675
I nternat. J. Math. & Math. Sci.
No. 4 (1980) 675-694
Vol.
TRANSLATION PLANES OF EVEN ORDER
IN WHICH THE DIMENSION HAS ONLY ONE ODD FACTOR
T. G. OSTROM
Department of Pure and Applied Mathematics
Washington State University
99164 U.S.A.
Pullman, Washington
(Received January 2, 1980)
Let
ABSTRACT.
G
be an irreducible subgroup of the linear translation complement
of a finite translation plane of order
in the kernel and
must divide
s
2 r
d
d
(qd +i)
H
where
q
d
q
where
is an odd prime.
r
is a power of 2.
GF(q)
A prime factor of
is
GI
(qi _i).
i=l
One possibility (there are no known examples) is that
group
W
has a normal sub-
which is a w-group for some prime w.
The maximal normal subgroup
i.
G
Cyclic.
elements in
2.
0(G)
0(G)
satisfies one of the following:
Normal cyclic subgroup of index
have order
KEY WORDS AND PHRASES.
r.
3.
0(G)
r
and the nonfixed-point-free
contains a group
W
Translation planes, collineation groups,
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
as above.
finite geometries.
50D35, 05B25, 20B25.
T. G. OSTROM
676
order
3
r
If
2.
W
W
r
3, s
If
S/0(S)
w =3
exists then
2
W
is elementary abelian and of
is nonsolvable and contains no affine
G
and if
or
be the subgroup generated by the 2-elements.
S
elations, let
or
and
PSL(2,u)
for
u--5, Ii, 13, or 19
0(S)
and
Then
S =SL(2,2
b)
is trivial unless
u--5.
We include some corrections to our paper "Translation planes of odd order
and odd dimension".
1.
INTRODUCTION.
For definitions of "spread", "translation complement", and other special
[9].
terminology with respect to translation planes, see
The reader is reminded
that if the spread is defined on a vector space of dimension
components of the spread have dimension
d
so that the
then it is customary to say that the
d (over its kernel).
translation plane has dimension
2d
The class of known finite
translation planes is very large but they tend to fall into a few classes and the
members of each class are very much like each other.
The class of linear groups
which are known to act as subgroups of the translation complement is very limited.
For translation planes of order
index which are subgroups of
Desarguesian planes).
p
r
(p is prime) many have subgroups of small
rL(2,p r) (the translation complement for
In a few cases, they are direct products of such groups.
In
This is the case in the irregular nearfield planes and some related to them.
the semifield planes of order
of order
pr
p
r
there are normal elementary abelian subgroups
(affine elations-or shears).
The Suzuki groups come in when
The only other cases which come to the
and
d =2.
and
SL(2,13).
The question arises:
2
SL(2,7)
Which groups can act as subgroups of linear
translation complements of translation planes?
a complete answer to this question.
author’s mind are
p
It may be unrealistic to expect
Nevertheless either there must be some more
translation planes around that are quite different from the known ones and that
677
TRANSLATION PLANES OF EVEN ORDER
admit different kinds of groups or there must be undiscovered restrictions on
translation complements.
Let
G
be a subgroup of the linear translation complement of a finite
translation plane.
A reasonable strategy would seem to be to identify certain
key normal subgroups (and how they act on the plane):
solvable subgroups if
IGI
is even.
G
the minimal normal non-
is nonsolvable, the group generated by 2-elements if
Group theoretic arguments often give us factor groups of these
subgroups with respect to normal subgroups of
G
which are maximal in some sense.
To understand the action of the minimal normal nonsolvable subgroup or the sub-
group generated by the 2-elements, we must determine what the subgroups at this
second stage are like.
For instance, we may need to know something about the
maximal normal subgroup of odd order.
In general, what happens is one of the following:
(2)
normal subgroup is cyclic
elementary abelian group
(3)
G
G
(I)
The
"Second-level"
is imprimitive and has a normal noncycllc
has a normal extra-special group.
Theorems that certain classes of groups cannot act on any finite translation
plane are hard to come by.
read:
In practice we shall have to reword our question to
For a given class of values of
translation plane of dimension
d
over
q
d, which groups can act on a
and of
GF(q)?
a long way towards answering this question when
(For instance, we can go quite
and
q
d
are both odd.)
we end up with is a set of theorems which screen out certain groups.
What
We should
like then to have results which say that the survivors, if they do act on certain
planes, must act in specified ways which are well enough defined for us to
specify explicit representations which may enable us to construct the planes or,
perhaps, to show that a plane admitting a particular group acting in a particular
way cannot, in fact, exist.
In this paper we investigate the case where
q
is a power of
2,
G
is
T. G. OSTROM
678
irreducible, and
d
=2Sr
r.
for some odd prime
We do not claim to have any spectacular results.
We believe that we use
methods which can be developed and used to investigate other cases. We do have
results as to how the groups act on the (possible) planes.
