ELATION GROUPS OF THE HERMITIAN SURFACE H{3.q2) OVER A
FINITE FIELD OF CHARACTERISTIC 2.
by
Robert L. Rostermundt
B.A., University of Colorado at Boulder, 1993
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
2005
This thesis for the Doctor of Philosophy
degree by
Robert, L. Rostermundt
has been approved
by
Eric Moorhouse
Rostermundt. Robert L. (Ph.D.. Applied Mathematics)
Elation Groups of the Hermitian Surface H(3. q2) Over a Finite Field of Char
acteristic 2.
Thesis directed bv Professor Stan Pavne
ABSTRACT
Let S = be a finite generalized quadrangle (GQ) having order
(.s. t). Let p be a point of S. A whorl about p is a collineation of S fixing all the
lines through p. An elation about p is a whorl that does not fix any point not
collinear with p. or is the identity. If S has an elation group acting regularly
on the set of points not collinear with p we say that S is an elation generalized
quadrangle (EGQ) with base point p. S. E. Payne posed the following question:
Can there be two nonisomorphic elation groups about the same point p? In this
presentation, we show that there are exactly two (up to isomorphism) elation
groups of the Hermitian surface H{3.q2) over the finite field of characteristic 2.
Moreover, we investigate some possible EGQ's that may be constructed as a
coset geometry from the new elation groups, and show that the EGQ we con
struct are isomorphic to the Hermitian surface.
m
This abstract accurately represents the content of the candidate's thesis. I
recommend its publication.
IV
DEDICATION
This is dedicated to my parents, whose constant support has helped to make
this thesis a reality.
ACKNOWLEDGMENT
There are many people who have been present with me throughout this
long journey and deserve thanks. I would first like to thank all the members
of my committee for their time and effort. In particular, I thank my advisor
Stan Payne, without whose patience I could have never reached as far as I have.
I can only dream of becoming a mathematician of his stature and reputation.
Furthermore, Bill Cherowitzo has been a constant source of assistance, espe
cially during the early stages of my career as a graduate student. Here I quote
Newton's letter to Hooke in 1676. in which he said. "If I have seen further than
others, it is because I was standing on the shoulders of giants."
I also would like to thank Tim Penttila for selflessly suggesting this problem.
Although we are separated by some distance, he has been extremely supportive
throughout the process.
I also extend my appreciation to the graduate committee for appointing me
a teaching assistant position. This financial support has made a great impact
on my success here at UCD. Moreover, I thank the Warren Bateman Family
for their generous funding of the Lynn Bateman Teaching Fellowship, and the
Lynn Bateman Teaching Award, both of which I was a recipient. These were
established in memory of Lynn Bateman, who had an outstanding reputation
as a teaching assistant in the department.
Finally, over the past, six years I have had many friends, both in and out
of the math department, who have encouraged me to push on during the most
difficult times. I can not name them all here, but I would like to thank a few
specific individuals: Marcia Kelly. Jennifer Thurston. Steve Flink. Scott Hace.
Ryan and Natalie McGrath, Christi Day, and also my family. Leo and Elizabeth
and Liz Rostermundt. To all of you. I am in your debt.
Rob Rostermundt
CONTENTS
Figures ................................................................ ix
1. Introduction and Review............................................... 1
1.1 Preface............................................................. 1
1.2 Historical Background............................................... 2
1.3 Basic Definitions arid Combinatorics .............................. 3
1.4 Elation Generalized Quadrangles..................................... 5
1.5 Introduction to ijclans............................................ 8
1.6 Some examples of gclans........................................... 10
1.7 gClans and Flocks................................................. 11
1.8 4Gonal Families from gClans...................................... 14
1.9 A Construction of the Hermitian Surface H(3. q2)................... 19
1.10 Regularity in H(3. q2) 20
2. Main Results ...................................................... 21
2.1 Forming a Sylow2 Subgroup in the Group of Whorls....................21
2.2 Elation Groups of H{3, q2) As Subgroups of 52.................... 27
2.3 The Commutator Subgroup............................................ 31
2.4 The Quotient Group S/S'............................................ 35
2.5 Looking for Hvperplanes............................................ 36
2.6 Lower central Series .............................................. 38
2.7 Building Subgroups A(t) in the Exotic Elation Group E.............. 51
vii
2.8 The Function S(t)............................................... 53
2.9 The Function gt................................................. 55
2.10 Property(G)..................................................... 67
2.11 A Theorem of Matt Brown......................................... 80
2.12 The Final Push.................................................. 81
2.13 Suggested Problems ............................................. 81
Appendix
A. Theorems......................................................... 83
B. The Hermitian Surface: A Unitary Representation.................. 98
References.......................................................... 103
viii
FIGURES
Figure
1.1 The coset geometry S(oc^........................................... 6
1.2 Construction of F and F* from (S^p\ G).............................. 8
2.1 Property(G)........................................................ 68
2.2 3x3 Grid........................................................... 71
2.3 3x3 Grid........................................................... 73
2.4 Near 4x4 Grid...................................................... 75
A.l P2 : A* C Ai = {id} for all j.................................... 87
A.2 P3 : AjAi H Ak = {id} for all distinct i, j. k................... 87
A.3 P3 : AjAi H Ak = {id} for all distinct i. j. k................... 88
A.4 PA : A*Aj = G for all i ^ j........................................ 89
A.5 P5: A* = Ai U {Atg : Ai9 C Q = 0}............................... 91
A.6 P5 : A* = Ai U {Atg : Atg n Q = 0}............................... 91
IX
1. Introduction and Review
1.1 Preface
The goal of this thesis is to answer a question posed by S.E. Payne: Given
an elation generalized quadrangle and a point p. can there be nonisomorphic
elation groups having the base point pi In this thesis we will give a proof that
the Hermitian surface H(3, q2) has two nonisomorphic elation groups. Initially.
Tim Pentila completed a search for the cases q = 2 and q = 4, using the Magma
software package. In each case, he discovered q2 1 new elation groups that
were nonisomorphic to the known elation group. For all cases q = 2e, we will
construct the q2 1 elation groups of H(3, q2). Moreover, we will show that they
are nonisomorphic to the known elation group, while being pairwise isomorphic.
We start with a representation of the standard elation group due to S.E.Payne,
and adjoin an involution d> to form a Sylow2 subgroup (denoted S2). having or
der 2q5. of the group of whorls about the point (oc). We then show that the
only nonelations in S> are contained in the coset of the commutator subgroup
containing the involution d. We proceed to show that the factor group S2/S'2 is
elementary abelian; i.e., a vector space over GF(2). and then using the complete
set oflinear functionals from S2f S2 onto GF{ 2). we construct the complete set of
elation groups about (oc). This set has size q2. We then show, using nilpotency
class, that q2 1 of these groups are nonisomorphic to the standard elation
group. Furthermore, we show that these q2 1 groups are pairwise isomorphic.
