a desarguesian projective plane will be called projective, if it is induced by a linear mapping of the underlying vector space. 42.7' Lemma. Let P, PI' P 2, P 3, Q and R be the points of a 6-arc in a projective plane ~ over a commutative field. Let a and T be projective collineations of ~ with the properties: a) P/J = Pi = P/, for i = 1,2,3. b) Qa = P and R l' = P.
Then there exists a conic c with Cd {P,PI>P 2,P3,Q,R}, if and only pa and p1' are collinear.
If P,
PROOF. Assume that P, pa and p1' are collinear. As P =1= Q, Rand Qa = P, R = P, we have P =1= pa,p7'. Thus ppa = pp1' = I is a line. As a is projective, it induces a projectivity of the pencil of lines through Q onto the pencil of lines through P. Hence 7'
c = {x n x a
1X
is a line through Q }
is a conic (Steiner's construction of conics). Moreover P,Q,PI ,P2,P 3 E c, as QPi n (QPit = QPi n PPi = Pi' Furthermore 1= ppa = (QPt and hence I is a tangent of c. Similarly b = {y n ya I y is a line through R} is a conic with P, R,P 1 ,P2 ,P3 E b and I is a tangent of b. As c and b have the four points P,
218
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
P ,P2,P3 and the tangent I in common, c = b. This proves one half of the lemma. To prove the converse let c be a conic containing P,Q,R,P I ,P2,PJ' Then J
c = {x n x ° Ix is a line through Q }
42. Ovals in Finite Desarguesian Planes
°
equation (PIXI - P3X3) - a(P2x2 - PJx J) = in the original coordinates. In the new coordinates, SI ,S2,s,t l ,t2,t have equations Y2 - KIYJ = 0,
= {y n Y I Y is a line through R }
Y2 - A'YJ
T
by Steiner's theorem. Hence ppo and pp T are tangents of c through P, whence ppo = PPT. 0 ~ be the projective plane over GF(q), q > 5 being odd. If 0 is a q-arc in ~, then there exists exactly one point P in ~ such that 0 U {P} is an oval. In particular, there exists exactly one conic c with o C c.
42.8 Theorem (B. Segre). Let
Put (0,0,1)
b2x J - bJx 2 = 0, (b 2 - b3)x I
°
bJx I - blx J = 0,
+ (b J
-
°
b l x 2 - b2x I
= 0,
b l )X 2 + (b l - b2)x J = 0,
Pick X E QJ' QJ + Q,. Hence all the coordinates of X are distinct from 0. Thus X = (PI' ,P2-" pJ-I)K. Any line through X has an equation of the form u(PIXI - PJx J) V(P2X2 - PJx J) = 0. As neither of the points QI ,Q2,QJ is on a tangent of 0 through X, the tangents through X may be represented by the equations (PIXI - PJx J) - a(P2 x 2 - PJx J) = and (PIXI - PJx J) - (3(P2 X2 - PJx J) = with a =1= =1= {3. Since the points QI,Q2'X are not collinear, the vectors (l,O,O)=e l , (0,1,0) = e2 and (PI- I ,P2- I,PJ- I) = f form a basis of the underlying vector space. Denote by (Yl 'Y2 'Y3) the coordinates with respect to this basis and let sand t be the tangents of 0 through X. We may assume that S has the
°
°
°
YJ - A2 YI = 0,
YI -
K JY2
= 0,
YI - A3Y2 = 0.
the above equations, one finds that SI 'S2,s,f l ,f2,t have the equations (KI + p;I)X3 - PJ-'x 2 = 0, (1 + K2 PI- I)X I - pJ-IXJ = 0, PI-'(PIXI - P3 XJ) (AI + P2- I )X 3 - P3- IX 2 = 0,
I
K3P2-
(1
(P2 X2 - P3X3) = 0,
+ A2 PI I)X I -
PI-I(PIXI - PJx J) - A3P2- I(P2 X2 - P3x3)
P3-IX3
= 0,
= 0.
Comparison with the original equations for these lines yields I AI -- (b 2P3- I - b3P2-I)b3'
Put c I = (a 2P2 + a 3P3)(b 2P2 - b3P3)' c 2 = (a 3P3 + a I PI)(b 3PJ - bIPI)' c 3 = I (aIPI + a2P2)(b I PI - b2P2)' Then a{3 = C2CI- , as -1 = KIK2K3AIA2A3 by
42.4.
Next we use the basis e l ,eJ,j. Denote by (YI 'Y2'Y3) the coordinates with respect to this basis. Then SI ,s,SJ,f l ,f'!3 have equations Y2 -
+ (a 3 + a l )x 2 + (a l + a2)x J = 0. o\c. Then X is on neither of the lines QI + Q2' Q2 +
= 0,
+ Y2 e2 + YJf= (YI + YJPI-I)e l + (Y2 + Y3P2- I )e 2 + Y3P3- le 3' 1 I Hence YI + Y3PI- = XI' Y2 + YJP2- = x 2, Y3P3- 1 = x 3· From this we get 1 YJ = XJP3' Y2 = x 2 - XJP3P2- , YI = X I - X3P3PI- I. If one puts these into
and sl ,S2,S3 ,S4 have equations (a 2 + a3)x I
K 2Y,
= eJ • Then
P= Q.
°
= 0,
YJ -
Ylel
PROOF. To prove uniqueness assume that there are two such points P and Q. Then c = 0 U {P} and b = 0 U {Q} are conics by 42.6. Moreover c and b have at least q points in common. As q > 5, we see that c = b, whence To prove the existence of such a P we have only to show that there exists a conic which contains o. By 42.3 there exists a point Q in ~ which is on r > 5 tangents of o. Let t I' . . . , tr be these tangents and put Qi = ti n o. By 42.5 there exists a conic c with Qi E c for all i = 1,2, ... , r and such that the tangents Sj of c through Qi are also tangents of o. Moreover Sj =1= ti for all i. We may assume that QI = (l,O,O)K, Q2 = (O,l,O)K, Q3 = (O,O,I)K and Q4 = (l,I,I)K. Let Q = (b l ,b 2,b 3)K. Then c has the equation a l x 2x J + a2x Jx I + a Jx l x 2 = with a l + a2 + a J = and aj =1= for all i. Moreover t l ,t2 ,t 3 ,t4 have equations
219
K I Y3
= 0,
K2YI
Y3 -
= 0,
YI -
K3Y2
= 0, = 0.
Y2 - AI Y3 = 0, Y3 - A2 YI = 0, YI - A3Y2 (The K'S and A'S have, of course, a new meaning.) The coordinates (YI 'Y2'Y3) are connected with the coordinates (x I ,x2,x3) by the equations YI = XI - PI-~2X2' Y2 = X2P2' Y3 = x J - PJ-~2X2' Putting these into the above equations, and comparing with the original equations for the lines S I' ... , t 3 yields KJ
K3
= -
a2( a2P3- 1 + aJP2- 1) -I, 1
= -(a I P2- +
a2PI I )a 2-
1 ,
K2 = PI(P3(1 - a))-I,
Al = b2( b3P2- 1 - b2P3- 1),
I A3 -- (b IP2-I - b2PI-I)b2' From - 1 = KI K2 KJA,A 2A3 we obtain after a little computation (1 - a)(1 (3) = C3CI- I. Hence a + {3 = (c l + c 2 - C3 )C I- I , as exf3 = C2 C]-I.
220
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
Finally we play the same game with the basis el' f, and e4 = (1, I, I). Then SI ,S,s4,t I ,t,t4 have equations
Y2 -
IY3 Y2 - Al Y3 K
= 0, = 0,
2YI = 0, Y3 - A 2 YI = 0, Y3 -
K
Y2 = 0,
YI -
K3
YI -
A3 Y2
= 0.
Moreover the coordinates (YI'Y2'Y3) and (X I ,X 2,x3) are connected by the 1 1 1 equations XI = Yl + Y2Pl- + Y3' X2 = Y2P2- + Y3' X3 = Y2P3- + Y3' Put these values in the original equations; then comparison with the above equations yields
1
KI
= -(a 2 + a3)(a2P3- +
K3
=
Al A3
1 a3P2- ),
K2
= -PI (PI - P3
43. Twisted Cubics Let ~ be a projective space of rank 4 over a commutative field K. A set of points of ~ which can be mapped by a collineation of ~ onto the set <£ = {(S3,S2t,st 2,t 3)Kjs,t E K, (s,t) =F (O,O)} will be called a twisted cubic. The set (£ is contained in the intersection of the two cones
~I = {(Xl 'X 2 'X 3 ,x4)KjXI X3 - xi = O} and ~2 = {(Xl 'X 2 'X 3 'X4)Kj X2X4 - xj =
+ ex(P2 - P3»-t,
+ f3 = (c l + C2 - c3)ci I ,
pi
°
we
P2-
l Assume alPI + a2P2 + a3P3 = 0. Then a 1P2- P3- I + a 2P3+ a3PI= 0. But this is not the case, as X is not on c. Hence b 1Pi(P3 - p0 b2Pi(PI - P3) + b3pj(P2 - PI) = 0, since 2 =F 0. This is equivalent to
O}.
The. centre of ~I is the point (O,O,O,I)K and the centre of ~2 is the point (1,O,O,O)K. Moreover the line L = {(x,O,O,y) j x,Y E K} is contained in ~I n ~2' The tangent plane Tl at ~l containing L has equation x3 = and the tangent plane T2 at ~2 containing L has equation X 2 = 0. Conversely we have
-(a 2 + a 3)pi l + (a 3 + al)p;I + (a l + a 2)p;I)(a 2 + a 3)-I, = (b 3 - b2)(b2P3- 1 - b3P2- I), A2 = -PI(PI - P3 + f3(P2 - P3»-I, = -(b 2 - b3)PI- 1 + (b 3 - b l)P2- 1 + (b l - b2)P3- 1)(b 2 - b3)-I.
As -1 = KIK2K3AIA2A3' exf3 = C2 C I- 1 and ex obtain after some computations that
221
43. Twisted Cubics
1
+
43.1 Theorem. Let ~I and ~2 be two cones of ~ with distinct centres PI and P2. Assume that PI + P2 is contained in ~I n ~2 and denote by Ti the tangent plane of ~j with PI + P 2 c: Tj. If TI =F T 2, then ~\ n ~2 is the union of PI + P 2 and a twisted cubic <£. Moreover P I ,P 2 E (£. PROOF. We may assume PI = (O,O,O,I)K and P 2 = (1,O,O,O)K. As the stabilizer of L = PI + P2 acts doubly transitively on the set of planes through L, we may assume that TI has equation X3 = and that T2 has equation x 2 = 0. Let ~i be represented by the equation
°
-0 aI;ixl2 + a2;jx22+ a3;jx 32 + a4;jx42 + a l2 ;jx lX2 + ... - . Then al; i = a4; i = 0. As L c: ~I n ~2 we have a 14 ; jX IX4 = for all XI ,x4· Hence a 14 . i = 0. Furthermore (Xl' 1,0,x4)K f/:. ~I' since TI is a tangent plane of ~;. Hence a2; 1+ a 12 ; IXI + a 24 ; IX4 =F for all XI ,X 4· Thus a2; I =F and a I2 ; 1= a24 ; 1= 0. We may assume a 2 ; 1= 1. Similarly (x l ,0,I,x4 )K E~2' Hence a3;2 + aI3;2xl + a34 ;2 x 4=FO for all X I'X 4. Therefore a 13 ;2 = a34 ;2 = and a3;2 =F 0. Again we may assume a3;2 = 1. Thus ~I is represented by
°
Thus the points Q, X and Y = (PI- 2,P2- 2,P3- 2)K are collinear. Let a be the projective collineation induced by
° ° ° Then Qio= Qi for i= 1,2,3 and Q;=X and XO= Y. Using QI' Q2' Q3 and Qs instead of Q\,Q2,Q3,Q4 we obtain a projective collineation T with Q/ = Qi for i = 1,2,3 and Q; = X such that X T is on the line X + Q. Therefore Q\ ,Q2,Q3,Q4,Q5 and X are on a conic D by 42.7. Since QI' ... , Q5 E c n D, we have c = b whence X E c, a contradiction. This contradiction proves 0 C; c. 0
°
°
°
xi and
~2
+ a3; IXj + a 13 ; IX IX3 + a23 ; I X 2X 3 + a34 ; IX3x 4 = 0.
is represented by
a2;2xi
+ xj + a I2 ; 2XI X 2 + a23 ;2X2X3 + a24 ;2X2X4 = 0.
The plane T2 contains two lines of ~I' namely the line L and the line represented by the two equations X2 = and a3; IX3 + a 13 ; IXI + a 34 ; IX4 = 0. Since the group of all elations with axis T J and centres on L fixes T2 and acts transitively on T2 \L, we may assume that (O,O,I,O)K is on ~I' This yields a3; I = 0. As (O,O,l,O)K is on T 2, we may also have
°
222
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
(0, 1,0,0)K E s:r2 • This yields a2; 2 = 0. As ~ I is a cone, (O,O~O, l)K + (0,0,1,0) K c- ~ I' Therefore a34,. I = 0. Likewise a 12 .• 2 = 0. Thus ~I IS represented by 2
°
xi + (a I3 ; IXI + a23 ; I X2)X 3 = and ~2 is r~presente? by X3 + (a 23 ;2X3 + a24 ;2X4)X2 = 0. As a cone is not contained m th~ umon of two planes, we have a 13 ; I =1= and a 24 ;2 =1= 0. Therefore the matnx
°
- a 13;
I
°° °
-
a23 ; I
1
°°
°° - a123 ;
2
-
°°° a 24 ;
43. Twisted Cubics
Then cp is a homomorphism from GL(2,K) into GL(4,K), as is easily seen. Let If; be the canonical homomorphism from GL(4,K) into PGL(4,K) and put l' = If;cp. If (~ ~) is in the kernel of 1', then b3 = c 3 = 0, whence b = c = and hence a 3 = a 2d. This yields a = d. Thus the kernel of l' is equal to the set of all scalar matrices. Hence GL(2,K)/Kern(1') is isomorphic to PGL(2,K). Put s' = as + bt and t' = cs + dt. Then
°
a3 a 2c ac 2 c3
2
is regular. Introducing new coordinates by means of this matrix, we. see that we may assume that ~I and ~2 are represented by the equatIOns x~ - X IX3 = respectively x~ - X2X4 = 0. . Let P = (XI ,X2,X 3,x4)K be a point of ~I n ~2' If X3 = 0, then x 2 = 0, whence P is on L. Hence we may assume x3 = 1. This yields X 2 =1= and
°
°
P = (x; 'X 2 ,1,x 2- )K = (x~ ,x; ,X2 ,1)K. Hence P is on G1 = {(S3,S2t,st 2,t 3)KI s,t E K, (s,t) =1= (O,O)}. Finally I
223
3a 2b 2 a d + 2abc bc 2 + 2acd 3c 2d
3ab 2 2 b c + 2abd ad 2 + 2bcd 3cd 2
a 12
G1 = {(S3,S2t,st 2,t 3)Kls,t E K, (s,t) =I=(O,O)}. Let a,b,c,d E K be such that ad - bc =1= from GL(2,K) into GL(4,K) defined by
.(ac ~)
cpo
--?
a3 a 2c ac 2 c3
°
3a 2b 2 a d + 2abc bc 2 + 2acd 3c 2d
and consider the mapping cp
3ab 2 b 2c + 2abd ad 2 + 2bcd 3cd 2
b3 b 2d bd 2 d3
(3
a 13
1 a43
a 42
PROOF. We may assume that
S'(2
1
The proof of 43.1 shows that the following is also true.
43.3 Theorem. Let G1 be a twisted cubic and let G be its stabilizer in the projective group of~. Then G is isomorphic to PGL(2,K) provided IKI > 4. Moreover, G acts sharply triply transitively on G1.
S,3 S,2t ,
° °°° ° ° °° + °+
all
o
Next we prove:
S3 s2t st 2 t3
This shows that G contains a subgroup H isomorphic to PGL(2,K) which acts sharply triply transitively on G1. Let y E G fix three distinct points of G1. We may assume that y fixes (l,O,O,O)K, (O,O,O,I)K and (l,l,l,l)K. Then y fixes ~I and ~2 and also TI and T 2 • (Here we use IKI > 4.) This yields that y is induced by a matrix of the form
(0,0,0,1 )K, (1 ,O,O,O)K E G1.
