COLLINEATION GROUPS OF TRANSLATION PLANES OF SMALL DIMENSION

A subgroup of the linear translation complement of a translation plane is geometrically irreducible if it has no invariant lines or subplanes. A similar definition can be given for "geometrically primitive". If a group is geometrically primitive and solvable then it is fixed point free or metacyclic or has a normal subgroup 2a+b a of order w where w divides the dimension of the vector space. Similar conditions hold for solvable normal subgroups of geometrically primitive nonsolvable groups. When the dimension of the vector space is small there are restrictions on the group which might possibly be in the translation complement. We look at the situation for certain orders of the plane.

mitive and solvable then it is fixed point free or metacyclic or has a normal subgroup 2a+b a of order w where w divides the dimension of the vector space.Similar conditions hold for solvable normal subgroups of geometrically primitive nonsolvable groups.
When the dimension of the vector space is small there are restrictions on the group which might possibly be in the translation complement.We look at the situation for certain orders of the plane.KEY WORDS AND PHRASES.Translon plan, translaon complement, linear groups. 1980MATHEMATICS SUBJECT CLASSIFICATION CODES.51A40, 20H30.A translation plane of order q with kernel GF(q) F can be represented as follows: Let V be a vector space of dimension 2d over F. A spread defined on V is a class of d-dimensional subspaces (called the components of the spread) such that each nonzero element of V belongs to exactly one component.The points of are the ele- ments of V, the lines of are the components of the spread and their translates.The group of nonsingular semi-linear transformations of V which permute the components is called the translation complement of .The subgroup consisting of linear transform- T. G. OSTROM atlons is the linear translation complement.
We are interested in finding information as to which abstract groups can act as subgroups of the linear translation complement; what the nature of the action is and also what the nature of the plane is.The action is fully as important as the (abs tract) group.
A particularly simple kind of action is for the group to be flxed-polnt-free (f.p.f.).A linear group is fixed point free if no nontrlvlal element fixes any nonzero vector.The translations and a fixed-polnt-free group generate a Frobenlus group with the f.p.f, group as Frobenius complement.A normal subgroup G 1 of a non-f.p.f, group is a minimal non-f.p.f, group with respect to G if it is non-f.p.f.but every normal subgroup of G properly contained in G 1 is f.p.f.It can happen that a minimal non-f.p.f, group with respect to G is also a minimal normal non- solvable subgroup.This situation has been analyzed in previous papers and some of the results are given in (2.4) below.
In (2.6) we show that if G is solvable then, subject to certain irreducibility requirements at least one of the following holds: (I) G is f.p.f.(2) G is meta- 2a+b cyclic (3) G has a normal subgroup W which is a w-group of order w for some a and b and w a divides the dimension 2d of the vector space.The nature of G and its action is much easier to analyze when case (3) does not occur.A similar situation occurs in the nonsolvable case.See (2.3) and (2.4).
In (2.7), (2.8) and (2.9) we develop some circumstances under which case (3) cannot occur.In Section 3 we develop some lower bounds on the value of d if the plane is to admit SL(2,u) or PSL(2,u) for a given u relatively prime to q. (Recall that the plane is defined on a vector space of dimension 2d over GF(q)).This is a slight sharpening of some standard results. (See Harris and Hering [2].)How- ever this sharper result is useful in looking at particular cases.
We confess to a poor background in group representation theory.All of this is representation theory in some sense and may be implied by results in classical representation theory.We would be pleased if some expert could show us how to get our results from standard representation theory if this could be done in substan- .tially less space than we have used to get them directly by fairly elementary means.
What we seem to be ending up with are results which show that when d and the characteristic of GF(q) are suitably restricted, the number of groups to be con- sidered is reasonably small.
In Section 4 we illustrate how this works for some specific values of d with q odd or even.
DEFINITION I.i.If G is a group of nonsingular transformations, V(G) denotes the vector subspace consisting of all vectors fixed by G.If is an element of G, V(o) means the same as V(<o>).
DEFINITION 1.2.The prime u is a q-primltive divisor of qd-I if u divides d-I q but u does not divide qa-I for 0 < a < d NOTATION 1.3.If G is a group, Z(G) denotes the center of G. Whenever we are considering subgroups of a given group G, C(H) will denote the centralizer of H in G.If G O is another subgroup of G, the centralizer of H in G O will be denoted by G O N (H).