1
Next we proceed to find conditions for the existence of 4gonal families in
these new elation groups. Given these conditions, we will construct a family
of EGQ and and prove that these EGQ are flockGQ. Moreover, we will be
able to show that all the new GQ we have constructed are classical, and hence
isomorphic to H(3.q2).
I. 2 Historical Background
The notion of a finite generalized quadrangle (GQ) is a fairly recent one.
J. Tits first introduced generalized polygons in 1959 [29]. Generalized polygons
are the building blocks of Tits buildings, and are the precursors of more general
geometries such as partial geometries, partial quadrangles, semipartial geome
tries. and near polygons [30]. In actuality, generalized quadrangles have been
around for some time as line systems corresponding to symplectic polarities in
threedimensional projective space over a field. But the explicit study of such
geometric objects is due to Tits. Initially, progress was made in the study of
what are now called classical generalized quadrangles (see [Deni68] and [FH64] )
that is. those quadrangles that can be imbedded in projective space. It was in
the late 1960's that other researchers began looking deeply at these geometric
objects, and since then, many new examples and results have been discovered.
The focus of this thesis is the classical GQ known as H(3.q2). the Hermi
tian surface in threedimensional projective space over the field GF(q2). Here
we will study a certain class of collineations called elations. More specifically,
we answer the question posed by S.E. Payne: Can there be two nonisomorphic
2
elation groups about the base point /;?
In this thesis we show that there are two elation groups of H(3.q2), up to
isomporpliisni. We also look into the construction of elation generalized quad
rangles (EGQ) arising from the newly described elation group of the Hermitian
surface. Although we do not have a complete classification of these EGQ. we
show that all of our constructions are isomorphic to the classical H(3,q2).
1.3 Basic Definitions and Combinatorics
We start with some basic definitions. Let V and B be two nonempty sets,
called points and lines, with an incidence relation T such there are two positive
integers .s and t satisfying
Gl) Each point is incident with t + 1 lines; any two points are mutually
incident with at most one line.
G2) Each line is incident with s + 1 points: any two lines are mutually
incident with at most one point.
G3) Given a line L and a point x not incident with L there is a unique point
y and a unique line M such that x 1 Ally!L .
Such a collection S = (V.B.l) is called a generalized quadrangle of
order (s,t) written GQ(s.t); when s = t the GO is said to have order s. The
dual of a GQ(s. t) is the GQ(t. s) obtained by interchanging the roles of points
and lines. Furthermore, any theorem or definition given for a GQ can be dualized
by interchanging the words points and lines. It will therefore be assumed that
whenever a definition or theorem is given, its dual has also been given.
3
Two points incident with a common line are said to be collinear and two
lines incident with a common point are concurrent. If :r and y are collinear
we use the notation x ~ y. Similarly, if L and M are concurrent we denote this
L ~ M.
If X is a set of points (respectively, lines) of S, then X1 denotes the set
of all points collinear (resp., lines concurrent) with everything in Ah A J is also
called the trace of Ah If X {:r} is a singleton set. it is common to write A1
as x1. The span of Ah written ATJ, is the set of all points collinear (resp..
lines concurrent) with all of AiA By convention x E xL.
Straightforward counting arguments demonstrate the following:
P = (l + s)(l + st).
\B\ = (l+t)(l + st).
For x E V. .r = 1 + s + st.
For L E B. \LL\ = 1 + t + st.
For two noncollinear points :r. y, {:r, y} =t + l
and 2 < {.r./y}11
For two nonconcurrent lines L, M, \{L. h/}J  = s + 1
and 2 < \{L. d/}1! < s + 1.
Let x, y be two noncollinear points of a GQ(s.t). We say that {a., y} is a
regular pair provided {a,. .(/} t + 1. If x is a point such that for every y.
with x 7^ y, we have{.r, ;
{.r. /y.s} of pairwise noncollinear points is called a triad of points. If {x.y.z}
is a triad of points, then all points in {:r, y. z}* are called centers.
The next important theorem is known as Higman's inequality.
Theorem 1.3.1 (D.G. Higman) [13] Let S = (P.B,L) be a GQ of order
(s.t). Then s < t2 and dually t < s'2. Furthermore, t = s2 if and only if for
some pair (:r. y) of noncollinear points every triad (x, y, w) has exactly s + 1
centers if and only if every triad of points has exactly 1 + s centers.
Proof: We give the proof in Theorem A.l.
Corollary 1.3.2 Let S (P. B. I) be a GQ of order (q2, q). Then every set of
three pairwise nonconcurrent lines has exactly q + 1 transversals.
1.4 Elation Generalized Quadrangles
The focus of this thesis is a certain class of GQ called elation generalized
quadrangles. Here we explain this class of quadrangles.
Let S = (V.B.T) be a GQ(s.t). s > 1. t > 1. and let p E V be a point
of S. A whorl about p is a collineation of S that leaves invariant each line
incident with p. If there is a group of whorls acting transitively on the points
not collinear with p we say that p is a center of transitivity. Let 0 be a whorl
about p. If 0 = id or if 0 fixes no point of V \p. then 0 is an elation about p.
If there is a group G of elations about p acting regularly on V\pL. we say S is
an elation generalized quadrangle (EGQ) with elation group G and base
point p. We will often denote this quadrangle as (S^KG). or simply .
We now describe a standard construction of elation generalized quadrangles.
Let G be a group with order s2t. Then let F = {Aq, A\.......4r} be a family
of r + 1 subgroups of G, each with order s, and let F* = {.4q. ,4)\ .... A*} be
another family of r + 1 subgroups of G, each having order st. where At < A* for
each 0 < i < r.
Our geometry, which we denote
types of points: (i) elements g G G. (ii) cosets A*g. (iii) a symbol (oc). There are
two types of lines: (i) cosets ALg. (ii) symbols [.4;]. Incidence is as follows; the
symbol (oc) is incident with the r + 1 lines of type (ii), the s cosets of ^4* are the
other s points on a line [Ai\. each point A*g is incident with lines corresponding
to the cosets AJi that are completely contained in the coset A*g. the remaining
points on a line Aih are the group elements contained in the coset Aih. The
diagram in figure 1.1 may be helpful.
AJ9
\ i /'
A*g
[Aj]
(oo)
Figure 1.1: The coset geometry
6
Theorem 1.4.1 Let G be a group of order s2t and let F = {T0. ___, .4,} be
a family oft+1 subgroups, each unth order s. and. let F* = {.4q. A\.A}} be
another family of t + 1 subgroups, each having order st. where Aj < A* for each
0 < i < t. Then if we build the coset geometry
is a GQ, having order (s.t), if and only if properties AT and I\2 hold, where
AT : AjAi n Ak = {id} for all distinct i.j. k.