We need more knowledge
of representations to carry out the last phase of the program.
Indeed the
representation theory we need is probably quite different from that included in
standard representation theory.
Section II deals with the nonsolvable case without restriction on the
Much of it is just a summary of previous results except that the
dimension.
connection with extra-special groups is clarified.
In Section III we do restrict the dimension (and characteristic of the
field) as mentioned above.
Our main result in this section is Theorem (3.7)
which gives the possibilities for the maximal normal subgroup of odd order
in terms of
0(G)
(I), (2), and (3) above.
In Section IV we look at planes still more explicitly, especially for the
case
r =3.
With respect to the general question as to which subgroups of
GL(2d,p) can be subgroups of the translation complement for a translation plane
of order
pd,
Theorem (4.8) says that certain primes cannot divide the order of
IGL(2d),P) I-
the translation complement even if they divide
[5]
some results of Herlng and Ho
GF(q), where
r
is odd and
q
for translation planes of dimension
is even.
S
Let
S/0(S).
about the possible nature of
exist admitting a group
S
With
0(S)
r
over
Herlng and Ho
restricted to be prime we can say more
and how
in Hering and
2r
be the subgroup of the linear
translation complement which is generated by the 2-elements.
determine the nature of
We also make use of
S
Ho’s
acts on the plane if a plane does
list.
We have not been able to
construct new planes furnishing examples but we can narrow down the possibilities.
Some, but not all, of the other papers in the same spirit and the cases with
679
TRANSLATION PLANES OF EVEN ORDER
odd order, odd dimension
which they deal are:
[7];
istic 2 with all involutions Baer
SL(m,q)
admitting
[15], [14], [8].
[12];
dimension 2 and character-
[Ii];
odd order, dimension 2
GF(q)
kernel
We hope that these results are explicit
enough to suggest to someone either how other types of planes may be constructed
or how to obtain stronger nonexistence theorems.
This research was supported in part by the National Science Foundation.
The rest of this Introduction consists of preliminary definitions and
results, as well as some matters of notation.
(i.I)
DEFINITION.
on a vector space
Let
V.
G
Let
be a group of nonsingular linear transformations
G
V.
G
minimal
subspace
G-space if
V
I
G, V
With
G
leaves
is a homogeneous
of the minimal G-spaces in
V
G
if
properly contained in
(1.3) DEFINITION.
G
Then a normal subgroup
G
non-f.p.f, with respect to
G
(f.p.f.) if no
is said to be fixed-point-free
fixes any nonzero vector.
acting on a vector space
of
G
Then
nontrivial member of
(1.2) DEFINITION.
be a group of nonsingular linear transformation acting
G
1
O
O
is said to be minimal
is not f.p.f, but every normal subgroup
is f.p.f.
O
V
as (1.2) a subspace
V
0
0
of
V
is called a
invariant and acts irreducibly on it.
G-space if
V
I
is invariant under
are isomorphic as G-modules.
doubt as to which group is being referred to, the
G
G
A
and all
(When there is no
may be omitted.)
REMARK. An f.p.f, group is a Frobenius complement.
It is well known that
the Sylow subgroups of odd order in a Frobenius complement are cyclic and the
Sylow 2-groups are kit.her cyclic or generalized quaternion.
Notation:
We shall be looking at vector spaces over the field
power of the prime
p.
GF(q)
where
q
is a
T. G. OSTROM
680
The order of the group
0(G)
We use
H
If
(H)
by
G
o
If
G
If
GI.
is denoted by
to denote the maximal normal subgroup of odd order.
GI,
is a subgroup of
G.
group
G
1
(H)
(2.1)
G
will be denoted
1
1
V, V(o) denotes
is a linear transformation acting on a vector space
G
.
G.
is the Fitting subgroup of
LEMMA.
Let
G
be a nonsolvable group of linear transformations acting
on a finite dimensional vector space
V
G
of
included in
Z(G0)
(2.2)
LEMMA.
G
Z(G0).
is in
G
O
PROOF.
(2.3)
LEMMA.
G
O
Then
G0/H
O
is equal to
H=Fit G
See the proof of Theorem (2.3) in
be a normal subgroup
be maximal with respect
O
=Z(G0).
[12].
In the notation of the previous Lemmas, suppose that Fit
Then
G
O
includes a subgroup
G, where
is a maximal normal subgroup of
W0
H
and let
O
H
is a direct product of isomorphic simple groups.
non-f.p.f, group with respect to
0
G
but not equal to
is fixed point free then
not fixed point free.
then
G
[12].
In the notation of the previous Lemma, let
included in
Fit
W
Then every normal cyclic subgroup
The Fitting subgroup of
See Lemmas (2.1) and (2.2) in
to this property.
If
O
and let
if it is fixed point free.
