TRANSLATION PLANES OF DIMENSION TWO AND CHARACTERISTIC TWO

This article discusses translation planes of dimension two and characteristic two. Let G be a subgroup of the linear translation complement of such a plane π. The nature of G and its possible action on π are investigated. This continues previous work of the authors. It is shown that no new groups occur.


INTRODUCTION
Part I.A translation plane of dimension two over GF(q)  F may be represented by a vector space of dimension four over F. The lines through the zero vector form a spread--a class of two-dimensional subspaces which are pairwise independent.The col- lineations fixing the zero vector (the translation complement) constitute a subgroup of FL(4,q); the intersection with GL(4,q) is the linear translation complement.
Finite translation planes of dimension two have been studied more than those of other dimensions.A complete classification may not be possible if this is understood to include an enumeration of all existing planes.Yet we have reached the stage where we can give a reasonable list of classes when q is even.
Hering [6] has determined the possible groups generated by the elations in the translation complement and one can at least make general statements as to the nature TRANSLATION PLANES OF DIHENSION TO AND CHARACTERISTIC TO 3 REMARK.The above two definitions require no restrictions on dimension or characteristic.We wish to remind the reader, however, that throughout this paper G is assumed to be a group of linear translations acting on a vector space of dimension 4 over GF(q), where q is a power of 2. Indeed G is assumed to be a subgroup of the linear translation complement of a translation plane of dimesion two defined on this vector space.
LEMMA 2.3.Let G be primitive and solvable and let be a maximal abellan normal subgroup.Assume that contains Z, where Z is the group of scalars.Then PROOF.The proof is by contradiction, so suppose Z is the unique maximal abelian normal subgroup.Since G is irreducible, G Z. If G is solvable G has a normal subgroup A such that A/A Z is abelian and characteristic in G/Z.(Let A be the full pre-image of A/A Z.) Suppose that A is f.p.f.Then A must be odd, since no involution is f.p.f, at characteristic 2.An f.p.f, group is a Frobenius complement.See [9] Lemma 2.2.
A Frobenius complement which is not cyclic has a characteristic abelian group not in its center.See [I0] Theorem 3.7.This contradicts the condition Z. Hence A is not f.p.f.
Then A contains a subgroup G 1 which is minimal non-f.p.f, with respect to G.
Note that A/A Z is abelian implies that A is nilpotent and hence that the Sylow subgroups of A are characteristic.Hence the minimal property of G 1 implies that G 1 must be a u=group for some prime u.If H is the maximal normal subgroup of G included in G 1 but not equal to G I, then GI/H is elementary abellan. (See [I0], Theorem 3.3.) If H is trivial, G 1 is abelian and by our assumption that Z is the unique maxi- mal abellan normal subgroup, we could have G 1 Z.This cannot happen since Z is f.p.f.Thus H is an f.p.f, u-group.Two-groups are not f.p.f, at characteristic two and the Sylow subgroups of odd order in a Frobenius complement are cyclic.Hence H is abelian and by our present assumeption, H Z.
N. L. JOHNSON AND T. G. OSTROM components of the spread and q + 1 2 mod u implies that a fixed 2-space must include at least two fixed l-spreads.Cliffords Theorem (see [3] page 70) implies that if G is primitive, all of the minimal Gl-spaces must be isomorphic as Gl-mOdules and hence that G 1 must be faithful on its minimal Gl-spaces.Since this does not happen, we have a contradiction to the assumption that Z.
THEOREM 2.4.Let G be primitive and solvable.Then G is isomorphic to a sub- group of FL(1,q4).
PROOF.Let / be a maximal abelian normal subgroup of G such that contains Z.By the previous Lemma, # Z.
If is not f.p.f., it leaves invariant the subspace pointwise fixed by one of its elements and hence is not faithful on its minimal subspaces.Thus Cliffords Theorem implies that is f.p.f.