K2 : A* H Ai = {A/} for all i =4 j.
Proof: See Theorem A.2 in Appendix A.
In the previous theorem, we call F a 4gonal family of G. and {G. A, A*} is
called a Kantor family.
Let (S^.G) be an elation GQ with base point p (and group G of elations
about p). Then we can obtain a 4gonal family in the following way. To obtain
the set T* we choose a point q not collinear with p and consider {p, q}L. For
each of the t + 1 points x G {p. q}L define a subgroup A* to be the stabilizer of x
in G. Then define .4, to be the stabilizer in G of the line through q and x. The
set F = {ToT!......At} will be a fourgonal family for G with accompanying
set F* = {4*.Tt.... .T;}.
Theorem 1.4.2 [28]. Suppose that S = (fP.B.T) is a GQ with order (s.t).
s.t > 1. with s and t powers of the same prime p. Suppose (oc) is a regular
Figure 1.2: Construction of F and F* from (S^.G)
point that is a center of transitivity, and let H be the full group of whorls about
(oc). Let G be a Sylowp subgroup of H. Then we have
1. (C = s2t. or
2. p = 2, G = 2s2t, and S contains a proper thick (s, t > 1) subGQ of order
t isomorphic to \V(t); consequently, s = t2.
1.5 Introduction to ^clans
There is a strong connection between flocks of a quadratic cone and general
ized quadrangles. In fact, there is a large clas of GQ known as flock generalized
quadrangles. The connection between the two structures arises from particular
sets of 2 x 2 matrices.
8
,4 =
is anisotropic if and. only if tr
the absolute trace function.
A =
ad
2).. the: . the matrix
/ A
a b
\C d)
5) = d AaT = 0 if and only if a
where q = 2P. Then the matrix
( a
V V
(b + c)2
I =1. where tr : GF(q) > GF( 2) is
Put q = 2e and let A' : x i xt. Y : y
functions from F to F. Put
(
At =
ijt. and Z
zt be three
Ut
0 zt
and define the set C = {At : t G F}.
Definition 1.5.3 The set C is a gclan provided all pairwise differences As At
(s. t G F, s 7^ t) are anisotropic.
Note: The anisotropic condition on At .4S. with At.AN Â£ C. is precisely
that
{ {:rt + xs)(ztl + Zs)
v (ijt + VsY
= i
9
1.6 Some examples of (/clans
Example 1.6.1 (Classical) Let At = t
To see that this is a (/elan look at
/
T T
vd)
where tr{d) = 1.
At As
t s t s
0 (t s)d j
i i (t s)2rf\ ,,
and notice that tr ( T ) = tr(d) = 1.
{t, sy1
Example 1.6.2 (Kantor) At =
eodd.
(, A
t r
V J
, where t E F = GF(q). q = 2e. and.
To see that this is a (/clan look at
=
t  s t '2  s
0 I  A
/
Then we have
. (t s)(t:i A) \ ... ( st
tr  V A ..... 1 1 = tr(l) + tr
A2A)2 J v ; \F A
The equation x2 + x + 1 = 0 has no solutions in GF(2) and so its roots lie
in a quadratic extension of GF{2). But e is odd and so GF(4) is not a subfield
of F. Therefore x2 + x + 1 is irreducible in F and hence tr{ 1) = 1. Furthermore.
10
the polynomial sx2 + (s + t)x + t has the root x = s/(s + t). It follows that
st
tr
t1 + 81
 I = 0. and Kantors example is indeed a gclan.
Example 1.6.3 (Payne) [17] At =
and e odd.
ft fO
1 n
. where Â£ G F = GF(q), q = 2e.
Consider the difference
/
At .I.,
/ .S /:i .s:i
Then
tri (t:f)(tT,s5)W(i)+fT 3t^+f2)
(t3 ,s3)'2 J \ (t + st + S2)2 t
In this case tr{ 1) = 1 because e is odd. Furthermore, the polyno
mial (st)x2 + (s'2 + st + t2)x + (s2 + t2) = 0 has the root x = L and so
tr[st(s2 + t2)/(t2 + st + .s2)2] = 1. It follows that Payne's example is a q
clan.
1.7 qClans and Flocks
Let K {(xq.Xi.x2, .T3) G PG(3.q) : x\ x0x2}. Then I\ is a quadratic
cone in PG(3.q) with vertex (0.0.0.1). Recall that all quadratic cones of
PG(3,q) are equivalent under the action of PTL(i.q), hence WLOG we can
choose our favorite cone.
11
Definition 1.7.1 Let K be a quadratic cone with vertex V. .4 flock of K is a
partition F = {Ct : t, G F} of I\ \ {V} into q pairwise disjoint conics Ct.
Each conic is a plane intersection Ct = rt n K. where nt = [xt. Ut zt, l]r, and we
often consider the flock to be the set of planes 7p.
Theorem 1.7.2 (J.A. Thas) [27]. If F = GF{ 2e), let X : t ^ xt; Y :fn yt.
and Z : t i> Zf be three functions on F. For each t Â£ F. put nt = [xt.yt.zt. l]T
(
Ct = fi K. IF = {Ct : t e F}. Also put At =
and C = {At:te F}.
Then C is a qclan if and only if F is a flock.
0
Proof: The set F is a flock provided every two distinct conics Ct and Cs
with s different from t are disjoint. If ~t = [;r(. yt. zt. 1]T and tts = [;rs. ys. zs, 1}T
are planes defining the conics Ct and Cs, we want that 7it and ns meet at a line
external to K. Therefore, the following system:
XtXo + ijtX i + ztXo + A3 0
xsXq + ysX i + ZgXo + X% = 0
A'i2 = A'0A'2
can have only the trivial solution. Subtracting the second equation from the
first we get
(xt xs)X() + (yt ys)Xi + (zt zs) X2 = 0
Not all of A'o.Ad.Ab are zero. So assume WLOG that X> 0 and divide
12
through by X2 to get the following equation.
0 (xt xs)tt + [ijt
A 2
= (xt xs) + (yt
A 2
Vs) TW + {Zf Zg)
A 9
Vs,
vxoxi
x,
Letting Y =
\P^X~2
X2
we get
(xf xs)Y 2 + (ijt Vs)Y + (zt
0.
Since we are in even characteristic this equation has no solutions if and only
if iias trace eclual to 1 for any s. t E F. s ^ t. This is exactly the
condition that C be a 5clan.
Note: Two flocks are projectively equivalent when there exists a pro
jective semilinear map of PG(3. q) leaving the cone invariant and mapping one
flock to the other.