PROOF.
of
G
F--GF(q)
over a finite field
be a minimal nonsolvable normal subgroup.
O
If
in
in the full
THE NONSOLVABLE CASE.
2.
G
H
is a subgroup, the centralizer of
the subspace consisting of all fixed points of
Fit
H
is the centralizer of
_c Z(G0)
and
W/W0
G
W
W
G
O
which is minimal
is a w-group for some prime
included in
is elementary abelian.
is
W
but not equal to
w.
W,
681
TRANSLATION PLANES OF EVEN ORDER
(2.4)
LEMMA.
0
If W is nonabelian
W
0
w
W
is a fixed-point-free w-group.
0
We already have that
must be cyclic.
If
W
Then
0
is
Z(W).
Note that
PROOF.
of the previous Lemma is nonabelian.
W
Suppose that
cyclic and equal to
W
[12].
See Lemma (2.8) in
PROOF.
Z(W)c_ W
is even, then W
W
0
! Z(G0)
W
must be included in
Thus if
by (2.1)
so
so W --Z(W)
if
0
0
W
0
c__ Z(W).
w
is odd.
If
is either cyclic or generalized quaternion.
0
is cyclic we can use the same argument that we used for odd
W
generalized quaternion, then
group of automorphisms on
is odd,
w
W
0
i Z(G0)
0
G
and
w.
If
W
is
0
must induce a nonsolvable
O
contrary to the fact that the automorphism group
of a generalized quaternion group is nonsolvable.
(2.5)
LEMMA.
W
If
W/W
(a)
(b)
is a symplectic space;
W
(c)
3
and
PROOF.
(2.6)
W
W.I
G
by
w
Suppose that
W
abelian of order
w
0,
1
2
W
w
WI, W2,... Wm,
be a subgroup of order
Let
W =<
0
>
Since
W
1
Let
W
G
[6].
w
can, without loss of generality, assume
0
W
where
W =< 0,>
W
is nonabelian but
-i
IWI
=w
2m+l
w-1
=v(o)
G0 =o.
v(o)
m
w
1
is elementary
1
let
0,
be pre-
Without loss of generality,
is elementary abelian we
Then
oi
pointwise and
v(o) + v() +...+ v()
Then
is irreducible.
< 0,o, >.
1
where
2
is nonabelian and that
respectively so that
is nontrivial.
2m+l
is a subgroup of Sp(2m,w).
is elementary abelian of order
Let
PROOF.
V()
IWI
if
w.
has order
0
is a central product of subgroups
divides the dimension of the vector space if
images of
W
This is essentially Satz 13.7 in Huppert
LEMMA.
Furthermore:
is extra-special.
has square order;
0
the automorphism group induced on
..IWil =w
W
is nonabelian then
...e v(o)w-i
fixes
i
T.G. OSTROM
682
Thus we have a vector space of dimension
W
V
Call this vector space
1
each of
V() ,V()0, etc. invariant.
subspace of dimension divisible by
WIW2
W
Now
I.
dim V()
w
which is invariant under
W
centralizes
2
V(),
In fact, each of
w
multiple of
W
m.
If
In
[12]
w
REMARK.
W
which is invariant under
w
Thus
I.
2
has invariant subspace whose dimension is a
=WIW2...Wm
G
leaves
etc. contains a
has invariant subspace whose dimension is a multiple of
By induction
ence
and
m
V, then w divides the dimension V.
is irreducible on
we obtained results similar to (2.5) and (2.6) by a more
complicated argument.
In Lemmas (2.1)- (2.6) we assumed that
G
was nonsolvable.
requires no assumption as to whether or not
G
is solvable.
(2.7)
Let
LEMMA.
on a vector space
G
V
normal subgroup of
be an irreducible group of linear transformations acting
of dimension
’G.
V=V
1
[4]
V
LEMMA.
w-group
()
G
in
and C (
G
Then
is prime.
V I, V 2,.. such that each
Each homogeneous
are isomorphic as W-modules.
if
V
as a direct sum
leaves
V
V
and the stabil-
I
invariant.
1
divides
dim V
By Hering
I.
is irreducible and has a normal abelian non-f.p.f.
G
is imprimitive with subspaces of
Vi
pointwise fixed by some nontrivial subgroup
PROOF.
n
divides
n--k dim V
where
in the stabilizer of
Suppose that
W, where w
imprimitivity
k
has index
the index of
(2.8)
M-spaces,
of homogeneous
k
G
be a cyclic
1-spaces.
By Clifford’s Theorem, we can write
...@ V
izer of
in
Let
F.
over the field
n
Then the index of C ()
is faithful on the homogeneous
PROOF.
The next Lemma
is a homogeneous
W(V i)
of
W-space and is
W.
W-space is a direct sum of minimal W-spaces that
If
W
is elementary abelian
is not f.p.f, on any of its invariant subspace.
