I nternat. J. Math. & Math. Sci.
(1978)447-458
Vol.
447
THE GEOMETRY OF GL(2,q) IN TRANSLATION PLANES OF EVEN ORDER q2
N. L. JOHNSON
Department of Mathematics
The University of lowa
lowa City, lowa 52242
U.S.A.
(Received August 5, 1977 and in revised form April 6, 1978)
ABSTRACT.
In this article we show the following:
tion plane of even order
group.
Then
u
q
2
GL(2,q)
that admits
Let
II
be a transla-
as a collineation
is either Desarguesian, Hall or Ott-Schaeffer.
KEY WORDS AND PHRASES.
Translation planes, Desarguesian, Hall, Ott-
Schaeffer planes.
AMS(MOS) SUBJECT CLASSIFICATION (1970) CODES.
i.
50D05, 05B25.
INTRODUCTION.
The author and Ostrom recently studied translation planes
order that admit
SL(2,2 s),
s>l.
of even
In [i0], we considered the case where
has dimension 2 over its kernel and in
_
II
[9]
no assumption was made concern-
ing the kernel.
Schaeffer
[13]
whose kernel
has shown that if a translation plane of even order
GF(q)
admits
SL(2,q),
q
2
then the plane is Desarguesian, Hall
448
N. L. JOHNSON
or 0tt-Schaeffer.
In [9], it is shown that if the assumption on the kernel is dropped, the
planes admitting
SL(2,q)
must have properties quite similar to the Desar-
guesian, Hall or 0tt-Schaeffer planes but it is an open question whether such
planes must, in fact, be in one of these three classes.
In the dimension 2 situation, if the plane admits
GL(2,2 s)
also admits
(see (2,2)).
Schaeffer planes of even order q
2
SL(2,2 s)
then the plane
So, in particular, the Hall and Ottmust admit GL(2,q)
as a collineation group.
In this paper we observe that this situation can be reversed (see Theorem
(.)).
We assume the reader is familiar with the papers [9] and [i0].
2.
THE MAIN THEOREM.
(2.1) NOTE.
GF(2 s)
is a translation plane of even order with kern
If
and admitting
SL(2,2s),
s >i,
then
T
admits
GL(2,2 s)
as a
collineation group.
GL(2,2 s) =SL(2,2 s)
PROOF.
SL(2,2 s) K
admits a group
versely, if T
then, by the obvious isomorphism,
Since the kern homology group
K
(2.1)),
(2.2) COROLLARY.
whose kernel contains
we have a group
If
w
GF(q)
2S-l,
GL(2,2s).
is isomorphic to
2s-i comnmtes with elements
and SL(2,2 s) is linear by [10] (see
SL(2,2 s) K (since SL(2,2 s) is simple).
of order
is a translation plane of even order
admits
Con-
where K is cyclic of order
SL(2,2 s) K
of the linear translation complement
the proof of
GL(2,2 s).
where $ is the center of
SL(2,2 s)
s>
I,
then
w
22r=q2
admits
GL(2,2s).
PROOF.
group
s
(of
lr
by [i0]
((2.2)
the group of kern
have the proof to
(2.2).
and
(2.4)).
Thus, there is a cyclic sub-
homologies) of order
2S-l.
By
(2.1),
we
449
GEOMETRY IN TRANSLATION PLANES
We also note that (2.2) is not valid for translation planes of odd order
(see
[4]).
Foulser
If a translation plane is of
dim 2
then the group
in the linear
SL(2,2 s)
translation complement generated by all Baer involutions is always
if there are no elations and the group is nonsolvable
(see [i0], (3.27)).
However, essentially nothing is known oncerning the group
assuming that the Baer subplanes are disjoint
(2.3) THEOREM.
Let
Q
or
Then either
Q
(3)
the suDgroup generated by the Baer involutions is
)
Suppose
Foulser
,
8.