Suppose that C = {At
(
xt
t E IF} is a 5clan. It is easy to
\ '7
check that C = {A't = At AQ : t E F} is also a 5clan. The 5clan C has
an associated flock F(C) = {irt = [xt,yt.zf. l]r : t E F}, and C has the asso
ciated flock F(C') = {/tf = [xt xQ. yt ijQ.Zt Zq. l]r : t E F}. Then since
T : [,r. y, z, l]r [,r .r0l y yo z l]r is a projective linear map on PG(3, q)
that leaves the cone invariant, without loss of generality we can assume that each
gclan C contains the zero matrix, which by convention we will label as Aq.
13
1.8 4Gonal Families from qClans
Let F = GF(q) where q
define a o j3 by
27 Put P
o o 3 =
 (
V1 7
qP3t
and for q 7 E F2
F x F
Then (a, 0) i> ct o 3 is a nonsingular, alternating, bilinear form with the prop
erty that a o 3 = 0 if and only if {o, 7} is Fdependent.
On the set = F2 x F2 x F = {(a, ,7. c) : a. 3 E F2,c E F} define the
binary operation
(a. 3. c) (a 7', c) = (o + o', 7 + 77 c + c + 7 o')
This operation makes G into a group of order q with center Z = {(0, 0), (0. 0). c)}.
This group has two important families of elementary abelian subgroups of order
For O/76 F2, put = {(7 Z o. c) E G : a E F2.c E F}, and for
0 7 o E F2. put Ua = {(7 0 o. c) E G* : 7 E F2. c E F}.
Theorem 1.8.1 [8], Each Â£, and each 1Za are elementary abelian groups of
order q3. And for nonzero a. 7 E F2, C1 = Ca (resp.. 7T,, = 7Za) if and only if
{a. 7} are Fdependent, so we may think of the groups Â£7 and 7Za as 'indexed by
the points of PG( 1. q).
Note: We use the elements of F = F U {oc} to index the points of PG{ 1. q)
as follows: 7^ = (0.1) and pt = (1. t) for f E F.
14
/
I't Vt
Let C = {At
: t E F} be a 5clan with Aq
. Also
/
0 0
put Ax.
. and 7y
(0,1). Then define g(cx.t) = aAtaT for
\0 0
t eF and a E F2. It will be useful to recognize that
g(a + ;3.t) = (a + f3)At(a + ;3)T = g(a. t) + g((3, t) + yt(a o (3)
For each t E F there are subgroups A(t) and A*(t) of G* defined in the
following wav:
A(t) = {(7yt S a.g(aJ)) E G2 : a E F2} < A*(t) := A(t) Z = Clyt < G3.
It is easy to see that A{t) is a subgroup when noticing that
(7vt A a\g{ai. t)) (7yt 0 02. t)) = (7ut S (qi + a2).g{ai + a2, t))
Observe also that A(t) is a subgroup of order q2 and A*(t) is a subgroup of
order qi. It is also helpful to see that for t E F. a typical element of A{t) has
the form
Theorem 1.8.2 [8j. Put J(C) = {A(t) : t E F}; J*(C) = {A*(t) : t E Â¥}.
Then the triple (G*, J(C)< J*{C)) is a Kantor family, i.e.. J{C) is a 4gonal
family for G The associated GQ is denoted GQ(C) and is referred to as a
flock GQ.
(o. yta. ct\:rt + e\ia2yt + a2zf), where a = (c*i, c\2) E F2.
Further, an element of A(oc) looks like (0, a.0).
15
Proof: We know that Y : t i> yt is a permutation and so for distinct
t.u G F we have yt p yu. Showing property Ah is easy when looking at two
elements from Apt) and A*(n) with u t. If
(a, yta, Ot\xt + QiQoijt + aid) = (b\ yu6c, c)
then q = q which forces both to be ecpial to (0, 0) which then forces c = 0.
Therefore, A(t) fl A*(u) = (0, 0. 0) = {id}.
Showing property Ah is easy in one case. Look at A(oc)A(t)nA(u). Because
At Au is anisotropic we get this intersection being the indentity.
What we need to show is that A(s)A(t) n A{u) is the identity when none of
s.t. u equal oo.
Suppose that for elements (~!yt 7 a.g(a.i)) G A(t) and (~nju 7 a.g(a, v)) G
.4(u) we have the product
(W Fn. g(a. t)) (qyu 0a,
is contained in A(v). Then the following two conditions must be satisfied:
1 !Jta + y 1 = yc(a + p). or (yt + yu)a = (yu + y,)P = y for some 7 G F2:
2. g(a, t) + g(p. u) + yt{a o 6) = g{a + p. v).
From condition 1. we get a = (yt + :(/tt)_17 and p = (yu + yv)~l"i Then
16
using the definition of i;o we now get
aod= {{yt + yu) 1'))P{{yu + yv) li)T
= (yt + Uu)~l{yu + yv)~1iPiT
= (Vt + !Ju) 1{>Ju + Vv) 1 0
= 0
where the third equality holds since o is an alternating form. Now using condi
tion 1. a o (3 = 0. and the equation labeled (*) for condition 2 we get
0 = g(a. t) + g((3, u) + g(a. v) + g((3, v)
= (yt + yu)~2(9{l, t.) + g(~.v)) + (yu + yv)~2(g( 7, u) + g{ 7. v))
= 7fl7r
where B =
fxP
Voz/
is the matrix
/ .Tf + xv xu + xv
B =
(yt + Uv)2 (yu + VvY
\
0
yt + Vc Vu. + Vv ^
(yt + yv)'2 (iju + yv)2
(yt + yv)'2 (yu + yv)2 )
17
Easy computations show that
XZ xtzt(yv + JJu)4 + xHzu(yt + yv)4 + xvzv{>Jt + J/t,)4
+ (: rtzu + x uzt)(yv + IJtfitVv + IJu)2
+ (xtzv + xvzt)(iju + yt)'{yv + yu)~
+ {xuzv + xL.zu)(yu + yt)~(yv + Ut)2
and
Y2
+ Vv
2
Hence
tr(XZ/Y2) = tr
tr
(xt + Xu)(Zt + zu) ^ (xt + Xv)(zt + zv) (xu + Xv)(zu + S')\
{Vt + Uu)2 {'!Jt + !Jv)2 (lju + Vv)2 )
(xt + Xu)(zt + Zu)\ / (xt + Xv)(zt +
(yt + Vu)2 ) \ (yt + y)2 )
\tv
(xu + xv)(zu + zv) \
(iju + yu)2 J
= l
since At + Au, At + Av, and Au + are anisotropic matrices.
We have just shown that B is an anisotropic matrix and so yByT = 0 if
and only if 7 = (0.0). Using 7 = (0.0), condition (i) forces a = 3 = (0,0). So
the only element. (~yt 0 a, g(a.t)) G A(t) and (zyu S' a. g(a. ti)) G A(u), whose
product is contained in H(u) is (0.0.0). which is the identity and property K'2
holds. Since both K1 and K2 hold. J{C) is a 4gonal family for G .