If
W
(not cyclic) then
and
V(o) N V
1
is
W
TRANSLATION PLANES OF EVEN ORDER
nontrivial then
V()
W
is invariant under
and includes a minimal W-space.
Thus @ fixes pointwise all minimal W-spaces in V
and hence fixes V
[12].
Here we would like to make certain corrections to
(2.9) in
[12],
normal in
G
where
W
pointwise.
1
In the proof of
was a non-f.p.f, w-group
was the maximal fixed point free normal subgroup of
0
We wished to show that, under the hypotheses,
We disposed of the case where the subgroup of
trivial.
W
W
and
W.
included in
G/C(W) by
we denoted
683
G
O
.
W
had to be
0
which centralizes
was nonsolvable but did not include the argument where this group might be
solvable.
i
Let
0
be the subgroup of
0/I
then
which centralizes
0
must be nonsolvable since
I
If
must be nonsolvable.
is
GL(2,w).
hence to a nonsolvabie subgroup of
G/Z(G)
and
By hypotheses, the Sylow 2-groups
0/HI
are dihedral and this implies that the Sylow 2-groups of
GL(2,w) do
But the nonsolvable subgroups of
are cyclic or dihedral.
This disposes of the case where
cyclic or dihedral Sylow 2-groups.
solvable,
0/i
Then
is isomorphic to a nonsolvable subgroup of the automorphism group of
of
G
not have
1
is
solvable.
LEMMA (3.3) and Theorem (3.4) of
appear to be correct.
THEOREM.
where
and
q
complement.
d
a translation plane of order
Let
are odd.
Suppose that
G
either
G
Let
0--SL(2,u)
G
O
G
q
d
with kernel
GF(q),
be a subgroup of the linear translation
is irreducible and is faithful on its subspaces
of imprimitivity or is primitive.
distinct primes.
should be deleted-the proofs do not
The statement of (3.5) should be altered to read:
H be
Let
[12]
Suppose also that d is the product of
be a minimal nonsolvable normal subgroup of
for some odd
u
or
G
O
A6 or A
7.
Here
G
O
G.
Then
G0/Z(G0).
T. G. OSTROM
684
3.
ORDER;
GROUPS OF ODD
In this section
space
V
(3.1)
Definition.
DIMENSION HAS ONLY ONE ODD PRIME FACTOR.
G
is again assumed to be acting irreducibly on a vector
of dimension
over
n
GF(q) =F.
The dimension will be assumed to be a power of 2 times an
odd number
Specifically the dimension
integer and
r
t
t>= I.
(3.2)
LEMMA.
then
C (I)
q =2
IGI
If
is odd and
G
has index in
I.
which divides
r
V
is an
where
G,
is faithful on
as a direct sum of homogeneous
must divide both
As in the proof of (2.7), the index of
r.
the stabilizer of a homogeneous
1-space must also divide
remainder of the proof is also similar to the proof of
is cyclic if
C ()
provided
W-spaces
The number of homogeneous
LEMMA.
F =GF(q)
Also
s
-spaces.
and hence must divide
(3.3)
where
is a maximal cyclic normal subgroup of
Use Clifford’s theorem to write
PROOF.
s
2 r
is equal to
is an odd prime or is equal to
the homogeneous
1-spaces.
n
IG]
Suppose that
G
is f.p.f.
G
f.p.f, elements in
G
is odd and that
If Fit G
have order
is f.p.f, but
(n,
n
GI
and
C(G)
GI).
in
The
(2.7).
is irreducible.
G
G
Then
is not, then the non-
r
and Fit G
H
is fixed point free and thus is a
is a cyclic normal subgroup of
r.
index
PROOF.
Let
H =Fit G.
Frobenius complement.
cyclic.
Let
H
The Sylow subgroups of odd order in a Frobenius comple-
H
ment are cyclic and
is the direct product of its Sylow subgroups so
C(H).
I
Then
Then Fit H =H.
I
contradiction so
index in
G
H
I
Z(H I)
if
is cyclic and
which divides
r
H
This would
were not cyclic.
H =H.
I
is
The Fitting subgroup of a solvable
nonabelian group is strictly greater than its center.
is strictly greater than
H
imply, that
This would be a
Hence (3.2) implies that
H
has
H
TRANSLATION PLANES OF EVEN ORDER
G
Thus if
order
is non-f.p.f., the non-f.p.f, elements in
G
If
r.
685
IGI
f.p.f, and
is
G
is odd, then
G
must have
is a Z-group-i.e., all
of its Sylow subgroups are cyclic.
G
Thus,
<
>
is generated by two elements
G
is normal in
be an element of order
G
is cyclic.
Passman
centralize
(3.4)
if
LEMMA.
If
G
then
prime
c_
>
IGI
<0 >.
<>
< 0>
W
IWI
(a)
Either
=r
then
<>
p. 106.
< G
in this case.
