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L(nyI,
if p does not divide n. This yields
(3) If P does not divide n, if p
.
> 3 and if L(n) > 3, then L(np) > pL(n).
Next we prove:
If P
(4)
> 5, then L(2p) > L(p) > 2p.
As 2 does not divide p, we have 2p(a, b) = L(p).
Since we are considering only those a and b for which b. Thus holds, a
*
Put x
lal > Ibl + I > 2
lp(a,b)1 = laP  bPlla  bl I > (lal P lbIP)(lal + Ibl)I. = lal and y = Ibl. Then x > y + 1 > 2. Assume
Then
(x P  yP)(2y + 1)
< (y +
It  yP)(x + y).
This yields
x P(2y + 1)  yP(y + 1)  yp+1 < (y + It(x + y)  yPx  yp+l. Therefore,
x P(2y + 1) «y + It(x + y) + yP(y + 1 x). As x
>Y +
> x P (2Y + I) 
(y + l)P x .
+ 1 < x that  (y + lyl >  x P  I and, therefore, + I)x p . This yields (y + Ity > x P(2y + 1)  (y + l)x P = xPy. (y + IY > x P and hence y + 1 > x. Thus x = Y + 1. This
It follows from y (y + IYx > (y
This implies shows that equality holds in (*). Therefore, we have
> (y + It  yP)(2y + 1)1 for all x, y such that x > y + 1 > 2. As Y > I and p > 5, we have py > p, pypI > p, (Dyp2 > 2py (p!2)y2 > 2py. Hence (x P  yP)(x + y)I
Assume that p does not divide n. Then each primitive (pn)th root of unity is the product of a uniquely determined primitive pth root of unity and a uniquely determined primitive nth root of unity. Therefore, we have in this case
L(2p)
27
6. Some Number Theoretic Tools
pl
(y + 1t  yP
= i~O ( ~ )y i > P + 2py + 2py + P = 2p (2Y +
+ 1) 2p and hence (4). We assume now that 6.1 is false. Let n be the smallest integer distinct from 1, 2, 3 and 6 such that L(n) < ITp/n p. Let q be a prime divisor of n and put n = qm. If q is also a divisor of m, we have L(n) > L(m) by (1). Furthermore, IT /m P = ITp/n p. Therefore, m = 1, 2, 3 or 6 by the minimality of n. This yields q = 2 or 3. If q = 2, then m = 2 or 6 and n = 4 or 12 which is impossible by (2). If q = 3, then m = 3 or 6 and n = 9 or 18 which is likewise impossible by (2). Thus q does not divide m. This proves that n is square free. From this we infer that we may assume q > 5, since otherwise n = 1, 2, 3 or 6. Moreover, m 1 and 2 by (4). Thus m > 3. If m 3,6, then L(m) > ITp/m P = m > 3 and hence L(n) > qL(m) > ITp/n P by (3). Therefore, m = 3 or 6. By (3) and (4), L(3q) > 3L(q) > 3q. Therefore, m = 6. Again by (3) and (4), we reach the final contradiction L(n) = L(6q) > 3L(2q) > 6q. Hence 6.1 is proved. 0
*
*
6.2 Theorem (Zsigmondy 1892). Let a and n be integers greater than 1. Then there exists a prime p which divides an  1 but not a i  I for any i E {I, 2, ... , n  I} except in the cases where n = 2 and a + 1 is a power of 2 or n = 6 and a = 2. PROOF. If n = 2, then the theorem follows from (a  1, a + 1) < 2. Thus we may assume n > 3. Let n be the nth cyclotomic polynomial. Then 2 and assume that every prime divisor of n divides q  1. If 1 2. Since every prime divisor of n divides q  1, we have q = 4 and n = 3. Hence (qn  I)n  I = 21. As 21 divides 2i  1 only for i = 6, the lemma is proved in this case. In the former case, as p is sprimitive and n ;;;;. 2, we have that p does not divide q 1 and so it also does not divide n. Hence p divides (qn l)n I and thus s iI. This yields i = tn. 0
7. Finite Nearfield Planes Let F be a set with two binary operations + and nearfield, if the following conditions are satisfied: 1) F( + ) is an abelian group. 2) If a,b,c E F, then (a + b) 0 c = a 3) (F\ {O})( 0) is a group. 4) a 0 = for all a E F.
°°
0
c + b 0 c.
o.
We. call F( +, 0) a
j
I
Every nearfield is obviously a weak quasifield. Therefore, by 5.3, the finite nearfields are exactly those finite quasifields for which (Q \ {o})( 0) is a group. Let F be finite nearfield. Then W(F) is a nearfield plane by 5.1 and 5.5, and one obtains every finite nearfield plane in this manner. An analogous result is true for infinite nearfield planes. But in this case the nearfields which are admissible are those which at the same time are quasifields. A nearfield which is also a quasifield is said to be planar. Thus every finite nearfield is planar. 7.1 Theorem (Andre 1955). Let F and F' be planar nearfields. Then W(F) and W(F') are isomorphic, if and only If F and F' are isomorphic. PROOF. F ~ F' implies W(F) ~ W(F'). In order to prove the converse we assume W(F) ~ W(F'). If W(F) is desarguesian, then W(F') is desarguesian. It follows from 1.11, 1.15 and 5.4 that F and F' are isomorphic in this case. We may assume henceforth that W(F) is nondesarguesian. It is easily seen that there is up to isomorphism only one nearfield of order 9 which is not a field (see also section 8). Therefore, we may assume IFI > 9. Let a be an isomorphism from W(F) onto W(F'). By 3.19, we have {P, Qt = {P', Q'}, where P = V(O) n 100 , Q = V(oo) n 100 , P' = V(O') n 1'00, Q' = V(oo') n 1'00. Using 3.11 and the fact that W(F) is a translation plane we see that we may assume V(Or = V(O') and V( 00 t = V( 00'). As the stabilizer of V(O') and V(oo') in the collineation group of WeFt) operates transitively on {(x,y)lx,y E F'\{O}}, we may also assume that (1, It = (1', 1'). We infer from V (0)" = V (0') and V ( 00)" = V ( 00 ') that there are bij ections (J and y from F onto F' such that (x, 0)" = (x.B,O') and (0, x)" = (O',x Y ) for all x E F. As (1,1)" = (1', 1'), we have V(1)" = V(1'). Hence (J = y. Furthermore, f3 is additive, since a is. Let V(m)" = V(m'). Then x.B 0 m' = (x 0 m).B for all x E F. Putting x = 1 we obtain m' = m.B. Therefore, x.B 0 m.B = (x 0 m).B. This establishes that (J is an isomorphism 0 from F onto F'. The proof of 7.1 also establishes the following result. 7.2 Theorem. Let F be a planar nearfield and let a be a collineation of W(F) with (0,0)" = (0,0), (1, 0t = (1, 0) and (0, 1)" = (0, 1). Then there exists an automorphism (J of F such that (x, y)" = (x.B, y.B) for all x, y E F. Conversely, if (J E Aut(F), then a defined by (x, y)" = (x.B, y.B) is a collineation of W(F) fixing (0,0), (1,0) and (0,1). A nearfield F( +, 0) is called a Dickson nearfield, if there is a third binary operation . defined on F such that F( +, .) is a division ring and such that the mapping x ~ (x 0 a)a  I is an automorphism of F( +, .) for all a E F\{O}. We shall determine now all finite Dickson nearfields. Let F( +, 0) be a finite Dickson nearfield. For x E F and m E 7L we
32
II. Generalized Andre Planes
~enote by x)m the mth power of x in F( 0) and by xm the mth power of x ~n F(). FurtherI?ore, we define pea) by xp(a) = (x 0 a)a I. The mapping p
a homomorphism from F*( 0) into Aut(F( +, by the following computation.