We shall make repeated use of the fact that the Sylow subgroups of a Frobenius complement (and hence of an f.p.f, group) are cyclic or generalized quaternion.
This research was supported by the National Science Foundation.
2 SOLVABLE NON-F.P F. GROUPS DEFINITION 2.1.Let G be a subgroup of the linear translation complement of a translation plane w.Then G will be said to be geometrically irreducible either if G is irreducible as a group of linear transformations or if none of the invariant vector subspaces is a proper subplane of or is a component of the spread defining w.If G is geometrically irreducible, G will be said to be geo- metrically primitive if (as a vector space) cannot be written as a direct sum of proper subspaces which are subplanes or components of the spread and are permted by G.

REMARK. Recall that if
is a nontrlvlal member of the linear translation complement then the subspace V(O) pointwise fixed by is either a proper subplane or is a (not necessarily proper) subspace of a component.A subgroup G of the linear translation complement is not geometrically irreducible if G has a normal subgroup H# 1 such that V(H) is nontrivial.
LEMMA 2.2.If G is geometrically primitive then G every normal elementary abelian subgroup is cyclic of prime order.
PROOF.Suppose that G> W where W is an elementary abelian w-group of order a w where a > i and w is prime.Then W cannot be fixed point free and for some element W, V(o) is nontrivial.Furthermore V() is invariant under W.
Let V be a minimal invariant G-space.By Clifford's Theorem V--VI... V k where VI, V2, etc. are homogeneous W-spaces--i.e. the minimal W-spaces in V i are isomorphic as W-modules.If an element o of W is non-f.p.f, on Vi, it fixes a minimal W-space pointwise.Hence every minimal W-space in V. is pointwise fixed by o so V i itself is pointwise fixed by .Clifford's Theorem also says that the V. are subspaces of imprimitivity for G. be Let W 1 be the subgroup of W which fixes V I pointwise.Let V I, V2,..., V k distinct images of V(W I) under G.Note that V(W I) need not be a subspace of V. * * .. -and that V is non- Suppose that V + V + +Vk_ I V I V 2 Vk_ I * * etc. are invariant under W and each is pointwise fixed trivial.Note that V I, V 2, * V.* is nontrivial for some by some conjugate of W I. It follows that V k l -IwI%I -IwI%2 i--I, 2, h-l.Thus there are two conjugates, say %1 and %2 both fixing V k*N V*i pointwise.Hence there is some homogeneous W-space Vj which is -IwI% IW I pointwise fixed by %1 I and % %2" But the subgroup of W which fixes V.
is a direct sum of V(WI) and its distinct images under G contrary to the condition that G is geometrically primitive.
THEOREM 2.3.Let G be a nonsolvable subgroup of the linear translation complement of a finite translation plane .Let G O be a minimal nonsolvable normal subgroup of G and let H be a maximal normal subgroup of G included in G O but not equal to G O Then either H is fixed point free or G O contains a subgroup W which is minimal nonfixed point free with respect to G. Furthermore W is a w-group for some prime w and W/W 0 is elementary abelian, where W 0 is the maximal normal sub- group of G included in W but not equal to W.
THEOREM 2.4.In (2.3) if H is fixed point free then H--Z(G0).If H is not fixed point free so that W exists, then either W is elementary abelian or 2a a IW/W01 w where w divides the dimension of the vector space on which is defined and the group of automorphlsms of W/W 0 induced by conjugation with respect to G is isomorphic to a subgroup of Sp(2a,w).
The above two theorems are contained in Lemma (2.2) and (2.8) of our paper on planes of odd order and dimension [9] and Lemma (2.5) of our paper on planes of even order in which the dimension has one odd factor [I0].The key to (2.4) above is Huppert's Satz 13.7 Chapter III in his book [5].
We now return to the solvable case LEMMA 2.5.Let G be a solvable group of linear transformations acting on a vector space V of dimension 2d over GF(q).Then at least one of the following holds: (I) G is fixed point free; (2) G is metacycllc; (3) G has a normal sub- group W with the properties of W in (2.3) and W 0 is cyclic; (4) G has a normal sub- group Q isomorphic to the quaternion group of order 8 and if G 1 is a minimal non-f.p.f, group with respect to G then either the non-f.p.f, elements in G 1 have order 2 or 3 or G I centralizes Q.