43.2 Theorem. Let ~I '~2'~; ,~; be two pairs of cones in ~ with centres Assume PI =1= P 2 and P{ =1= P~ as well, as P I ' P2' P'I ' P'2 resnectivelv. r J PI + P2 \: ~I n ~2 and P{ + P~ C~; n ~;. Denote by T j respectively T j the tangent plane of ~j resp. ~; with PI + P2 t:: Ti and P{ + P~ C T/. If TI =1= T2 and T{ =1= T~, then there exists a projective collineation a of ~ with ~~ = ~; and ~~ = ~;.
b3 b 2d bd 2 d3
a44
with all + a 12 + a 13 = 1 = a42 a43 a44 • Now the intersection of ~I with the plane having equation x 4 = consists of the points (S2,st,t 2,0)K. The 2 line (S2,st,t ,0)K + (O,O,O,I)K has a point distinct from (O,O,O,I)K in common with G1, if and only if s =1= 0. This yields that the plane with equation XI = which is tangent to ~I is fixed by y. Therefore al2 = = a l3 and all = 1. Similarly a42 = a43 = and a44 = 1. Hence H= G. 0
°
°
°
°
The plane 0 with equation x I = is said to be the osculating plane of in the point (O,O,O,l)K and the line {(O,O,x,y)KI x,y E K} is called the tangent of (i£ in (O,O,O,l)K. Using the group G, one sees that the osculating plane in (s3,S2t ,st 2,t 3)K is the plane with the equation t 3x I - 3t 2sX2 + 3 tS 2X3 - S3X4 = 0. (i£
43.4 Theorem. Let ~ be the projective 3-space over the field K and assume that the characteristic of K is not 3 and that IKI > 2. If G1 is a twisted cubic in ~, then there exists exactly one symplectic polarity l' of ~ such P'" is the osculating plane of G1 in P for all P E G1. PROOF. That there is at most one follows from the fact that any subset of four points of (i£ generate ~. In order to prove that there is one, we may assume G1 = {(S3,s2t ,st 2,t 3)K I s,t E K, (s,t) =1= (O,O)}. Then the skew sym-
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
224
metric form f defined by f(x,y) such a polarity.
= Xl Y4 - X4Yl - 3(X 2Y3 - x 3Y2)
defines 0
Case I: U is not a line. Then we may assume that U is a point. Let U = (Xl ,X2,X 3,x4)K. Then (XI 'x 2'x 3 ,x4) is fixed by
The tangent at ~ in (l,O,O,O)K is the line {(x,y,O,O)lx,y E K). This line has no point in common with the tangent at ~ in (O,O,O,l)K which is the line {(O,O,x,y) I x,y E K}. Also, G acts triply transitively on the set of tangents by 43.3. Thus we have: 43.5 Theorem. The set of tangents of a twisted cubic of ~ is a partial spread
of 145. Let
(6 ?)
_?)
43.6 Theorem. If IKI > 2 and if the characteristic of K is not 3, then S acts irreducibly on V = K EB K EB K EB K. If the characteristic of K is 3, then S fixes the line {(O,x,y,O) I x,y E K}. If IKI = 2, then S fixes the plane spanned
by f£. PROOF. The image of
(6
225
43. Twisted Cubics
1 A
= 0
o o
b I
b2 0
0 0
I 0
b3 2
b b I
eigenvalues of2 this matrix are I . Hence (x I' x 2' x 3' x) - (x I + as all the 2 3 4 bX2 + b x3 + b X4 ,x2 + b x 4,X 3 + bx 4,X4) for all b. Putting b = I yields x 4 = 0 and x 2 = x 3 • Hence bxil + b) = 0 for all b. This yields x 2 = 0, as IKI > 2. Hence U = (1,O,O,O)K is on ~, a contradiction. ~ase 2:
f! is a li~e. Consider again the above matrix A. As U is fixed by
S, It con tams an eIgenvector of A. Each such eigenvector has the form (XI 'X 2'X 3'0) with x 2 + bX3 = O. As U contains at most two points of f£ and as S acts doubly transitively on ~, we see that U does not contain a point of ~. Hence x 2 -=1= 0 -=1= x 3. We may therefore assume that x3 = I, whence (.Xl,X2,X~,0)=(Xl,b,I,0) .. If b-=l=l, then (xpb,I,O) and (x;,I,I,O) are lmearly mdependent. ThIS shows that U is contained in the plane with equation x 4 = O. Using the transpose of A which is also in S one sees that U = {(O,x,y,O) I x,y E K}. Then U TT = {(x,O,O,y) Ix,y E K}; hence U TT contains the points (1,O,O,O)K and (0,0,0, I )K, which is impossible. This contradiction proves our theorem. 0
:) under
1 0 0 0
3
1 0 0
3 2
1 0
:]
43.7 T~eo~em. Let q be a power of an odd prime and assume q > 3. If 145 is the projective 3-space over GF(q) and if ~ is a set of q + 1 points of ~ such
1 . I
that no four points of ~ are coplanar, then
Hence
(~- [~ 1)'=
0 0 0 0
6 0 0 0
n
(~_
1)3=
[0~
0 0 0 0
0 0 0 0
~]
and (-r - It = O. Thus if Char(K) -=1= 2,3, then (x - 1)4 is the minimal polynomial of 1'. Hence V is a cyclic primary l'-module. Therefore, {O}, {(x,O,O,O) I x E K}, {(x,y,O,O) I x,y E K}, {(x,y,z,O) Ix,y,z E K} and V are the only l'-invariant subspaces of V. From this we infer that S acts irreducibly, if Char(K) -=1= 2,3. If Char(K) = 3, then S fixes {(O,x,y,O) I x,y E K}, as is seen immediately. Thus assume Char(K) = 2 and IKI > 2. Let '7T be the polarity TT defined in 43.4. If S fixes the subspace U, then it also fixes U , since S cen tralizes
'7T.
~
is a twisted cubic.
PROOF. Pick P E~. Then ~/ P is the projective plane over GF(q). The q chords through Pare q points of ~ / P no three of which are collinear. Hence they are contained in a unique conic of ~ / P by 42.8. Thus there is a unique line through P such that this line together with the q chords through P form a cone Si'l' Pick another point Q E f£. Then we obtain likewise a cone Si'2 such that the q chords of f£ through Q are on Si'2' In order to prove the theorem, we have only to show by 43.1 that the tangent planes T; of Si'; containing P + Q are distinct. Suppose Tl = T 2. Then we may assume P = (l,O,O,O)K, Q = (0,0,0, I)K and that TI is represented by the equation x 2 + X3 = O. We may further assume that (O,I,O,O)K and (O,O,l,O)K are contained in the intersection of Si'l and Si'2' This implies Si'2 is represented by ax I(x 2 + X3~ + X 2X3 = 0 and Si'l is represented by bX 4(X 2 + x 3) + X 2X 3 = O. Therefore, If X = (x) ,x 2 ,X 3 ,x4)K is in the intersection then (ax - bx ) (X2 + x 3) = O. Thus if X =1= P,Q, we have aX I - bX 4 = O. Hence ~t lea:t
226
VII, Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
q - I of the points of aX 1 - bX4 = 0 whence
~ are contained in the plane with equation
Let 0=1= b E K and put M(b) =
q - 1 < 3, i.e., q = 3, a contradiction. This proves
43.7.
M(b) =
baa da a bc a dc a
aao: = caO: d [ acO:a cc
b)
aba cbO: adO: cdO:
1
1
As ,all eigen~alues of M(b) are equal to the fixed points of M(b) consist entIrely of fixed vectors. Let (x) ,x 2 'x 3 ,x4 ) be such a fixed vector. Then x 4 = ,and bX2 + b aX3 = 0. Therefore the Sylow 2-subgroup L = {M (b) Ib E K} mduces a, group of elations in the plane T(L) with equation X 4 = such that each lme through (1,O,O,O)K contained in T(L) is the axis of an elation except the lines L)(L) = {(x,O,y,O) Ix,y E K} and L 2(L) = {(x,y,O, O)lx,y E K}. Denote by ~ the set of chords of ~(a) through (I,O,O,O)K. Then {L)(L)',Li L )} u ~ is a (q + 2)-arc in V /(1,O,O,O)K. As a (q + 2)-arc is dheterml{nLed(~)Y an~ q of its points, it follows that PGL( V)Q:(o:),(),O,O,O)K fixes e set ) ....., ,Li""")} and hence T(L). Let a E PGL(V)Q:(a) fix the points (1,O,O,O)K, (O,O,O,I)K and (1,I,I,I)K. Then a fixes T(L) by the remark made above and, by the same token, also the plane with equation x) = 0. Hence a is induced by a matrix of the form
°
°
t
a
°
0
Oed
e
Then
bb a hO: b
bO:
° ° °° °° ° I, 1
[
Theorem 43.7 is not true if q = 2', as we shall see now. In fact, if r > 2, then there are exactly ~ cp(r) projectively inequivalent classes of point sets ~ with the property that no four points of ~ are coplanar and such that ~ admits a group induced by collineations isomorphic to SL(2,q) operating in its natural representation on ~. Throughout this section let q = 2' > 4, K = GF(q) and V = K(4). Furthermore, for a a generator of Aut(GF(q», denote by ~(a) the set of points (ssa,tsa,sta,tta)K with s,t E K and (s,t) =1= (0,0). Define the mapping from GL(2,q) into GL(4,q) by
f). Then
b
I
0
44. Irreducible Representations of SL(2,2r)
227
44. Irreducible Representations of SL(2, 2 ')
° J °°° e
°0 ° g
As (1,I,I,l)K is fixed by a, we have a = c + d = e + f= g. We may assume a = 1. Assume L)(Lt = L)(L). Then LiLt = LiL). Since
~ ~ ~ ~ [~] [i] =
44.1 Lemma. a) No four points of \£(a) are coplanar. b) If q > 4, then G*(a) = PGL(V)Q:(a)' c) If q = 4, then PGL( V)Q:(o:);;;; Ss and IPGL( V)Q:(a): G*(a)1
and
= 2.
PROOF. In order to prove the first assertion, we may assume that the four points are (1,O,O,O)K, (O,O,O,I)K, (1,1,1, I)K and (ssO:,sa,s, I)K with s =1= 0,1, as G*(a) acts triply transitively on ~(a). Then sO: + s =1= 0, since a generates Aut(K). Thus
°° °° ~1 1
This proves the first assertion.
1
1 =s+sa=l=O . 1
°°° ° ° ° °° ~] [~ O° ed ° e f ° y fy' ° f°= ° ° ° 1
1
=
we have e = d = 0 and c = L)(~t = L 2(L) and Li~rJ Now
=
1 1, whence a = 1. Thus, if a fl G*(a), then L)(L). This yields c = J = 0 and e = d = 1.
°°1 ° ° °o 1
228
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
Therefore there exist p,A,p, E K with pssa = AAa, ps = p,A a, psa = p,a A, 2 p = p,p, a. Putting v = Ap, - I yields s = va, sa = v. Hence s = sa for all s E K. Thus a 2 = 1 whence r = 2. This proves b) and IPGL( V)Q:(a): G*(a)1 ,.;; 2, if q = 4. On the other hand, assume q = 4 and let a be the collineation induced by
Then a E G*(a) and @:(aY @:(a). As a restricted to @:(a) is a transposition and since G*(a) acts triply transitively on @:(a), we finally see that PGL( V)Q:(a) ~ Ss. 0 44.2 Lemma. Let a and [3 each generate Aut(GF(q)). Then the following statements are equivalent: a) There exists a collineation mapping @:(a) onto @:( [3). b) There exists a projective collineation mapping Q.:(a) onto Q.:([3). c) a E {[3,[3 -I}.
PROOF. a) implies b): Let p be a collineation mapping @:(a) onto @:([3). Then p is induced by a semilinear mapping a with companion automorphism v. Let l' be the mapping defined by (XI' ... y = (X(I, ... ). Then l' leaves invariant 0::( [3), since Aut(GF(q)) is abelian. Moreover, a1' is linear. Thus aT induces a projective collineation mapping 0::( a) onto 0::( [3). b) implies a) trivially. b) implies c): We Fay assume that the projective collineation which maps @:(a) onto 0::([3) fixes (l,O,O,O)K, (O,O,O,I)K and (l,l,l,l)K. It follows that it also fixes the planes with equations XI = and x 4 = 0. Hence it is induced by a matrix of the form
°
° °a b°O° °° 0 0 1° 1
+b=
c
d
{{ (x,y,O,O) I x,y E K}, {(x,O,y, 0) I x,y E K}}. If it fixes each of these lines individually, we get c = b = whence a = [3. If it switches the two lines, then a = d = Now
° ° °° °° ° o 0 °°° 1
I
°°
and a = d = 1, and c = b = 1.
~l
0
°
Call the plane with equation X 4 = the osculating plane of Q.:(a) in (l,O,O,O)K. Using the group G*(a), one sees that the osculating plane of @:(a) in (ssa,tsa,sta,tta)K is the plane with equation ttax 1 + st ax 2 + ts aX3 + ssax4 = 0. 44.3 Lemma. There exists precisely one symplectic polarity the osculating plane of @:(a) in P for all P E 0::(a).
'TT
such that p7T is
PROOF. The uniqueness follows at once from 1@:(a)1 = q + 1 > 5 and 44.1 a). Define f: V X V ~ K by f(x,y) = X1Y4 + X 4 Yl + X 2Y3 + X3Y2' Thenf is a non-degenerate skew symmetric form which obviously defines a polarity of the kind required. 0 Denote by Sea) the image of the restriction of to SL(2,q). Then Sea) ~ SL(2,q). Also, we have 44.4 Theorem. Sea) acts irreducibly on V. PROOF. If Sea) fixes U, then Sea) fixes U 7T , as G*(a) centralizes the polarity 'TT defined in 44.3. Case I:· U is not a line. Then we may assume that U is a point. Let U = (X 1 ,X 2,X 3,x4)K. Then (X 1 ,X 2,X 3,X 4) is fixed by 1 b b a bb a
M(b) = [
+ d = 1. Moreover it fixes the set
I
1 0 0 0 1 0 1 0 0 0 maps 0::(a) onto 0::([3), as is easily seen.
°
=° ° °
with a
Hence there exist p,t E K with pssa = tt(3, ps = t(3, psa = t, P = 1. Thus s = s af3 for all s, whence a = [3 - I. c) implies b): We may restrict ourselves to the case a = f3 - I. Then the projective collineation induced by
°
100 001 010
c
229
44. Irreducible Representations of SL(2, 2')
~
g~
y
for all b E K, as all eigenvalues of M(b) are 1. Hence XI = XI + bX 2 + b ax 3 + bb ax 4, x 2 = x 2 + b ax 4, X3 = X3 + bx 4· Putting b = 1 yields x 4 = and x 2 = x 3 • Therefore (b + b a)x2 = for all b. As a is a non trivial automorphism, there exists b with b + b a =1= 0. Hence x 2 = X3 = 0. Thus U = (l,O,O,O)K, a contradiction. Case 2: U is a line. Then U contains an eigenvector of M(b). This eigenvector has the form (X 1 ,X2 ,X 3,O) with bX 2 + b ax3 = 0. Since Sea) has no fixed point on {£(a) and also no subgroup of index 2, we see that no
°
°
230
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
point of U is on G£(a). Hence X3 =1= 0. We may therefore assume X3 = b, if b =1= 0. This yields (xI ,x 2 ,x3 'O) = (XI ,b!X,b,O). If b =1= 1, then b!X =1= b. Hence (x] ,bIX,b,O) and (x] ,1,1,0) are linearly independent. This shows that U is contained in the plane with equation X 4 = 0. Using the matrices
°1 ° ° b!X ° °I °0' bb!X b!X ° 1 U is also contained in the plane with equation XI = 0. 1 b
one sees that Therefore U = {(O,x,y,O) Ix,y E K}. Now un = {(x,O,O,y) Ix,y E K} which is impossible, as un carries the points (l,O,O,O)K and (O,O,O,I)K. This contradiction proves our theorem. 0
44.5 Theorem. Let G£ be a set of q + 1 points of the vector space V such that no four points of G£ are coplanar. If S is a subgroup of G L( V) isomorphic to SL(2,q) such that S leaves G£ invariant and acts triply transitively on ~, then there exists an automorphism a of GF(q) generating Aut(GF(q» and a collineation K such that G£K = G£(a). PROOF. Pick P E ~. Then there exists IT E SyllS) with pIT = P. It follows that IT fixes a plane T(P) through P, since the number of planes through P is q2 + q + 1. As IT acts transitively on ~\{P}, it follows that P is the only point common to G£ and T(P). Assume that there is a a E IT\ {l} fixing T(P) pointwise. Pick Q E ~\{P}. Then QO =1= Q and (Qo + Q) n T(P) is the centre of the elation a. Since I~I = q + 1 ,> 5, there exists R E ~\{P,Q,QO}. The plane Q + QO + R passes through the centre of a. Therefore it is fixed by a. Hence RO = R, as no four points of ~ are coplanar. This yields a = 1, since IT acts regularly on ~\{P}. This contradiction shows that each a E IT\ {I} induces a non trivial elation in
Let Q be a point of
distinct from P. Then Z = MQ is cyclic of order T(P) n T(Q) and has two fixed points on it, namely Ni(P) n T(P) n T(Q). As all involutions in IT are conjugate under Z, the group Z acts transitively on the set of axes through P and hence transitively on the set of the q - I points on T(P) n T(Q) which are distinct from Ni(P) n T(P) n T(Q). Assume N](P) n N](Q) =1= {O}. Then N](P) n N](Q) is a point on T(P) n T(Q) which is fixed by Z. It follows that NiP) n NiQ) is the second fixed point of Z on T(Q) n T(P). There exists an involution a E S mapping Ponto Q whence NlCPt = Nl(Q) and N2CPt = N 2(Q). It follows that a fixes 2 and hence all points on T(P) n T(Q). But a cannot be an elation of the space. Hence all fixed points of a are on T(P) n T(Q). From this we infer that T(P) n T(Q) meets ~ in a point which is distinct from P, a contradiction. Hence N](P) n N\CQ) = {a} for all Q E ~\{P}. Thus S'r\ consists of pairwise skew lines. Likewise ~2 consists also of pairwise skew lines. On the other hand ~
q - 1. The group Z fixes the line
{Ni(P)
n
T(P)
n
T(Q)li = 1,2} = {Ni(Q)
n
T(P)
n
T(Q)li = 1,2}.