(i)
PROOF.
in
be a translation plane of even order admitting a
be a sylow 2-subgroup of
is normal in
for
QX
(as subspaces).
whose sylow 2-subgroups fix Baer subplanes pointwise.
collineation group
Let
without
s > I.
Suppose neither (I) nor (2).
for
SL(2,2 s)
(QX) NQ
Consider
(normalizer of
xE.
gE(QX) NQ
([2], (2.5)),
g
so that
QXg=Qx
for
gQ.
Since
must centralize some involution
of
h
Igl=2
by
QX
Thus
(g,h>
is an elementary abelian 2-group and so is contained in a sylow 2-sub-
group
Q.
Qx,Q,
Let
gQNQ so
Then
respectively fix the Baer subplanes
g
fixes
i =w3" Thus,
(Theorem 2), <QX,Q> is a
the involutions of <QX,Q>
similarly
and
x
2 and 3
<QX,Q> fixes
pointwise.
wI
(QX) nQ
(i>
Thus,
pointwise.
pointwise.
w2=3
and
By Foulser [3]
subgroup of a one-dimensional affine group.
belong to the same sylow 2-subgroup.
(Q).
Thus,
l,W2,w 3
for
x-(Q).
Thus
So,
QX= Q
N. L. JOHNSON
4.50
.
We can thus apply Hering’s main theorem of [7].
closure of
Q
PSU(3,2s),
or
in
Then
SU(3,2s).
S
is isomorphic to
SL(2,2 s)
But only
S
GL(2,q)
SL(2,2s),
denote the normal
Sz(2S)
s>l,
.
([2], (2.5)).
is the group generated by the Baer involutions of
(2.h) THEOREM.
Ostrom [5].)
S
has elementary abelian sylow
2-subgroups which must be the case by Foulser
Clearly,
Let
(Special case of the
Let
main theorem of Foulser-Johnson,
be a translation plane of order
and where the
(pr=q)
p-elements
q
2
which admits
are elations.
w
Then
Des-
is
arguesian.
(2.5) THEOREM.
admits
GL(2,q)
Let
where the sylow
Then
pointwise.
PROOF.
be a translation plane of order
r
p =q,
.
is even then by Johnson and Ostrom
subplanes fixed pointwise fall into a derivable net
plane is Desarguesian by
and thus
(2.6) THEOREM.
admits
GL(2,q)
Let
(2.h) since
GL(2,q) must
[9], (h.1),
and
fix the net.
1
Let
G
under
Iwl
If
[2]
if
q# 3
(2.h) applies
u be a translation plane of even order
Then
q
q
2
which
that are
is an Ott-Schaeffer plane.
By [9], (.6), we have orbits on
q(q-1)
the Baer
Thus, the derived
where the sylow 2-groups fix subsets of order
contained in components.
PROOF.
which
fix Baer subplanes
is odd, the Baer subplanes fall into a derivable net by Foulser
(Theorem (5.1))
or 9
is a Hall plane.
II
If
p-subgroups,
q2# h
of lengths
q+l,
)2q_q_l_,l
SL(2,q).
be in the center of
GL(2,q).
Then
G
must fix each Baer sub-
plane which is fixed pointwise by an involution and must fix each line of the
orbit of length
The
SL(2,q)
q-1
q+l.
Baer subplanes fixed pointwise by elements of a sylow 2-group of
cover the components other than the components in the orbit of length
Let
q+l.
ilPi
ment of order
G.
be the prime decomposition of
.
Ien, since
. .
Thus,
2/kern
G’I
be an ele-
indicated
n’.
G.
G
u
so that
q+l,
and so fixes each compo-
fixes each component of
i=l
dim
and let
fixes each Baer subplane
fixes a component from each Baer subplane
nent of w.
q-i
shares precisely one component with the orbit of length
w.
above and
_
451
GEOMETRY IN TRANSLATION PLANES
Therefore,
has
is an 0tt-Schaeffer plane by Schaeffer.
We can now state our main theorem.
(2.7) THEOREM.
GL(2,q)
admits
Let
be a translation plane of even order
q
2
which
Then the fixed point space of each
as a collineation group.