PROOF.Suppose that G is not fixed point free.Let be a maximal normal subgroup of G.If is not cyclic then is not f.p.f, and contains a subgroup W which is minimal non-f.p.f, group with respect to G.
By Corollary (3.3) of [8] if W is a solvable minimal non-f.p.f, group with respect to G and W 0 is the maximal normal subgroup #W of G included in G, then W/W 0 is elementary abelian.
Hence we have conclusion (3) if is not cyclic.Suppose that is cyclic.
If is its own centralizer in G, then G/ is cyclic.In this case we have conclusion (2) since the outer automorphism group of a cyclic group is cyclic.
If / is not its own centralizer in G, then the centralizer of / in G contains a minimal normal nonabellan subgroup W* of G.The conditions of Huppert's Hilfsatz I are satisfied [4] and the conclusions include the following: a I. W* has prime power order, say w 2. W* rood its center is elementary abelian.
3. If w> 2 and W then w= I. 4. If w= 2 and W then 4= I.
Thus if w 2, we may take W*= W and we have our conclusion (3).We also have conclusion (3) if w=2 and W* is not f.p.f.Otherwise w= 2, W* is f.p.f, so W* is generalized quaternion.But if W* has exponent 4 this implies that W* is quaternion of order 8. Let us change our notation and let Q be this normal quaternion group.
Suppose that G has a minimal non-f.p.f, group G 1 which does not centralize Q.
Since G 1 must be generated by its non-f.p.f, elements, it follows that G 1 contains an element which induces a nontrivial automorphism on Q by conjugation.But the outer automorphism group of Q is isomorphic to S 4 so its order must divide 24.This gives us case (4) of the Lemma.d THEOREM 2.6.Let be a translation plane of order q and kernel GF(q).Let G be a solvable subgroup of the linear translation complement which is geometrically irreducible and geometrically primitive.Then at least one of the following holds: (I) G is fixed point free.
(2) G is metacyclic.(3) G has a normal subgroup W such that W is a w-group for some prime w.W mod its center W 0 is elementary 2a a abelian of order w for some a >i where w divides 2d.The group induced by G on W/W 0 by conjugation is isomorphic to a subgroup of SP(2a,w).
The situation at this stage can briefly be described by saying that if G is geometrically irreducible and geometrically primitive then every solvable normal subgroup is fixed point free and if G has a minimal nonsolvale normal subgroup G O then G O modulo its center is a direct product of isomorphic simple groups.
Note that G O is its own derived group and (see Huppert [5] Hilfsatz 5.23.3) in this case Z(G0) is a subgroup of the Schur multiplier of G0/Z(Go).The situation where G O contains a (noncyclic) metacycllc normal subgroup of G does not arise when G is nonsolvable due to the fact that G O is a minimal normal subgroup of G.
We can now drop the irreducibility considerations and consider the cases where Go/Z(G0) is simple.As we shall see the possibility frequently arises in trans- lation planes that G0/Z(GO) is PSL(2,u) for some u.For u # 9, the order of the Schur multiplier divides 2 so that G O turns out to be SL(2,u) or PSL(2,u) in these s i tuat ions.
We might remark that we have not made much use of the geometry and that primitivity as a linear group is at least as strong a condition as geometrical primitivi ty.
In any case it is desirable to have conditions in which every solvable normal subgroup is fixed point free.
We find it convenient to use the expression "geometrically primitive" instead of the more complete "geometrically irreducible and geometrically primitive".d COROLLARY 2.7.Let be a translation plane of order q with kernel GF(q).Suppose that both q and d are powers of the same prime u.Let G be a solvable geometrically primitive subgroup of the linear translation complement.Then no minimal non-f.p.f, group with respect to G is a w-group for a prime w.
PROOF.At characteristic u a u-group has a nontrivlal subspace which it fixes 2a a pointwise.If W exists with W/Z(W) of order w where w divides 2d it follows in this case that w= u.Hence W cannot exist.The possibility that W might be elementary abelian is excluded by the geometrical primltivity.