Therefore N1(P) n NiQ) =1= {a} which is also true if P = Q. This proves that ~l and ~2 are reguli which are opposite to each other. Let n be the quadric carrying S'r \ and S'r2. As all ruled quadrics are projectively equivalent, we may assume that n is defined by the equation x\x 4 - X 2 X 3 = 0. As no two of the points P = (l,O,O,O)K, Q = (O,O,O,I)K and R = (l,l,l,I)K are on a line belonging to ~\ U ~2' we may assume P,Q,R E~. It follows that T(P) has equation x 4 = and T(Q) has equation Xl = 0. Consider the groups Sl and S2 consisting of all the matrices
°
° °° ° °° °° 8f3' = a y
T(P).
Again let Q be a point of ~ other than P. Then it follows from our assumptions that {T(P) n (X + Q) IX E ~\ {Q} } is a q-arc containing P. Pick a E IT\{I}, then T(P) n (Qo + Q) is a fixed point of a in T(P) other than P. Hence P + [T(P) n (Qo + Q)] is the axis of the elation of T(P) induced by a. Since {T(P) n (X + Q) I X E ~\{ Q}} is a q-arc, we see that exactly q - 1 of the lines of T(P) which pass through P are axes of elations induced by elements of IT\{l}. All these lines are fixed by IT, because IT is abelian. Since IT consists only of projective collineations and since q - 1 ,> 3, it follows that IT fixes all the lines of T(P) which pass through P. Denote by NI(P) and NiP) the two lines through P which are not axes. Then {NI(P),NlP)} is fixed by M = 9C s (IT). As IMI = q(q - 1), it follows that NI(P) and NiP) are fixed individually by M. Denote by S'r i the orbit of NiCP) under S. Then /S'r i / = q + 1. Moreover choose the notation so that NiCQ) E S'ri for all Q E G£. We shall show that S'r] is a regulus and that S'r2 is the regulus opposite to S'r l .
231
44. Irreducible Representations of SL(2, 2')
f3
8
a
A
respectively
° °
° A
B
°
COD
° B 0'
COD y I = AD - BC. Then Sl ~ S2 ~ SL(2,q). As S is simple, we
with a8 - f3y have S ~ SIX S2' Furthermore, S n Si = 1. Hence S is a diagonal of S\ X S2; i.e., there exists an automorphism p from Sl onto S2 such that S = {aa P / a E S d. Let T be an element of order 2 of S fixing (l,0,0,0). Then there are a,f3,y,8,A,B,C,D E K with a8 - f3y = 1 and AD - BC = 1 and
° ° A A° B B° aAyA 8Af3A aByB 8Bf3B °a °f3 C° D° f3D f3C aD ° D° aC y 8 yC 8C yD 8D C ° ° ° ° Hence aA = I, yA = aC = yC = 0, whence y = C = 0. As all eigenvalues T=
a y
f3 8
0
0
of Tare 1, we have SA = aD = SD = 1. Hence a = S and A = D. We infer
232
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
233
45. The Hering and the Schaffer Planes
2
from 1 = ao = a and 1 = AD = A2 that a = 0 = A = D = 1. Therefore
1" =
1
{3
0
0
1
0
o o 1 o o 0
0110 o
o
B
o
1
o
B
This yields
{3 -1
{3 0 o 1 o 1 0 o o Since p is an isomorphism, B is uniquely determined by {3. Therefore B = {3f where f is an automorphism of the additive group of K. There exists an involution f.L E S with pp. = Q and R p. = R. Again the mapping f.L has the form
0
f3
0 0
0 0
f.L=
It follows a = A = O. o= D = O. This yields
{3A oA {3C
aD
OC
yD
aB
{3B oB {3D oD
yB
From this we obtain that all the involutions of S fixing Q are of the form
f.L1"f.L
=
o o o o
o
o
{3{3f
{3f
Write 1"( {3) instead of T and put a( {3)
=
{3
f.LT( {3)f.L. Now for {3 =1= 0 we obtain
o
0
0
0 0
0
0
{3f
0
0
0
0
{3f
0
K, (s, t) =1= (0,0) } .
0=1= det
~ ssf
o o
1 sf
0
0
0 1 s
1 1
= sf + s.
This proves that f generates Aut(GF(q)).
o
o o
{3{3 - f
{3 -1{3f
o
{3{3f
o
o
In this section we shall describe the planes discovered by Hering 1970, 1971, Ott 1975, and Schaffer 1975 arising from irreducible representations of SL(2,q). 45.1 Theorem. Let q be a power of a prime satisfying q == - 1 mod 3 and let V be the vector space of rank 4 over GF(q). Assume that Q: is either a twisted cubic of V or a point set of the type Q:( a) described in section 44 and let S be the subgroup of GL( V)a: isomorphic to SL(2,q). Then we have: a) If q is odd, then there exists exactly one spread of V which is left invariant by S. b) If q is even, then there exist exactly two spreads of V which are left invariant by S.
o
o o
There exists f3 =1= 0 such that pa(f3) = R. Hence (1,{3,{3f,{3{3f) E (l,l,l,l)K which yields If = 1. Hence i\(l) = f.L. Therefore
{3{3 -f
o o
o
{3 -1{3f
o o
o
o
{3{3f
o
0 {3
0 {3 -f
45. The Hering and the Schaffer Planes
=
1
{3 - f
= {( ssf, tsf,stf, ttf) K Is, t E
1
~ [~ ~ ~ H o
0
0 0
Moreover, if s =1= 0,1, then P, Q, Rand (ssf,sf,s,I)K are not collinear. Hence
from pp. = Q that aA = yA = aC = 0 and yC =1= O. Hence Moreover Qp. = P implies {3B =1= 0 = 8B = {3D = oD whence Since f.L2 = 1 and R p. = R, we see that (1,1,1,1) is fixed by f.L. {3B = yB = {3C = yC = 1. Thus
1 {3 {3f
0 0 {3 -1
From this we infer that f is an automorphism of K. As Q: = {P} U {QT(f3)1 {3 E K}, we obtain Q:
aA yA aC yC
p
0
o
PROOF. Let pES be of order 3 and pick a point P E Q:. As q == - 1 mod 3, the mapping p acts regularly on Q:. Hence E = P + pp + P p2 is a plane left fixed by p. By Maschke's theorem, E is a completely reducible p-module. We infer from P =1= pp that p does not act trivially on E. Hence E = Q EB I with Q a trivial p-module and 1 a line on which p acts irreducibly. Again by Maschke's theorem V / I is a completely reducible p-module. As (Q + 1)/ I is a trivial p-submodule of V / I, we deduce that V /1 is a trivial p-module. Invoking Maschke's theorem a third time, we see that V = I EEl m with m a trivial p-module.
234
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
Let n be a third line fixed by p. Then Inn = {O}, since p has no fixed point on I. If m n n =1= {OJ, then p fixes the plane m + n and hence the point (m + n) n I. Thus m n n = {O}. Therefore I, m, and n are pairwise skew. But then any transversal of {/,m,n} is fixed by p contrary to the fact that I is an irreducible p-module. Hence I and m are the only fixed lines of p. It follows from this fact that the planes fixed by p are exactly the planes containing I. The normalizer N of (p) in S is a group of order 2( q + 1) which is at the same time a maximal subgroup of S. As S acts irreducibly on V by 43.7 and 44.4, we have S, = N = Sm' Hence Ilsl = 1m sl = ~ q(q - 1). Moreover IS n m S = 0 by the same token. Now let 'TT be a spread of V left invariant by S. Case 1: q is odd. Let P E: C£ and let ~ be a Sylow p-subgroup of S fixing P. Furthermore, let X E: 'TT be such that P ~ X. Then X~ = X. As we have seen in the proof of 43.7, every 7' E: ~\ {l} has minimal polynomial (x - 1)4; thus every element 7' E: ~\{l} fixes exactly one subspace of rank 2 of V. Hence X is the tangent of C£ in P. Thus 'TT contains the set ~ of all tangents of C£. Moreover ~ n IS = 0 = ~ n mS. We infer from I'TT\~I = q(q - 1) == 2 mod 3 that p fixes at least two lines belonging to 'TT\~. Hence I,m E: 'TT\~ whence 'TT = ~ U IS U m s. This proves the uniqueness part in this case. Case 2: q is even. Let uti and ut 2 be the two reguli belonging to C£. (See section 44.) Pick P E: C£ and X E: 'TT such that P ~ X. Then X is fixed by Sp. This yields X E: ~I or X E: ~2' From this we infer as above that S or 'TT = ~2 U IS U mS. Hence we have proved the 'TT = ~l U IS U m uniqueness part in all cases. By 43.4 and 44.3 there exists a symplectiv polarity (J such that po is the osculating plane in P for all P E: C£. Pick a point Q on m. Then QO is a plane fixed by p. Hence I ~ Q 0 by the remark made above. This yields m 0 = I and 1° = m. Let t E: ~, respectively t E: ~l U ~2' Then I n t =1= {O} if and only if m n t =1= {O}, as to = t. But if m n t =1= {O}, then t n t P =1= {O}, whence t = t P, a contradiction, or ut) = ut 2 and uti = ut 1 which yields the contradiction that 2 divides o(p) = 3. Let p' be an element of order 3 of S which is not in (p) and let I' and m' have the meaning for p' that the lines I and m have for p. We want to show that {/,m,I',m'} is a partial spread of V. If I n I' =1= {O}, then (l n I't = m + m' yields m n m' =1= {O}. If I n m' =1= {O}, then I n m' = (l n mit = IP' n m' yields I niP' =1= {O}. Therefore, replacing pi by p,-lpp' if necessary, we may assume m n m' =1= {O}. The group H = (p,p') fixes m n m ' vectorwise. If H has odd order then we infer from 14.1 that \H\ divides q(q - 1) or that IHI divides q + 1 and H is cyclic. The first case cannot hold, as q == - 1 mod 3 and 3 divides IH I. The second case cannot hold either, as p' fl (p). Hence IH I is even. If q is odd, then S contains only one involution. If 0: is this involution, then 0: E Hand x a ;::: x for all x E m n m'. On the other hand va = - V for all v E V. This
235
46. Three Planes of Order 25
contradiction shows that q is even. As H is generated by two elements of odd order, H cannot contain a subgroup of index 2. Therefore 4 divides IHI. We infer from this fact that there exist two distinct commuting involutions 0: and f3 in H. Since two distinct commuting involutions of S have exactly one fixed point in common and since this fixed point is on C£, we finally obtain the contradiction m n m' E C£. This contradiction shows that {l,m,I',m ' } is a partial spread of V. As Iisl = Imsi = ~ q(q - 1), we finally get that 'TT = ~ U IS U m s, respectively 'TT = uti U IS U m s, is a 0 spread. The planes of the form (~ U IS U m s)( V) are called Hering planes and the planes of the form (uti U IS U m s)( V) are called Schaffer planes. If q = 2' > 4, then there are ~ cp(r) inequivalent C£(o:)'s, as we have seen above. We shall see in section 50 that the cp(r) Schaffer planes determined by them are pairwise non-isomorphic.
46. Three Planes of Order 25 Let K
=
GF(5) and denote by 7'(s) the matrix
s 3s 2 S3
0 1
0 0
S
3s 2
0 0 0
s
and by p the matrix 0 - 1 0 0
1 0 0 0
0 0 0 - 1
0 0 1 0
Then {7'(s) Is E K} is a group of order 5, as is easily seen. Let S be the group generated by the 7'(s)'s and p. Then S fixes the subspace L = {(O,O,x,y) I x,y E K}. Moreover S induces groups isomorphic to SL(2,5) on L and V / L, where V = K(4). The kernels of the two homomorphisms coincide: it is the group N fixing L and V / L vectorwise. We want to show N={l}. Put Vo={(x,y,O,O)lx,yEK}.ThenVo= {(x, - sx + y,O,O) I x,y E K}. Since 0
s 3s 2
S
S3
3s 2
0 0 S
0 0 0
x -sx + y 0 0
x
Y + sy - 2s 3x + 3sy 2S2X
236
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
we see that Vo is mapped by 1"(s) onto
Vs
= {(x,y,2s
2
237
46, Three Planes of Order 25
46.1 Lemma. If l' is a spread of V left invariant by S, then L E 3
x + sy,-2s x + 3sY)lx,y E K}.
1'.
An easy computation shows that p maps Vs onto V 2s3. Therefore {Vs Is E K} is an orbit of S. Pick JJ EN. Then JJ is induced by a matrix of the form
PROOF. The minimal polynomial of 1"(1) is (x - 1)4. Hence there exists exactly one subspace P of rank 1 and exactly one subspace X of rank 2 with p'T"(I) = P and X'T"(I) = X. Hence L = X. Moreover there exists exactly one Y E l' with P ~ Y. Therefore P <: Y n Y'T"(I) whence Y = Y'T"(I). This 0 yields L = Y E 1'.
100 100 b I c dOl Applying this to Va yields the subspace
Put ~ = {8(s) Is E K*}. Then ~ is a group of order 4 which IS normalized by p, as p -18(s)p = 8(s -I). The group ~ fixes the two subspaces
° °
°a
V'
=
+ by,cx + dy) I x,y E K}. we get a = 2S2, b = s, c = - 2s 3 , and d = 3s 2. Now
{(x,y,ax
';s,th,is is one of the Vs ' JJ IS mduced by the matrix
1
1 2a 2b 1 2c 2d 1 2 Hence there exists t E K with 4s = 2a = 2t 2 and 2s = 2b = t. Hence t 2 = 2t 2, i.e., t = 0, whence s = 0. This proves N = {I}, i.e., S ~ SL(2,5). S acts reducibly on V, as L is left invariant by S, but S does not act completely reducibly, since the minimal polynomial of 1"(1) is (x - 1)4. Put a(s) = P1"( - s)p. Then
a(s) =
Moreover, for
°
s
°
3s 2
S3
S
3s 2
°° °° I
s
Finally, if
8(s) =
l'
PROOF. We infer from rk(Xj n L) = 1 and 46.1 that Xi rt. 1'. a) implies b) and c): It follows from Maschke's theorem that ~ has two fixed points on X. As Xi rt. 1', we deduce that ~ has exactly two fixed points on X. Hence rk(X n X;) = l. b) as well as c) imply a): Pick 8 E~. Then X n Xi = (X n Xi)S <: X n X 8 whence X = Xs. 0 1',
then
~
fixes exactly six
0
PROOF. This follows immediately from 46.2.
° S
-S-I 0
0
° ° ° 0 S
° °° o s ° ~ ° °° ° ° °
0 -S-I 0
As P normalizes ~ and fixes L and since p2 is the identity on the projective space belonging to V, it follows from 46.3 that p fixes either one, three or five of the components other than L of l' which are fixed by ~. 46.4 Theorem. There exists exactly one spread 1'1 of V left invariant by S such that p fixes all the components of 1'1 fixed by ~. PROOF. Let P be a point of Xl which is not on L. Then P
s
Cx
OS-I
then 8(s) = p'7'(s)a( - s - 1)'7'(S) E S for all s E K*. N ext we want to determine all spreads left invariant by S. Fir'st of all we note:
the
a) X A = X. b) rk(X n XI) = l. c) rk(X n X 2) = l.
46.3 Corollary. If S leaves invariant the spread components of l' including L.
0
=F s E K we find
1"(s)a( - S-I)1"(S) =
pointwise. Moreover Xi = X 2 and Xi = XI' 46.2 Lemma. Let l' be a spread of V left invariant by S. Then for X E following statements are equivalent:
° ° °° °° °
1
Xl = {(x,O,y,O) \ x,y E K} and X 2 = {(O,x,O,y) \ x,y E K}
Then
C: =
Cx and Cx
n
=
= (l,O,x,O)K.
{(k,l,xk,xl)\k,l E K).
Cy = {a} for x =F y. Moreover
C:(s)= {(k,l,2s 2k + sl + xk,-2s 3k + 3s 21 + Ix) I k,l E K}.
Put
238
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
239
46. Three Planes of Order 25
Assume c~, n C;(s) =1= {a}. Then there exist k,l E K with (k,l) =1= (0,0) and 2S2k + sl + xk = yk, - 2s 3k + 3s 21 + xl = yl. Putting z = y - x yields
°°
Hence
(25 2 - z)k + 51 = 0, - 25 3k - (25 2 + z)! = 0. Therefore = - (25 2 + z)· 2 4 4 (25 - z) + 2s = - 2s + z2. As 2 is not a square, we get s = and x = y. From this we infer that 7T1
=
U
+ (35 2 + v) I) I k,l
{C:(s)ls E K} U {L}
xEK
is a partial spread. Moreover 17T II = 52 + 1. whence 7T I is a spread of V. This proves the uniqueness part of the theorem. In order to prove the existence, we have to show that 7T I is left invariant by S. As 7T1 is left invariant by all the 7T(S)'S, it suffices to prove that 7TI is 3 left invariant by p. But C;(s)p = C;(2s ) as is easily seen. This proves 46.4. D
° °° °° °°° ° ° 1
1
b
1 Then a(b) centralizes S and fixes XI and X 2. Moreover Cxo:(b) = Cx + b'
7T I (
det(
V).