Sylow 2-subgroup is a component, Baer subplane, or Baer subline and
(i)
is Desarguesian if and only if the Sylow 2-subgroups fix corn-
ponents pointwise,
(ii)
w
is Hall if and only if the Sylow 2-subgroups fix Baer subplanes
pointwise,
(iii)
is Ott-Schaeffer if and only if the Sylow 2-subgroups fix Baer
sublines (sublines of order
q)
pointwise.
By (2.3), (2.4), (2.5), and (2.6), it remains to show that the
PROOF.
fixed point spaces are as asserted.
are Baer.
We can thus assume that the involutions
We can assume that a Sylow 2-subgroup
Q
does not pointwise fix
Let
Q
fix
a component, Baer suplane, or Baer subline.
Let
denote the center of
=(o>. en,
o
a line fixed by
X
GL(2,q).
PEX,
if
then
Q
fixes
GIPEX.
mq
If
X
intersects, nontrivially,
points of a component.
does not nontrivially intersect any component fixed by
2
s
If
cannot lie on a component of
X
Obviously then
of order
pointwise.
The center is fixed-point-free.
Case i:
Let
X
p
Q
and fixes the Baer subplane
So, assume
G.
and is thus a subplane
p
pointwise then
X
N. L. JOHNSON
452
is a subplane of
up
Thus
X
be a component of
l(Gi>l<2s-1
length
q-1 <
2
(Gi>
and
fixes
s+l since
(s2-1) (2S+l) 2s-1 K q-i
< 2
is
Q
maximal subgroup of
II =2(,
groups then
But,
(,x)
<
X
and
so that
QIu
2)),
j>
That is, if
=--SL(2,2-)
SL(2,q)
Let the normalizer of
so there is a subgroup
Q
SL(2,a)
in
of
since
is transitive on the points of
fixed by
Q,
such that
(G)
Gg
fixes a point of
X
+
i
Q.
-1
X
Suppose
(_.,.>
(Gg>
Q
is a component not in
X
pointwise.
Hence,
So there are
and fixes
u.
and
X
.
q+l)(q-)
If
.
[9](2.1)
by
u
that fixes
u
pointwise
Moreover,
permutes the points
g
then there is an element
That is,
fixes
II >
fixes
and
be
distinct subgroups
Q-<i>.)
be the
I,I>, 2>
Q
X
of
pointwise.
n.
Qi
of
pointwise and also fixing pointwise a Baer subplane
the involutions of
Let
SL(2,q)
or
and thus fixes
Let the fixed point subplane of
Now there are
Then
Since
be
of order
we have that if
<
By [9](2.9), we know that
q.
or
and the orbit
are in distinct Sylow 2-sub-
->l.
for some
2-=Jq
so
x
and
Thus,
So,
components
has order
SL(2,)
Let
has
.
_ .
X
pointwise.
(,x)
2,
2 s+l
has
(G)
in
i
-orbit of
has order
u
which fixes
If
x6 GL(2,q).
for some
X
.
by assumption ).
X#up
the stabilizer of
+i.
By Foulser ([3] (Theorem
then
(since
@ is fixed-point-free and the
since
length of i
q=4
s
Q
u..
1
(iiQi
-\ (I>
common to
u.
each fixing
(
X
is regular on
(i>.
and
..
Then
pointwise and thus fixes a Baer subplane
share no components outside of
components that are permuted by
X.
(G>.
453
GEOMETRY IN TRANSLATION PLANES
u.
X
and also leaves
Since
fixes each subplane
and
u.-X
1
thus permutes the same set of components.
(Gg>
So,
-
X=Y
QX
X
then fixes
Q
fixes
X
pointwise,
(By [9] (h.3),
each of the
Y
maps
components and permutes the remaining set
2-primitive.
for otherwise
There are
k*
where
components.
k
of these
1
q (q-l)
is odd
(the
must fix one of these
k
Unless,
T
=2,
planes of order
Thus,
X
and
<,x>
p
components).
wT
pointwise.