Then (see Herlng [3] Hilfsatz 5) G is isomorphic to a subgroup of FL(I,q4) or FL(2,q2) since G is primitive. , FL(I we are done, so suppose G is a subgroup of FL(2,q ).We may 2 also assume G GL(2,q centralizes G > G SL (2,q2).The solvable subgroups of SL(2,q2) PSL(2,q2) (for q even) are cyclic or dihedral except in cases where there is a characteristic 2-group.G has no normal 2-group if it is primitive so G SL(2,q2) is cyclic or dihedral and hence contains a characteristic abelian group I" Since is maximal and is centralized by I' /I is abelian so 1 But is actually included in the center of GL(2,q2), so SL(2,q2) is trivial.Hence G SL(2,q 2) is trivial.At characteristic 2. GL(2,q2) is the direct product of SL(2,q2) and the center of 2 q2) 2 GL(2,q ).Hence, in our case, G N GL(2, In that case G G N FL(2,q can- not act irreducibly on the vector space of dimension 4 over GF(q).Thus the only possibility left is G is isomorphic to a subgroup of FL(I,q4).
THEOREM 2.5.If G is irreducible but imprimitive then G has a reducible sug- group of index 2.
PROOF.The theorem is trivial if the vector space is the direct sum of two subspaces of imprimitivity.
Hence assume that the basic vector space V is the direct sum of subspaces of imprimitivity of dimension I. Let G O be the normal subgroup of G which fixes each of VI, V 2, V 3, V 4.
the minimal G0-spaces in W.I for fixed i are all isomorphic on G0-modules and each minimal G0-space belongs to one of the W.1's; the W i all have the same dimension.
Note that k I, 2, or 4. If k 2 we are done.
If k 4 then there are no minimal G0-spaces besides V I, V 2, V 3, V 4. G01 can- not be even, since an involution in G O would have to fix VI, V 2, V3, V 4 polntwlse.
If G O is odd, then G O must fix some other 1-space on the component of the spread which contains V I. We conclude that if k 4, there must be two components, each containing two of the V In this case G has a subgroup of index two fixing these two components.
We have left the case where k I, so that all of the minimal spaces are iso- morphic as G0-modules.We may assume that G and G O contain all of the scalars. (If they do not, adjoin the scalars.)If the minimal G0-spaces are isomorphic as G 0- modules, then G O must be faithful on its minimal subspaces.With the minimal subspaces being 1-dimenslonal, it follows that if G O contains the scalars 2, then G O Z.
But G/G 0 is a subgroup of the symmetric group S 4 and its order is divisible by 4. IS41 24 and any subgroup whose order is divisible by 4 is one of the following: (a) a two-group, (b) A 4, (c) S 4. In each case there is a normal 2-group (not necessarily a Sylow 2-group).
Let G G/Z G/G 0 Let S be a 2-group i G, where S is normal in G.If is an involution in S and % G then %-I I for some I in S and some Z. Hence 2 22 2 and I. Then the subgroup of S generated by its involu- Hence the subspace which is pointwise fixed by this 2-group is invariant under G and G is reducible, contrary to hypothesis.Hence the case k I does not happen and the Theorem is proved.

4)
THEOREM 2.6.Suppose that (G) is irreducible.Then G c FL(I,q PROOF.If (G) is irreducible, it may be primitive by the previous Theorem.
THEOREM 2.7.Suppose that G @ (G) contains the scalars and that G is redu- cible.Then G is cyclic or G is not faithful on one of its minimal subspaces.
PROOF.If G is reducible, it has some invariant 2-space V 1.If G is not faith- ful on V we are done.Otherwise G is a subgroup of GL(2,q).The subgroups of 2 r PSL(2,q) q of odd order are cyclic.See Dickson [I].But PSL(2,q) SL(2,q).Thus G SL(2,q) is cyclic, with order dividing q-1 or q + I.As a subgroup of GL(2,q) acting on a vector space of dimension two, G must either be cyclic or have a pair of invariant 1-spaces and not be faithful on at least one of them.

G GENERATED BY ELATIONS
In this section we are concerned with the case where the translation complement contains elations.Actually, we will limit ourselves to the case where G is the sub- group of the translation complement generated by the elations.