46.7 Corollary. Let G be the subgroup of GL( V) fixing and G acts transitively on 7T] \ {L}.
7T].
7T1
Then L G
=
L
o
If and
PROOF. It is
{(k,l,uk,vl)lk,l E K}
=
{(k,-sk
+ Qr+ It'
I
S2) = s
3
s2(3s 2 - (u - v») D
a E S} is a partial spread consisting of 10 lines.
+ l,uk,-vsk + vl)lk,l
then (2S2
+ I)k + sl =
°
and (- 2s 3
+ s)k +
This shows that
{(P o + QI)r(s) Is E
K)
U
{(PI
+ QoY(S) Is E K)
is a partial spread of 10 lines. Moreover, this partial spread is left invariant by p, as is easily checked, and hence by S. This proves the lemma. D
46.10 Theorem. There exist exactly five spreads 7T 2 ,r (r E K) of V left invariant by S such that p fixes exactly four of the six components of 7T 2,r fixed by Is. The five translation planes 7T 2,r( V) are all isomorphic. PROOF. Assume that there exists a spread 7T of this type. Then it contains a component of the form Pr + Qs with r =1= s. It follows from 46.8 that r - s E {I,-I}. As (Pr + Qsl = Ps + Qr' we may assume s = r + l. Thus 7T
E K} is a partial spread,
=
3+2s(2u-v ) s
4
= (l,O,u,O)K and Qv = (O,I,O,v)K.
Pu + Qv
s
+ QoY(s),
= 1= 0.
for all
PROO'F. It follows from 46.5, 46.6, and the proof of 46.4 that GL acts transitively on 7T] \{L}. If G =1= GL , then G contains a subgroup isomorphic to SL(2,S2) by 35.10 and 46.6. Hence 7T I (V) is desarguesian by 38.12. But then all subgroups of G isomorphic to SL(2,5) are conjugate within G, whence S cannot act in the above way. This contradiction proves G = GL .
{( Pu + QV>'T(s) Is
2
2S2 + 1 - 2s 3 + s
we infer k
PROOF. This follows at once from 46.5 and the fact that Cx E x E K.
46.8 Lemma. ® = u-vE{O,I,-l}.
-
for all s E K*, i.e., if and only if u - v E {O,I, -I}.
If (k,l,O,l) E (PI 3s 21 = I. From
PROOF. This follows from straightforward computations.
Put Pu
0=1= det(
= {a}
PROOF. We may assume by 46.5 that r = 0. It follows from 46.8 that {(Po + QIY(s) Is E K} and {(PI + QoY(s) Is E K} are partial spreads. Now (PI + QoY(s)= {( k,l, (2S2 + l)k + sl, (-2s 3 + s)k + 3s 2/) I k,l E K}.
1
46.6 Corollary. a(b) induces a shear with axis L on
E K }.
As ® is a partial spread, if and only if (P u + Qv) n (P u + QvY(S) for all s E K*, we get that ® is a partial spread if and only if
46.9 Lemma. {(Pr
46.5 Lemma. Denote by a (b) the matrix
b
(P u + Qvr(S)= {(k,l,(2s 2 + u)k + sl,( -2s 3 + (u - v)s)k
E
K).
only
if
= {L}
U
{(Pr
+ Qr+l)ala
E
S}
U
U {(cra+ i=2
i
)Ia E S} =
7T 2,r'
Hence, by 46.5, the theorem is proved, as soon as we have proved that 7T2 ,O is a spread. We infer from {L} U U;=2{Cia la E S} <;;;:; 7T] and 46.9 that {L} U U;=2{ Cia Ia E S} and {(Po + QIY' Ia E S} are partial spreads. Furthermore L n (Po + Qlt' = L n (Po + QI) = {O}. 3 As c/(s)p = C/(2s ), we have {Cia I a E S} = {C/(s) I s E K}. Assume (0,0, 0,0) =1= (x] 'x 2'x 3 'x 4 ) E C/(s) n (Po + QI)' Then we infer from Po + QJ =
240
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
{(k,l,O,f)lk,I,E K) and C/(s) = {(k,I,2s 2 + sl + ik,-2s 3 + 3s 21 + il)lk,1 E K) that x3 = and X 2 = x 4. Hence 2S2k + sf + ik = and 1= -2s 3k + 3s 21 + if. Therefore
°
°
°
(2S2 + i)k + sl = 3 -2s k + (3s 2 + i - 1)1 = 0. 2 Hence 0= (2S2 + i)(3s + i-I) + 2s 4 = 3s 4 + 3s 2 + 6i(i - 1). Thus = S4 + s2 + 2i(i - 1) = S4 + 2S23 + 32 + 1 + 2i(i - 1) = (S2 + 3)2 + 2i(i - 1) + 1. We infer that 2i(i - 1) + 1 is a square, as -1 is a square. Now i = 2, 3 or 4 whence 2i(i - 1) + 1 = 0, 3 or 0. This yields S2 = - 3, a contradiction. Hence '772 ,0 is a spread. 0
°
46.11 Theorem. There exist exactly five spreads '77 3,r (r E K) of V left invariant by S such that P fixes exactly two of the six components of '77 3" fixed by 6. The five planes '77 3 ,,( V) are all isomorphic. Assume there exists a spread '77 of this type. Then it contains exactly one component of the type C,+4' It then follows from 46.8 that PROOF.
1
'77
= {L}
U {
CrG+ 41
a E S} U
U {( P, + 2i + Q, + 2i + I) G Ia E
S } = "'3" .
i=O
Hence, by 46.5, the theorem is proved as soon as we have proved that "'3,0 is a spread. Using the spreads '772 , constructed above, it suffices to show that {(Po + Q1YIa E S) U {(P2 Q3Yla E S) is a partial spread. Assume (0,0,0,0) =1= (XI ,x 2'X 3'X 4) E (Po + QI) n (P 2 + Q3Y' Then x3 = and x 2 = x 4· Moreover, there exists s E K such that (P 2 + Q3Y = (P 2 + Q3Y(s) or (P2 + Q3Y = (P 3 + Q2Y(s). The first case yields
+
o=det(
2S2+2 - 2s 3 - S
°
= det(
2s2 + 3 - 2s 3 + S
restriction. of G to L is a non-trivial subgroup of OL(L) ~ OL(2,5). Hence G acts faithfully on L. Similarly, G acts faithfully on V / L. . As all automorphisms of SL(2,5) are induced by inner automorphlsms of OL(2,5), we have that the elements of G may be represented by matrices of the form a b c d e f a b g h c d with ad - be = 1. From
°°
°°
° ° ° °° ° °°
s) = _ 2( S2 _ 3)2 -
3s 2 + 2
which is also impossible. This proves 46.11.
2
3
e g
2 f h
3
2
° ° ° °e - 1 - °I ° ° - °1 °that e-h= °= °h. Replacing
-1
we infer in this particular case the basis (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1) by the basis (l,O,O,g),(O, 1, - f,O),(O,O, 1,0), (0,0,0,1), we see that we may assume g = f = 0. Hence the element Pa represen ted by a
° °
°° °
a
°
belongs to G. Consider the element T(S) defined by
s es gs
s )=-2(s2- 1)2- 1 3s 2 + 1
which is impossible, as 2 is not a square, whereas the second case yields
°
241
46. Three Planes of Order 25
Jl
°1 °° °°
Is
°
s
hs
I
Then we have es + I = es + tis + el whence tis = sj;, as T(S)T(t) = T(l)T(S). Hence is = sfl = sk. As (x - 1)4 is the minimal polynomial of T(S), we get k =1= 0. Conjugation by
1
o
°
100
°° °1 k-° °° ° ° Ok-I 1
r
46.12 Theorem. Let V be the vector space of rank 4 over OF(5) and let G be a subgroup of OL( V) isomorphic to SL(2,5). If G fixes a subspace of rank 2 of V and if all elements of order 5 of G have minimal polynomial (x - 1)4, then there exists y E OL( V) with y -lGy = S, where S is the group defined at the beginning of this section. PROOF. Let L be a subspace of rank 2 of V fixed by G. If T is an element of order 5 of G, then V is a cyclic T-module since (x - 1)4 is the minimal polynomial of T. Therefore, T does not induce the identity on L. Thus the
shows that we may assume k = 1. Hence eS + 1 = es hS+1 = hs + hI + st. Now
PaT(s)Pa-
1
=
a2s es a 2g
s
°1 °0 °0 a 2s hs
0 2
as
+ e + st. l
Moreover
242
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
(Observe a -2 = a 2.) Therefore es = ea2s' hs = ha2s and ga2s = a 2gs ' In particular e =e4 and e2 =e3 as well as h =h 4 and h2 =h 3 • As o= e l +4 = e l + e4 + 4 = 2e l + 4, we have e l = 3. Similarly hI = 3. Furthermore, 0 = e2+ 3 = e 2 + e3 + 6 = 2e 2 + 6, whence e 2 = 2. Likewise h2 = 2. Therefore es = hs = 3s 2 for all s. 0 ) _ (a-I 0\( 0 I) h ave A s ( _ 01 0I)(a0 a-I 0 a) - 1 0' we J
"tV.
lurec:: t'lanes or vruc::r
From (:
1 0
U
V
x
Y
0 0 0 - 1
.. ~,
= diag(a-J,a,a-J,a)
Hence ua = ua Furthermore,
1 0
u x
Y
and ya - 1 = ya for all a
1
1 3 gl
0 0 diag( a,a - 1,a,a - I ) 1 0 0 - 1
v
-=1=
0 0 0 - 1
L4j
?)(6 -:)(: ?) = (? -J) we deduce therefore
1
J
0 - 1
L:)
0
0
0 1 1
0 0 1
0 1 0
- 1 0
-v
0
0 1 1 3
1 0 3 - 1
1 0
0 0 0 - 1
v
0
0 0 1 0
0 - 1 0 x
1 0
0 0 1 0
0 0 0 - 1
v
0
- 2v - gJ - 2v + 3
0 0 0 1
-v
0 0 1 0
-11
1 1 3 gl
0 1 1 3
0 0 1 1
0 0 0
0 0 -1
0
and hence -3g J -2v+3=O= -2g 1 -v-3. This yields g I = 1 and v = O. This shows that G is generated by the matrices
0 0 1 0
[~
O. This yields u = y = O.
0 I 1 3
0 0 1 1
~l
46.13 Corollary. The planes 0 - 1 0 x
-1 1
and
7T 1(
V),
isomorphic.
= diag( - 1, - 1, - 1, - 1).
0 - 1 0 0 7T 2,O(
1 0 0 0
V) and
0 0 0 -}
7T 3 ,O( V)
0 0 1 0
0
are pairwise non-
PROOF. Let 7Tx ''TTy E {'TT J ,'TT 2 ,O,'7T 3 ,O} and assume that 'TTxC V) and 7Ty ( V) are isomorphic. Then, by 46.12, there exists y E 91oL(vlS) mapping 7Tx onto 'TTy' Hence, obviously x = y. 0
This yields v = x. Also, Finally we show: 0 - 1 0
1 0
v
0
v
0 0 0 - 1
0 0 1 0
-I
- 1 0 -v 0
0 1 0 -v
0 0 0 1
0 0 - 1 0
46.14 Theorem. The Hering plane of order 52 is isomorphic to PROOF.
Thus we obtain 0 - 1 0
1 0
v
0
0 0 0 - 1
v
0 0 0
0 1
s 3s 2
S
gs
3s 2
1 0
-s 1
3s 2 -s
2vs - gs - 2v + 3s 2
0 0 1 0
0 0 S
0 0 -s
0 0 0
0 1 0
- 1 0
-v
0
-v
0 0 0
0 0 - 1 0
'TT 3 ,r( V).
Conjugating the group S of all the matrices a3 a 2c ae 2 e3
3a 2b a d + 2abe be 2 + 2acd 3e 2d 2
3ab 2 b e + 2abd ad 2 + 2cdb 3ed 2 2
b3 b 2d bd 2 d3
with the matrix diag(1,1,3,1) yields a group S* with one of its Sylow 5-subgroups being the group of all matrices
o
0 0 100
s 3s 2
S
S3
3s 2
0 S
244
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
Moreover, ""'<=e.-'
0 0 0 - 1
0 0 3 0
~]
0 3 E S*. 0 0 The group S* leaves invariant a twisted cubic whose set of tangents is the set /J = {L} U {Co(s) Is E GF(5)}. Furthermore, 8 leaves invariant 'lT3 I\/J as is easily checked. The theorem now follows from 45.1 and 46.1l. '
0=
0
. The planes 'IT I(V) and 'IT 2,O( V) are called the exceptional Walker planes, smce they were discovered by Walker 1973.
47.1 Lemma. Let a be a quasitransvection of the vector space V and let U be a subspace of V such that a induces the identity on U and on V / U. Then C(a) c: U c: A (a). Furthermore (a - 1)2 = 0 and a is a bijection. PROOF. Put 0: = a-I. Then va E U for all v E V; as a induces the identity on V / u. Hence C(a) c: U. Moreover, u a = U O - U = U - U = 0, as a induces 2 the identity on U. Hence U c: A (a). As va E C( a) c: A (a), we have va = 0, whence 0: 2 = O. Finally (1- 0:)(1
va
'T(a)'T(f3) = (1
+ a)(1 + f3) = 1 + 0: + f3 + af3 = 'T(a + f3).
Moreover 'T(a) = 1, if and only if a = O. Hence 'T is a monomorphism. Finally, if a E T(U, W), then a = a-I E Hand 'T(O:) = a. Thus 'T is an isomorphism from H onto T( U, W). 0
PROOF. This follows from the fact that a has no eigenvalue other than 1, as
Let V be a vector space over K and let U be a subspace of V. If a E End K ( V) is such that it induces the identity on U and on V / U, then we call a a quasitransvection of V. If a is a quasitransvection of V, then we o I call C(a) = v - the centre and A(a) = kern(a - 1) the axis of a.
=
PROOF. Put H = {o: 10: E Hom( V, U), W c: kern( o:)}. Then H is a subgroup of Hom( V, U) which is isomorphic to Hom( V / W, U). For 0: E H we put 'T(a) = I + a. Then ('T(a) - If = a 2 = 0, as U ~ W ~ kern(a). Therefore 'T(a) is a quasitransvection by 47.2. Furthermore C('T(a» = ~ U, whence 'T(a) E T( U, W). For 0:,f3 E H we have af3 = O. Hence
47.4 Lemma. Let a be a quasitransvection of V. Then a fixes no subspace of rank 1 outside of A (a).
47. Quasitransvections
(1- o:)a
245
47. Quasitransvections
+ 0:) =
1- 0: 2 = 1 = a(l- 0:).
Hence a is bijective. 47.2 Lemma. LeI a E GL( V) be such that (a - 1)2 silransvect ion.
o =
(a - 1)2 = O.
0
47.5 Lemma. If V has finite rank n and rk(C(a» 1n. PROOF. This follows from rk(C(a))
47.3 Lemma. Let V be a vector space and let U and W be subspaces of V with U c: W. If T( U, W) is the set of all q1;lasitransvections a with C( a) c: U and W c: A (a), then T( U, W) is an abelian group which is isomorphic to Hom( V / w, U). In particular, T( U, W) is an elementary abelian p-group, if the characteristic of the underlying skew field is p > O.
nand C(a)
c: A(a).
0
PROOF. Assume char(K) = 2. Then (aT - 1)2 = (aT? - 1 = a 2'T2 - I = 0, whence aT is a quasitransvection by 47.2. Assume conversely that a'T is a quasitransvection. Put A (a) n A (T) = Xo' There exist subspaces X I ,X2 ,X3 with rk(X,.) = nand A(a) = XoEB XI' A(T) = Xo EB X 2 and V = Xo EB Xl EB X 2 EB X 3 • Using this decomposition, a and 'T are represented by matrices I
o
+ rk(A(a» =
of V, then
47.6 Lemma. Let V be a vector space of rank 4n over K and let a and 'T be two commuting quasitransvections of V with rk(A(a» = rk(A('T» = 2n and rk(A(a)nA('T»=n. Then aT is a quasitransvection, If and only If char(K) = 2.
O. Then a is a qua-
PROOF. Put U = kern(a - 1). Then U O = u for all u E U • Moreover ( 2 o I V 0-1) = 0 implies v E U for all v E V. Hence a is a quasitransvection.