T
k
16
in
and
Y
[ll]
Let
components of
so
k
none of which can fix any
fixes a
of length
1 (q-1)k*
q
T
component
which
of
Clearly, the Sylow 2-subgroup
must have a subgroup of order
we have a contradiction.
j+l
components.
contains an involution
So
Thus,
cannot fix a common
So there is an orbit under
p
components.
i and a Sylow 2-subgroup
,
conjugates of
Thus,
fixes each of
2
q-4
(x>
fix
other Sylow
acts on these sets
+l
of
q
2-subgroups.)
SL(2,q)
fixes
Let
normalizer of
and fixes a Baer subplane
containing
(p)
Then
fixes a Baer subplane pointwise.
is odd.
-
as it acts on the Sylow
fixes each of these components.
,
GL(2,q)
it follows that
components.
other pairwise disjoint
q
subplaues is on the same set of
q+l
is
onto
4+1
which are pointwise fixed by the
of subplanes of order
component
[9] (h.3).
pointwise, contrary to
Q
(QX,Q>
But,
Y.
and
share a component then they share
2-subgroups.
IPl
X
and fixes
X NY#(} then
If
pointwise.
Y
subplanes of order
p6 9
4
and
X
(Q,QX>
q2+l-((+l)(q-) ++l))
of
of order
X-O
is regular on
SL(2,q)
Since
Y
fix a subplane
since
If
fixes
q(-l).
components
Let
permutes the remaining set
invariant,
fixing
uT
pointwise.
However, we have considered
and this possibility does not occur.
do not share a component.
Thus,
(Gg>
fixes
Y
2
N. L. JOHNSON
454
(
X
must fix
Y
X
plane
w
fixing
of
<g>
I < -i.)
l<g>
There exist elements
Baer subplanes
(<og>
We assert that
Y.
+i.
faithfully on
tain
of
Y -n.
@
Let
T
X
and
n
and
Y.
Y.
XNY
Since
of order
be the prime
(og>
p.
in
Each
n
0i
which fix
ms fixed point-
k
@.
So
@.@
g
<g>
fixes
Y-[0],
-i
<g>
fix a component
and thus cannot act
Y
u
Thus
Y
and
can fix
component.
share a
(X
=UT
and acts faith-
Y
fixes a component of
O,
pointwise.
fixes
no element of
X Y.
Let
n pointwise.
fixes
i=l I
u and Y must
That is,
So
(by uniqueness of subgroups of
be an involution that maps
uT
both con-
r-dim
subspaces
and
are
nT).
(o-lg)
So
p
of order
If not, then
Q __QX
<> =
2
g 6
fixes
,
2(q-l)
g
2
i
=Dq_ I.
and thus
pointwise.
<,>
Thus,
Thus
pointwise, then
also fixes
Since
X--Y.
since
and is thus dihedral of order
pointwise.
-i
pointwise.
Thus,
<g>
must
fixes pointwise a Baer sub-
Since the elements of order dividing
Note that
Since
@.
is regular on
we must have that
Y,
of
Since
Y.
a point of
w@ =.
Thus,
<g>
Let
n@ i
+i
of order
be the element of order
fully on
Y)
fixes
wise by a subgroup
order).
.*
and fixes
I-i
there is a subgroup
(That is,
decomposition.
of a given
QX)
and
of prime power order
[3] (Thm. 2)
pointwise.
so
Q
normalizes
if
and thus fix a Baer subplane pointwise
and by Foulser
+I
Y
and
So the elements in
pointwise.
fix a component of
>
X
fixes two subplanes
g
Let the
GT=G-I
(-ig)(g)=g2
X i,Xj
and
[Q,QX]
(Gg)T" =G -i g.
fixes
i.
q+l
subplanes of order
fixed pointwise by the Sylow 2-subgroups be denoted by
be the Baer subplane containing
stabilizes
XI,X2,..-
,Xq+ I.