18
1.9 A Construction of the Hermitian Surface
Let F = GF(q). where q = 2e. and fix a 8
for each t 6 F = F U oo, define the subgroup A(t) as
A(t) = {(71 S a, g(a, t)) : a e F2}
Clearly, A(t) < A*(t) where
A*{t.) = {(;fZo,c) : a gF2.cGF}
Let J{C) = {.4(f) : t 6 F} and J*(C) = {A*{t) : t G F}}.
Theorem 1.9.1 (S.E. Payne and J.A. Thas) If (G w J(C). J*{C)) a Kan
tor family, as prescribed above, let S GQ(C) be the corresponding EGQ. Then
S is a GQ(q2.q) isomorphic to the Hermitian surface H(3.q2).
Proof: See Theorem A.3 in Appendix A.
tr
(s1/28 tl!2d)(sll'28 t1'r2d) ,
r fAL =tr(d
(s 1/2 tA2)2 J 1
= tr{5)
= 1
and C = {At : t G F} is a gclan. Put Ax
. Then as in section 1.8.
19
1.10 Regularity in H(3.q2)
Let Q{5.q) be an elliptic quadric of PG(5.q). It can be shown that Q(b.q)
is a GQ of order (q. q2).
Theorem 1.10.1 The quadrangle Q(5,q) is isomorphic to the pointline dual
0fH(3.q2).
Proof: See Theorem A.4 in Appendix A.
Theorem 1.10.2 Any pair of lines in Q(5,q) is a regular pair.
Proof: The 3space defined by any pair of nonconcurrent lines of Q(5. q)
intersects Q(5.q) in an hyperbolic quadric (or regulus), and so it is clear that
any pair of lines of Q(5,q) is a regular pair.
Corollary 1.10.3 Every point of H(3, q2) is a regular point.
Let S be H(3.q2) with elation point (oc). as contracted by Payne. If we
can find an whorl about (oc) which is not an elation and is an involution, it will
follow from Theorem 1.4.2 that a Sylow2 subgroup of the group of whorls about
the point (oc) in H(3.q2) will have size 2q.
20
2. Main Results
From here on we assume that q = 2e and that S is the Hermitian surface
H( 3.
whorls about the point (oc). We note that any elation group must be a 2group,
and furthermore, all Svlow 2subgroups are conjugate in IT. We aim to find a
Sylow 2subgroup of IT which we call ST Then contained in So we are looking
to find a subgroup E < S2, of elations about (oc). such that E is not isomorphic
to the regular elation group (which we will denote G) of S. Clearly, such a group
will have index [E : S2] = 2, and thus be normal in So.
2.1 Forming a Sylow2 Subgroup in the Group of Whorls
For all (q. S. c) Â£ G*. define the map [n. I. r : G 1> G* so that
(o'. S', c') = (a'. /3', c') (a. 3, c) = (a1 + 0. 6 + S, d + c + 3 o a). Let
G = {[a,/Tc] : (a.fj.c) Â£ }. Then G is the regular elation group of S.
Next, define the involution o : G G* so that for (a.d.c) Â£ G we have
(u. 3. c) = (aP. 3P. c). The map o is a whorl of IT about the point (00). Next,
consider the following computations.
21
(a, 3. = (ap. ;jp c)K/3'.c']oo
= (qP + a', pP + 3'. c + c' + pP o {a'P ff
= {aP + a. pP + P'.c + P + pPo PTalTf
= (aP + a', PP + 3'. c + d + p o <\')
= ((aP + <>')/>. (pP + b';/W + c' + i o o')
 in o'/.d p'P. c + c' + p o a')
= (a.3.c)VpPR*
Note: Because (aP.pP.c) = (a, p. c)A we will denote [a'P.p'P.c1] by
[o'./f.c'f.
We have shown that o [o'. The'] o p = [a.T.o]^ G G. That is. the map p
normalizes G in the group of whorls about (oc). and we can define the semi
direct product S2 = G xi (p).
Remark 2.1.1 The group So is a Sylow2 subgroup of IV.
A typical element in S2 can be identified as [a. 3. c]o p\ where i = 0.1. That
is. an element of S2 should be thought of as the composition of the maps [a. p. c]
and 6l. We now investigate products of elements in S2, where multiplication of
elements is simply composition of functions. We have the following possible
products:
22
1. It is trivial to see that
[a. 3. cl fa'. 3'. c'l =
a + a'. 3 + 3'. c + c + a' o 3
Therefore, it makes sense to define the following:
(a, 3, c) (ah 3', d) := [a. 3. c] [c/, 3' c'}
2. It is also easv to see that
[a. 3. c] [ah /T. c'] o 0 (a. 3. c) (a', ,3\ c')
o o
3. It takes a bit more work to show that
[a'.S'.c'joo
[a. 3. c] o
We start with
(x. y. z)[a'lix]Q
Then we compute
(x + o)P. (y + 3)P, z + c + y o a^
(x + a)P + ah (y + 3)P + 3'. z + c + yoa
+c' + (y + B)P o a'
x + a + a' P. y + 3 + 3' P. z + c + c
+y o a + (y + 3)P o a'
= 3'y) 9*
23
where g* = (a. 3. c) (a'P. 3'P, c').
That is.
[a. i3. c] o 0 [a'. 3'. c'\ o 0 = (a. i. c) (ah 3'. c')
4. Using the computations above we can also show the following:
[a. 3. c] o0 [ah 3'. c'\ = (a, 3 c) (ah 33 c')
o 0
To see this we reunite the element
+ a)P + o', (y + 3)P + 33 z + c + ya + c' + (y + B)P o a'j
in the following form.
.
t \
(x + a 4 a'P, y + 3 + 3'P, z + c + y o a + c' + (y + B)P o a'^j
This can be done since 6 is an involution.
Using these results we can now rewrite products of arbitrary elements of S
in the typical representation of elements of S as follows.
[a, 3. c] o [a', p3 c'\ o 6' = (a, ft, c) (ah 33 c)0>
o c3
+i
It is now easy to obtain the following theorem.
24
Theorem 2.1.2 Let F = GF(q) where q = 2e. Put P
. and let
0 1
1 0
G be the usual group of elations of H (3, q2) about the point (oc). Iff is the
involution such that (a,p,c) (aP. 3P,c), then every nonidentity element in
S2 = G x (f)) has order two, or four.
Proof: Choose an arbitrary element id g = 7r(a, fi. c) o o' E S. If i is
even then
g2 = [a. 3. c] o f>l [. L c] o ol
= [a. i3. c] (a, 13, c) o o2'
= 0.0. dPa1^
Then g2 = {id} iff {a. 3} is Fdependent, and we always have gA = {id}. Now
suppose that i is odd.
g2 = [a. 3. c] o 0' [q. 3. c] o p'
= [q. 3. c\ [a. 3, c] o oh
a + aP. 3 + 3P. iP<3
~.a.ai3o + Ct2 31
where p (a. a), o = (b, b), and 0 = 0! + o2 F and b = 3i + 32 = ra fr some
c E F. So {7,0} is an Fdependent set and we easily see that g4 = id.