If a is a quasitransvection
a=
o
o I o o o o
0
o o
o
o
I
and
T
=
o
I
As a and T commute, we have A (ay = A (a) and A (TY = A ('T). Hence A3 = 0 and B3 = O. Computing aT and 'Ta and using aT = 'Ta yields AIB4 = BIA4' Since aT is a quasitransvection, (aT - I? = O. A simple computation then yields 2B ,A4 = O. As a-I and'T - 1 both have rank 2n by assumption, we conclude that B I and A4 are regular matrices, whence BIA4 =t= O. Therefore char(K) = 2. 0
246
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
47.7 Corollary. Let p be an odd prime and let V be the vector space of rank 4 over G F(p r). Suppose that 2: is a group of quasitransvections with rk( A (a)) = 2 for all a E 2: \ { 1}. Then 2: is an elementary abelian p-group and A(a) = A(T)for all a,T E 2:\{l}. PROOF. As aP - 1 = (a - lY = 0 for all a E 2.:, we see that 2.: is a P-group of exponent p. Let 1 =1= a E 3(2.:) and 1 =1= T E 2.:. Then A (aY = A (a). Since A (a) is a p-group, we find that A (a) n A (T) =1= {O}. Hence A (a) = A (T) by 47.6. This proves the last statement of the Corollary. Finally, 2.: ~ T(A (a),A (a)), whence 2.: is abelian by 47.3. D
247
48. Desarguesian Spreads in V(4,q)
K'-rank 1, we have 7T O = 7T'. Moreover, we deduce from GF(q) ~ K n K' that a induces a semilinear mapping on the G F( q)-vector space V, i.e., a E fL( V) = fL(4,q). Putting 7T = 7T' in the above considerations, one sees that the group of all semilinear mappings of the GF(q)-vector space V which leave invariant 7T is isomorphic to fL(2,q2). Hence the number of desarguesian spreads is N = \fL(4,q)\\fL(2,q2)\-I. As IfL(4,q)1
= IAut(GF(q))l q 6(q4 -
1)(q3 - 1)(q2 - 1)(q - 1)
and
48. Desarguesian Spreads in V(4,q) In this section we shall study the geometry of reguli and desarguesian spreads in the vector space V of rank 4 over GF(q). We shall call a spread 7T of V desarguesian, if the plane 7T( V) is desarguesian.
sn be the projective space of rank 4 over GF(q). If n is a sn, then there are exactly t q2( q - 1)2 lines in sn which do
48.1 Lemma. Let ruled quadric in not meet n.
sn
PROOF. n carries exactly two reguli ))11 and ))12 and each line of which is completely contained in n belongs to ))1 I U ))12' Hence there are 2( q + 1) lines which are completely contained in n. Pick PEn. Then there is exactly one Ii E ))1i with P ~ Ii' The lines through P contained in II + 12 and distinct from 11 ,12 meet n only in P. The q2 remaining lines through P carry exactly 2 points of n. Hence there are exactly (q + I)2(q - I) lines having exactly one point in common with n, whereas the number b of secants satisfies (q + I)2q2 = 2b, whence b = tq2(q + Ii. As (q2 + 1)(q2 + q + 1) is the total number of lines in the number of lines which do not meet n is
sn,
( q2
+ 1)( q2 + q + 1) - 2( q + 1) - (q + 1)2( q - 1) =
t q2( q - I?,
t q\ q + 1)2 D
48.2 Lemma. Let V be the vector space of rank 4 over GF(q). Then the number of desarguesian spreads of V is t q4(q3 - l)(q - 1). PROOF. Let 7T and 7T' be desarguesian spreads of V and denote by K resp. K' their kernels. Then \K\ = \K'\ = q2, whence K';;;£ GF(q2) ';;;£ K'. Moreover, GF(q) is contained in K as well as in K'. As K and K' are isomorphic and as rkK( V) = 2 = rk K,( V), there exists a bijective semilinear mapping a from the K-vector space V onto the K' -vector space V. Since 7T consists of all subspaces of K-rank 1 of V and since 7T' consists of all subspaces of
and IAut(GF(q2))1 = 2IAut(GF(q))I, we finally obtain N = (q - 1).
t q4(q3 -
1)·
D
48.3 Lemma. The number of reguli in a projective space of rank 4 over GF(q) is q4(q3 - 1)(q2 + I). PROOF. It is easily seen and, in fact, well known that all reguli are in one orbit of PGL(4,q). Moreover the stabilizer of a regulus in PGL(4,q) is isomorphic to PGL(2,q) X PGL(2,q). Hence the number of reguli 1S IPGL(4,q)IIPGL(2,q)I- 2 = q4(q3 - 1)(q2 + 1). D Let U be a vector space of rank 2 over GF(q). Then V = U® GF(q)GF( q2)
is a vector space of rank 4 over GF(q) and a vector space of rank 2 over GF(q2). Pick u E U with u =1= O. Then lu = {u 0 x Ix E GF(q2)} is a subspace of rank 2 of the GF(q)-vector space V. Moreover lu n lu' =1= {O}, if and only if uGF(q) = u'GF(q). In this case we have lu = lu" Consider ))1 = {X IX = lu for some u E U}. Then ))1 is a regulus the opposite regulus being the set of subspaces {y 0 k lyE U} for 0 =1= k E GF(q2). Moreover each X E))1 is a subspace of rank 1 of the GF(q2)-vector space V. This shows that each desarguesian spread of V contains a regulus. More precisely: 48.4 Lemma. Let V be a vector space of rank 4 over GF(q). Then each desarguesian spread of V contains exactly q( q2 + 1) reguli and each regulus is contained in t q( q - 1) desarguesian spreads. PROOF. Let 7T be a desarguesian spread of V and let G be the subgroup of GL( V) leaving 7T invariant. Then G acts triply transitively on 7T. Hence each set of three distinct components of 7T is contained in a regulus which consists of lines of 7T by the remark made above. As three pairwise skew lines are in exactly one regulus, 7T contains (q2~ I)(q~l)-l = q(q2 + 1) reguli.
248
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
49. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
249
Consider the incidence structure whose points are the reguli of V, whose lines are the desarguesian spreads of V, and whose incidence relation is the inclusion. Then this incidence structure is a tactical configuration with parameters v = q4(q3 - 1)(q2 + 1), b = ! q4(q3 - 1)(q - 1), k = q(q2 + 1) and r. From vr = bk we obtain r = ! q(q - 1). 0
a) If S fixes 91 linewise, then a(V) is a Hall plane. b) If S fixes the opposite regulus of 91 linewise, then a( V) is desarguesian.
48.5 Lemma. Let 91 be a regulus of the vector space V of rank 4 over GF(q) and let G ~ GL(2,q) be the subgroup of the stabilizer of 91 in GL( V) which fixes the opposite regulus of 91 linewise. Then G leaves invariant each desarguesian spread which contains 91.
U= U~~?)q(q-I)(1Ti\I) contains tq2(q - 1)2 lines which do not meet the ruled quadric defined by I. As these are all the lines of V which do not meet this quadric by 48.1, there exists 1T E {1TI"'" 1T(1/2)q(q-I)} such that (a\I) n (1T\I) is not empty. Therefore a\I = 1T\I. Hence a = 1T, if I = 91, and a = (1T\91') U 91, if I = 91'. 0
PROOF. Let 91 be represented by means of a tensor product as before. Then a ---7 a 0 1 gives an imbedding of GL( U) into GL( V)?7 where 1T consists of the subspaces of rank 1 of the GF(q2)-vector space V. Moreover, this imbedding is such that it leaves the opposite regulus invariant linewise. This proves the lemma, since GL( V) acts transitively on the set of pairs (91',1T/) where 91' is a regulus, 1T' a desarguesian spread and 91' ~ 1T'. 0
48.6 Lemma. Let 91 be a regulus, 91' its opposite regulus, and assume that 1T is a desarguesian spread containing 91. If S is a subgroup of GL( V) isomorphic to SL(2,q) fixing 91' linewise, then 1T S = 1T and S acts transitively on 1T\91. S PROOF. It follows from 48.5 that 1T = 1T. We represent 91 and 1T by means of a tensor product as in the proof of 48.5. We also imbed SL(2,q) into GL( V)?7 ~ GL(2,q2) by the mapping a ---7 a 0 1. Interpreting this in the projective space belonging to V, we obtain an imbedding of S* ~ PSL(2,q) into G = PGL(2,q2). Let Z be a cyclic subgroup of S* of order 1(q + 1) or q + 1 accordingly to whether q is odd or even. Then Z is contained in a cyclic subgroup C of order q2 - 1 of PGL(2,q2). Hence Z fixes two lines X and Y of 1T, since PGL(2,q2) acts in its natural representation on 1T. As Z operates regularly on 91, we see that X and Y do not belong to 91. Now s; ~ Gx and \Gx \ = q2(q2 - 1). Moreover the Sylow p-subgroups of S* act regularly on 1T\91. Therefore S; is cyclic. As Z ~ S; and as Z is a maximal cyclic subgroup of S*, we get Z = S;. Thus \X s *\ = q(q - 1). This proves the last assertion of the lemma. 0
48.7 Lemma. Let 1T and 1T' be two desarguesian spreads of V and assume that 91 is a regulus contained in 1T n 1T'. If 177 n 77'1 > q + 1, then 1T = 77'. Let 91' be the opposite regulus of 91 and let S be a subgroup of GL( V) isomorphic to SL(2,q) which fixes 91 linewise. Pick X E (77 n 77') \91. Then 1T'\ffi = X S = 1T\91 by 48.6. Hence 1T = 1T'. 0 PROOF.
1
48.8 Lemma. Let a be a spread of V containing a regulus 91 and let S be a subgroup of GL( V) isomorphic to SL(2,q) leaving invariant 91 and a.
PROOF. Let I' E {91,91~ be the regulus fixed linewise by S. Then the opposite regulus I of I' is contained in 1q(q - 1) desarguesian spreads 1T I' . . ., 77 t q( q - I) . Moreover 77·I n 77·) = I by 48.7, if i =1=)'. Hence
49. Translation Planes of Order q2 Admitting SL(2,q) as a Collineation Group In this section we shall determine all translation planes of order q2 whose kernels contain GF(q) and which admit a group of collineations isomorphic to SL(2,q). The results in this section are due to Walker and Schaffer. We assume throughout this section that p is a prime and q = p r. Moreover V will always denote the vector space of rank 4 over GF(q) = K. 49.1 Lemma. Let S be a subgroup of GL( V) isomorphic to SL(2,q) with the following properties:
1) If 1 =1= r E 8(S), then v r = -v for all v E V. 2) If P = 2, then S fixes no subspaces of rank 1. 3) If 1 =1= T is a p-element of S, then T is a quasitransvection with rk(A (T» = 2. 4) Ifp = 2 and L E Sy12(S), then A(a) = A(T) for all a,T E ~\{l}.
Then the following is true: a) If~ E Syl (S), then A(a) = A(T) for all a,1' E ~\{1}. b) If F(~) d:notes the subspace of V left fixed vectorwise by ~ E Sylp(~), then ffi = {F(~)\~ E Syl (S)} is a partial spread. Moreover, there eXist at least two distinct trans:ersals of 91 and each transversal of ffi is fixed by S. PROOF. If P = 2, then a) is just the assumption 2). If P > 2, then a) follows from 47.7. Assume that F(~) n F(T) =1= {O} for two distinct Sylow p-subgroups ~ and T. Then S fixes a point P t;;;;; F("i.) n F(T) vectofwise, as <"i.,T) = S. By 2) we have therefore p > 2, whence \3(S)\ = 2. Let 1 =1= f E 3(S). Then v = v r = - v for all v E P by 1). This contradiction proves F(~) n F(T) = {O}. Thus ffi is a partial spread.
250
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
Let V be a transversal of ffi and pick 2: E Sylp(S). Then va is also a transversal for all 0 E L. Moreover UnA (0) = U a n A (0). This yields U = U a, as /91/ = q + I > 3. As S is generated by its p-elements, we have S = U. In order to prove that ffi has at least two transversals we first consider the case p = 2. Pick F, G E ffi with F =J= G. Then Z = SF,G is ~yc~ic of order q - I and acts transitively on ffi\{F,G}, since S acts In Its natural representation on ffi. It follows from Maschke's theorem that F is a completely reducible Z-module. Hence Z has at least two fixed points on F, as Z cannot act irreducibly on F. Let P be such a fixed point. There exists an involution 0 E S with Fa = G. Since 0 normalizes Z, it follows that pa is a fixed point of Z on G. Moreover, A(o) meets P + pa non-trivially, because P + pa is fixed by 0 and P + pa is a 2-group. Thus p + pa is a transversal of {F, G,A (a)}. As Z acts transitively on ffi\{ F, G} and fixes P + pa, we conclude that P + pa is a transversal of ffi. Hence ffi has at least two transversals in this case, since Z has at least two fixed points on F. Now let p > 2 and pick 2:,T E Sylp(S) with 2: =J= T. Put A = F(2:), B = F(T), and Z = SA,B' Then Z is cyclic of order q - l. It follows from Maschke's theorem that Z has at least two fixed points on A, as Z cannot act irreducibly on A. Case 1: Z has at least three fixed points on A. Then Z fixes all the points on A, as Z induces projective collineations on the line A. Since there exists an element in S interchanging A and B, it follows that Z fixes also all the points on B. Let X be a plane containing A. Then X n B is a point, whence X Z = X. Moreover L induces a group of transvections on X. Since all subgroups of order p of L are conjugate under Z and since Z fixes A pointwise, it follows that all the transvections induced by L on X have the same centre P. Hence P + (X n B) is a line fixed by L. It then follows from the transitivity of L 'on ffi\ {A} that P + (X n B) is a transversal of ffi. As different planes yield different transversals, ffi has q + I transversals in this case. Case 2: Z has exactly two fixed points on A. There exist 0 ELand l' E T such that (a,r) acts transitively on ffi (see e.g. Gorenstein [1968, p. 44]). Let a be a linear mapping such that a fixes A vectorwise, leaves B invariant, and maps B a onto AT; furthermore, let f3 be a second linear mapping such that [3 fixes B vectorwise, leaves A invariant, and maps AT onto B a • Then af3 = f3a and af3 leaves A, B, AT, and B a invariant. It follows that (af3) - laaf3 = a' is a quasitransvection with A (a') = A which maps B onto B a • Pick b E B. Then f3 b a' = bf3-1!y'-laaf3 = b a- 1a!Y.f3 = (b!y'- l + a/ = (b + al= b + af3. On the other hand, there exist b' E B and a' E A with b a' = b'a = b' + a'. Hence b - b' = a' - a E A n B = {O}. Therefore b a' = b a. This shows that
u
49. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
251
a and af3 commute. Similarly raf3 = a[3r. Hence af3 centralizes (a,r). As (o,r) acts transitively on lR, it follows that af3 fixes each line of lR. Similar arguments then show that a[3 centralizes S. Let P and Q be the two fixed points of Z on A. Then {P, Q }!Y.f3 = {P, Q }. Assume that F is a fixed point of af3 on A. If F E {P, Q }, then af3 has the two fixed points P and Q. If F fi {P, Q }, then F is not a fixed point of Z, whence a[3 has two fixed points in this case, too. Let L be the unique transversal of A,B,A through F. Then L = F + (L n B). As Ff3 = F!Y.f3 = F and (L n B)f3 = L n B, we have L = Lf3. Thus L is a transversal of A,B,A Tf3 = B a. Therefore L is a transversal of A,B,A T,Ba. This implies that L is fixed by 0 and by l' and hence by (0,1'). Since (0,1') acts transitively on ffi, we see that L is a transversal of ffi. Therefore ffi has at least two transversals, if af3 has a fixed point on A. If af3 fixes a point X not on A, then X + A is a plane which is met by B in exactly one point. As this point is fixed by af3, we also find a fixed point of af3 on A by the transitivity of Son ffi. Hence we may assume that a[3 has no fixed point. But P and Q are fixed by (a[3l Hence (af3)2 has at least two fixed points on each line belonging to ffi. Assume (af3)2 =J= l. Then (a[3)2 has exactly two fixed points on each line of ffi, as is easily seen using the fact that (a[3)2 is linear. P + B is a plane fixed by (af3f This plane is not fixed pointwise by (af3)2, as B is not fixed pointwise by (af3f Hence the set of fixed points of (af3)2 which are contained in P + B does not contain a quadrangle. As X n (P + B) is a fixed point of (a[3)2 for all X E ffi\{B}, we see that (a[3)2 has at least q + 2 fixed points on P + B. This yields that q + I of them are on a line and the remaining one is the second fixed point of (a[3)2 on B. This yields a transversal of ffi through P. Similarly we find a transversal through Q. It remains to consider the case where (af3)2 = 1 and af3 has no fixed point. Then the set @5 of lines fixed by a[3 is a spread with ffi C @5. Moreover the kernel of @5 is GF(q2) and S is contained in the group fixing @5, as S centralizes a[3. This means that S acts as a subgroup of GL(2,q2) on V considered as a vector space over GF(q2). Since all subgroups of GL(2,q2) which are isomorphic to SL(2,q) are conjugate in GL(2,q2), we infer that ffi is a regulus in this case. 0 T
49.2 Lemma. Let S be a subgroup of G L( V) isomorphic to SL(2, q) and let U and W be subspaces of rank 2 of V such that the following is true: a) V= U EB W. b) US = U and W S = W. c) U and Ware irreducible S-modules.