But
We conclude that
G
is odd, if
3
eassman [13],
Let
T
G=<T, H>
and
so
By the argument on the top of p. 197 in
G
W, where
w=r, W/Z(W)
G
is elementary abelian and
< >
must also
is cyclic if
G
is irreducible and Fit
has a minimal non- f.p.f, group
w.
or (b)
<
See
<T> c < o>
c__
<T>
must centralize
O
II01"
If
r.
Otherwise
<T>
GI
and
of relatively prime order,
and
O
W
G
is f.p.f.
is not f.p.f.,
is a w-group for some
is elementary abelian of order
is imprimitive with
r
homogeneous
r
W-spaces as subspaces of imprimitivity..
PROOF.
Let
W
W
if
G
is irreducible.
W
Note that
is nonabelian.
C (W0)
If
C(W0)
C (W0)
W
wm.
Then
W
in
G
C (W0)
C(W0)
Note that if
must divide
then
is a minimal non -f.p.f.
W
q
r.
in
G
If
IWI
divides
m +I
=w
and is not f.p.f.
0)
W= C(W
W, i.e.
If
W
0
included in
is a power of 2 and
W
is faithful on the homogeneous
C(W0) < G
f.p.f, with respect to
must be odd if
w
follows from (2.7) that the index of
index of
G
be the maximal f.p.f, normal subgroup of
0
W
0
2Sr
so the
then
is min. non-
! Z(W).
is not f.p.f., then it is equal to
W, since
W
In this case we again have
G.
group with respect to
spaces, it
w0 ! z(w).
If
C(W0)
W
is f.p.f., we must have
is faithful on the minimal
W
0
(W0) Q W has index
r
W0-spaces
in
W.
C(W0)
W =W
0.
Then
C(W0) Q W
so we can apply (2.7) to conclude that
Suppose that
W
0
is cyclic of order
b
w
2
T. G. OSTROM
686
.
generated by
i
=a
o
a
mod w
a=l.
and
and an element
r =b
= -w
and
#i
b
a
We must have
w
w
a
so
W
That is,
w
a
in
b
-=I mod w
such that
This implies
W
must centralize
W
0
and we again have
w0 ! z (w).
The rest of the proof is essentially the same as for the nonsolvable case.
See (2.3)-(2.6) and (2.8).
(3.5)
COROLLARY.
If the
is a power of 2 and
n
irreducible then
G
is cyclic.
PROOF.
r
i in (3.3) and
G
then
(3.6)
Put
(3.4).
If
IGI
G
is odd and
is
in conclusion (b) of (3.4)
r
i
G
generated by its 2-elements
is reducible, contrary to hypothesis.
Let
DEFINITION.
0(S), 0(G)
and let
S
be the subgroup of
the maximal normal subgroups of odd order in
S, G
respectively.
(3.7)
THEOREM.
Let
be an irreducible group of linear transformations acting
G
2Sr
on a vector space of dimension
r
either
(a)
in
G
1
If
divides
(b)
If
or
is an odd prime.
r
0(S)
unless C
[0(S)]
order
r
is a power of 2 and
q
C [0(S)]
is cyclic and the index of
0(S)-spaces.
is not faithful on the minimal
is f.p.f, then
0(S)
has
and the non-f.p.f, elements in
0(S)
have
is not f.p.f, but Fit
a normal cyclic subgroup of index
GF(q), where
Then the following hold:
0(S)
is f.p.f., then
2Sr
0(S)
over
0(S)--Fit G
r.
(c)
If Fit 0(S)
is not f.p.f, then
G
has a normal subgroup
W
sat-
isfying the conditions of (3.4).
(d)
S
1
If
0(S)
S
is f.p.f, and
is nontrivial then
S
contains a subgroup
which is generated by its 2-elements and is a minimal (normal) non-f.p.f.
is nonsolvable then
SI/0(S)
direct product of isomorphic simple nonabelian groups and
0(S) S
subgroup with respect to
G.
If
S
1
S
is a
=Z(Sl).
687
TRANSLATION PLANES OF EVEN ORDER
0(S)
In each of the above
PROOF.
G
If
is irreducible then the dimension of a minimal invariant
2Sr,
subspace for any normal subgroup must divide the
Lemmas apply to
S
G
pointwise if
and
so that (a)- (c)
0(S)
so that the previous
are direct results of the
Note that a normal subgroup of
previous Lemmas.
0(G).
can be replaced by
G
does not fix any subspace
is irreducible.
In case (d) note that no group of even order is f.p.f, when the charF
acteristic of
to
G, then
Hence if
S
is a
non-f.p.f, group with respect
must be a minimal nonsolvable normal subgroup of
I
G so that (2.2)
TRANSLATION PLTqES
(4.1)
DEFINITION.
In this section
G
complement of a translation plane
and
minimal
The details are left to the reader.
applies.