Let
and fixed pointwise by the
.
cyclic subgroup
GEOMETRY IN TRANSLATION PLANES
SL(2,q)
of
l,j
(Recall
455
SL(2,q)
that
[Xi].
on its Sylow 2-subgroups and thus on the subplanes
w contains precisely X and Y,
Note that
contain all
# X
planes
subplanes of order
q+l
or
X
I
PROOF.
subplane
for otherwise
is regular on remaining sub-
XI,X2
and
J 2,
XI,Xj’
share only the components
If the subplanes share a component
(smallest subplane properly
J #2’
XI ,X2=WXI ,j
X
w would
Y.
We assert that
of
as
is 3-transitive
X
but then
X
containing
I,X2,Xj
(,XI>
then
()2)
has order
I
are in
is a Baer
so
contrary to the
XI ,X2
above.
We also assert that
Let
PROOF.
’1,j
since
X
I
UXI,X2
share no common components
2
(q+l)(q-l)ll(l,2,C3,4lfollows that
So
subgroups
each
pointwise.
1,2 is regular
3,4 is regular
q>4, 21, 2 A23,
4
Since
i.
and
<1,2,3,4)
SL(2,q)
Qi
fixes
share no common component.
UX3,X4
be a common component.
fixes
X
and
XI,X2
Then
Note that
q+l>
<1,2,3,4>
4.
q >
2(j+l).
fixes
That is, since
So
on
IX3’’’" ’Xq+l] 1,2
on
IXI,X2,x 5,--.,Xq+I} #3,4"
so that there is an orbit of length
-
By examination of the subgroups of
SL(2,q).
and acts faithfully on
fix subspaces of
i pointwise.
.
q+l.
Thus,
SL(2,q),
it
Then the Sylow 2-
X
Thus
i
intersects
But, this is a contradiction.
We consider
X.I
X
j
-[Xi’Xj]
and
WXk X
m
statements, there are no common components if
count the components of
i,jU (Xi,X.- [Xi ,Xj
),
-[Xk’Xm]"
[i,j]
By the above two
[k,m].
for the
q(q+12
We thus may
unordered
for
N. L. JOHNSON
456
(i,j),
pairs
(q+l) (+i)
have at least
q2+l
q(q+l)
>
@
The center
Case 2:
GL(2,q).
is normal in
Case 2a:
2 s+l
permutes
2.
2
SL(2,q)
That is,
2S<q
fixes a Baer subplane
#
must fix
2S+l>q+l
So,
(since some
is faithful on
by
So, if
2 s+l > q+l,
SL( 2 ,q
2S+l<q+l
then
2 pointwise, which is a
2S=q.
or
@
Thus,
fixes it
SL(2,q)
pointwise).
is Desarguesian by
so
SL(2,q)
[ii] Satz 4 and the Sylow 2-subgroups of
and
affine points on each of the
element of
[9] (4.2)
fix
s.
214 [8]).
p.
p6@-(i>
2.
components and unless
(8.23),
[9] (4.3).
contradiction to
Let
is characteristic in
fixes
must fix pointwise each set of
components of
SL(2,q)
(p)
##O.
SL(2,q)
So
must fix each component (see
SL(2,q)
which is clearly a contra-
So Case i is completed.
2 is a subplane of order
SL( 2, q)
q>4
and
Therefore,
is not fixed-point-free.
pointwise a set of affine points
so we must
i=l
additional components.
q+l-2+l<2q=2.
diction unless
U X.i
We also have
(q+l-2(+l)+ (q+l(+l)
2
So
q+l
q(q+l)(q+l-2(+l)).
2
as
L[[neburg
act as elations on
That is, the 2-groups either fix Baer subplanes or Baer sublines pointwise.
Thus, we must have
2 is not a subplane so
Case 2b:
IP
Assuming
and
y= 0.
also fix
Let
2
is prime,
e
and
x=0
and
y= 0
and
GL(2,q)
x
0
must
y=0.