Hering 6 has investigated this situation.In one of his cases, G has a normal subgroup of index two and odd order.In unpublished work he has shown that, under much broader circumstances than we have here, G must be dihedral.Since this work is not published, we will have a proof that G is dihedral for a plane of dimension two over GF(q), where q is a power of 2.
LEMMA 3.1.Suppose that (a) G has a normal subgroup of index and odd order; (b) G is primitive; (c) The involutions in G are elations; (d) G contains more than one elation.Then G is dihedral.
PROOF.Let O be an involution and T an element of odd order such that <T> has i index 2 in <,>.Then the involutions have the form r for those i such that i -i Hence the group generated by the involutions is dihedral.

4).
By (2.4) G is isomorphic to a subgroup of FL(I,q Then G GL(I,q4) is a cyclic normal subgroup of odd order.The fact that G is a group of linear trans- 4 formations over GF(q) implies that the index of G GL(1,q in G must divide 4. Because of (a) this index is 2. Hence (d) and the above argument implies that G is dihedral since it is generated by its elations.
TRANSLATION PLANES OF DIMENSION TWO AND CHARACTERISTIC TWO 47 LEMMA 3.2.G is dihedral if G is reducible and the conditions other than (b) of (3.1) are satisfied.
PROOF.If there are two elations with the same axis, 4 divides IGI.If there are two elations with different axes, no component of the spread is invarlant.If G has an invarlant 1-space, the component containing this 1-space is invarlant.Hence if G is reducible, it has an invariant subspace which is a Baer subplane 0" If G is faithful on 0' the hypotheses are sufficient to guarantee that G is dihedral.
Otherwise, let G 1 be a maximal dihedral subgroup of G acting faithfully on 0" Specifically, let G I <o,T> where 0 is an involution and <T> is the cyclic stem.If G is not dihedral, there is some involution o such that <,GI> is not dihedral and some element fixes 0 polntwse.
Suppose that 0 fixes n 0 polntwise.Then the restrictions to 0 of the two involutions o and 0T are identical.In particulat, they fix the same points of 0 so that the axes of the elations o and 0 must be the same.But this requires that either 0T or that G has a subgroup of order 4.
Neither of these conditions is satisfied.Suppose that TI fixes 0 polntwlse. -i

-2i
Then the restrictions of o and to n 0 are identical, so must fix 0 pointwise.
But G I <0,T> is faithful on 0 so i 0. In this case, fixes n 0 polntwlse, con- trary to the assumptions that is an elation.
LEMMA 3.3.Suppose that G is irreducible but imprimitive and the conditions of (3.1) excluding (b) are satisfied.Then G is dihedral.
PROOF.By (2.5) G has a reducible subgroup H of index 2.We may assume that V2, where V 1 and V 2 have dimension 2, are interchanged by each involution G and some invariant under H. Lemmas (2.3), (2.4), (2.5), (2.6) for a translation plane of dimension 2 over its kernel and a vector space of dimension 4 over GF(q) apply as well as to a vector space of dimension two, i.e., to a Desarguesian affine plane over GF(q).
Hence if H is irreducible and faithful on V 1 and has odd order, then H is cyclic as in (2.6).
If H is faithful on V I and is reducible on V I then it has at least two invarlant 1-spaces and is a subgroup of the direct product of two cyclic groups.Hence H is abelian.If H is f.p.f, and abelian, then H is cyclic.
Since H has two fixes 1-spaces in V2, V must be the direct sum of 4 H-invariant l-spaces.If H has an additional l-space in V 1 (or V2) then H fixes each l-space in V 1 (V 2).If H is not cyclic, it is not f.p.f, on V 1. Thus if H fixes each l-space in VI, then H is not faithful on V I.
If H has precisely 2 invariant 1-spaces in V and precisely 2 invariant l-spaces in V 2 then therefore 1-spaces must occur in two orbits of length 2 under G.In this case, the 2-space containing an invariant pair is invariant so G is reeucible and we can return to the previous case.
We have left the case where H is not faithful on V I or V 2.
Let G I <O,m> be a maximal (dihedral) subgroup where <T> !H acts faithfully on V 1 and O is an involution.If G # G I and G is generated by its involutions, there is an involution such that <o,GI> contains an element which fixes V pointwise.