4.
S
is 2.
2
d
s-I
is a subgroup of the linear translation
G
s >=2.
is an odd prime and
r, r
q
of order
d
where
is a power of 2
q
is irreducible and no component
has an orbit of length 2; no proper subplane has an
of the spread defining
orbit of length 2.
REMARK.
2Sr
2d
The dimension of the vector space on which
so that
(3.7) applies to
G.
G
is defined is
2
Translation planes of order
frequent possible exceptions to theorems, so the case
q =s =2, r =3
6
are
has special
interest.
(4.2)
LEMMA.
Then
W
then
V()
Suppose that
G
has a normal subgroup
has at least three invariant components;
if
W
W, where
W
is a 3-group.
is non-f.p.f.,
is a subplane; the partial spread consisting of those elements of the
spread fixed by
W
is invariant under
PROOF.
q
is a power of
If
of the spread is
d
q +i
least two components.
2, then
so that, if
If
W
G.
d
is normal,
q
2
-1 mod 3.
The number of components
is even, a 3-group in
G
G
must fix at
must permute the components fixed
T. G. OSTROM
688
by
W.
is
lft
By (4.1),
W
LEMMA.
(3.4)
for some prime
If
r
3
w
then
W
has a normal nonabelian group w-group
G
and
w=3
and
W
as in
satisfies the conclusions of (4.2).
See (3.4)and (4.2).
PROOF.
W
The rest of the proof
to the reader.
(4.3)
LEMMA.
(4.4)
has at least three fixed components.
If
2
s =q
in (4. i)
and if
is an elementary abelian w-group of order
G
W
has a normal subgroup
> w, then
w
where
3 or 5 and the
dimension of a homogeneous W-space is 2 or 4.
PROOF.
minimal W-space has dimension 1,2,4,r,2r, or 4r.
W-space pointwise, then
has dimension
divides
2
4
i, then
G
w
If
could not be reducible.
must divide
4r, so a
is defined has dimension
The vector space on which
2 i- i.
Thus
W
should.fix a minimal
Thus if a minimal W-space
i
1
-I =3.5 if’i =2 or 4.
If a minimal W-space has dimension which is a multiple of
r
then a
homogeneous W-space must also have dimension which is a multiple of
W
w
is impossible and
is not faithful on homogeneous
W-spaces the case 4r
r.
Since
does not happen (the
whole vector has dimension 4r).
A subplane in this context is a vector space of even dimension.
Thus a
homogeneous W-space is either an r-dimensional subspace of a component or has
dimension 2r.
In the latter case there will be just two homogeneous W-spaces,
contrary to the assumptions in (4.1) that no component or subplane has an orbit
of length two.
We are left with the possibility that
W-spaces and that
they
the.re
are precisely 4 homogeneous
are r-dimensional subspaces of four distinct lines.
Furthermore, Clifford’s theorem implies that every minimal W-space is in one of
these 4 homogeneous W-spaces.
689
TRANSLATION PLANES OF EVEN ORDER
o
Let
be any element of
must be a Baer subplane.
of the
r
2 +i
(2r+1)-4.
W
V(o)
W
are invariant under
2r-I
But also we must have
mod w.
and
must divide
w =2
This implies
The only possibilities left are
w
which is a
w=3 or 5.
We conclude that each homogeneous W-space has dimension 2 or 4.
must be faithful and hence
(4.5)
LEMMA.
w
is an elementary abelian
24-I
must divide
d =6 in (4.1) and if
If
V(o)
Then
is nontrivial.
If there are precisely 4 minimal W-spaces then four
components of
contradiction.
V(o)
such that
G
IWI
then
W
5.
3
W
has a normal subgroup
>w
w-group of order
But
=w
2
IWI
or
W
where
3
m
with
m-<4.
By (4.4), either w=3
PROOF.
or
w=5
w=5
and if
homogeneous W-spaces (each of dimension 4 in this case).
there are just 3
VI, V2,
Let
three homogeneous W-spaces whose direct sum is invariant under
of
W
which is non -f.p.f.
hence fix
V
be f.p.f.
Hence if
Suppose
w
m-2
m
3.
0(G)
IWI =wTM,
V
V
1
and
V
2
q .)
order
If
w=3
An element
0(G).
W-space pointwise and
m
w
m >i
cannot
m-I
pointwise.
V
pointwise has a subgroup of order
A similar phenomenon works for each
space; the result is that there is a subgroup of order
whose fixed point space has dimension greater than
d
be
3
is pointwise fixed by a subgroup of order w
l
Then the subgroup fixing
which fixes
minimal
must fix a minimal
1
elementary abelian group of order
pointwise.
1
V
on
V
This cannot happen so
m-2
w
(Recall that
d.
has
must be the identity.
and there are 6 homogeneous W-spaces, again take
reasoning to the above there is a subgroup of order
four homogeneous W-spaces pointwise.
m-2
w
m-4
w
IWI =wTM.
