TRANSLATION PLANES OF ODD ORDER AND ODD DIMENSION

The author considers one of the main problems in finite translation planes to be the identification of the abstract groups which can act as col lineation groups and how those groups can act. The paper is concerned with the case where the plane is defined on a vector space of dimension 2d over GF(q), where q and d are odd. If the stabilizer of the zero vector is non-solvable, let GO be a minimal normal non-solvable subgroup. We suspect that GO must be isomorphic to some SL(2,u) or homomorphic to A6 or A7. Our main result is that this is the case when d is the product of distinct primes. The results depend heavily on the Gorenstein-Walter determination of finite groups having dihedral Sylow 2-groups when d and q are both odd. The methods and results overlap those in a joint paper by Kallaher and the author which is to appear in Geometriae Dedicata. The only known example (besides

(when dimension and order are odd); it is possible that the key non-solvable group is always SL(2,u) for some u or, perhaps, is a pre-image of A 6 or A 7. This is suggested by the Gorenstein-Walter Theorem [5]. The author [14] has previously shown that this is the case for minimal non fixed-point-free groups (see below) which are non-solvable. However, a non-solvable linear group need not have a non-solvable minimal non-f.p.f, subgroup.
Throughout this paper G is a non-solvable group of linear transformations and G O is a minimal normal non-solvable subgroup.
Our most important result is Theorem (3.5) which states that, if d is the product of distinct primes, a minimal non-solvable normal subgroup of the linear translation complement either has the form SL (2,u) or is a pre-image of A 6 or A 7. The present paper is similar in method, spirit, and results to the joint one. Here there are more restrictions placed on d and weaker initial restrictions on the group.
The notation and language are more or less standard. Some of the terminology and even some of the facts, may not be familar to every potential reader of this paper. We finish this Introduction with a brief discussion of thee matters and some remarks on notation. A minimal invariant subspace of a reducible group G will sometimes be called a minimal G-space. If V 1 is a subspace such that all of the minimal G-spaces in V 1 are isomorphic as G-modules, then V 1 will be called a homogeneous space.
The order of G is denoted by IGI. If G is the full group and G O is a subgroup, 0 (Go) is the centralizer of G O in G.
The subgroup of G which fixes is G(C).  [15].) Suppose that the 2-group in Fit G O is a generalized quaternion group Q. Then GO/O (Q) F G O is isomorphic to a group of automorphisms of Q.
The automorphism group of Q is a 2-group or S 4 (see Passman [15].) and hence is solvable. Hence (Q) F G O is non-solvable; by the minimal property of  [5], G-0 PSL(2,u) for some odd u or is equal to A 7. Furthermore, if G-0 PSL(2,u) and u # 9 then G O SL(2,u).
One of the key assumptions of Theorem (2.3) was that Fit G O is fixed point free. The next few Lemmas develop the machinery for examining the case where Fit G O is not fixed point free. Dixon [2] gives a similar development for vector spaces over an algebraically closed field. We have attempted to modify Dixon's argument to apply to vector spaces over a finite field. Let $ be a minimal G-invariant subspace of . By Lemma (2.7) in Dixon [2], the dimension of is n and there is a vector v in V F such that <v> V F. Let {u I, u2' n } be a basis for $. The {Vl 1, vl 2, v n} is a ba$i for Vf. Label these vectors v I, v n respectively.
Let w be an arbitrary non-zero element of V and let . a,v,j J V . ail JJ where the aiF.j Let (w) be an eigenvalue wl of the matrix (aij). If F, let E be an extension which contains ,.
Note that wl 1 vu v[. aijj li]. The determinant of the matrix (aij) },I is zero, so the vectors (w v)i are dependent and the dimension of <(w-v)> is less than n.
We had vl v i. Let v w i. Now v and w are in V F (which is embedded in VE). Let the mapping T be defined by wit >,v i. Then T becomes a linear transformation on V E by defining (ClW I + + CnWn)T Cl(WlT + + Cn(WnT) for c 1, c n E. (Note that T also acts as a linear transformation on V F.) Let x be an arbitrary member of V F. Then x + xT belongs to <(wv) > Suppose that XlT x2T. Then (x I + XlT) (X 2 + X2T) x I x 2 is a vector in V F which belongs to <(w-) >. Now <(w -v)> I"1 V F is a G-invariant subspace of V F and, since G is irreducible, is either the null space or all of V F. Suppose that V F <(wv) >. A basis for V F is also a basis for V E so in this case <(w->,v)$> V E, which is a contradiction.