IS
.)). This will be established
xp(a)p(b)( a b) = xp(a)p(b)aP(b)b = (xP(a)a)P(b)b = (x a) b 0
0
=
f
divides nr 
0
0
1,
r1
qn1 =(qnr 1 _l)
=
V(·).
IV.
(ws)a= ws(qQI)(qI)I.
This is true for a = 1. Assume that (3) is true for some a :> 1. Then
As a consequence, (3) is true for all a. We infer from (3) that s, S(q2  1)(q  1)1, ... , s(qn  1)(q  1)1 is a complete residue system mod n. This implies in particular (s, n) = 1. Therefore wSV generates F* / V. Hence we may assume s = 1. Furthermore we have
(4)
(qn  l)(q  1)1= 0 mod nand (qa  1)(q  I)It:O mod n for 0
< a < n.
Next we prove: (5)
If P is a prime divisor of n, then p divides q  1.
Assume that p does not divide q  1. As n divides qn  1, the prime p also divides qn  1. Therefore (p, q) = 1. Let pd be the highest power of p dividing n. Then qpdI(pl)
=
q ... , CXn E Am and kl' ... , k n E K such that n
Hence x = O. Thus {O} is a maximal right ideal. As L is a commutative ring with 1, this implies that L is a field. Because every Klinear mapping which centralizes A also centralizes K[A] = L, we have Gl:GL(V,K)(A) \: GL(V,L). On the other hand GL(V,L) \: GL(V,K), as K C; L. Furthermore, A C; L* = 3(GL(V,L)). Therefore Gl:GL(V,K) (A) = GL(V,L). Let cxi E. A, k i E K and v E :RGL(V,K)(A). Then v I 2:7_1 kicxiV = 2:7= I kiv ICXiV. Therefore, v li\v E L for all II. E L. Hence II. ~ v li\v is an automorphism of L which fixes K pointwise. This establishes v E fL( V, L : K), i.e. :R GL ( v, K) (A) \: fL( V, L : K). If L is finite, A is characteristic in L *, as L * is cyclic. This yields :RGL(V,K/A) = fL(V,L: K). The statement about the ranks is trivial. 0 9.2 Theorem (Uineburg 1976a). Let X be a vector space over the field K and let 'TT be a spread of V = X EB X with K C; K( V, 'TT) and V(O), V(1), V( (0) E 'TT. Furthermore, let X( +, 0) be the quaslfield constructed with the aid of 2:( 'TT) as in section 5. If A is an abelian collineation group of 'TT( V) with A C; G L( V, K) and V (O)A = V (0), V ( oo)A = V ( (0) and if for all a E 2:( 'TT), the subspace V(a) is an irreducible K[Aa]module, where Aa denotes the stabilizer of V( a) in A, then X( +, 0) is a generalized Andre system and 'TT( V) is a generalized Andre plane. The division ring X( +, .) belonging to X ( +, 0) is a field.
(x,xom)=
2: (u,uomtiki' i= 1
Therefore
(x,O)
+ (O,x m) = 0
n
n
i=1
1=1
2: (u,O)a iki + 2: (0, u
0
mtiki'
As A fixes V(O) and V(oo), we obtain (x,O) = 2:7=1 (U,Otiki' Thus V(O) is generated as a K[Am]module by (u,O). Since this is true for all u E X\{O}, the K[Am]module V(O) is irreducible. That V(oo) is an irreducible K[Am]module is proved similarly. 0
9.5 Lemma. V(O) and V(oo) are isomorphic K[Amlmodules. PROOF. Define the mapping f.L from V(O) onto V(oo) by (x,O)/L = (O,x 0 m). Then f.L is Klinear. Furthermore, (x,O) + (x, O)Jl E V(m). Let cx E Am' We infer from (x,O) + (x, O)/L E V(m) that (x,o)a + (x,O)Jla E V(m). On the other hand, (x,o)a E V(O). Hence (x,o)a + (x,Ot/L E V(m). Therefore (x,O)Jla  (x,o)a/L E V(m) n V(oo). Thus (x,O)/L a = (x,Ot/L. 0
9.6 Lemma. Km = K[Aml is a field and V is a vector space of rank 2 over Km' Furthermore, Gl:GL(V,K)(A m) = GL(V,Km);;;;; GL(2,Km)' PROOF.
o
This follows from 9.1, 9.4 and 9.5.
For X E KI we define cp(X) by (cp(X), 0) = (1, O)x. Then cp is a bijection of
K J onto X. As Before we prove 9.2 we state and prove the following: 9.3 Corollary. Let
collineation group of
(cp(X
m be a translation plane and let A be an abelian mwhich fixes two points P and Q on 100 and an affine
point O. If A has the property that the stabilizer
Aw
of any point
WI
too
+ 11.),0) = (l,O)X+A= (l,0l+ (1,0/= (cp(X) + cp(i\),O),
the mapping cp is additive. Define a multiplication in X by xy isomorphism from Kl onto X( +, .).
=
I(Xq:>(X»= (xqJ(X),O). Define tV by tV(x, 0) = (0, x). Then tV is an KIisomorphism from V(O) onto V(oo). Hence
(O,x)x= (tV(x,O))x= tV((x,O)X) Thus (x, y?