I@I
be an element such that
(which exists by [i]
q-i
fixes exactly two components, say
must fix both
Thus,
x=0
p
2 is contained in a component.
this case does not come
unless q=
up).
Then
@
8
is a prime 2-primitive divisor of
and since
q
is assumed to be square
is irreducible on any component
fixes.
That is,
@
either fixes
i
pointwise or is fixed point-free on i
GEOMETRY IN TRANSLATION PLANES
e
(since
is completely reducible on
pointwise then
Thus,
0
e
implies
must fix
J.
0
if
So we may take
That is, either
element fixes
SL(2,q)
fixes
cannot exist by
If
=
or
of
pointwise).
fixes
J to be
x= 0.
Thus,
is
pointwise.
and every prime 2-primitive divisor
SL(2,q)
Since
is faithful on
[9] (4.2).
SL(2,q)
is f.p.f, on
So
y= 0
SL(2,q)
y= 0
,
must fix a subspace
and fixes
then
SL(2,q)
Thus,
II
IBI
Sylow 2-subgroup fixes a subplane
IBi)
N(x=0)so
12s-i
w
induces on
is represented by
SL(2,q)
of order
IBll2s-is"
which implies
either
y= 0,
=.
we may
Q
of
is f.p.f, on
contrary to our assumptions.
q
(IAI) if
q-I=LCM(IAI,IBI). Since
fixes
fixes
Since
pointwise.
Thus, our argument shows that the group
semiregular of order
and thus
That is, a Sylow 2-subgroup
of
II
,
The first alternative
is faithful on
then since
X
.
is simple and fixes
J pointwise or
apply or previous argument of Case i.
LCM(
fixes a subspace
0 ,x=0 )-homology.
a
But,
457
x= 0
fixes
2
s
x
(y 0)
is
(x,y) (xA,yB).
and
pointwise
and similarly
0
IAll2s-i
y= 0,
where
a
2s< q.
so
q< 2
Thus, we have the proof to our theorem.
ACKNOWLEDGMNT.
The author gratefully acknowledges the helpful suggestions
of Professor D. A. Foulser in the preparation of this article.
REFERENCES
i
2.
Birkhoff, G D
Ann. Math.
Foulser, D.
and H
S
Vandiver
On the integral divisors of
an-b n
5(1904) 173-180.
Baer p-elements in translation planes, J. Algebra 31(1974)
354-366.
3.
Foulser, D.
Subplanes of partial spreads in translation planes, Bull.
London Math. Soc. 4(1972) 1-7.
N. L. JOHNSON
458
4.
Foulser, D. Derived translation planes admitting affine elations,
Math. Z. 131(1973) 183-188.
5.
Foulser, D. A., N. L. Johnson, and T. G. 0strom.
the Desarguesian and Hall planes of order
6.
7.
Hering, Ch.
Hamburg
Characterization of
by SL(2,q), submitted.
q
On shears of translation planes, Abh. Math. Sen. Univ.
37(1972) 258-268.
Hering, Ch.
On subgroups with trivial normalizer intersection, J.
Algebra 20(1972)
622-629.
8.
Huppert, B. Endliche Gruppen I, Springer-Verlag, Berlin-HeidelbergNew York, 1967.
9.
Johnson, N. L. and T. G. 0strom.
The geometry of
planes of even order, Geom. Ded., to appear.
i0.
Johnson, N. L.
ii.
Johnson, N. L.
12.
Lineburg, H.
The translation planes of order
Ostrom, T. G.
Schaeffer, H.
16
that admit
SL(2,4).
Charakterisierungen der end_lichen Desarguesschen projek-
85(1964) 419-450.
Linear transformations and collineations of translation
planes, J. Algebra
14.
in translation
Translation planes of characteristic two in which all
involutions are Baer, J. Algebra, to appear.
tiven Ebenen, Math. Z.
13.
SL(2,q)
14(1970) 21-217.
Translationekenen, auf denen die Gruppe
Diplomarbeit, Univ. Toingen, 1975.
SL(2,p n) operiert,