Hence Xl x2T, x I x 2 e V F implies x I x 2 so T is a non-singular linear transformation on V F. If p e G and x V F, then (x + XT)p Xp + XpT SO commutes with G. Hence z is a scalar transformation on V F. We had W.T V.. Hence ; c F and E F.
In particular this holds for w v 1, v 2, v n. Let   But n must be a power of u and divide the dimension of V. Hence n u.
(2.8) LEMMA. In the notation of (2.1) ( PROOF. The theorem holds if W of (2.8) is non-trivial and W 0 is trivial, so suppose W 0 is non-trivial. We wish to apply (2.7). For this purpose, we can restrict our attention to a minimal Go-space V 1. Note that the dimension of a minimal Go-space is also the product of distinct primes. Now all of the minimal Z(G O) spaces in a Go-space are isomorphic as Z(Go)-modules. As in Hering [7] Hilfssatz 5, there is a field K so that the additive group of V 1 is a vector space over K and the elements of Z(Go) become scalars. Now consider the action of W on a minimal W-space in V 1. Again, the 2 dimension is a product of distinct primes, so (2.7) implies that IW/WoI w and w is one of the primes dividing the dimension of V over F. Let G G/C (w). Then G induces, by conjugation, a group of automorphisms of IV. That is, there is a homomorphism from G into GL(2,w), since is elementary abelian of order w2. The kernel of the homomorphism is the subgroup of G which centralizes W. If the subgroup of G O which centralizes W were non-solvable, then G O would centralize W. We wish to show that this cannot be the case. But, except for the w-groups, the Sylow sugbroups of G O are included in G 2, so Go/G 2 is a w-group. But if G 2 is solvable, Go/G 2 must be non-solvable, so we again get a contradiction. Thus W 0 must be trivial if W exists and W must be abel ian.
(2.10) LEMMA. Suppose that the Sylow 2-groups in G/Z(G) are dihedral.   PROOF. If W 0 is trivial, W is elementary abelian by (2.9). Thus if and the subspace V() pointwise fixed by >, is non-trivial, then W leaves V(>,) invariant. Hence W is not faithful on its minimal spaces.

Let H be a maximal normal subgroup of G included in G O but not equal to G O Then Go/H is simple and either H W[c'(W) FI G O or W Z(Go
Since each homogeneous space is a direct sum of minimal spaces that are isomorphic as W-models, W is not faithful on its minimal spaces. If V.* c for some component C, then is invariant under W. If W leaves just one or two components invariant then G must fix or interchange these two and cannot be non-solvable. If W has 3 invariant components every non-f.p.f, element of W must fix a subplane pointwise. Hence V is a subplane and has even dimension. This implies that k is odd, since 2d k dim V 1. PROOF. Suppose that W 0 is trivial, so that (3.3) holds. Let G(V 1) be the stabilizer of V 1 in G 1. The index of G(V 1) in G is equal to k, so a Sylow 2-group of G(V 1) is a Sylow 2-group of G. Let G(V 1) be the induced group on V 1 i.e. G(V 1) may be identified with the factor group obtained by taking G(V 1) dulo the subgroup fixing V 1 pointwise. Then W is a normal subgroup of order w in G(Vl), and all of the minimal W spaces in V 1 are isomorphic as W-modules. As in Hering [7], G(V 1) rL(s,qt) and the subgroup centralizing W is isomorphic to G(V 1) 1 GL(s,qt) for some s, t such that st dim V 1. Thus the index of e (W) (I G(V I) divides t and is not divisible by 4.
Hence the index of G(V 1) rl (2 (W) in G(V 1) is not divisible by 4. Let S be a Sylow 2-group of G(Vl). As pointed out at the beginning of the proof, S is then a Sylow 2-group of G. Hence S/S rl e (W) is a Sylow 2-group of G/C (W) and its order is I or 2. This implies that G/e (W) is solvable.