= tV (xqJ (X), 0) = (O,xqJ(X)).
with a,b E X\{O}. A C ~GL( v, K) (A I) = GL( V, K 1) by 9.6. Furthermore, V(O)A = V(O) and V( oo)A = V( (0). This yields that to every a E A there are elements X, A E KI such that (x,O)'x = (x, oy and (0, yt = (0, yi' for all x, y E X. Hence (x, y)a = (xqJ(X), yqJ(A)). PROOF.
0 m)mI. Then a is a mapping from X\{O} into Aut(X(+, .)). Moreover, Km consists of all the mappings (x, y) ~ (xa, ya a(m») with a E X \ {O}. Finally, Fm is isomorphic to
.).
As m =1= 0, the mapping a(m) is a permutation of X. Obviously (1, m) E V(m). Let a E Q. Then (a, a 0 m) E V(m). Since V(m) is a Kmsubspace of rank 1, there exists X E Km with (a, a 0 m) = (1, mY. By 9.8, there exist u, v E X with (x, y? = (xu, yv) for all x, y E X. Thus (a, a 0 m) = (u, mv) and hence a = u and v = aa(m) (remember that X(+,') is commutative) proving that the mapping (x,y)~(xa,yaa(m») belongs to Km' Conversely, if X E Km, then (l, m)X = (a, a 0 m) for some a E X and hence (x, y? = (xa, yaa(m»). This proves that Km consists of all the mappings (x, y) ~ (xa, ya a(m»). Now PROOF.
=
(x 0 m + yo m)mI
= (xo m)mI + (yo m)mI Moreover, if (x, y? = (xa, yaa(m»), (x, y)A = (xc, yc a(m»), then
which
yields
c = ab
and
=
(ca(a)  ca(b»)a
for all c EA. Thus a(a) = a(b), as a =1= O. If a + b =1= 0, then
ca(a+b\a + b) = co (a + b) = co a + co b
= ca(a)a + ca(b)b.
= c a(a)d a(a)a 2 + ca(a)da(a)ab + ca(b)da(b)ba + c a(b)d a(b)b 2.
9.9 Lemma. For mE X\{O} define a(m) by xa(m) = (x
=
+ co b = ca(a)a + ca(b)b
(cdr(a+b)(a + b)2= (ca(a)da(a)a + ca(b)da(b)b)(a + b)
This follows from 9.7 and Am CA.
(xab, yaa(m)ba(m»)
If a + b = 0, then 0= co(a + b) = co a
PROOF.
Replace c by cd. Then
9.8 Lemma. Km \{O} C B for all m E X\{O}.
(x + yt(m)= ((x + y) 0 m)mI
The next lemma is essentially Andre's. 9.10 Lemma. Let Q( +, 0) be a generalized Andre system and assume that the division ring Q( +, .) belonging to it is a field. If a, b E Q \ {O} and if co (a + b) = co a + cob for all cEQ,· then a(a) = a(b).
= (xcp(X), YqJ(X)).
9.7 Lemma. A is a subgroup of the group B of all mappings (x, Y) ~ (xa, yb)
X( +,
45
9. Generalized Andre Planes
=
xa(m) + ya(m).
= (xb, yba(m»)
and (x, y)XA
(x, y)XA= (xc, yca(m»)
aa(m)ba(m)
=
(abt(m).
Hence
a(m) E
Aut(X(+, .)). Finally, the mapping (a,aa(m»)~a provides us with an isomorphism from Km onto X(+, .). 0
As Q( +, .) is commutative
(cdt(a+b)(a + b)2= ca(a+b\a + b)da(a+b)(a + b) = (ca(a)a + ca(b)b)(da(a)a + da(b)b) = c a(a)d a(a)a 2 + ca(a)da(b)ab + ca(b)da(a)ba + c a(b)d a(b)b 2. As ab =1= 0, we infer
0= (ca(a)  Ca(b»)( da(a) _ da(b»). Since this is true for all c,d E Q, it follows that a(a)
= a(b).
0
9.11 Theorem (Foulser 1967a). Let Q( +, 0) be a generalized Andre system. If the division ring Q( +, .) belonging to Q( +, 0) is a field, then the
following statements are equivalent: a) Q(+, 0) = Q(+, .). b) Q( +, 0) is a division ring. c) There exists bEQ\{O} such that co(a+b)=coa+cob for all a,c E Q. PROOF. It suffices to show that c) implies a). As a consequence of 9.10 we have a(a) = a(b) for all a E A \{O}. In particular a(b) = a(l) = 1. Thus a(a) = 1 for all a E Q\{O}. Hence x 0 a = xa for all x,a E Q. 0
Remark. Let I2l be a desarguesian plane over a noncommutative division ring. Then 9.2 and 9.11 imply that 2[ does not satisfy the assumptions of 9.2. Hence 9.2 does not characterize the generalized Andre planes. I do not know whether a plane over a generalized Andre system for which Q( +, .)
40
11.
UCI1t;laIlLc.:U .t"\.ll\. J.ll"..
.I.
It..1I1 ..... ~
is a field always admits such a large abelian collineation group. However,
IV. r II1lle veneralizeu Anure t'lanes
41
and
if Q( +, 0) is finite, then the existence of such an abelian collineation group can be proved, as we shall see in section II. In this context see also
a 0 (x 0 y) = a(xa(yy) = aa(Y)xa(y~ = (ax) 0 y = (a
0
x) 0 y.
o
R. Rink [1977]. The following corollary is essentially due to Foulser [1967a, Lemma
4.1].