If qd-I has a q-primitive divisor u (see Definition (1.2)) and if the stabl- lizer of some component of the spread is transitive on nonzero points, then the order of the translation complement is divisible by u.Kallaher and the author made much use of this idea [6].
LEMMA 2.8.Let G be a geometrically primitive subgroup of the linear trans- d latlon complement of a translation plane of order q with kernel GF(q).Let d* be the largest prime power factor of 2d and let d be the largest prime power factor of d.Suppose that u is a prime factor of IGI such that u >+ I if q is even or 1 , d* u ,> (d + I) and u# if q is odd.Suppose that G has a normal subgroup W as in (2.3) and (2.4) or (2.5).Then all elements of order u in G centralize W.
PROOF.The order of Sp(2a,w) is (w 2a-l)(w2a-2-I) (w 2l)w a 2a a Under the hypotheses if IW/W01 w where w divides 2d, the prime u does not divide ISp(2a,w) l.Thus if W exists and is some element of order u in G and % 6 W we must have that -1%@ l for some v in W 0 since must induce the trivial automorphism of W/W 0. Then I o-u%u lu so that 1 since (u,w) 1 and is a w-element.
THEOREM 2.9.Suppose that the prime u is a factor of IGI which satisfies the hypotheses of (2.8) plus the extra condition that either u is a q-primitlve d divisor of q -I or u is a q-primitive divisor of qt-I where d 2t and q is even.
Let G be a solvable geometrically primitive subgroup of the linear translation complement.Then no solvable minimal non-f.p.f, group with respect to G is a w-group for a prime w.
PROOF.Suppose that W does exist with the usual properties.By Huppert, Satz 13.7 [5] W is a central product WlW2..W k where W./Z(W)I is elementary 2 abelian of order w If W then w Z(W)=W 0. Hence if V() is nontrivlal, then II =w since W 0 must be fixed point free.Furthermore the conjugate of with respect to an element of W is either or the product of with an element of W 0. It follows that ( has exactly w conjugates with respect to W and that z as a vector space is the direct sum of w copies of V(o) so that dim V(@) 2d / w.If q is even, w# 2. If dim V(o)= e, then the prime u does not divide qe-I if u is a q-primitive divisor of qd-I or if qt-I with q even.By (2.8) each element of order u in G must leave V() invarlant; in the present situation such an element must fix V(O) pointwlse.
Thus if W exists the normal subgroup generated by the u-elements fixes some nontrivial subspace pointwise and G cannot be geometrically irreducible.Hence W cannot exist if G is geometrically irreducible and geometrically primitive.
REMARK.A prime q-primitive divisor of qd-I is not ncessarily larger than d or even larger than the largest prime power factor of d.However this is "usually" the case at least for small q and d.Note that 26-1 has no 2-primltive divisors but 23-1 has a primitive divisor larger than 6.
3. BOUNDS ON THE DIMENSION.
It is probably well known from classical representation theory that the 1 dimension of the smallest complex representation of PSL(2,u) is (u-l).For a group acting on a vector space over a finite field with characteristic prime to u this is part of the results of Harris and Hering [2]even if the representation cannot be obtained from a complex representation.This section represents a slight sharpening of this part of the results of Harris and Hering [2] for the case where the group is part of the translation complement of a finite translation plane.
Furthermore our methods are relatively elementary.We consider this to be an asset.
The group W in this section does not play the same role as the group W of Section 2.
LEMMA 3. i.Let W be a group of prime order w acting on a vector space of finite dimension d.Suppose that (I) does not fix polntwise any proper subspace of V. ( 2) There exists a nonsingular linear transformation on V which normalizes and induces by conjugation a regular automorphlsm group of order h 0 on W (i.e centralizes no nontrivial element of W for i < 0 < h 0 but h i centralizes W for i=h0).
Then h 0 divides d.
PROOF.We shall show that permutes the eigenvalues of , where W =< c >.
Let K be an extension of GF(q) which contains all of the eigenvalues of c.