Theorem 2.1.3 There are q4 1 involutions in S2.
Proof: Suppose that tt(q. 3. c) o o' ( 5 has order two. Then
[a. 3. c] o (p1 [a. 3. c] o
r(o. 3. c) (a. /3, cf
o o
2;
0.0.0
Case 1. i = 2k:
(a, /?, c) (a. 3. cf (a. 3. c) (a 3, c)
= (0. 0. 3 o a)
which equals zero if and only if {a. 3} is an Fdependent set; i.e.. 3 = ta for
some t G F. So for each fixed a f 0 there are q choices for each c and t. This
gives us a total of (q2 1 )q2 = q4 q2 elements of order 2.
Case 2. i = 2k + 1:
If q = (a,i, 02) anh T = (&i. b2). then we get
[a. 3, c] (a. 3. cf  ^(u 1. (i2) (61, 62). fj \ ia1,a2)P, (b1.b2)P. c
(iai a2) {b\. b2),csj {[a2.ai),{b2.by).cj
(a 1 + a2. a 1 + ci2). (bi + 62^1 T b2). a\b2 + a2b 1
26
This equals [0. 0. 0] if and only if cq = a2 and hi = b2. So we get q2 1 choices
for a and p, both not equal to (0. 0).
Adding the two cases we get y4 1 involutions in S2.
Corollary 2.1.4 There are 2q y4 elements of order four in S2.
2.2 Elation Groups of H{3. q2) As Subgroups of So
We first look into the group S2 and determine which elements are not ela
tions about (oc).
Theorem 2.2.1 The only elements in So that fix any points not collinear with
(oc) are the conjugates off.
Proof: Suppose that Q is a point opposite (oc) that is fixed by d>. As S2
is a group of whorls about (oc) and G < So, the size of the orbit of Q under So
is exactly y5.
From the orbit stabilizer theorem we also know that
I So I
^ = size of the orbit of Q under So
We immediately get (*S<2)q = 2. But since 0 Â£ {So)q we must have (So)q =
{id. o}.
27
Now choose any point Q' opposite (oc). Because G acts regularly on points
not collinear with (oc), there is a unique g G G such that Q'9 Q. So
[Q')9^9 1 = Q'. Thus g4>g~l G Sq> and using the orbitstabilizer theorem again
we get SQ> = {id.gog1}.
Corollary 2.2.2 A subgroup E < S2, with \E\ q5. is an elation group of
H(3.q2) if and only if E contains no conjugates of o.
Observation 2.2.3 See that gog~l = yog '<;> 'o = [g,
all nonelations will be in a coset of the commutator subgroup containing d>.
First we determine what the conjugates of o look like in the group S2
Let g = (a. i3. c) G G. Its easy to show that g~l = (a. /3. c + 3 o a). Then
[a. 3. c] [q. 3, c + 3 o o] = (a. 3. c) (a, [3, c + 3 oq) = [0. 0. 0] = id G S'?. So
conjugates of o can be written as follows:
[a. 3. c] 4>
a, 3.c + 3 o a
= [a. 3.c\
0.0.0
O 0
< i. 3. c + 3 o a
= [o. 3. c] o o
a. 3.c + 3 o a
(a. d, c) (oP. 3. P. c + 3 o a)
o O
a + aP. 3 + 3P. c + c + 3oa + 3o aP
o 0
a + aP, 3 + 3P. 3 o a. + 3 o aP
O 0
It is easy to see that o + aP = (a. a) and 3 + 3P = (b. b) for some n, b G F.
28
Furthermore, we also have /joa loo/' ab. so we can simplify this last term
to the following
(a. a). (b, b). ab
o o
It follows that there are at most q2 elements in S2 that are not elations.
Before we determine all elation groups contained in S2 we will need some
results from the theory of groups.
Theorem 2.2.4 Given a group G. the commutator subgroup. G' [G.G], is a
normal subgroup ofG. Moreover, if H
G' < H.
Proof: For completeness we include the proof from [26].
A subgroup G' is normal in G if and only if for every g G G. all conjugates
hgh~l remain in G. Therefore, if G' < G, then G' < G if and only if y(G/) < G'
for every conjugation y.
Let / be a homomorphism, f : G G. Then f[a.b\ = [fa.fb]. It follows
that f{G') < G'. But conjugation is a homomorphism from G to G. So the
commutator subgroup is a normal subgroup in G.
Next, suppose that H < G. If G/H is abelian then HaHb = HbHa for all
a.b G G. So Hob = Hba. So. ab{ba)~l = aba~ib~l = [n, 6] G H and G' < H.
Conversely, suppose that G' < H. Then ab(ba)~l G H and Hab = Hba which
29
implies HaHb = HbHa.
We know that [S2 : G] = 2 and so G <1 Si Furthermore, the quotient group
S/ G has order 2 and so must be cyclic and abelian. It follows that the commu
tator subgroup is contained in G.
It turns out that all elements that are not elations will be in the coset of
the commutator subgroup containing
group S2/S2 is an elementary abelian group of order 2q2. We will then be able
to employ the following theorem.
Theorem 2.2.5 (Correspondence Theorem) Let K < G and let v : G 1>
G/K be the natural map. Then S > i/(S) = S/K is a bijection from the family
of all the subgroups S of G which contain I\ to the family of all subgroups of
G/K. Moreover, if we denote S/K by S*, then:
1. T < S if and only ifT* < S*. and then [S : T] = [S'* : T*\:
2. T <1 S if and only ifT* < S*. and then S/T = S*/T*.
This ensures that finding a subgroup of index 2 that does not contain the
nonelation elements will be the same as finding a subgroup of S2/S'.2 that does
not contain the elements corresponding the commutator subgroup coset contain
ing o. First we form the commutator subgroup.
30
2.3 The Commutator Subgroup
Our main goal in this section is to show that the commutator subgroup
equals the Frattini subgroup, denoted T^). which is the intersection of all
maximal subgroups. Then given this result, the quotient group S2/S2 is elemen
tary abelian and therefore a vector space over GF(q).