Let pES be of order t where t divides q + 1 but not pS + 1 for all s < r. If p fixes a subspace X of rank 2 other than V and W, then ffi =
252
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
{F(~)I~ ESylp(S)} is
a regulus and S fixes the regulus opposite to fR
linewise. PROOF. Obviously, fR is a set of q + 1 pairwise skew lines of V, and U and Ware transversals of fR, and all transversals of fR are fixed by S. Hence it suffices to prove that fR has a third transversal. Let 1 =1= r E L E Sylp(S). Then r induces a quasitransvection on U and on W. Hence r is a quasitransvection of V. Thus r fixes all the planes containing F(2:.). Let us consider the case r = 1 first. Then IL:l = p. Hence L: = r) for r E ~\ {I}. Pick a plane E containing F(L:) other than F(L:) + U and F(L:) + W. Then L: induces a group of elations on E. Therefore there exists a point P on F(2:.) such that r fixes all the lines through P which are in E. Pick T E Sylp(S)\{L:}. Then P + (E n F(T» is a line in E. We infer from 2:. =
<
49. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
253
Remark. As the proof of 49.2 shows, we can drop the assumptions on p without affecting the conclusions of 49.2, if r = l. 49.3 Lemma. Let ~ be a finite projective space. If'TT is a spread of~, then is also a spread of the dual of~·
1T
PROOF. This follows from the fact that the number of points in ~ is equal to the number of hyperplanes in ~. 0 The finiteness assumption in 49.3 is essential. See Bruen & Fisher 1969.
49.4 Lemma. Let r be a linear mapping of order p of V leaving invariant a spread. Then the minimal polynomial of r is (x - Ii or (x - 1)4. If it is (x - 1)4, then p > 5. PROOF. We have 0 = 1 - rP = (1 - rY. Hence the minimal polynomial fL of r divides (x - lY. Thus fL = (x - lY. As r =1= 1, it follows that 2 ~ i ~ rk( V) = 4. Moreover i ~ p, whence p > 5 if fL = (x - 1)4. Assume fL = (x - 1)3. Then r has the rational normal form
[ ~6
6 ~ ~1
- 3 3 o· 001
Hence V, considered as a p-module, is the direct sum of a cyclic p-module U with rk( U) = 3 and a cyclic p-module W with rk( W) = 1. Let (J be the spread left invariant by r. By 49.3, there exists exactly one X E (J with X C U. Moreover, there exists precisely one Y E (J with W Y. Therefore X7'-= X and Y7' = Y. This yields U = X EB (Y n U) where X and Y n U are r-modules. This contradicts the fact that U is a cyclic r-module isomorphic to the K[x]-module K[x]/ fLK[X]. 0
c:
49.5 Lemma. Let r be a linear mapping of order p of V leaving invariant a spread (J. Then the following statements are equivalent: a) The minimal polynomial of r is (x - Ii· b) The subspace A(r) = {vlv E V, v7' = v} has rank 2. c) r is a shear or a Baer collineation. PROOF. a) implies b): Obviously, r is a quasitransvection. Hence rk(A(r» > 2 by 47.5. Assume rk(A(T» = 3. Then A(r) contains exactly one X E (J by 49.3. The remaining q2 components of (J meet A (r) in a point. Hence Y7' = Y for all YEa. This yields that T is in the kernel of (J whence o( r) = p divides q2 - 1, a contradiction. This proves rk(A (r) = 2. b) implies c): If A(r) E G, then r is a shear. If A(r) rJ. 0, then A(T) consists of the points of a Baer subplane of a( V), since rk(A (T» = 2. Hence T is a Baer collineation in this case.
49. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
254
c) implies a): If (x - I? is not the minimal polynomial of T, then T has minimal polynomial (x - 1)4 by 49.4. Hence V is a cyclic T-module. As a result rk(A (T» = I, whence T has exactly q affine fixed points in a( V), a contradiction. 0
q
The next theorem was proved by Walker for even.
q odd and by Schaffer for
49.6 Theorem (Walker 1973, Schaffer 1975). Let St be a translation plane of order q2 such that GF(q) is contained in the kernel of St. If St admits a group S of collineations isomorphic to SL(2,q), then one of the following holds: a) S operates completely redUCibly but not irreducibly on the translation group of St and St is either desarguesian or a Hall plane. b) S operates irreducibly on the translation group of St, and q == - I mod 3, and St is either a Hering plane or a Schaffer plane. c) S operates reducibly but not completely reducibly on the translation group of St, the order of St is 25, and St is either the Hering plane of order 25 or one of the two exceptional Walker planes described in section 46. PROOF. It follows from our assumptions that St ~ 7T( V) where V is a vector space of rank 4 over GF(q) and 7T is a spread of V left invariant by S. As all translation planes of order 4 and 9 are known, we may assume
q
255
VII. Translation Planes of Order q2 Admitting SL(2, q) as a CoJlineation Group
> 3.
1) Let 1 =1= ~ E 3(S). Then v.l = - v for all v E V. We have S2 = I whence V= V+ EB V- where V+ = {vlv E V, v.l = v} and V- = {vlv E V, v.l = -v}. Assume V+ =1= {O}. As is either a homology or a Baer collineation of 7T( V), we see that I V + I = q2. As S fixes V + we obtain a nontrivial homomorphism of S into GL( V +) ~ GL(2, q). As GL(2, q) does not contain a subgroup isomorphic to SL(2, q)/ 3(SL(2, q», we deduce that S acts trivially on V +. (Remember q > 3.) If V + E 7T, then S consists only of perspectives with axis V + and centres on (XJ' This yields that S is a Frobenius group which is absurd. Hence V + ti:. 7T whence V + is a Baer subplane. But then S acts regularly on the q(q - 1) points of 100 which do not belong to V +. As a consequence q(q2 _ 1) = IS I divides q(q - 1), a contradiction. Therefore V + = CO} which proves 1).
s
PROOF OF a). We infer from 2) that there is no irreducible S-submodule of rank I or 3 of V. It follows therefore that there exist two irreducible S-submodules U and Wof V of rank 2 with V = U EB W. From this and I) we infer that S satisfies the assumptions of 49.1. Hence ~ = {F('L) I 'L E Sylp(S)} is a set of q + 1 pairwise skew lines and U and Ware transversals of ~. Since S acts transitively on ~, we have ~ ~ 7T or
91 n
7T
=
0.
Case I: 91 n 7T = 0. Pick 'L E Sylp(S) and let P be a point on F('L). Then there exists exactly one X E 7T with P ~ X. Hence X~ = X. The plane E = F('L) + X is fixed by 'L. Moreover all elements of 'L induce elations in E with axis F('L) and centre P = F('L) n X, as X~ = X. Also 'L acts transitively on the set of points of Yother than P for all lines Y with P ~ Y <: E and Y =1= F('L). Pick T E Sylp(S) with T =1= 'L. Then F(T) n E Q: F('L). Hence P + (F(T) n E) is a transversal of ~ by the above remarks. Thus ~ is a regulus and the opposite regulus ~' of ~ is fixed linewise by S. We deduce from this fact that 9C s ('L) fixes F('L) pointwise. Hence X is fixed by 9C s ('L). Let T again be a Sylow p-subgroup of S distinct from 'L. Pick 8 E (9C s ('L) n 9C s (T»\3(S). This is possible since q > 3. Then 8 'fixes F(T) pointwise. As the collineation induced by 8 is not the identity, all fixed points of 8 are on F(2:,) or on F(T). As X is fixed by 8, Maschke's theorem implies that 8 has a fixed point Q on X other than P. Hence Q = X n F(T). Thus X E ~/, i.e., ~' ~ 7T. It follows now from 48.8 that 7T( V) is a Hall plane. Case 2: ~ ~ 7T. If r = 1, then 9t is a regulus by the remark made after 49.2. As the opposite regulus is fixed linewise by S, it follows from 48.8 that 7T( V) is desarguesian in this case. Hence we may assume r > 1. Let u be a p-primitive prime divisor of p2r - l. Then u divides p r + 1 but u does not divide pS + I for s < r. If there exists no p-primitive prime divisor of p2r - I, then p = 2 and r = 3 by 6.2. In this case we put u = 2 3 + I = 32 . Obviously u is odd in all cases. Let p be an element of order u in S. As p leaves invariant 7T\~ and as 17T\~1 = q(q - 1), it follows that p fixes an X E 7T\~, since u is a power of a prime satisfying (q(q - 1), u) = l. Hence ~ is a regulus by 49.2. We infer from 49.1 that the opposite regulus of ~ is fixed linewise by S. Therefore, 7T( V) is desarguesian by 48.8. PROOF OF b). As the p-elements behave quite differently in this case according to whether p = 2 or p > 2, we have to separate the discussion into two cases. First we discuss the case p > 2. In order to make this apparent, we shall label our statements by 01), 02), ...
2) S fixes no point and no plane of V. 01) If Assume that 2) is Then S acts trivially follows that S fixes operates trivially on
false. By 49.3 we may assume that S fixes a point P. on P. There exists exactly one X E 7T with P <: X. It X. This yields that S acts trivially on X, since S P. Therefore S consists only of perspectivities with
axis X which is absurd.
T
E S has order p, then the minimal polynomial of T is (x - 1)4.
This follows from 49.4 and 49.1 by using 1). If T is an element of order p of S, then V is a cyclic primary T-modu1e by 01). Hence there exist exactly one point peT), exactly one line L(T),
256
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
and exactly one plane U(7') left invariant by L(7') ~ U(7'). 02) Let 'i. E Syl/S). Then P(7') for all 7',7" E'i.\{l}.
=
7'.
P(7"), L(7')
Moreover P(7')
= L(7"),
~
L(7') and
and U(7')
=
U(7")
This follows from the fact that 'i. is abelian. As P( 7') etc. depend only on 'i. we shall write P('i.) etc. instead of P( 7') etc. Pick 'i. E Sylp(S) and put H = 91 s ('i.). Then there exists a cyclic subgroup 11 of H such that H = 11'i. and 11 n 'i. = {l}. It follows that 1111 = q - 1. Moreover we have P('i./l. = P('i.), L('i.)~ = L('i.), and U('i.)~ = U('i.). The line L('i.) is a completely reducible l1-module. by Maschke's theorem. Hence there exists a point Q with L('i.) = P('i.) + Q and Q~ = Q. Moreover 'i. acts sharply transitively on the set of points on L('i.) other than P(2.). Let R be a point on L('i.) with R =!= P('i.), Q and assume that 0 E 11 fixes R. There exists exactly one 7' E'i. with R'r = Q. This yields Ro-I'ro = R'ro = QO = Q = R'r, whence 0- 17'0 = 7'. As 7' =!= 1, we obtain 0 E 3(S). Thus we have proved: 03) 11 splits the set of points on L(2.) other than P('i.) and Q into two orbits of length! (q - 1).
Using Maschke's theorem once more, we find a point A in U('i.) with U(L) = L(2.) + A and A ~ = A. Since 2. acts sharply transitively on the set of lines through peL) in U(2.) which are distinct from L(2.), we obtain similarly to 03) 04) t::. splits the set of lines through P(L) in U(L) which are distinct from A + P(2.) and L(2.) into two orbits of length !(q - 1). Let B be a point on U(2.) with B ~ L(2.) and B tl A H, Assume BTl = B for some YJ E H \ { 1}. Then YJ tl 2.. Hence there exists Il E H such that Il - I YJIl E 11. We infer that B J.L is a fixed point of 1l- IYJIl which is distinct from A. Therefore B J.L + P(2.) is a fixed line of Il - IYJIl . If B p. + P(2.) =!= A + P(2.), then (J - IYJIl E 3(S) by 04). This yields YJ E 3(S). If BIl + P(2.) = A + P(2.), then there exists 7' E 2. with A'r = BIl. This contradicts our assumption that B is not in the orbit of A under H. Thus we have proved IBHI = 1q(q - 1). Hence we have: 05) H splits the set of points of U(2.) into five orbits, namely {P(2.)}, L(2.)\{ P(2.)} , A H, and two orbits of length 1q(q - 1).
Let X E 'TT contain P(2.). Then XL = X whence X the only line fixed by 2.. This proves 06) L(L) E
'lT
for all L E Sylp(S).
Next we prove 07)
\.is
=
{peL) \ L
E
Sylp(S)} consists of q + 1 points.
= L(2.),
as L(2.) is
49. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
257
Assume P(2.) = peT). Then P('i.) is fixed by the group generated by 2. and T. This yields 2. = T, since S does not fix a point by 2). 08)
~
= {L(L) IL
E Sylp(S)} consists of q
+ 1 lines.
Assume L(2.) = L(T). Then L(L) is fixed by the group generated by L and T. Hence L = T, as S acts irreducibly. Put 'lTo = 'TT\~. Then l'lTol = q(q - 1).
09) No three points of ~ are collinear. Assume that P,Q,R are three collinear points of ~. Then the orbit of R under Sp,Q contains exactly 1(q - 1) points. Hence P + Q carries at least t(q - 1) + 2 = t(q + 3) points of ~. Pick two distinct points X, Y in ~ which are also distinct from these t (q + 3) points. Then X + Y also carries t(q + 3) points of ~, as S acts doubly transitively on ~. Now t(q + 3) + t(q + 3) = I~I + 2, whence P + Q and X + Y have two points in common. Therefore P + Q = X + Y. This yields that all points of ~ are collinear contradicting the fact that S acts irreducibly. 010) For L E Sylp(S) and X E
'lTo
we haveSp(L)
n Sx = 3(S).
Obviously 3(S) ~ SP(,Z) n Sx' As ~\{L(L)} is an orbit of H = Sp(L)' we have that L(T) n U(L) is in an orbit of length q of H provided T =!= 'i.. Hence X n U(2.) is in an orbit of length 1q(q - 1) of H by 05). Let YJ E H n Sx' Then YJ fixes X n U(L). It then follows that TJ E 3(S). 011) No four points of
~
are coplanar.
Assume that there are five coplanar points PI' ... , P 5 in ~. Let E be the plane containing them. As Spl,P z splits ~\{PI,P2} into two orbits of length (q - 1), there exists without loss of generality an YJ E S PI ,P2 with Pj = P4 . It follows from 09) that ETI = E. By 48.3, there exists X E 'lT with X ~ E. Hence XTI = X. By 1) and 010), we have X E 53, i.e., X = L(2.) for some L E Sylp(S). As YJ does not induce the identity on ~, we have P('i.) E {PI ,P2}. Starting with P 3 and P 4 instead of PI and P 2, we obtain the existence of aTE Sylp(S) with L(T) ~ E and peT) E {P 3 ,P4 }. As E contains exactly one component of 'TT, we finally reach the contradiction peL) = peT). Let P,Q,R be three points of ~ and let E be the plane containing them. Again there exists X E 'lT with X t;: E. Assume X = L(2.) for some L E Sylp(S). If E = U(2.), then all the points of ~ are contained in E whence E S = E, a contradiction. Hence IEkl = q. Moreover two distinct planes of Ek have only L(L) in common. Hence sq = q, where s + I is the number of points of ~ contained in E. This yields the contradiction s = 1. Hence X E 'lTo' In particular, X does not contain a point of ~. Let n be the number of planes in the orbit of E under Sx' The proof of 010) together with 05) yields \Sx\ = q + 1 or iSxi = 2(q + 1). Therefore Sx acts either
t
1.,)0
VII. 1 ranSlallOn
rldllC;' UI
VIUCI
I.{
n.UlIlIlLlIl~
.:>L\L.,
I.{J d~
d
,--UIIIII'-<",.vu
'V'VUj-'
transitively on ~ or splits ~ into two orbits of length ! (q + 1). Hence q + lor tn = !(q + 1), if I is the number of points of s:f3 in E. Assume t = 4. Then q == 3 mod 4. But then Sp acts transitively on the set of 2-subsets of s:f3\ {P}. Hence S E,P still acts 2-homogeneously on the set of the three points of s:f3 other than P which are in E. Hence 3 divides ISE,pl = ISp n Sxl contradicting 010). This proves 011). Moreover q == -1 mod 3. Therefore 7T( V) is a Hering plane by 43.8 and 45.1. Next we discuss the case p = 2. If a is an element of order 2 in S, then (x - 1)2 is the minimal polynomial of a. Hence rk(A(a» = 2 by 49.5. For L E SyI 2(S) put peL) = {v Iv E V, va = v for all a E L}. It follows from 49.1 that there exist a,T E L\{l} with A(a) ~ A (T). Thus peL) is a point. As S fixes no point by 2), we have therefore In =
El) ffi = {peL) I L E SyliS)} is a set of q
+I
points.
Put VeL) = A(a) + A(T) for a,T E L\{I}. As L is abelian, A(aY = A(a) and A (Tt = A (T). Hence VeL) is fixed by Land L fixes all the lines through peL) which are contained in VeL) individually. If L fixes a line through peL,) which is not contained in VeL), then L fixes all the lines through P(2:.), whence 2:. consists only of transvections of V which is not the case. Hence A (p) C U(2:.) for all p E 2:.\ {l}. Moreover U(L) is fixed by H = 91 s (L). Let Z be a cyclic subgroup of order q - 1 of H. It follows from Maschke's theorem that V(L)/ peL) is a completely reducible Z-module. Hence there are two lines Ll(2:.) and Ll2:.) through peL) and contained in V(2:.) which are fixed by Z. E2) Either
Ll~:) E 7T
or Ll2:.) E
7T.