9.12 Corollary. Let 2r be a translation plane, let P, Q be distinct points on 100 and 0 a point not on 100 , If A is an abelian collineation group of 2r with p A = P, QA = Q, OA = 0 and if Aw induces an irreducible group of automorphisms on T( W) for all W I 100 which are different from P and Q, then the following statements are equivalent: a) 2r is pappian. b) 6(P, OP) =!= {1}. c) 6(Q, OQ) =!= {l}. PROOF. It follows from 1.15 that a) implies b) and c). As the conditions are symmetric in P and Q, it suffices to prove that c) implies a). 2r is a generalized Andre plane by 9.3. Therefore, we may assume that 2r = 7T( V) where V = X EB X is a direct sum of a Kvector space X with itself. Furthermore, we may assume K ~ K( V, 7T) and V(O), V(l), V( 00) E 7T. Using 2.1, we may finally assume that Q = V(oo) n 100 and P = V(O) n 100 , Let X( +, 0) be the quasifield constructed with the aid of L( 7T). Then X ( +, 0) is a generalized Andre system and X ( +, .) is a field by 9.2. As 6(Q,OQ)=6((oo),V(00»=!={1}, we infer from 5.6 and 9.11 that X( +, 0) = X( +, .). 0
9.13 Lemma (Foulser 1967a). If Q( +, 0) is a generalized Andre system, then
10. Finite Generalized Andre Planes In this section we follow mainly D. Foulser [1967a, section 2]. Let p be a prime and q = pS. Furthermore, put K = GF(q) and F = GF(qd) and let p be the automorphism of F which is defined by x P = x q . Then p generates the Galois group of F over K. For kEN we put Ik = {O, 1, ... , k  I}. Let A be a mapping from Iqd_l into Id with A(O) = 0 and let w be a generator of F. We define an operation 0 on F by a 0 0 = 0 . A(i)· and a 0 Wi = a P Wi. Then we have: F( +) is an abelian group, a o l=l o a=a and (a+b)oc=aoc+boc for all a,b,cEF. We denote F( +, 0) by FA' Obviously, FA is a generalized Andre system, if and only if FA is a quasifield. To avoid tedious repetitions, we shall also write A(a) instead of AO), if
a
= Wi.
10.1 Lemma. FA is a quasifield, if and only if A satisfies the condition: If = (d,AU)  A(j», then i = j.
i,j E Iqd_l and i ==j mod ql  1, where t
PROOF. Assume that FA is a quasifield. Let i, j E Iqd_ I be such that i == j mod ql  1, where t = (d,AU)  A(j». We may assume AU) ;;;. A(j). As t = (d,A(i)  A(j», we have ql  1 = (qd  1,qA(i)A(j)  1). Since qA(j) and qd  1 are relatively prime, it follows that
ql 1 = (qd  1,qA(j)(qA(i)A(j)  1)). As ql  1 divides j  i, there is therefore an integer k such that
a) l\(Q) = {a I a E Q\{O}, a(x 0 a) = a(x)a(a) for all x E Q\{O}}. b) I\n(Q) = {a I a E Q\{O},a(a 0 x) = a(a)a(x) for all x E Q\{O}}. a) We have (x 0 y) 0 a = (xa(y~) 0 a = xa(y)a(a~a(a)a and x 0 (y 0 a) = xa(y a~a(a)a. Hence a E nr(Q), if and only if a(y)a(a) = a(y 0 a) for all y E Q \{O}. b) is proved similarly. 0 PROOF.
j  i == k( qA(i)  qA(j)) mod qd  1. Thus
0
9.14 Lemma. If Q( +, 0) is a generalized Andre system, then {a Ia E Q, aa(m) = a for all mE Q\{O}} c: k(Q).
As FA is a quasifield, Wi = wi and hence i = j. Conversely, let the condition on A be satisfied. Because of 5.3, we have only to prove that FA is a weak quasifield. In order to do this, consider the mapping x ~ a 0 x and let a 0 Wi = a 0 wi. As a = w k for some k, we have
Assume that aa(m) = a holds for all mE Q\{O}. Then aox = ax for all x E Q. Hence
Hence
PROOF.
a 0 (x
+ y) = a (x + y) = ax + ay = a x + a y 0
0
i + kqA(i)
== j +
kqA(j) mod q d  1.
48
II. Generalized Andre Planes
We may assume AU) ;;;, A(). Then )  i
== kqA(j)( qA(i) A(j)

1) mod qd 
== 0 mod q' 
l.
1,
whence i = ) by assumption. Therefore, the mapping x and hence surjective, as F is finite.
~a
0
x is injective
0
10.4 Lemma. N v and Nu are cyclic subgroups of F*(') of order v l(qd  1), resp. u l(qd  1). Furthermore, x 0 a = xa for all x E F and all a E N v ' In particular N v( 0 ) = N v(')' Moreover Nu C; N v '
10.2 Corollary. Let K = GF(2) and F = GF(2 d). If FA is a quasifield with k(FA) = K, then d = 1 or d is divisible by at least three distinct primes. PROOF. Let t = (A( i)  A(j), d) = 1. Then 2'  1 = 1 and hence i ==) mod 2'  1. By 1O.l, we have i = ) and thus AU) = A(j). Therefore 1 = (0, d) = d. Hence (AU)  A(j), d) =1= 1 for all i and), if d =1= l. In particular, (AU), d) = (AU)  A(O), d) =1= 1 for all i. Let p be a prime dividing d and assume that p divides A(i) for all i. Then
1. As GL(V) is cyclic and U 1(qd_l)
divides I G n GL( V)I, there exists a cyclic subgroup 2 of G n GL( V) of ?rder u l(qd  1). Since 2 is a characteristic in G n GL( V), it is normal III G. Let q = p r where p is a prime and let t be a pprimitive prime divisor of p:d : .r. Fur~hlerr:;ore, let t a be the highest power of t dividing q d  l. Then t diVides u (q  1). Let T be a Sylow tsubgroup of 2 and T. a Sylow tsubgroup .of. G which contains T. By 10.5, we obtain T = To~ As T is charactenstlc III 2 and 2 is normal in G, we have that T is normal in G. Hence T is the only Sylow tsubgroup of G. As T operates irreducibly on V by 10.5, Schur's lemma yields that C = [G(T) is cyclic. Furthermore, Z ~ C. Let ZI be an abelian subgroup of order u  \qd  1) of G. As Tis ulllqu.e, T C; Zl' Hence 21 C; C. This yields 2 = Zl' as C, being cyclic, contallls exactly one subgroup of order u l(qd  1). Assume that Z is not unique. Then there is no pprimitive prime divisor of prd  l. Therefore qd = 64 or q = p, d = 2 and p + 1 = 2b by 6.2. Let
12. The Andre Planes Let L be a field and r a group of automorphisms of L. Let M be a subgroup of L * which is invariant under r and assume that r operates trivially on L */ M. Finally, let f3 be a mapping from L */ Minto r with f3(M) = I and define the mapping a from L * into r by a(a) = f3(aM). Then we have in particular a(l) = 1. For a, bEL define a 0 b by a 0 b = 0, if b = 0, and by a 0 b = a a(b)b, if b =1= O. Obviously, L( +) is an abelian group, a 0 = and (a + b) 0 c = a 0 c + b 0 c for all a, b, c E L. Let a,b,rn E L*. Then