We can embed V in a vector space V* of the same dimension as V but where V* is a vector space over K. Let 8 be an element of K which is an elgenvalue of .Then the eigenspace (in V*) belonging to 8 is identical with V((IS-1), the subspace of V* pointwise fixed by -I.Now -I a for some integer a and -I (08-I) c[as-I -I so V(C8 ) v(ca8-I) if C ab--I, v(ca8-I) =V(c8-b) and is the eigenspace to @b The two eigenspaces are disjoint unless they are identical They are identical only if J8 -1= c@ -b so that 8= b.But then (I a--(J which cannot happen since )t induces an automorphism of order 1/2(w-l).In a simiiar fashion, )t pemtes eigenvalues #1 in cycles of length h O. Now V* is a direct sum of eigen- spaces of c and )t permutes these eigenspaces in orbits of length h0--1/2(w-1).
Hence h 0 divides dim V. THEOREM 3.2.Let G be a subgroup of the linear translation complement of a d translation plane with kernel GF(q), order q Let w be an odd prime relatively prime to q. Suppose that G has a normal subgroup W which is elementary abelian of order w and that G induces a cyclic automorphism group on W by conjugation so that the nontrlvial elements of W fall into exactly conjugate classes each of i a w a length -(w-1).Then either w a 4d+ 1 or -<_ 2d+ 1.If V(W) is trivial, then w -I divides 4d.
PROOF.Note that G is acting on a vector space of dimension 2d over GF(q).
Consider the case where a > I.There may or may not be some nontrivial subspace V(W) pointwise fixed by W. If V(W) is nontrivlal, W has some complementary space on which it acts faithfully.
Let V be a vector space on which W acts faithfully.(The following is suggested by the proof of Lemma 1.3 in Harris and Herlng [2] .) There exists an element h such that h induces a group of automorphisms on W of order 1/2(wa-l).Let G I <h, W> and let <h e 1 / 2 ,W>, where e (wa-l).
Since h e centralizes W, is abelian.Then Clifford's theorem implies that V is a direct sum VI... V k of homogeneous -spaces.If a > 1 some element o of W is not f.p.f, on V I.But W is abelian so o fixes a minimal -space pointwise and, since V 1 is a homogeneous W-space, V I is pointwise fixed by o.An f.p.f.elementary abelian w-group must have order w so W must induce a group of order w a-I on Wl--i.e.W I is pointwise fixed by a group of order w respectively are pointwise fixed by conjugate subgroups of W having order w a-I The number of subgroups of rder w in a conjugate class is the same as the number of subgroups of order w in a conjugate class and is either (wa-l)(w-l) -I or i a I (w -l)(w-l)-depending on whether a subgroup of order w has 2 or 1 conjugate classes of nontrivial elements.
i By the previous Lenmm h 0 (w-l) or w-i divides dim V 1 Hence ..V h is invariant under G and dim VI...VhI is a multiple of -(w -1).
Furthermore if V 0 V I. ..VhV' then the argument can be repeated to show that I a All V' contains a direct sum V I...Vh where (w -i) divides dim [V I V h of this holds for a--1 again by the previous Lemma.By induction, is the direct 1 a sum of V 0 and a subspace whose dimension is divisible by (w -I).If V 0 is trivial, I a then (w-i) divides 2d or 4d-0 mod wa-l.In this case either 4d--wa-I or 2d>__ wa-l.Now suppose that V 0 is not trivial.Here we make our first use of the fact that G is acting on a translation plane.If is any component such that V 0 is nontrivial, then is invarlant under W. Furthermore V 0 is a subplane iff at least three components are invariant under Wo With the proper basis, W can be (x,y) from a vector space of dimension d; the sets of points for which x= 0, y= 0, y x respectively are three invariant components.In this case if V i intersects an i a invariant component it intersects all of them.Hence h--(w-I) is less than d.
If W has precisely one invariant component we again conclude h < d. (This case doesn't really happen) If W has two invariant components and <%> leaves both invariant we again conclude h < d.If W has two invariant components in the same orbit under % then V 0 must intersect both of them nontrlvially and we are back to the case where V 0 is a subplane.
COROLLARY 3.3.Let G be a subgroup of the translation complement for a trans- d latlon plane of order q with-kernel GF(q).Suppose that G has some normal sub- group H such that G/H PSL(2,u for some odd prime u, (IHI, u) 1 and (u,q)--I.
a Then either ua--4d+l or u =< 2d+l.If ua-I does not divide 4d then a Sylow u-group in G fixes a nontrivial subplane pointwise.