We first need to form the inverse of a general element in S. Let g = [a, (5, c]o
p. Then if a = (ai.a2) and 3 (bi. b2) we get
9 =
(ci2. fli). (b2, 61), c T U\b2 + Q2b\
o p
= oP.BP.c + poa
o o
If g = [o'. 3'. c7] then
9 =
a3. d. + 3' o a'
We can use this information to create all commutators. Let g = [o, 3, c] o p'
and g' = [a7, 3' c7] opL and suppose that a = (a{.0,2). a7 = (a\. a'2). 3 = (Â£>1, b2).
and 3' = {b^.b'o) have a number of cases. First, if i is odd and j is even
then
[9.9'}
(n, o), (b. b). bpo\ + (?) + b2{o,2 + (?) + pci
31
where a = a\ + a'2 and 6 = b\ + b'2. Now see that if a = b = 0 we get
[fh 9 ] 0. 0. b^cii T 69(12 H 61 aL T 62(12
0. 0. 6^ (a1 + 02) + ci 1 (61 + 62)
But we can choose b[, blt b2. a[.ai. a2 to obtain any element in F. Therefore,
all possible elements of the form [0. 0, c], which are all in the center of S2. are
contained in the commutator subgroup. That is. if an element [a. 5, c] o
S!2. then [a, 0, c*] o
From now all our computations will neglect the third coordinate. If i is even
and j is odd we get
\.99'}= (a,a),(b,b).*
where a = a\ + a2 and 6 = 61 + 62.
The next case is for i and j even. Then
[99'] =
0,0.*
The final case is for i and j both odd. We get
\99'\ =
0, 0. *
All products of commutators will yield an element of the form (a. a). (6, 6). *
and we can then multiply this element by an element in the center of S2. We
have shown the following result.
32
Theorem 2.3.1 The commutator subgroup S'2 is the set of all dements of the
form
(a. a). (b. b). c
where a. b. c Â£ F
Furthermore, the commutator subgroup has size
To show that S!2 = $(S2) we need some additional group theoretic results.
Lemma 2.3.2 (Frattini Argument) Let I\ be a normal subgroup of a finite
group G. If P is a Sylow psubgroup of K (for some prime p). then
G = KNa(P).
Proof: For completeness we include the proof from [26].
If g Â£ G. then gPg~l < gKg~l = K. It follows that gPg~x is a Sylow
psubgroup of K. and so there exists a k 6 I\ such that hPk~l = gPg1.
Hence. P (k~lg)P(k~1g)~1. so that klg Â£ Nq(P). Therefore, we can factor
as g = k(k~xg).
Lemma 2.3.3 If G is a nilpotent group, then every maximal subgroup is normal
in G.
Proof: Let M be a maximal subgroup of G. Since M < Nq{M). we get
Na{M) = G.
33
Theorem 2.3.4 Let G be a finite group.
1.
2. If G is apgroup, then F(G) = G'GP where G' is the commutator subgroup,
and Gp is the subgroup of G generated by all pth powers.
3. If G is a finite pgroup, then G/$(G) is a vector space over Zp.
Proof: For completeness, we include the proof from [26].
1. Let P be a Svlow psubgroup of (G)
Frattini argument gives G = <&(G)Ng(P). But
and so G = NG(P) So P
low psubgroup we must have (G) the direct product of its Sylow psubgroups.
But all pgroups are nilpotent. and their direct product is then also nilpotent.
and (G) is nilpotent.
2. If M is a maximal subgroup of G, then M < G and [G : M] = p. Thus
G/M is abelian and G' < M; moreover. G/M has exponent p. so that xp e M
for all x e G. Therefore. G'GP e $(G).
3. G =pn. Since G'GP = (h(G). and hence G' < $(G). the quotient group
G/d>(G) is an abelian group. If M is any maximal subgroup it must have order
p',_1 and so G/il/ = p. So the coset Mx has order p and so M = (Mx)p = Mxp
and xp Â£ M. Now consider the coset
34
xp Â£ $(G) for all x Â£ G. It follows that ((G)x)p = $(G) and G/$(G) is
elementary abelian.
Theorem 2.3.5 The quotient group S2/S'2 is a vector space over GF{2).
Proof: So is a 2group, and so we show that for every g Â£ <5, we have
g2 Â£ S2. If g Â£ S2 has order 2. then g2 Â£ S'2. If g Â£ S has order 4, we have
already shown that g2 = [(a. a). (6. b), c] where a.b.c Â£ F. It is easy to see
that the subgroup generated by all squares of elements in S is contained in the
commutator subgroup. So by Theorem 2.3.4 we get S2 = and S2/S0 is a
vector space over Z2.
2.4 The Quotient Group S/S'
Elements in the quotient group will be representatives of the eosets of S2.
We can choose the representatives of the form [(0. a). (0, b). 0] o ol which would
give us 2q2 = representatives in S2/S2. It is easy to see that each
representative is in a different coset of S2. Furthermore, we will denote each
of these coset representatives as triples (a, b. i) where a, b Â£ GF{q) and i Â£ Z>
Then we can define the group operation in S2/S!2 by
(a, b, i) (c. d. j) = (a + c.b + d. i + j)
where addition is in the appropriate field. We note that (0.0.1) Â£ So/S2 corre
sponds to the coset containing p.
2.5 Looking for Hyperplanes
Let q = 2e and treat F = GF(q) as an edimensional vector space over
GF{2). Then for a fixed Q E GF(q)*, the map
e 1
tfc(x) =
1=0
is a linear functional; i.e., try : GF(q) i> GF(2). Letting Â£ vary over F* gives
us exactly q 1 nonzero linear functionals on F. Given a linear map T on a
vector space V we always have dim(V) = dim null(T) + dim range(T). But
range(tris a one dimensional subspace, and so miU{T) is a hyperplane in F.
So for each ( E F* there is a unique hyperplane in F. So we have q 1 distinct
hyperlplanes. But each hvperplane corresponds to a nonzero vector in F and
there are exactly q 1 such vectors in GF(q). Hence {tr^ : ( E F*} gives us all
linear functionals on GF(q).
Now choose a triple (a.b. i) E So/S!2 This is a vector space over GF{2).
Furthermore, the map
0^.cr(o. b.i) = tr\(a) + tra(b) + i
is a linear functional from S>/So onto GF(q). The kernel is a hyperplane and it
is easy to see that the vector (0. 0.1) is not in the kernel of This gives us
q2 hyperplanes of S2/S2 without the forbidden element (0.0. 1).
If we then define the map Qla(a.b. i) = tr^(a) + tra{b) we get the other
q~ hyperplanes of S2/S2, each one containing the element (0.0.1). We have
accounted for all 2q2 linear functionals on S2/S2. It follows that all elation sub
groups of S2 will correspond to the q2 hyperplanes from the kernels of the maps
0^. It is worth noting that when Â£ = <7 = 0 the elation group is the familiar
example.