°°
aa(m)bM = aa(m)MbM = aMbM = abM. Therefore, a(aa(m)b) = a(ab) for all a, b, rn E L *. In particular, a(a 0 b) = a(ab) for all a,b E L*. Let a,x,cEL with a=l=O and assume aox=c. If x =1=0, then a 0 x = a a(x)x =1= O. Therefore, c = implies x = 0. If c =1= 0, then x =1= 0 and aa(x)x = c. Hence x = aa(x)c. The remark made above yields a(x) = a(a IC). Thus x = a a(a1c)c is uniquely determined. Conversely, if a,c E L, then we put x = aa(a1c)c. A trivial computation shows a 0 x = c. Finally, x 0 I = x = lox, as is easily seen. Hence L( +, 0) is a weak quasifield. In general, L( +, 0) will not be a quasifield. However, if r is finite, then L( +, 0) is a quasifield: Let K be the fixed field of r. Then
°
[L : K] is finite. Furthermore, K is contained in the outer kernel of L( +, 0) as is easily seen. Applying 5.3 yields that L( +, 0) is a quasifield in this
case. If f is finite, then we define nr by nr(a) = IT'YEra'Y for all a E L *. Then nr(a'Y) = nr(a) = nr(a)'Y for all a E L* and all y E f. If a(a) = a(b) for all a, bEL * with ab I E ker(nr), then we shall call L( +, 0) an Andre system and every translation plane which is coordinatized by an Andre system will be called an A ndre plane. We have a'Y 1 E M n ker(n r ) for all a E L* and all y E f. Applying Hilbert's Satz 90 (see e.g., S. Lang [1971, p. 2l3]), we obtain that ker(n r ) c; M, if f is cyclic. Therefore, L( +, 0) is an Andre system, if L is finite. Not all the generalized Andre systems described above are Andre systems, as the following examples show which are due to P. Roquette. Let K be the field of rationals and let p be a prime with p == 3 mod 8 or p == 3 mod 8. Asp is odd, [j tl K(/2). Therefore, L = K(/2,[j) is an extension of K of degree 4. Furthermore, L is a Galois extension of K and the Galois group f is elementary abelian of order 4. The group f is generated by the elements a and T which are defined by (/2t = /2, Opt =lp and (/2y =/2, ([jy = [j. Put a = I +/2. Then ao+ 1 =  1 and lienee nrC a) = a 1 + ° + 1" + 01" = a (I + 0)( 1 + 1") = (  1) 1 + 1" = 1.
Let b, c, dEL and assume a = bO 1c1" ld°1"I. As d°1"1 = dO(1"I)do l , we may assume d = 1. Then we have 1 = aO+ 1 = b 02  IC(1"1)(0+ I) = C(1"I)(o+ I). As c E L*, there exist a, f3,y,8 E K with c = a + 13/2 +y[j +8/2 [j. This yields after an easy computation co+ 1 = a 2 + 2ay[j  213 2  4f38[j + y7;  2p8 2. Furthermore,
c1"(o+ I) = c(o+ 1)1" = a 2  2ay[j  2[32 + 4[38[j +
/p 
2p8 2.
As  co+ 1 = c1"(o+ I), we obtain from the two above equations 0= a 2  2[32 + p(y2  28 2). Since a, [3, y, 8 are rational, we may assume that they are integers. We may further assume that they are relatively prime, as c =1= O. Computing modulo p yields a 2 == 2[32 mod p. From p == 3 or  3 mod 8, we infer that 2 is not a quadratic residue modulo p. Therefore a == f3 == 0 mod p. Put a = qJ and [3 = tp. Then 0 = p2(f..2  2t 2) + p(y2  28 2) and hence y2 == 28 2 mod p which yields as above y == 8 == 0 mod p, a contradiction. This con tradiction proves that M = q, it follows that there is exactly one b E Q3 such that P, Q E b. Furthermore, Q3 consists of all the Baer subplanes of Ill: which contain D. The number of pointpairs (P, Q) with P =1= Q and PQ n 100 E D is q\q + 1)(q2  1) and the number of pairs (P, Q) with P =1= Q and P, Q E b where b E Q3 is q2(q + l)(q  1). Therefore, 1Q31 = q2(q + 1). We define a new incidence structure Ill:' as follows: a) The points of Ill:' are the points of Ill:. b) The lines of Ill:' are the lines I of Ill: with I n 100 E D* and the elements of
m.
c) E is the incidence relation. Then Ill:' contains q4 points and q2(q + 1) + q(q  1)q2 = q4 + q2 lines. Furthermore, each line carries exactly q2 points and each pair of points of Ill:' is on exactly one line. Thus Ill:' is an affine plane of order q2. We call Ill:' the plane derived from Ill: via D. Let D' be the set of points on 1'00 which are on lines belonging to Q3 and m' the set of affine lines of Ill: which carry a point of D. Then Q3' is a set of Baer subplanes of Ill:' and Ill:', \~3', D' satisfy 1) and 2): Let I E Q3' and let P and Q be two distinct points on I. Denote by (PQ)' the line joining P and Q in Ill:'. Then (PQ), Em and therefore I(PQ)' n II = q. A simple counting argument now proves that I is a Baer subplane of 21'. Furthermore 1 n 1'00 = D', This together with = q2(q + I) yields the desired conclusion. In particular, if 121" is the plane derived from 21' via D', then Ill:" =Ill:.
IIB'I
lJ,
13.1 Lemma. Let 2(,2(', D, D' be as above. If G is the collineation group of 2( and G' the collineation group oj m', then GD = G~,. This follows immediately from the remark made above that 'B and 'B' are the sets of all Baer subplanes of 2( and 2(' resp. which intersect 100 in D and 1'00 in D' resp.
13.2 Lemma. If mis a translation plane, then 2(' is a translation plane and the translation group of 2( is also the translation group of 2('. suffices to prove the second assertion. Let T be the translation group of 2( and P E D*. Then T(P) C GD = GlJ,. Furthermore, if T E T(P), then T is the central collineation in mas well as in %[' the centre of T being P. As T has no fixed point on 2( unless T = 1, we have that T is a translation of 2( and of m'. Therefore, the assertion of 13.2 follows from 1.3, as ID*I = q(q  1) > 2. 0 PROOF. It
13.3 Lemma. If
K
is a perspectivity of 2( with axis 100 , then
K
is a collineation
of 2('.
This follows immediately from 13.1. We remark that perspectivity of m'.
13.4 Lemma. Let
K
K
need not be a
be a perspectivity of 2( with D K = D and axis m 1= 100 ,
Then:
a) If m tf. 'B' , then b) If m E 'B ' , then
K K
is a perspectivity of 2('. is a Baer collineation of 2('.