(if m >4)
By
which fixes
But the pointwise fixed subspace of a non-
trivial element cannot have dimension exceeding 6 if
d =6.
Hence
m-<_4.
T. G. OSTROM
690
LEMMA.
(4.6)
W
Suppose that
(3.4).
as described in
r
and
3
Let o
W
s
2
V(o)
and let
has a normal subgroup
G
and that
Then
be nontrivial.
is a subplane.
PROOF.
W
If
IWl
implies that
is nonabelian apply
=w
2.
W
O
V(o)
and
included in some component
6
4
4
-q =q (q
i or 2,
q
2
2
(4.5)
W
which does not fix a homogeneous W-space pointwise
-i) -0 mod w
q
nents of the
nents so that
REMARK.
2
sprehd.
V()
If
Hering and Ho
2
-I mod w.
has dimension 4.
Likewise if
V(o)
is
Then
has dimension
6
+i E2 mod w, so
W
W
fixes at least two compofixes at least three compo-
must be a subplane.
2
q +i
0 mod w
[5]
we cannot guarantee that any components of the
W.
have investigated the case where
s =2
and
r
is odd
In their Theorem (5.7), they have given information
They come up with the following cases when all involutions are Baer
involutions.
G; the
q
V(o)
As in the proof of (4.2),
but not necessarily prime.
S.
Suppose that
and
--i mod w, then q
spread are invariant under
about
.
V(o)
is nontrivial but not a subplane then
--I mod w.
But if
in
is abelian then
-I --0 rood w.
If
q
W
The dimension of a homogeneous W-space must divide 4, so
must be f.p.f, on it.
4
If
Furthermore, the number of homogeneous W-spaces is equal
to 3 and each element of
q
(4.3).
They also include the situation where there are affine elations
nature of the group generated by affine elations is known from
earlier work
[3].
We shall leave this case out, although there is more informa-
tion than what is implied immediately by
no assumptions of
Hering’s
irreducibility; we
take if we assume (4.1).
Hering’s work on elations.
They make
state these results in the form that they
691
TRANSLATION PLANES OF EVEN ORDER
The Hering and Ho cases (with the restrictions mentioned above):
I.
S/0(S)
2.
S/0(S)
3.
ISI
4.
(S
is elementary abelian.
5.
S
SL(2,2 b)
4; C
is elementary abelian of order
u-
PSL(2,u)
[0(S)] O S
Z[0(S)].
+3 nod 8.
is not divisible by 4.
for some
is irreducible.)
b.
We would know more about the action of
if we knew more about 0(S).
G
This case is void if
G
on the plane in the first three cases
Here (3.7) of this paper is some help.
These results are sharpened by (4.3), (4.5) and (4.6) if
(The order of the plane
6
q .)
is
s =2.
and
We also have the following:
(4.7)
LEMMA.
Assume (4.1) and that
where
u
=u
Then
+3 nod 8.
r =3
u =3,
S/0(S) =PSL(2,u),
Suppose that
r=3, s=2.
5, II, 13 or 19.
0(S)
Furthermore
is
trivial unless u =3 or 5.
PROOF.
If
It follows from Harris and Hering
u-+-3 nod 8, and
Note that
S/0(S)
is simple unless u =3.
0(S)
has order 2 so
0(S)
0(S)
W
SL(2,3)
Thus if
is f.p.f., then
u #3,
S
0(S)
Z(S)
is a minimal
by (3.7)(d).
0(S) is f.p.f.
Suppose that
0(S)
has a normal cyclic subgroup
S/ C(/)
S
Then
G; by (2.7)
is solvable.
C (/)
If
has index in
S/0(S)
G
which
is simple this implies
This is a contradiction.
0(S)
is nonabelian then
abelian of order 9.
+ 1 =25.
(3.7).
Hence
If
12
and
divides 12.
Suppose that
2
u #3
But / is normal in
S =S.
=<
must be trivial if
of index 3.
C (/)
u
is a subgroup of the Schur multiplier; the Schur multiplier
satisfies case (b) of
that
0(S)
If
that
u=3, 5, II, or 13, or 19.
is a prime power then
u
normal nonsolvable subgroup.
Furthermore
[2]
Then
satisfies case (c) of (3.7) so that
(4.3) implies that
S
w =3
induces a subgroup of
is solvable, so the centralizer of
and
W/W0
W
of (3.4) exists.
is elementary
SL(2,3) on W/W0.
W/W0 must be nonsolvable.
But
T. G. OSTROM
692
S
Since
W/W0.
that
S
W/W0
if and only if it centralizes
so
S
in
S.
centralizes
W.
must centralize
W
If
G, this implies
must be a minimal nonsolvable normal subgroup of
The reader may verify that a 2-element centralizes
W.