Note that V 1 is not invariant under any element of the form oT i or 0T i.If i o0T fixes V I pointwise, we get a contradiction just as in the proof of the previous Lemma.We conclude that G must be dihedral.
THEOREM 3.4.Let G be a subgroup of the linear translation complement of a translation plane of dimension 2, characteristic 2. Suppose that G is generated by its elations.Then one of the following holds: (a) G is a Suzuki group.
(c) G is elementary abelian and is a group of elations all with the same axis.
PROOF.This is just a restatement of a Theorem of Hering [6, 3] except for case (d).In Hering's remaining case, G has a subgroup of odd order and index 2.
TRANSLATION PLANES OF DSION TO AND CHAPCTERISTIC TO

ALL INVOLUTIONS ARE BAER INVOLUTIONS
Here we are concerned with a subgroup of the linear translation complement which has even order but contains no elations.As pointed out in [8], the Sylow 2-groups are elementary abellan.The case where G is non-solvable is covered in [8], so we shall be mostly concerned with the solvable case.
LEMMA 4.1.Let G be a subgroup of the linear translation complement of a finite translation plane of even square order.Suppose that each Sylow 2-group fixes a Baer subplane pointwise and that there is more than one Sylow 2-group.Then no Sylow 2-group contains a non-trivlal element which normalizes another Sylow 2-group.
PROOF.Suppose that S 1 and S 2 are Sylow 2-groups and that S 1 contains an involution O which normalizes S 2. Then 0 centralizes some involution in S 2.
Hence <0,> is a group of order 4 contained in some Sylow 2-group S 3.
Then the Baer subplanes polntwlse fixed by , 0, S I, S 2, S 3 must all be identi- cal.By Foulser [2], Theorem 3, a Sylow 2-group of the polntwise stabllzer of a Baer subplane is normal and hence unique (in the polntwlse stabilizer).Hence S 1 and S 2 are not distinct.
LEMMA 4.2.If G is solvable and each Sylow 2-group fixes a Baer subplane point- wise, then either the Sylow 2-group is normal or IGI is not divisible by 4.
PROOF.See Hering [5].Theorem I.If S is an elementary abellan 2-group and is a Frobenlus complement then IS[ 2.
REMARK.Lemma (4.1) does not require that the plane have dimension 2 over its kernel.Foulser's Theorem 3 [2] implies that the Sylow 2-group must be elementary abellan in this case.Hering's Theorem 1 then implies that if G is non-solvable and generated by its Sylow 2-group it must be isomorphic to SL(2,2s) for some s.
If, in addition, G is solvable and generated by its involutions, then G/(G) is a Sylow 2-group.
PROOF.A Sylow 2-group must have at least one fixed component.If it contains no elations, it acts faithfully on a 2-space and is a subgroup of a Sylow 2-group of GL(2,q).Hence the Sylow 2-groups are elementary abelian.
N.L. JOHNSON AND T. G. OSTROM

GROUPS GENERATED BY INVOLUTIONS
In the last two sections, we have assumed that the involutions in the linear translation complement are all of the same kind---i.e., either they are all elations or all Baer involutions.As before G always denotes a subgroup of the linear trans- lation complement.
LEMMA 5.1.If G contains both elations and Baer involutions then the group gen- erated by the Baer involutions contains elations.
PROOF.If G contains involutions of both kinds, let G be the normal subgroup generated by all of the elations in G and let G 2 be the normal subgroup generated by all of the Baer involutions in G.
Let S be a Sylow 2-group and let S S 0 G I, S 2 S 0 G 2. By Herlng's results [6], G 1 eontalons no Baer involutions.If G 2 contains no elations, then SIS2=S1 x S 2.
Furthermore S 1 and S 2 are both elementary abelian.See (4.3) and again refer to Hering [6].It follows that SlX S 2 is elementary abelian and contains involutions which do not belong to either S 1 or S 2 (unless one of them is trivial).But every involution belongs to G 1 or G 2 so every involution in S belongs to S 1 or S 2. Hence G 2 must contain elations.