The proof is trivial.
13.5 Theorem. Let L = GF(q2) and V = L E:B L. Put w = { V(O), V( oo)} U { V(m) ImEL *} where V(m) = {(x, xm) Ix E L}. For t E GF(q)* define SI by SI = {x Ix E L*, x q+ 1 = t}. If DI is the set of points at infinity on the lines V(m) with m E SI and if 'BI = { V( 0 m) + x I m E Sf' x E V} where V(o m) = {(x,xqm)lx E L}, then w(V), DI' 'B{ satisfy conditions I) and 2) of page 57. It suffices to show that 'B;.o = {V(m) I m E SI} is a regulus in the projective space defined by the GF(q)vector space V and that 'BI 0 = { V( 0 m) 1m E SI} is the regulus opposite to Q3;, o' In this context we sh~ll write V instead of V if we consider V as a GF(q)vector space. Let X E 'B 1,0 U I, o' Then X is a line in the projective geometry belonging to V q . Furthermore, set PROOF.
i'
(} =
U X E'1.\,o
X= {(x,xm)Ix
E
L, mE SI}'
Il1t:
naJl t'lanes
59
Let mESt • Then (x,xqm)=(x,xxq1m). Moreover, (xq1m)q+1 = xQ2lmQ+1 = t, provided x1=O. Thus xq1m E St. Hence (x,xqm) EO. This shows V( 0 m) CO for all m E SI' Let m, n E SI' Then (mn I)q+ 1 = tt I = I. Applying Hilbert's Satz 90 yields an a E L with mn  I = a q  I • In particular a 1= 0 and hence (0,0) 1= (a, am) = (a, a qn) E V(m) n V( 0 n). This establishes 13.5, as the lines of 'Bt,o resp. 'B;,o are pairwise skew. 0
13.6 Theorem. Let L = GF(q2) and w( V) the desarguesian plane over L. If D is a set of q + I points on 100 and 'B a set of Baer subplanes such that 1) and 2) on page 57 are satisfied, then there is a collineation K of w( V) with D K= D 1 and 'BK = 'B 1 •
This follows from the fact that 'B' n w is a regulus in the projective geometry belonging to Vq , as is easily established with the aid of 'B n w.
13.7 Theorem. Let L = GF(q2) and K = GF(q). Let X and Y be subsets of K* with K* = X U Y and X n Y = 0. If w( Vh resp. w( V)y are the planes which are obtained by deriving simultaneously via D t for all t EX resp. all t E Y, then w( Vh and w( V)y are isomorphic Andre planes. Moreover, w( Vh is desarguesian, if and only if X = 0 or X = K*. PROOF. The construction shows that w( Vh and w( V) yare generalized Andre planes. As K is contained in their kernels, both planes are Andre planes by 12.5. The mapping cp: (x, y) ~ (x, yq) is an involutory linear mapping of Vq onto itself with 'Bt,o = cp'B;, o' As a consequence: w( Vh and w( V)y are isomorphic. The last assertion follows from 9.11. 0
It follows from 13.5 and 13.6 that there is up to isomorphism only one plane H(q) of order q2 which is constructed by a single derivation from the desarguesian plane of order q2. This plane will be called the Hall plane of order q2. These planes were first discovered by M. Hall Jr. H(2) is the desarguesian plane of order 4. This follows from 13.7, as 2  1 = 1. In all other cases we have:
13.8 Theorem. If q PROOF.
> 3,
then H(q) is nondesarguesian.
This follows from the last assertion of 13.7, as q  1 > 2.
0
13.8 and 8.4 yield that H(3) is the nearfield plane of order 9. The next theorem which is very useful is a special case of a theorem on Frobenius groups. As the proof of this special case is much easier than the proof of the general case, we shall give it here.
m
m.
be a finite projective plane and I a line of Let ~ be a group of perspectivities with axis I and denote by I.E the set of points P with
13.9 Theorem. Let
60
II. Generalized Andre Planes
!:l(P, l) =t= {I}. If T is the group of all elations with axis I which are contained in !:l, then \3 is an orbit of T.
PROOF (Andre). We have 1!:l1=ITI+
2:
(1!:l(P,/)Il)=ITIIs.BI+
PEm
2:
1!:l(P,/)I·
PEm
As 1!:l(P,I)1 ;;;;. 2 for all P E s.B, we obtain l!:ll ;;;;. ITI + Is.BI ;;;;. 1 + Is.BI· Let \31' ·.·,s.B r be the orbits of Ll with ls.Bil 3, whereas 8.1, 8.2, and 8.3 take care of the case q = 3. We state a few more properties of the collineation group of H(q). The proofs are left as exercises for the reader. 13.11 Theorem. Let I be a line of A (q2) with I n 100 ED. Then I is a Baer subplane of H(q) the pointwise stabilizer oj which has order q(q  1).
r
l!:ll
61
14. The Collineation Group of a Generalized Andre Plane
ILl( P, 1)1·
i= I PEm;
If Q E \3i' then 1!:l(P,I)1 = 1!:l(Q,/)1 for all P E s.B i . Hence
2:
This theorem gives another nondesarguesian if q > 2.
1!:l(P,I)I= l\3 i ll!:l(Q,I)1 = l!:ll·
P EIB;
proof
of
the
fact
that
H(q)
IS
13.12 Theorem. If P E D*, then there is a unique pi E D*\{ P} such that !:l(P, OP') contains a cyclic subgroup of order q + 1. Moreover, P = P.