But
S
is generated by its 2-elements
W
This is a contradiction since
(4.5) and (4.6) apply.
is elementary abelian then
has exactly 3 homogeneous subspaces, each of dimension 4.
be
V
V 2, V
I
The normal subgroup of
3.
must have index dividing
V I, V 2, V
3.
S
Then
6, since
H has
G
dimension at most 4.
Hence
u
(4.8)
THEOREM.
W
Furthermore
Let these subspaces
H, which fixes all three
G, call it
induces a transitive permutation group on
S/H
index dividing 6 so
is a nonsolvable normal subgroup of
is nonabelian and
G.
Hence
S
Applying Harris and Hering
c__
H.
[2]
S
Thus
V
fixes
of
I
4 + 1 =9.
u _-<2
again,
H0S
Hence
is solvable.
5.
Let
be a translation plane of order
subfield of the kernel.
lation complement with
d-i
d
Let
d
GF(q)
where
is
be an element of the intersection of the trans-
GL(2d,q)
such that
II
u.
is a prime
(qd-i_
q(q + i)
q
H
I). In particular if q is a prime
i=0
factor of the translation complement, then u divides
p
and
p(pd +i)
Then
divides
u
u is a prime
d-I
(pd-i-l)
H
i=0
and
u
PROOF.
dimension
We may think of
2d
GF(q).
over
z
If
of the spread.
V(o)
wise
Then
u
o
divides
If
o
fixes
for
q 2d_ I
does not divxde
u
o
In particular
Z pointwise, then
must act regularly on the
i
qd-i_l ).
But
u
d
q -q
i
is prime, so
establishes the first point of the theorem.
has fixed
then
fixes some component
qd-l.
must divide
u
is a vector space of some dimension
q
d <e <2d.
as being represented on a vector space of
points other than the zero vector.
so
pe-i
cannot be a p-primitive divisor of
i, where
points of
u
divides
Other-
0 < i < d.
not in
q
V(O),
(qd-i_ i).
The specialization to
q =p
This
is
693
TRANSLATION PLANES OF EVEN ORDER
[I]
It follows from
obvious.
u
pa-I
for
e
so that
e- 1
then
u
does not divide
b
p -i
then
b
is a multiple of
p
is a prime p-primitlve divisor of
that if
a < 0 < e
u
Furthermore if
does not divide
p2d
divides
u
1
if
d< e<2d.
(4.9)
COROLLARY.
PROOF.
If
q--2
in
(4.7), then u=3, 5, or 13.
The reader may verify that neither
(26 + I)
5
II nor 19 divides
(26_i_ I).
i=0
REFERENCES
n
i.
Birkhoff, G. D. and Vandiver, H. S. On the integral divisors of
Ann. Math. (second series) 5(1904) 173-180.
2.
Harris, M. E. and Hering, C. On the smallest degrees of the projective
representations of the groups PSL(n,q) Can. J. Math XXIII, (1971),
90-102.
3.
Hering, C.
a
b
n
On the shears of translation planes, Abh. Math. Soc. Hamb.
3__7(1972), 258-268.
4.
Hering, C. Zweifach transitive permutationsgruppen, in denen 2 die
maximale anzahl von fixpunkten von involutionen ist, Math. Z.
I04 (1968)
150-174.
5.
On free involutions in linear groups and
Hering, C. and Ho, C.Y.
collineation groups of translation planes, Fundac$o Universidade de
Brasilia, Brasilia, Brazil (1977).
6.
Huppert, B.
7.
Johnson, N. L. and Ostrom, T.G.
8.
Johnson, N. L. and Ostrom, T. G. The geometry of SL(2,q) in translation
planes of even order, Geom. Ded. 8(1979), 39-60.
9.
Ostrom, T. G. Finite translation planes, Lecture Notes
Endliche Gruppen I, Springer-Verlag, Berlin and New York (1967).
Translation planes of characteristic two
in which all involutions are Baer, J. Alg. 5__4(1978), 291-315.
in Mathematics
158,
Springer-Verla.g, Berlin and New York, 1970.
I0. Ostrom, T. G. Normal subgroups of collineation groups of finite translation
planes, Geom. Ded. 2(1974), 467-483.
II. Ostrom, T. G. Collineation groups whose order is prime to the characteristic,
Math Z. 156(1977), 59-71.
T. G. OSTROM
694
12.
Ostrom, T. G.
Translation planes of odd order and dimension,
Int. J. of Math and Math Sci. 2(1979), 187-208.
13.
Passman, D. S.
14.
Schaeffer, H.
operiert.
15.
Walker, M.
Permutation groups, W. A. Benjamin, New York, 1968.
Translationsebenen, auf denen die Gruppe
Dimplomarbeit, Universitt Tbingen 1975.
SL(2,p n)
On translation planes and their collineation groups.
Westfield College, University of London, 1973.
Thesis,