By 49.3, there exists exactly one X E 7T with X C V(2:.). Hence X H = X whence in particular peL) C X. If X ~ L I(2:.),Ll2:.), then Z fixes all the lines through P(2:.) which are in VeL), as Z consists only of linear transformations. But there exists ~ E Z with a r = T and hence A(ai = A ( T) ~ A (a). This contradiction proves E2). E3) Z has exactly three fixed points in U(L) one of them being peL) the other two lying on LI(2:.) respectively Li2:.). It follows from the proof of E2) that all fixed points of Z are on LI(2:.) or on Li2:.). Let Q and R be fixed points other than P(2:.) on Ll2:.). As A (p) =t= Li(2:.) for all p E 2:., there exists p E 2:. with QP = R. Let ~ E Z. Then Qr-1pr = QP~ = R~ = R = QP. Hence ~ -Ip~ = P for all ~ E Z. This yields ~ = 1, i.e, Q = R. By Maschke's theorem, there exists a fixed point of Z on Lj(2:.) other than peL). This proves E3). Lj(2:.) is fixed by L and by Z. Hence it is fixed by H. Thus we may label the Li(T) for T E SyliS) in such a way that the LI(T)'s as well as the LiT)'s form an orbit of S.
E4) Z has exactly four fixed points.
"'7.
I
JdllSJdllOn t'JdIleS or vroer q-
AdmittIng ~L(2,
q) as a Collineation Group
259
By Maschke's theorem, there exists a point F fixed by Z such that .V = ~(L) ffi.F. Pick X E 7T with F C X. Then XZ = X. Hence X n V(2:.) IS a fIxed pomt of Z on VeL) which is distinct from P(2:.) by E2). We may assume L](2:.) E 7T. Then X n V(2:.) is the second fixed point of Z on Ll2:.). In particular, all fixed points of Z which are not on L/2:.) are on X. There exists T E SyllS)\{2:.} with Z C; 8'(s(T). Assume pet) C; LI(2:.). Then LI(T) = LI(2:.), whence LI(L) is fixed by (L,T) = S, a contradiction. Hence pet) ~ LI(2:.) which yields pet) C; X by the remark made above. From this we infer X = LI(T). .If has more than four fixed points, then Z fixes LI(2:.) and LI(T) pomtwlse, as all the fixed points of Z are on LI(2:.) or on LI(T) and since there exists y E 91 s (Z) interchanging LI(2:.) and LI(T). The group L acts regularly on the set of points on LI(2:.) other than P(2:.). Hence L IS centralized by Z which is absurd. Thus Z has exactly four fixed points. Put ffii = {Li(L) I L E SyllS)}.
Z.
E5) ffi l and ffi2 are reguli opposite to each other.
If Li(L) = Li(T) for 2:. ~ T, then Li(2:.) is fixed by (2:.,T) = S, a contradiction. Thus Iffiil = q + 1. We may assume ffi l C 7T. Then ffi l is a partial spread. Pick LiL) E ffi 2 . We have to show that Li2:.) meets LI(T) for all T E SyI2(S), This is certainly true for T = L. Assume T ~ 2:.. Then Z = 8'(s(2:.) n 91 s (T) is cyclic of order q - 1. By E4), the group Z has exac~ly four fixed points, three of them being points of V(2:.), namely P(2:.), a pomt FI on LI(2:.) and a point F2 on Li2:.). Moreover, Z has two fixed points
o
e.
:II
I
260
VII. Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
semilinear mappings of the vector space K(2), we have that all elements of S may be represented by matrices of the form
° ° °
c
d
g
h
1
e
a'P c'P
b'P d'P
where cp is a fixed automorphism of K and ad - bc = 1. Moreover, using Maschke's theorem again, we may assume that Pa = diag(a,a - I ,a'P,a -'P) belongs to S. Consider elements a and T of order p defined by
s es gs
° °° °° °
1 Is hs
resp.
1 s'P
er
° °° °° ft1
1
gr
et
t'P
'0
°
*
1
s
°
1 1; h's
e's
g;
°° °° ° 1
s'P
Then es + r = es + tis + er , whence tis = sft, as aT = Ta. Hence Is = s11 = sk. Assume p = 2. Then 0 fixes a subspace of rank 2 vectorwise. Let (X I ,x 2,X 3,X 4) (0,0,0,x 4) be a fixed vector of o. Then SX 1 = 0, esxI + Isx2 = 0, gsx I + h 2x 2 + S'PX3 = 0.
one obtains a'P = a. Therefore a = a -3; i.e., a 4 = 1 for all a E K*. Thus q = 5. It now follows from 46.12, 46.4, 46.10, 46.11 that 7T( V) is isomorphic 0 to one of the planes 7T I (V),7T 2 ,O(V),7T 3,O(V).
This yields XI = 0. If X2 = 0, then x) = 0, as s'P 0. But then (XI 'X 2,x) ,x4 ) = (0,0,0,x 4 ), a contradiction. Hence Is = 0, i.e., k = and es + t = es + et • Conjugating 0 by Pa yields aa-'Pes = ea2s' In particular ael = a'Pea2. Pick a E K* with a + I 0. This is possible, as q > 4. Then
50. The Collineation Groups of the Hering and Schaffer Planes
*
*
°
*
+ e l = (a + 1)'Pe(a+1)2= (a'P + 1)(ea2 + e l) = a 'Pea + a'Pe l + ea + e l . a'Pe l + ea = (a'P + aa-'P)e l = a-'P(a 2'P + a)e l .
a 'Pea 2 + e l = ael
2
2
Hence 0= 2 Assume e l = 0. Then ea 2 = for all a and hence es = for all s. But then 2 0 = 1 yields hs = 0. Therefore the Sylow 2-subgroup of S consisting of the mappings
°
°
° °° °° ° °° ° gs s
1
1 s'P
1
fixes the line {(O,x,O,y) I x,y E K} vectorwise. This contradicts the fact that S is not completely reducible (Theorem 49.1). Thus a 2'P = a for all a E K and hence a'P = aa-'P for all a 0. Therefore ea 2 = a'Pe l = a 2'P2e) , whence es = s'P2e l for all s E K. Computing the coefficient in the lower left hand corner for 0 2 = I yields shs = s'Ps'P2e l for all s E K. Hence s2h} = s2'Ps2'P2ei = ss'Pei. Therefore 2 h}= s-Is
*
I
*
2 Since e l 0, we obtain S'P2 + 1 = S'P2 + s
*
abO
261
50. The Collineation Groups of the Hering and Schaffer Planes
(s
+ 1) 'P2 e I
=
(s
+ I) 'Phs + I
=
(s
+ I) 'P (hs + hi) = (s + 1)'1' (s - 'Ps 'P2 + l)e I
.
In this section we shall determine the collineation groups of the Hering and Schaffer planes as well as the number of isomorphism types of Schaffer planes of a given order. 50.1 Lemma. If 7T( V) is a Hering or a Schaffer plane, then 7T( V) does not possess a non-trivial shear. PROOF. Let S ~ SL(2,q) be the group used in the construction of 7T( V). Then S splits 7T into an orbit ffi of length q + 1 and two orbits of length ~ q(q - 1), as we have seen in section 46. Assume that a is a non-trivial shear with axis X. We may assume X E 7T. Assume X tl. ffi. The construction of 7T( V) shows that there exists a homology of order 3 in S fixing X. As this homology does not commute with 0, there are at least 3 shears with axis X. Let G be the group generated by all shears of 7T( V) whose axes pass through the origin. Then
G ~ SL(2,pS), G;;;;; S(22s+1) or 4 does not divide the order of G by 35.10. (Remember that q == - 1 mod 3.) In the latter two cases, the characteristic of 7T( V) is 2 whereas in the first case the characteristic is p. As there are at least 3 shears with axis X, we have that 4 divides IG I if the characteristic is 2. Hence G ~ SL(2,pS) or G ~ S(2 2s + I). The number n + 1 of axes of
1.61.
VII. Translation Planes of Order q" Admitting SL(2, q) as a Collineation Group
non-trivial shears is p S + I or 22(2s+ I) + 1 by 35.10. Therefore either n + 1 = q2 + 1 or n + 1 = q + 1 + t q(q - 1) = t q(q - 1) + 1 or n + 1 =tq(q+l)+tq(q-l)=q(q-l). But the latter two cases cannot occur, as n = pS. Hence n + I = q2 + 1 and G ~ SL(2,q2) or G ~ Seq). In the first case, 7T( V) is desarguesian by 38.12 which is obviously absurd. In the second case, 7T( V) is a Luneburg plane by 31.1. But 3 does not divide the order of the group of those collineations of the Luneburg plane which are induced by linear mappings of V. Hence X E ffi. Moreover ffi is left invariant by the stabilizer of the origin in the full collineation group of 7T(V) by what we have just seen. Pick Y E ffi\{S}. Then no element of Sx,y\3(S) fixes a third component belonging ~o ffi. Hence the only elements of SX,y centralizing a are the elements of 3(S). Thus there are at least t (q - 1) non-trivial shears with axis X. This yields that the group of shears with axis X has order q. We deduce from this fact and 35.10 that the group G generated by all shears with axes through the origin is isomorphic to SL(2,q). We infer from 49.5, 49.1, and 49.6 a) that 7T(V) is desarguesian, a contradiction. 0 50.2 Lemma. Let 7T( V) be a Hering or a Schaffer plane of order q2 and let S ~ SL(2,q) be the group used in the construction of 7T( V). If ffi C; 1T is the orbit of length q + 1 of S, then ffi is fixed by all the collineations of 7T( V)
which fix the origin. PROOF. Assume to the contrary that there exists a collineation y fixing the origin and an X E 7T\m such that XY E ffi. Then there exists a homology p of order 3 such that either XY is the axis or XY n 100 is the centre of p. The group Sx y then moves either the centre or the axis of p whence there exists a non-trivial shear by 13.9 contradicting 50.1. 0 50.3 Theorem (Walker 1973, Schaffer 1975). Let 7T(V) be a Hering or a Schaffer plane of order q2 with q =1= 5. Furthermore, let S ~ SL(2,q) be the group used in the construction of 7T( V). If G is the group of collineations of 7T( V) fixing the origin, then G = 9CrL(v/S).
PROOF. We first show that 9C rL(v)(S) C G. This is certainly true, if q is odd by 45.1. If q is even, then 9CrL(v/S) permutes the two spreads left invariant by S. Hence it permutes the two reguli ml and ffi 2 • Assume that there exists y E 9C rL (v)(S) with ffij = ffi 2. Then ffi~ = ffil' We may assume that y fixes the points (1,O,O,O)K, (O,O,O,I)K and (l,I,I,I)K on the curve ~(a). Put e 1 = (1,0,0,0), e2 = (0,1,0,0), e3 = (0,0,1,0) and e4 = (0,0,0,1) and let f3 be the companion automorphism of y. Then (2:;= Ieix)Y = 2:;= 1e/x!. Moreover e{ = e1a, el = e4 b and (2:;= 1ei)Y = (2:;= 1ei)c whence a = b = c. We may assume a = b = c = 1. The lines through elK belonging to ffi] resp. ffi2 are the lines {(x,O,y,O) I x,y E K} and {(x,y,O,O) Ix,y E K}; hence e l x f3 + eIy f3 = (e1x + e2y)Y = e]x' + e3 J' and e]xf3 + eJy f3 = (e1x + e3 y)Y = e1x" + e2y" for all x,y E K. Puttingy = 0 yields e1x fJ = e1x' + e30' and elx fJ = XIX" +
) I. 1 he 1 heorem
LbJ
01 Lolman-Prohaska
°
e20". Hence 0' = 0" = and x' = x" = x. Therefore, eIy f3 = e3 J' and eI = e2y"· Now 2:;=1 ei = (2:;=1 eJY = e 1 + e 3 1' + e21" + e4 • Thus I' = I" = 1 and eI = e3 and ej = e2 • Now (e1ss a + e2s a + e 3 s + e 4 )Y = e]sf3s f3 a + e 3 s f3a + e 2s f3 + e 4 , as Aut(K) is abelian. Hence there exists t E K such that s(3sf3a = ua, s(3 = ta, a2 s(3a = t. This yields s(3 = sf3 for all s E K whence a 2 = 1, i.e., q = 4~-1 mod 3. This contradiction proves 9C rL(v)(S) C G. Assume there exists 8 E G\9C rL(v)(S), Adjusting 8 if necessary by an element of 9CrL(v)(S), we may assume that 8 is linear. Then it follows from 43.3 and 44.l that there exists P E ~ respectively P E ~(a) with po ti. ~ resp. ~(a). Let IT E Sylp(S) fix P and let X E 1T contain P. Then XO is fixed by ITo. Moreover there exists Q E ~ resp. ~(a) with Q C; X by 50.2. Furthermore, po =1= Q. Let IT* E SylpCS) fix Q. Then S* =
> 4 and assume q == Then there exist up to isomorphism exactly
50.4 Corollary (Schaffer 1978). Let q = 2r
-
1 mod 3.
The proof is an easy exercise.
51. The Theorem of Cofman-Prohaska In section 13, we studied the construction technique called derivation in order to obtain the Hall planes. The finiteness assumptions made there are not really necessary. If we want to generalize this construction to infinite planes, we have to make clear what we mean by a Baer subplane of an infinite projective plane: A proper subplane n of a projective plane I.l3 is called a Baer subplane of 1.l3, if each point of I.l3 is on at least one line of n and if each line of I.l3 carries at least one point of n. In the finite case, this definition is equivalent to the one given earlier, as is easily seen. Obviously, Baer subplanes are always maximal subplanes. With this definition of Baer subplanes derivation carries over to arbitrary planes: Let %( be an affine plane, D a set of points on 100 and let ?B be a set of Baer subplanes of m: such that: 1) If 0 E?B, then 0 n 100 = D. 2) If P,Q are distinct points of %( such that PQ E ?B such that P and Q are points of
o
o.
n too
ED, then there exists
Let 0' be the plane generated by P, Q and D. Then 0' ~ o. Let X be a point of 0 which is not on PQ and does not belong to D. Then PX and QX
:> I. 1 ne 1 neun::JlI
are two distinct lines meeting 100 in points of D. Hence PX and QX belong to ,0' whence X belongs to 0'. Therefore all points of 0 not on PQ are pomt~ ~f 0' whence 0 = 0'. This implies that there is exactly one 0 E?B contammg P and Q. N ow we define 2(' in the same way we did in section 13. We want to sh~w that ~: i~ ,an affine plane. By the remark just made, each pair of pom~s of 2( IS Jomed by exactly one line of 2('. Let 1 be a line of 2(' and Q ~ pomt not on I. Assume that 1 is also a line of 2(. Then 1 n 100 ti. D. As 1 mtersects all the Baer subplanes in ?B in a point of the projective closure of ~ and ~ince 1 n 100 ti. D, we see that 1 intersects all the Baer subplanes in ?B m a pomt of 2('. From this we infer that there exists exactly one line of 2(' through Q which does not meet 1 in a point of 2(', namely the line Q(l ~ 100 ) of 2(. Assume that 1 E ?B. Then all the lines through Q which are also hnes of 2( meet 1 in a point of 2(', as we have just seen. . The point Q is on exactly one line h of the Baer subplane I, as Q is not m ~. Let g be a line of 1 p~rallel to but distinct from h. Furthermore, let j be a lme of 2( through Q wIth h =1= j an~j n 100 E D. Put X = j n g. Then X do~s not belong to I, as otherwise j would be a line of 1 contradicting the umqueness of h. There exists m E ?B with Q,X in m. Assume that m has a poi~t ~ in common with I which is not on 100 , Pick a line i of I through Y whIch IS not parallel to h. As h,g,i are also lines of m, it follows that i n h and i n g are distinct points common to m and I whence m = I contradicting the fact that Q does not belong to I. Hence m and I have n~ poin t in common. Let m' E?B be such that Q is also a point of m' and I and m' have no point of 2(' in common. Let Z be a point of m. Then Z is on a line z of I. As m and I have no point of 2(' in common, z is parallel to h. Hence all the lines of I parallel to h are lines of m. Likewise we obtain that all these lines a.re also lines of m'. As all the lines of m through Q are also lines of m', we fmally get that m = m'. Finally, it is obvious that there is a triangle in 2('. This proves that 2(' is an affine plane. The following theorem was first proved for finite planes by Prohaska 1972. The generalization to arbitrary planes and the proof given here are due to J. Cofman 1975.
51.1 Theorem. Let 2( be an affine plane, let D be a set of points on the line at infinity of 2(, and let ?B be a set of Baer subplanes of 2( such that the following is true:
1) If 0 E?B then b n 100 = D. 2) If P,Q are distinct points of 2( such that PQ o E ?B such that P and Q are points of o.
UI
a set of Baer subplanes of ~d with the following properties: If 0 E s.B then 100 is a point of band D consists of all the lines of 0 which pass through 100 , If I and m are lines of ~d not passing through 100 such I n m is on one of the lines of D, then there exists 0 E Q3 such that I and m are lines of o. Pick a E D and consider the following incidence structure Q.
a) The points of a are the points of ~d other than 100 which are on lines of D\{a}.
,8) The lines of a are those subsets of the set of points of Q which are intersections of lines of ~d other than a with Baer subplanes belonging to ?B. y) The planes of Q are the point sets 1\ {loo} with lED \ {a} and' 0\ {a} with 0 E ~. (Here we consider lines and planes as point sets.) We shall prove that
E D, then there exists
Then all the planes belonging to ?B are desarguesian.