Thus
/I
l!:ll = ITI  1\31 + rl!:ll Whence (I'  1)1!:l1 = IIBI  ITI. As I' ;;;;. 1, we obtain
o 0, as q > 2. Hence there is aPE D*\(D'Y U D ' ). Let 2 be a cyclic group of order q + 1 the elements of which are homologies with centre P leaving D* invariant. Then 2 also fixes D'. Similarly, let 2' be a cyclic group of order q + 1 which leaves Diy invariant, the elements of which are also homologies with centre P. Let a be the axis of the homologies in 2 and a' the axis of the homologies in 2'. We may assume 0 I a,a ' . If a =t= ai, then !:l(P,OP) =t= {I} by the dual of 13.7. Hence H(q) is desarguesian by 9.12, a contradiction. Therefore a = a'. By 11.7, we thus obtain 2 = 2'. As a consequence D' n D 'Y = 0 and 2 Y = 2 which yields pY = P and
13.14 Theorem. If q is odd, then each affine line oj H(q) is the axis of exactly one involutOlJ! homology.
14. The Collineation Group of a Generalized Andre Plane The groups PSL(2, q) and PG L(2, q) playa prominent role in the theory of finite projective planes. Thus a good working knowledge of them is important to every student of this theory. We shall state here without proofs a few theorems about their subgroups. Proofs may be found in Dickson (1958, Chapt. XII] and Huppert (1967, Chapt. II, §8]. 14.1 Theorem. If U is a subgroup oj PSL(2, p r), then U is one of the following groups: (1) An elementatJ! abelian pgroup of order pm with m 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 2subgroup 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
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 nontrivial 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 nontrivial 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 psubgroup 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 psubgroup of G which contains B IT B' By d), 1GB I = n 2k. Hence L IS. a Sylow psubgroup 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. Rank3Planes
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 rank3group 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 rank3plane, if it admits a group of collineations acting as a rank3group on the set of points of m. Any such collineation group will be called a rank3collineation group of 2:r. Every nearfield plane is a rank3plane. 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 rank3collineation group of the given nearfield plane. Every Hallplane is a rank3plane: 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 rank3group 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 rank3plane. The Luneburg planes which we shall construct later on are also rank3planes. The next theorem was proved independently by Johnson & Kallaher 1974 and by myself 1973.
G
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 2transitivity 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. Rank3Planes
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 2subgroup 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 2group. 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 rank3collineation group oj W, then one oj the Jollowing holds: a) G acts transitively on the set oj lines oj mand G{ acts as a rank3 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 rank3group 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 2subgroup of Gp , Q, h' Let ~ be a Sylow 2subgroup 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 2subgroup of Gp , R' 1 of Gp, as Gpv is 2transi 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 rank3 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 rank3collineation 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. Rank3Planes
17. Rank3Planes 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 rank3coflineation 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 pprimitive 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 nonnegative 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 'TTsubgroup of K. Then 2: n K(P, OQ) is a Sylow 'TTsubgroup 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:* 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 pprimitive 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 3subgroup 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 3subgroup 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 3subgroups, 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 pprimitive divisor of p r  I and if K(P, OQ) is soluble. We may assume from now on that there does not exist a pprimitive 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. Rank3Planes
the S~low 3s~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 3subgroup of K is isomorphic to a Sylow 3subgroup of OL(6, 2). Let ~ be a Sylow 3subgroup 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 3subgroup of OL(6,2)v is elementary abelian for a nonzero 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 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 7subgroup 2 in K(P,OQ). Fur~hermore, IAut(Z)1 = 2 . 3 and all Sylow 3subgroups 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. Rank3Planes 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 3subgroup 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 3subgroup of Bw' Then ~ has order 9. Hence ~ is abelian. Moreover, S n ~ = {I}. Let SI be a Sylow 3subgroup of K(P,OQ) and S2 a Sylow 3subgroup 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 3subgroup of K. This yields that (S~)* is a Sylow 3subgroup of K*. As I( S~)* I = 33 and K* C;; fL(l, 64), we see that (S ~)* is a Sylow 3subgroup of fL(l, 64). This is a contradiction, since (S ~)* is abelian but the Sylow 3subgroups 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 2subgroup of G(P,OQ) is either 25 or 25 + I, where 25 is the highest power of 2 dividing p
+ I. Moreover
Ill.
KanKJnalle~
if the order of a Sylow 2subgroup of G(P,OQ) is 2s, then the Sylow
2subgroups ofG(P,OQ) are cyclic, unless 25 + 1 divides IG(Q,OP)I. PROOF. Let 2: be a Sylow 2subgroup 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 2subgroup of G(Q,OP). If IL*I = 2 1 , then t:< s + 1, as IG(Q,OP)I divides p2  1. Furthermore 22s+2+b1 = 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 2subgroup 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 2subgroup 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 2subgroup of G(P, OQ) is 2s  I , then the Sylow 2subgroups 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 pprimitive 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 2subgroups 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 Zsubgroup 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 Zgroup, the Sylow asubgroup A of N is normal in N
17. Rank3Planes 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 2subgroup 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 2subgroup of K(P, OQ). Then LO is a Sylow 2subgroup 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 2subgroup of K(P, OQ)/ N and let L) be a Sylow 2subgroup 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 2subgroups 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. Rank3Planes
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, ... , IZoll)· 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 noncyclic 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 ipartial spread, if UYE'lT Y= Bi U {(O,O)}.
18.10 Lemma. For each i there exist
exactly exactly exactly exactly
five ipartial spreads if p = 11, three ipartial spreads if p = 19, two ipartial spreads if p = 29, one ipartial spread if p = 59.
92
III. Rank3Planes
PROOF. By 18.7, it suffices to show that [77 n ffi[ = 2 for all ipartial 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 ipartial 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 nontrivial 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 ipartial spread. This proves that 0 is the union of ipartial 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 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+I02,kl02,k02)
= (1, 1, 1)7J(k
I ).
As wYJ(k)wYJ(k1)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) = (kll,ko+l/ol,ko+2/02)
= k.
E K} If b =!= •
= (1,/ 0+1,/ 0+2).