DEFINITION 5.2.In the rest of this section G 1 is the group generated by the elations; C (G I) is the centralizer of G 1 in G and H is the subgroup of C(G I) gen- erated by the involutions.
LEMMA 5.3.The center of G 1 is trivial unless G 1 is a 2-group.Thus G IC(GI) G 1 C(G I) LEMMA 5.4.Suppose that G 1 is not trivial, that C(G I) contains involutions and neither G 1 nor C(G I) has a non-trivlal normal 2-group.Then the axes of the elations in G can be embedded in a derivable net which contalons the Baer subplanes pointwise fixed by involutions in C (GI).If C (G I) is non-solvable, this derivable net is embedded in the given translation plane.
TRANSLATION PLANES OF DSION TNO AND CHARACTERISTIC TNO 53 PROOF.If C (G I) is non-solvable this follows from (2.7) in 8 ], since C(G I) must be reducible.Each axis of an elation in G I is invariant under C (GI).If G 1 has no normal 2-group, there are at least three such axes.. Hence the non f.p.f.elements in C (G I) including the involutions, must fix Baer subplanes polntwise.
The Lemma is now a direct consequence of the classical theory concerning a regular determined by three skew projective lines, etc.
In the rest of this section, the subgroup G 1 is assumed to be non-trivial.
LEMMA 5.5.If G 1 is not a 2-group, (G I) is isomorphic to a subgroup of GL(2,q).PROOF.By (5.2), G 1 C (GI) is trivial, so C (G I) contains no homologies.If C (G I) contains an affine homology whose axis is the axis of an elation in G 1 then must be invariant under G 1 and all of the elations in G I fix .Hence G I is a 2-group, contrary to hypotheses.
Thus, C (G I) must be faithful on each elation axis.This implies that (GI) _ GL(2,q).LEMMA 5.6.If G 1 is not a two-group, G/G 1 C(G I) is abelian.If H is not a 2-group (or trivial) G/H (H) is abelian.
-1 PROOF.If E G, the mapping G 1 % GII is an inner automorphism of G 1 if and only if belongs to G I C(GI).Hence G/G I C (G I) is isomorphic to a subgroup of the group of outer automorphisms of G I. By (3.4), G I is dihedral, a Suzuki group, or SL(2,2 s) for some s.
If G I is dihedral, the cyclic stem has odd order.Let G I <0,T> where O is an involution and <> is the cyclic stem.All of the involutions in G are equivalent under the inner automorphisms so the outer automorphisms fix 0 and induce automor- phisms of <T>.The automorphism group of a cyclic group is abelian, so the automor- phism group of G I is abelian in this case.
The outer automorphisms of SL(2,2 s) are essentially the field automorphlsms; this group is cyclic.
According to Suzuki [15], Theorem II, page 139, the same remark holds for the H is normal in G; if H is not a 2-group then H is dihedral or SL(2,2s) for some s since H is a subgroup of GL(2,q) generated by involutions.(Note that H G 1 must be trivial unless H and G 1 are both 2-groups.)Thus the arguments just used also apply to H. THEOREM 5.7.Except in the cases where H is trivial and G 1 is dihedral or where G has a normal 2-group, G 1 C (G I) contains the group generated by the involutions in G and if G 1 is not a Suzuki group G G 1 C(G I).PROOF.As in the proof of Lemma (5.6), G/G 1 C(G I) is isomorphic to a group of outer automorphisms of G I.
Thus the first part of the Theorem is true unless G contains some involution not in G 1 C (G I) which induces an outer automorphism of G I.
If G 1 Sz(2s) then s must be odd.The Theorem of Suzuki referred to before implies that the outer automorphism group has odd order.
If G 1 SL(2,2s) then the net whose components are elation axes is Desarguesian.This net is invariant under G.
By Ostrom [Ii ], the induced permutation group on the affine elation centers is isomorphic to a subgroup of PFL(2,2s) for some s.