Hence Go/G 0 FI e (W) is solvable. This is a contradiction since G O 1 I2 (W) is solvable and G O is non-solvable. We conclude that W 0 must be non-trivial. d (3.5) THEOREM. Let 1I be a translation plane of order q with kernel GF(q), where q and d are odd. Let G be a subgroup of the linear translation complement. Suppose that G is non-solvable and irreducible with a minimal normal non-sOlvable-G 0 and that d ks the product of distinct primes. Then either SL(2,u) for some odd u or G-0 A 6 or A 7. Here G-0 Go/Z(Go)-PROOF. This is a consequence of (3.4), (2. It may be worth while to take a look at some aspects of the ways that G O SL(2,u) can act on a translation plane. One possibility is that u is a power of the characteristic p and that the p-elements are affine elations.
If G O contains affine homologies of prime order greater than 5 then a result of the author [13] shows that G O contains affine elations. The group generated by these elations will be normal in G O and, in fact, equa) to G O (3.7) COROLLARY. If r divides u + I, the conclusions of (3.6) hold even if u is not prime.
REMARK. The cases where u is not prime or where G contains affine homologies of order 3 or 5 were handled in the Kallaher-Ostrom paper [12] under the assumption that a certain p-primitive divisor of qd I (which turned out to be u) divided the order of the group induced on by G().
Note that when (3.6)  PROOF. lL(l,qd), in its action on a vector space of dimension d over GF(q) has a cyclic normal fixed point free subgroup of order qd I and index d.
REMARK. In the Kallaher-Ostrom paper [12], Theorem 6.1, it turned out that d divides u I. A subgroup of a Frobenius complement whose order is the produce of two distinct primes must be cyclic. SL(2,u) has a subgroup of over u(u I) which is not fixed point free for u > 5. Putting this together with (3.6), it appears that an important subcase for the possible action of SL(2,u) is the one where the orders of the non-fixed point free elements divide u 1. In the context of (3.8), especially if d is prime, we again arrive at a situation where d divides u 1.
(3.9) LEMMA. If u is not a power of p, if p and d are both odd, and if d divides u -1 then d 1/2(u 1) or d 1/4(u 1).
PROOF. If (IGoI, p) 1 and if G O is absolutely irreducible, it has a complex representation of dimension 2d (u 1) and the representation we are using can be obtained from the complex representation. (See Dixon [2].) In the other case, G O is reducible over a field extension but G O can be obtained from a complex representation of dimension u 1.
Suppose that u + 1 has an odd prime factor r. Let be an element of order r in G O Under the hypotheses, it follows from (3.7) that k is fixed point free.
Let (!) be a complex rth root of 1. Then the character of has the a 0 + ale +-..+ ar_l Er-1 where a is the multiplicity of the eigenvalue form Ei, so that a 0, a I, are non-negative integers. If e is the dimension of u 1 the complex representation, a 0 + a I +-..+ ar_ 1 e. First, suppose e 2 From character table (see Dornhoff [3]) the character of is equal to -1. Recall that, by (3.9) d 1/2(u 1) or (u 1). If r exists, so that 1/2(u + 1) is odd, then u 1 0 mod 4 so that if d is odd, d cannot be equal to 1/2(u i). Thus if 1/2(u + 1) is prime, d (u I).
If r does not exist, so that u + 1 is a power of 2, then (u I) is not integer so that d 1/2(u 1) in this case.
We can use (3.11) to make a slight improvement in Theorem 6.1 of [12].
In the following corollary, u is a prime p-primitive divisor of qd I where p is prime and q pko G has the usual meaning of this paper and G O is a minimal non-solvable normal subgroup of G. Here and earlier in this paper when reference is made to the group induced on by G(), it should be understood that the subgroup fixing pointwise has been factored out. We continue to assum q and d are both odd.
(3.12) COROLLARY. Suppose that, for each component , the order of the group induced on by G() is divisible by u and that G O exists and is non-trivial. Suppose that 11 is non-Desarguesian. Then at least one of the following holds.
(c) u 2d + I, q p, p divides d and G O SL(2,u).
PROOF. In Theorem (6.1) of [12], it is shown that, under the present hypotheses we have case (a) or (d) or G O SL(2,u) q p, u 2d + 1.