(_;:'i:~=:~ II;
23. The Luneburg Planes
°
/ \:)~ r;\'~~\,1"
, 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 2subgroup of G. We infer from Q:G(g) n Q:G(g') =1= {I} that Q:G(g') is also a Sylow 2subgroup 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(q21) +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 nonreal 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 2subgroup LO distinct from {I}. Then the same argument works by taking for X a Sylow 2subgroup of Seq) containing Lo. By 24.9, it remains to consider Case 3: The group W contains more than one Sylow 2subgroup. Let Lo be a Sylow 2subgroup 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 2subgroups 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 2element, then ~S(q)(Y) is a 2group by 24.2 d). If Y is not a 2element, 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 2subgroup 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 = qoI. 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)(qol)(qol) = 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 = Uwrw, then 7" since Q = U W= Uwr = 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, Uwr(O, b)WT(O, c) = (0, b, b atT(O, c) = [( b a, 1, b, O)GF( q) reo, c)
= [(1, b a, b Ia, O)GF( q) reo, c) =
= 1,
= (bla,c,b a + c a + cbla). This implies in particular that the ycoordinates 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)wr(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
=
(gla, g, ga + ga + g2a)w
= [(1, ga + ga + g2a, gla,
=
(I,A, gA, glaA)GF(q)
= (gA, glaA,A), where A = (ga + ga + g2a)I. (It is easily seen that ga + ga + g2a i= 0.) By the remark made above glaA E gGF(qo). Hence gaA E GF(qo) whence ga(ga + ga + g2a) 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 characterfree 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
= (ga + ga + g2a, 1, g, gla)GF(q)
(f) (g)
(b Ia, 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) pointpairs 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 nondegeneracy. 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 pointpair on k, then bye) there exist exactly n circles intersecting k in the given pointpair. 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 nonempty point set 0 of a projective 3space 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 3space 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 nonconcyclic points in 9)((0). 0 A Mobius plane 9)( is called egglike, if there exists an ovoid 0 in a projective 3space 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 pcpI. 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 2subgroup 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 2subgroup 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 2subgroups 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
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 nontrivially 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 2subgroup 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 2subgroup 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 2subgroup 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 nonidentity 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 2subgroup 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 2subgroup 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 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
nondegenerate 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 nondegenerate 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: = AIXA = 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 vI = 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 nonabsolute 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 nonabsolute 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 nontrivial 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 psubgroup of G containing E(P), then E(P) = 3(TI). (h) Let II be a Sylow psubgroup 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 pgroup, 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 psubgroups 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 ~(q21) 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 p1T(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 nontrivial 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)IpIT(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 yIHp,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 XI = (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 = yI. 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(yIxIX) = O(yI) = 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) 4. He also proved:
34.3 Theorem. If r 3 and 1 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 vshear, 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 vshear, every W E v being an axis of it. 35.1 Lemma. If 1 1= a E G and a
if a
is a vshear, 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(al)2
2r = rk( V) = rk( vaI) 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(al)(Tl) = V(Tl)2 = {o} = V(TI)(a I). This yields (a  1)(T  I) = 0 = (T  1)(a  1). Hence aT = a + T  1 = Ta. Moreover, aT  1 = aI + T  1, whence vaTI 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 'ITshear.
+ rk(Kern( aI))
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 Dlinear, 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 'ITshear. Hence a) 0 implies b). The converse is trivial. PROOF.
+ rk(Va) = rk(VO 1) + r.
Hence rk( vaI) = r = rk( Va) proving v o 1 = Kern(a  1) = VO' b) If P > 0, then a P  1 = (a  lY = (a  ly2(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 vshears 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
ni
(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 WpI ~ Kern\p  1). As rk(Wpi) + 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 pgroup.
?}.
1. . [I
'.,.:,r
ij
'!
Finally, L
= (®).
By 3S.2, the Sea) are elementary abelian pgroups.
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 aI 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 pgroup of (R, S) containing R. As all subgroups of order p in SL(2, q) are conjugate, R* consists only of vshears. This yields R* = R, i.e., Rand S are Sylow psubgroups 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 = XI 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 vshear, (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 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 3subgroups, 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 = {(xal,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 nontrivial vshears 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 = 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 pgroup 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 pgroup 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 pgroup 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 pprimitive 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 Rprimitive 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 pprimitive 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 7subgroup 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 3subgroup in H. Therefore, if IT is a Sylow 3subgroup 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 nonabelian. Let N be a minimal nonabelian 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 3group. Moreover, N I SeN) IS an elementary abelian 3group 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 7subgroup 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. 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 pprimitive 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 pprimitive 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 nondesarguesian 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 psubgroup 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 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 psubgroups. 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 psubgroups. 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 nonincident point line pair (P, I), that the pelements 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 9, the group D..j 8(D..) is simple. Therefore 11* ~ PSL(2, q). The number of points respectively of lines of 0 is 52 + 5 + 1 = q + 1 + s. Note that q = s2 is distinct from 5, 7, and 11. If q =t= 9, it therefore follows from 14.1 that each non trivial orbit of 11 has length at least q + 1. Moreover, S2 + s + 1 does not divide 1111 = 1 S2(S4 1). Hence D.. fixes a line I of 0, as the number of lines is q + 1 + s < 2q + 1. From s + 1 < q + 1, we infer that D. fixes all the points of 1 which belong to O. Hence each point orbit of 11* which is contained in o has length .;;;; q + s + 1  5  1 = q < q + 1, a contradiction. Thus
°
°
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 2transitive 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 2subgroups of SL(2, q) are generalized quaternion groups, it follows that the Sylow 2subgroups 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 psubgroup 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 psubgroup 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 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 psubgroups. 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) 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 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 1, we see that p does not divide n. Thus a psr(p2s  1)· (p2r _ 1)1  1 divides as(p2r  1). Therefore
psr[ 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)prs
i=l: Thenp S(p2SI) 11, thenp(p21»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 nontrivial 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 pgroup. 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 pgroup 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 FeitThompson theorem. As Gp acts transitively on 100 , the group G acts flag transitively on m. Hence G p 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 qarc 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" pJI)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  pJIXJ = 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 
PII(PIXI  PJx J)  A3P2 I(P2 X2  P3x3)
P3IX3
= 0,
= 0.
Comparison with the original equations for these lines yields I AI  (b 2P3 I  b3P2I)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 + YJPII)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 IP2I  b2PII)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 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 3space 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
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 nondegenerate 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 qarc 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 qarc, 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 pmodule. We infer from P =1= pp that p does not act trivially on E. Hence E = Q EB I with Q a trivial pmodule and 1 a line on which p acts irreducibly. Again by Maschke's theorem V / I is a completely reducible pmodule. As (Q + 1)/ I is a trivial psubmodule of V / I, we deduce that V /1 is a trivial pmodule. Invoking Maschke's theorem a third time, we see that V = I EEl m with m a trivial pmodule.
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 pmodule. 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 psubgroup 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 nonisomorphic.
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
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 pgroup, 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 01) = 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 aI 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 pgroup 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 Pgroup 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 pgroup, 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 Kvector space V onto the K' vector space V. Since 7T consists of all subspaces of Krank 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(qI)(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(qI)} 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 psubgroups 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 pelement 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 psubgroups ~ and T. Then S fixes a point P t;;;;; F("i.) n F(T) vectofwise, as 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 2subgroup 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}= sIs 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 psubgroup 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 3group 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 psubgroup 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 nontrivial 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 nontrivial 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 = 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 psubgroup 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 pelements 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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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 karc 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
'ITadmissible 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
rank3group 76 rank3plane 76 real 116
Zgroup
II
I
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I)~Sbg2k
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•
I
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i
I
1 • :
I
Y
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I
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1\,,'1,,\ It'" i i I
I
r
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II J
23
33