If IGII and IC(GI) are both even we can apply (5.4) to infer that the elation axes are embedded in a regular (derivable) net invariant under G. (We do not need the condition that all components of this derivable net are components of the spread which defines the plane.)Again the induced group on the affine elation centers is a subgroup of PFL (2,2s) An element of G fixes all of the elation centers if and ), GF(2s) must be a subfield of GF(q2).If 2 s the plane must be Desarguesian.See Prohaska [14].If the plane is not Desarguesian, GF(2s) will be a proper subfield of GF(q).This will also he the case where H is non- trivial and G 1 is dihedral by (5.4).Since we are looking at linear transformations over GF(q), it follows that an element of G which fixes three elation centers must fix all of them and thus be in C (GI).Hence is in PGL(2,2s) PSL(2,2s), not merely in P FL (2,2s)      PROOF.By (5.7) G G IC (G I) G 1 C(G I) where G 1 is, in this case a Suzuki group.If G contains Baer involutions, then C (GI) will contain Baer involutions.
This implies that G 1 induces a Suzuki group on a Desarguesian plane of order q; that is, on a Baer subplane pointwise fixed by some element of C (GI).
LEMMA 5.9.Suppose that has a derivable net 71.Let 7 2 be the net whose components are the Baer subplanes in 7 I" (I and 7 2 correspond to a regulus and its opposite regulus.)Then G cannot contain 2-group Q1 and Q2 with the following properties (a) Q1 and Q2 centralize each other.
(b) I and 2 are invariant under QIQ2" Q1 consists of elations with a given axis in I; Q2 fixes pointwise some Baer subplane V(Q2) of l(line of 2 ). (c)

Q1
Q2 q0 > 2. Q1 has an invariant subplane of order q0 embedded in V(Q2).In its action on 72, Q2 has an invariant subplane of order q0 embedded in 4, where now is a Baer subplane of 7 2 We can choose a representation so that the components of 71 are x 0 and y xW, where W runs over the various scalar matrices over GF(q).Furthermore, we can choose a basis so that Q1 is represented by the various matrices of the form 1 o g 0 1 0 0 g GF(q0) and V(Q2) is the set of points (Xl, x2, YI' Y2 such that x 2 Y2 0. This representation for Q1 is unchanged under conjugation with respect to any matrix [ ] where A is 2 by 2 over GF(q).Hence Q2 can be represented by matrices of the form where varies over GF(q0).But the difference of distinct matrices in must be non-slngular.Thus if matrices in have identical first columns (second rows) they must be identical.Hence m I + 2m 3 + m 2 + m 4 for all in GF(q0).Thus [(m I + m4) + m3] 0. This 2 can only happen if m 3 0, m I 4.But if the q distinct matrices in have distinct second rows every ordered pair of elements of GF(q) must appear as the second row for some matrix in .In particular, there must be matrices in which m 3 belongs to GF(q0) but is not equal to zero.We have a contradiction which establishes the Lemma.REMARK.A similar proof works for odd characteristic.THEOREM 5. I0.If G 1 and H (see (5.2)) are both non-solvable, then G SL(2,q), G 2 SL(2,q2) where GF(ql) and GF(q2) are subfields of GF(q) such that GF(ql)N GF(q2)ffi GF (2).

CONCLUSION
We have tended to ignore subgroups of the translation complement which are both solvable and reducible or even solvable and irreducible but imprimitlve.We have TRANSLATION PLANES OF DIMENSION TWO AND CHARACTERISTIC TWO 57 generally assumed that there is no normal 2-group.Here the subspace polntwlse fixed by the normal 2-group is invariant.
In all cases the induced group (the factor group modulo the subgroup fixing a minimal invarlant subspace polntwlse) is cyclic or dihedral in the solvable case.
Thus the group theoretic structure is relatively uncomplicated.Yet "most" of the known planes of even order and dimension two are in this class.This includes most semi-field planes, generalized Hall planes and generalized Andr planes.We would conjecture that there are many more planes with an invariant component or Baer sub- plane, and we will not attempt to enumerate all of the possibilities.
Our investigations have led to two interesting possibilities for which we do not know whether or not examples exist.The first is described in (4.4); the second in (5.10).
TRANSLATION PLANES OF DIMENSION TWO AND CHARACTERISTIC TWO 55 If should have a normal 2-group, so would G. Recall that G 1