PROOF. Let ~ be the projective closure of 2(. Then D is a set of lines in the dual ~d of ~ all of them passing through the point 100 of ~d. Moreover ~ is
Q
is a 3-dimensional affine space.
A) Let A,B,C be three distinct points of a which are not collinear in ~d. Then there is exactly one plane of Q containing them. At most one of the lines AB,BC,CA is in D. Hence there is at least one b E s.B containing A, Band C. By the remarks made before 51.1, the plane o is uniquely determined. B) If two planes band c of ?B have two distinct points A and B of common, then AB n b = AB n c.
Q
in
Case 1: AB ti. D. Then AB n b = {l n AB\I E D\{a}} and AB n c E D\{a}}. Case 2: AB ED. Suppose AB n b =1= AB n c. Then we may assume without loss of generality that there is a point C E AB n b with C fl d AB n c. Pick a line h of the Baer subplane b of l-l3 through C other than AB. Then h carries exactly one point H of c. As H is on a line kED, we see that H = h n k is a point of b, too. Now AH,BH fl D. Therefore, there is exactly one plane in s.B such that AH and BH belong to this plane. Hence we reach the contradiction c = b.
= {l n AB\I
(1) Any two distinct points of
Q
are joined by a unique line of
Q.
It follows from B) that there exists at most one line of Q joining the two distinct points A and B. As there is always a point C of Q such that A, B and C are not collinear in ~d, it follows from A) that there is at least one line of a joining A and B. (2) Each line of
n 100
'J""""-""
\""-UlIl1o.U-' •
Q
contains at least two points.
This is trivial. (3) The line of Q joining two distinct points of a plane of a is completely contained in this plane. This follows from B).
266
VII, Translation Planes of Order q2 Admitting SL(2, q) as a Collineation Group
(4) Any three distinct points of unique plane of a,
Q
which are not collinear in a are on a
Let A ,B, C be three points of a which are not collinear in a. If A ,B, C are not collinear in SUd, then (4) is true by A). Hence assume that A,B,C are collinear in SUd and let I be the line carrying A ,B, C. Then it follows that I belongs to D. From this, B) and the fact that A,B,C are not collinear in a, we infer that I is the only plane of a containing A,B,C. (5) Each plane of a contains three points which are not on a line of a. This is trivial. (7) If two planes of a have a point in common, then they have a line in common. This is certainly true if at least one of the planes is induced by a line of D. We may therefore assume that the two planes are induced by elements ex,f3 E?S. By assumption ex and f3 have a point P of a in common. Suppose there is a line I of SUd through P other than looP which is a line of ex and of f3. Then {I n h [h E D \ { a}} is a line of a common to the two planes of a which are induced by ex and f3. Hence we may assume that there is no such line. Let Q be a point of a on looP belonging to f3 but not to ex. Pick a line m of f3 through Q which is distinct from looP. Then m carries a point R of ex which is distinct from P and 100 , As R is on a line of D, it also belongs to f3. Hence P R is a line of SUd which belongs to both ex and f3 in contradiction to our assumption. Hence f3 n IooP ~ ex n looP. By symmetry ex n looP ~ f3 n looP. This proves (6). C) The points and lines contained in a plane b of a form an affine plane. This is certainly true if b is induced by a f3 E ?S. Hence we may assume that b U {loo} = lED \ { a}. Pick y E ?S and let X be a point on a which is not a point of y. Denote by c the plane of a induced by y. For all Y E c define yr by yr = Xy n I. Then y is a bijection from c to b (as point sets). Using A) and B), it is easily seen that y induces a mapping of the set of lines of a contained in c onto the set of lines of a contained in b. This proves C). Two lines I and m of a are said to be parallel if they are contained in a plane of a and if either 1= m or I n m = 0.
52. Prohaska's Characterization of the Hall Planes
267
not in c, we have by (6) that p is a line of a. Assume p =/=- n. Then p meets ~ by C). As p n I ~ c n b = m, we have p n I ~ m n p n I = 0. Hence P, IS
parallel to I. Therefore m and p are paralle~s ,of I passing through the pomt
p n m. As there is exactly one pla~e ~ontammg I and p n m, we reach the contradiction p = m. This contradIctIon proves p = n. If I and n are ~ot
parallel, then they meet by C). Therefore I and n a~e ?arallels of m havmg a point in common, whence 1= n, again a contradIctIOn. (9) If I is a line of a, then there exists a line of a parallel to and distinct from I. This is trivial. From (1) to (9) we conclude that a is an affine space which is not an affine plane (see e.g. Lenz [1967, p. 136]). Moreover, the special form of (7) yields that the dimension of a is 3. . ' Now, if f3 E?S then the plane of a induced by f3 IS de,sargueslan, as all affine spaces are desarguesian. Hence f3 itself is desargueslan, 0 As an immediate consequence we have: 51.2 Corollary (Prohaska 1972). If 21 is a finite derivable plane, then the order of mis a power of a prime.
52. Prohaska's Characterization of the Hall Planes We shall finish this book by proving the following theorem due to· Prohaska 1972: 52.1 Theorem. Let 21 be a finite affine plane of order q2 and let D be a set of q + I points on 100 such that 21 is derivable with respect to D. If G is a rank-3-group of 21 leaving invariant D and if 211 is the plane derived from 1 with respect to D, then one of the planes 21,21 is desarguesian and the other one is a Hall plane.
m
Before we can prove this theorem, we have to establish a few lemmas.
(8) Parallelism is an equivalence relation. We have only to show that parallelism is transitive. Let I,m,n be three lines of Cl such that III m and m "n. If I, m and n are all contained in the same plane of a, then III n by C). Hence we may assume that they are not contained in one plane of a. This implies in particular that I, m and n are pairwise distinct. Let b be the plane containing I and m and let c be the plane containing m and n. Pick a point X on n. It follows from (4) and (3) that there exists a plane d of a containin!! X and l. T.et n = rf n r A" 1 j"
52.2 Lemma. Let 21 be the desarguesian affine plane of order q = p r and let II be a p-subgroup of the collineation group of 21. If [III> q and if II fixes a point of 21, then II contains a shear of 21. PROOF. Let 0 be the fixed point of II and let r be the group of all collineations of 21 fixing O. Then ~ fL(2,q). Hence
r
In = rQ(Q
-
l)(Q2 - 1).
As the number of lines through 0 is q + 1, there is a line I through 0 fixed by IT. Let 2: be the group of all shears with axis I. Then ILl = q. Moreover 2: is normalized by IT, whence IT2: is a p-subgroup of r. Let pS be the highest power of p dividing r. Then IITLI divides pS+'. As pS ~ r < p' = q, we see that IIT 2: I < q2. Therefore lIT n ~IIIT~I = IITIILI > q2 yields ITn2::f={l}. D 52.3 Lemma. Let 2( be an affine plane of order q2 > 4 and let D be a set of q + 1 points on 100 such that 2( is derivable with respect to D. Denote by 2('
the plane derived from 2( with respect to D. If 7J is a collineation of 2( fixing a point 0 of 2( and each line through 0 which does not carry a point of D, then 7J is either a homology of 2( or of 2(', PROOF. First of all we note D7J = D. Hence 7J is a collineation of 2( and of If 7J fixes a point of 2( other than 0, then 7J fixes a quadrangle and hence a subplance of the projective closure of 2( pointwise. As 7J has at least q( q - I) fixed points on 100 , the order of this subplane is at least q(q - 1) - 1. As q > 2, it follows that the order of this subplane is greater 2('.
{c1.
than q = Hence 7J = 1. Thus we may assume that 7J has no fixed point other than O. Assume that 7J is not a homology of 2(. Then 7J :f= 1. Moreover, there exists a line I through 0 with 17J:f= I. Let 0 be a Baer subplane of 2{ containing 0 and D. Then we have to show o7J = o. Assume o7J :f= O. Then 0 and o7J have only one affine point in common, namely O. Moreover, I is a line of O. Hence there exists a point P of 0 other than 0 which is on I. As I:f= 17J, we have P:f= p7J. If the line pp7J carries a point of D, then pp7J is a line of 0 as well as of o7J, since P is a point of 0 and p7J is a point of O7J. But all affine lines common to both 0 and o7J pass through O. Hence 0 I P p7J whence 1= OP = OP7J, i.e., 1= 17J. This contradiction proves that pp7J does not carry a point of D. Let F = 100 n pp7J. Then pp7J = FP = Fp7J and hence (Pp7J)7J = pp7J. This yields that 7J has a fixed point other than 0 in 2( whence 7J = 1, a contradiction. D 52.4 Lemma. Let p be a prime and q = p' > p. Furthermore, let ?B be a set of length q(q - 1) and S a permutation group of?B isomorphic to PSL(2,q). If
no element of order p of S fixes an element of ?B and if all orbits of Shave the same length, then there exists a partition tJ of?B such that IXI = 2 for all X E tJ which is left invariant by S and such that Sx is a dihedral group of order 2ISI(q(q - 1»-1. This is proved easily with the help of 14.1. PROOF OF 52.1. As G is a rank 3-group of 2( as well as of 2(', the planes 2( and 2(' are translation planes by 16.3. Moreover G contains the translation group T of 2( and T is also the translation group of 2('.
Let 0 be a point of 2( and let ~ be the set of lines through 0 which carry a point of D and denote by we the remaining lines through O. Furthermore, let ?B be the set of Baer subplanes containing 0 and D. Then Go acts transitively on the sets ~,we and ?B by 16.2. Moreover I~I = q + 1, IWCI = q(q - 1) and I~I = q + 1. Put a = {P I 0 :f= P I 1 and I E ~} and b = {P I 0 :f= P I I and I Ewe}. Then a and b are orbits of Go = H. Moreover lal = (q + 1)(q2 - 1) and Ibl = q(q - 1)(q2 - 1). Let d = (2,q - 1). Then d -lq(q2 - 1)2 is the least 1 common multiple of lal and Ibl. Hence d- q(q2 - I? divides IHI· N ext we prove that H contains a collineation which is a shear of 2{ or of 2{'. Assume not. Pick a Sylow p-subgroup II of H. Then q divides IHI· Moreover there exists 1 E ~ with lIT = I, as I~I = q + 1. Also, there is 0 E ?B with oIT = O. The group IT acts faithfully on 0, since there exists no shear of 2{' in H. Furthermore, 0 is desarguesian by 51.1. Hence there exists 'IT E II which induces a non-trivial shear on 0 by 52.2. As 1 is a line of 0, it must be the axis of 'IT considered as a shear of 0, i.e., each point in 1 n 0 is fixed by 'IT.
Assume that HI does not act faithfully on I. Then HI contains a non-trivial homology with axis 1 of 2(. Let C be its centre. Then C is on the line at infinity of 2(. It follows from 13.9 that C is fixed by HI' Hence HI ~ He. Likewise He ~ HI' whence HI = He· This proves ~ ED. Hence C is a point of the projective closure of O. But 'IT E HI YIelds C'TT = C, whence 'IT induces the identity on I, a contradiction. As the assumptions on 2( and 2(' are symmetric, we also see that H 0 acts faithfully on O. It follows from 16.2 that H 0 acts transitively on the set of points of 0 other than O. From this we infer that Ho acts transitively on D. Put S = <'IT a 1'lT E II, a E H 0' 'IT induces a shear in b). Then either S ~ .SL(2,q) or p = 2 and S is dihedral of order 2( q + 1), since 0 is desargueslan and since H 0 acts faithfully on O. As 52.1 is certainly true if q = 2 or 3, we may assume q > 3. Assume that S is dihedral of order 2(q + 1). Obviously, S is normal in Ho' Thus IH I divides 2r(q + 1) where r is determined by q = 2'. But IHI = (q + 1)· IH:I. Hence q(q - 1)2(q + 1) divides IHbl, whence q(q - I? divides 2r. This yields q = 2, a contradiction. Therefore S ~ SL(2,q).. . Let 'l' = II n S. Then 'l' is a Sylow p-subgroup of S, as S IS normal III H . The line I is a Baer subplane of 2(' and 0 is a line of this plane. As 'l' acOts faithfully on I, it induces a group of shears of order q on I. This yields that 'l' acts transitively on ~\{l}. Hence S acts transitively on ~\{l}. Therefore S contains a subgroup of index q. It then follows from 14.1 that q = 3, 5, 7, or 11. Hence GF(q) is in the kernel of 2(. Moreover the p-elements of S have minimal polynomial (x - 1)4 by 49.5. H~nce. 2{ is one of the exceptional planes of order 25 by 49.6. But thIS Yields the contradiction that b is a line of 2L Hence H contains a collineation which is a shear of 2{ or of 2{'.
We may assume that H contains a shear of 2(. Let S be the group generated by all these shears. It follows from 1)31 = q + 1 that the axes of these shears belong to D. Hence S ~ SL(2,q) or p = 2 and S is the dihedral group of order 2( q + 1). Assume that S is the dihedral group of order 2(q + 1). Since S is normal in H, all orbits of S in we have length t. Hence t divides (2(q + 1), q(q - 1» = 2. (Remember that q is even.) Hence t = 2, as a shear does not fix a line in we. Let 2 be the cyclic stem of S. Then 121 = q + I and 2 fixes all the lines of we. As 2 acts transitively on )3, we have by 52.3 that 2 is a group of homologies of 2('. This yields that GF( q2) is the kernel of 2(' whence 2(' is desarguesian. Moreover 2( is a Hall plane in this case. Assume S ~ SL(2,q). If q = p, then 52.1 is a consequence of 49.6. Hence we may assume q > p. Then S induces a group S* isomorphic to PSL(2,q) on we. Since H acts transitively on we and since S is normal in H, we have the assumptions of 52.4. Let .p be the partition of we whose existence is guaranteed by 52.4. Furthermore, let 2 be the centralizer of S in H. Then 2 fixes .p elementwise, as 2 centralizes each dihedral subgroup of S*. Moreover H /2 is isomorphic to a subgroup of prL(2,q), as this group is the automorphism group of S. Hence IHlt = rq(q2 - 1)121. On the other hand IHI = sd- 1q(q2 - 1)2, i.e., std- 1q(q2 - Ii = rq(q2 - 1)121. This yields std- l (q2 - I) = r121. From dr < q - I we infer 121> q + 1. Let A be the subgroup of 2 fixing we linewise. Then IAI > 1(q + 1). By 52.3, each A E A is a homology of 2( or of 2('. As A is not the union of two proper subgroups, A consists entirely of homologies of 2( or of 2('. This implies that the kernel of 2( or of 2(' is GF(q2), whence one of the planes is desarguesian and the other one is a Hall plane. D
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1965
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Index of Special Symbols
1==.":
A+
165
2f(Q) 22 A(o) 244 "L(V,K) 5 '_i{a) 226 C(o) 244 t::.(P) 5 t::.«0) , V(oo» t::.« (0), V(O» @G
E(P)
10
10
Seq)
156 95
24 K( V, 17) 3 9](0) 126
124
nr(Q) 24 nm(Q) 24 17(G),17(V) 2 POU(3, q2) 155 PSU(3, q2) 156 R*p 88 S(K, 0) 108
6
F*p H(P), G(P) H(q) 59
k(Q)
9](P)
68
Index
~l
III 10
~r
11
~(7T) 7 T,T(U) Trans( V, G) 6 lli(V) 6 V(O), V(l), V(oo), V(o)
7
admissible 139 Andre plane 54 Andre system 54 arc 214 associative specialization 166 avoiding 124 axis of a quasitransvection 244
generalized Andre plane 41 generalized Andre system 41 geometric partition 139
Baer collineation 19 Baer subplane 19 Bol plane 210
intersecting
carrier 124 centre of a quasitransvection
derived plane 57 desarguesian 6 Dickson nearfield Dickson pair 33
Hall plane 59 Hering plane 235
124
Jordan algebra 165 Jordan division algebra Jordan inverse 170 244 k-arc kernel kernel knot
31
214 of a quasifield 24 of a translation plane 127
Lilneburg plane egglike 126 exceptional Wa1ker plane exterior line 214
276
170
244
3
115
middle nucleus 24 Mobius plane 123
277
278
nearfield
Index
reflection
30
nearfield plane
order of a Mobius plane 124 osculating plane 223, 229 outer kernel 23 ovoid 126
Schaffer plane 235 secant 214 semi dihedral 62 spatial partition 140 special Jordan algebra 166 special universal envelope 166 spread 2 strongly real 116 Suzuki group 111
'IT-admissible 139 pappian 10 partial spread 91 pencil 124 planar nearfield 31 plane of type F *P 95 plane of type R*p 88 (P, I)-transitive 14 primitive divisor 28 projective collineation 217
quasifield 22 quasi transvection
125
reversal operator 168 right nucleus 24
14
tangent 214 tangent plane 126 touching 124 translation plane twisted cubic 221 type of a Dickson nearfield
244 weak quasifield
rank-3-group 76 rank-3-plane 76 real 116
Z-group
II
I
"1
I)~Sbg2k
.,;';\ -
-
\
:'\ L._ -_.
•
I
~.
i
I
1 • :
I
Y
~
I
I•
. ,
\
1\,,'1,,\ It'" i i I
I
r
\ \
(
_ _ ~_
\ ,\
; \.
~._.-
II J
23
33
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