PROOF.The group G of the previous theorem exists.If ua-I does not divide 4d, the Sylow u-group has at least one fixed component.There is more than one fixed component since (u,q)= i and each fixed component must contain fixed points 1 a different from 0 if (u-i) does not divide 2d. 4 THEOREM 3.4.Let be a translation plane of order q and with kernel GF(q) where q is even.Suppose that the translation complement G of contains no afflne elations.Then the Sylow 2 groups of G hve nilpotency class at most 2.
PROOF.Let S be a Sylow 2-group.Since the number of components of the spread is odd, S must fix some component and act faithfully on .Furthermore each involution in S is a Baer involution and hence fixes a 2-space on pointwise.
Furthermore if is any nontrivial element, we must have that V() has dimension at most 2.
Let us restrict ourselves to the representation of S on by 4 x 4 matrices.
We can choose a basis so that the elements of S are represented by upper triangular matrices with l's on the diagonal Furthermore we may assume that some involution in the center is represented by a matrix of the form Hence either c3=c4 0 or c=0.In the former case the points (0,I,0,0)(0,0,I,0) and (0,0,0,I) are all fixed contrary to the condition that the dimension of V() is equal to 2

Hence c=O
In abbreviated form o (o I C where the capital / letters are 2 by 2 matrices.By a further change of basis we can take ; 0 Now the condition that ( E) I commute with is that A= B. We may take (I a) The reader may verify that the commutator of Il I C ) where E DII DI2 + DII CII C12 DII + CII DII DI2 CII + CII C12 (using the fact that Cll Dll are commuting involutions).
If we set DII ( ) and CII (01 ) it turns out that the element/T =\ in the left hand corner of E is 0.
The commutator of with respect to a general element then has the form ( )where K has the formCl l\u ] 0 f 1 2 1 But then ( KI)fixes the vectors (0,i,0,0)(0,0,i,0)and (0,0,0,I).It follows that K= 0 and thus the commutator of three elements is the identity.

APPLICATIONS.
As before G is understood to be a subgroup of the linear translation comple- d ment of a translation plane with kernel GF(q) and order q We shall assume that G is geometrically primitive and G O will denote a minimal nonsolvable normal sub- group of G if G is nonsolvable.
As we have pointed out in several previous papers when q and d are both odd and G is nonsolvable G O contains a normal subgroup H of G such that G0/H is PSL(2,u) for some odd u or G0/H is A 7.
Consider the case q--d=5.Applying (2.7), (2.3), (2.4) and the remarks preceding (2.7) we can say that H is a Schur multiplier for G O By (4.8) of [I0], each odd prime factor of G must divide 5(55+ 1)(55-1)(54-1)(53-1) so A 7 does not apply.Hence GO= SL(2,u) for some odd u.If (u,5) 1 then by (3.2) either u=4.5+I =21 or u -< II.Since 21 is composite we cannot have u= 21.Thus u= 11,9, or a power of 5.If we put in the condition that G has a subgroup fixing some component E and transitive on nonzero vectors of E, then IGI is divisible by the prime factor 71 of 55-1.(This condition arises naturally in the investigation of rank three planes.)We leave it to the reader to verify that in this case the 52 only possibility is u so that the plane is Desargueslan.
A similar argument works for q 3, d--5 if we again assume G has a subgroup fixing and transitive on some component E of the spread.These are special cases; our results appear to be the most useful in narrowing down the possibilities for groups in particular situations.
Consider the case where q is even and d 4. If G contains affine elations (shears) we can apply Hering's results on the groups generated by elations [3], so assume that G has no affine elations.By (3.4) the Sylow 2-groups have nil- potency class at most 2.
According to Walter [Ii] the nonsolvable simple groups with abelian Sylow 2 s 2-groups are in the following list: PSL(2,u), u > 3 u 3 or 5 rood 8 or u J(ll), Ree type (R l(u)).
To illustrate the methods of this paper, we restrict ourselves to the cases where Go/H=PSL(2,u) u-3 or 5 rood 8 or u-7 or 9 rood 16 or G is solvable.Note that if q is even and G is geometrically primitive then G can have no normal 2-groups.By (2.6) and (2.7) every solvable normal subgroup will be fixed point free or metacyclic.