We can now give an explicit description of all elation groups in the Svlow
2subgroup S2 of the group of whorls of H{3.q2). The hyperplanes in S2/S2 are
the kernels of the maps 0^.CT. In other words, for a fixed pair. Â£. a Â£ GF(q)
(both not zero) one hyperplane is the set of all (a.b.i) Â£ S2/S2 such that
Â£Â£() 1 6a(b) + i = 0. To pull back to the original group S2 we simply ask
which coset of S!7 contains the general element
(oi.a2) {bi,b2).c
o 0. It is the
coset S!2
theorem.
(0. a2 + i),(0, b2 + f>i).0
Theorem 2.5.1 Let S2 be the above mentioned Sylow 2subgroup of the group
of whorls of H(3. q2). Fix two elements Â£, a Â£ GF(q) and put ^(a, b. i) =
ti\{a) + t.ra(b) + i. Then there is an elation group E^a < S2 where
E^.a 
(cp. a2). (bi. b2). c o (jf : Qq tj(ai + a.2. bi b2. /) 0
For each case where at least one of Â£ or a is nonzero, we will call these cor
responding q2 1 elation groups of H(3, q2) "exoticr elation groups. When
Â£ = <7 = 0 we will call the group the "familiar" elation group of H(3. q2).
37
We will say that the ordered pair {(",
to simplify the notation. Fix f. a G GF(q). and for each pair a. 3 GF{q) x
GF(q). define the function a + 3 : GF(q) x GF(q) GF(2) to be a + 3 =
tr^(a) + tra(3). If we then use the notation
we can redefine the group as
Ec;.a <> 3. (
with the group operation
J C.cr
:a.3e GF(q) x GF{q):e e GF(q)
[a, /?, c]c.CT [o'. 3'. c']^ = [Q + n"^. 6 + 3,a+3. c + c + a'^Pl3T}c.0
Next we show that these exotic elation groups are not isomorphic to the
familiar elation group.
2.6 Lower central Series
Definition 2.6.1 Set T^G) = G. and T2(G) = [G. G]. Inductively define T,, =
[rn_i(G),G]. The lower central series of a nilpotent group G is the norma,l
series
G = F^G) D r2(G) D F3(G) r(G) = {id}
The length of the lower central series is the number of strict inclusions in
the series. If the length of the series is n we say the group G has nilpotency
class n. If two groups have different length central series then the two groups
38
are nonisomorphic. This follows since for any group automorphism a we have
a[n.b] = [a(a).a(b)}.
Every pgroup is nilpotent and has a lower central series. The group G has
class 2 since G = [G,G\  Z(G) and every element in the commutator has
order 2. We note that the group S2 has nilpotency class greater than 2, since
since S'2 is not abelian. We are interested in showing that each of these exotic
elation groups has nilpotency class 3.
Observation 2.6.2 Let E = Eqm < S2 be an exotic elation group of H(3. q2).
Then
E' = [E. E] =  (a, a), (b. b).c : c. G GF(q) and tr<(a) + tra{b) = o
Proof: We have already verified that S!2 = { [(a, a), (b. b). c] : a, 6, c G F}.
and clearly E' C S!2.
Put g = (ai. a2), (bi, b2). coo1 and g'  (a[. a'2), (b[, b'2). c' o &>. If i and
j are both odd. or both even, then [g, g'] = [D. 0. *]. If i is odd and j is even,
then [g.g'] = [(*. a*), (b*, b*). *] where a* a[ + a'2 and b* = b[ + U2. Since
g' G Eq.v we have tr^(a*) + tra(b*) = 0. The case with i even and j odd is the
same. This completes the proof.
Note: It is easy to see that \Ela\ = q*/4.
39
Theorem 2.6.3 Let E = E^ < S2 be an exotic elation group of f/(3.
Then E has nilpotency class 3.
Proof: We show this for E = T'u. The proof follows since {id} T3(Â£7) =
[E'.E] Â£ Z(E) whence T^E) = {/}. To show this consider T3{E) = {[g.g1] :
g Â£ E\g' Â£ E). Let g = {a.a).(b.b),c and g' = (qi. q2). (di. /i2), d
Then
o o.
99 9 V i =
(a. a). [b, b). c (a 1, a2), {3i. ih). d
(ct2. o ; [32. 3i). d + 3 GO OO
(a + Q], a + a2). [b + 3ib + L). *
(a + 02. n + 03), (6 + /?2, & + Ti). *
0.0, *
O 0
o 0
O 0
(a. a). (6, 6), c
Â£Z{E)
Similar computations show that when g' = tt (03. o2), (di, d2).d
we still get
^y1 e ^(^).
So we get the lower central series
E = T3E) D T2{E) d T3{E) D Ta(E) = {/}
The proof for all Eqm ^ E0 0 will follow once we show that = Eqi^i when
at least one of Q. o are not equal to zero.
40
Theorem 2.6.4 All of the q2 1 exotic r elation groups of H(3. q2) are pair
wise isomorphic.
Proof:
We saw in Appendix B that if 0 = q, then
/
a (3 p
112 =
0 10/3
0 0 1 d
o ol G PTU(4. q2) : p + p + a3 + 3a 0 >
IV 0 0 0 1 /
is a unitary representation of a Sylows subgroup of the group of whorls about
the point p = (0. 0. 0,1) in H{3. q2). We will use the following representation of
this group.
1I'2 = {[o, p. p] o ol : ap + Pa + p + p. = 0}
with group operation
[o. 3. p] o 0W [o'. 3'. p'} o & = [o + a'*'. i3 + 3''. c+e'* + a{P'f + 3{Wf] o oi+j
Let x = (1. a. b. c) G P \ {p1}. Then, for any id g G 114 we have
x9 = {l.(a + a)q'.(b +p)q'.(c +p + a3 + ba)q>)
If / = 0. then x9 7^ x. If i = 1. then x9 = x only if a = a + aq. 3 = b + bq, and
// = 0. That is. g is not an elation only if a, 3 G GF(q), p = 0. and i = 1.
41
When we compute the commutator subgroup of U2 we get
H 2 = {[a. b. c] : a, b, c Â£ GF(q)}
So all nonelations in 11'2 will be contained in the coset of the commutator
subgroup containing
We have 2q2 distinct coset representatives of lib /11 listed below (split into
two sets of size q2).
{[a. 3. 0] : a = (0. h), 3 = (0, k). h.ke GF(q)}
{n. ()' : O : <\ = (0. h), 3 = (0. k)h. k E GF(q)}
Since GF(q2) = GF(q) x GF(q), there are no problems when considering
each matrix entry to be in GF(q) x GF(q). Suppose that S E GF{q) with
tr(S) = 1, and let i be a root of the polynomial x2 + x + S. Then iq = i + 1 is
also a root of the polynomial. We can let any element in a Â£ GF{q) x GF(q)
be represented as a = a + bi. where a.b E GF(q). It then easily follows that
tr(b) = tr(a + aq).
Furthermore, we already know that 1T2/11 2 hs a vector space over GF(2).
So if o = (ai,a2) Â£ GF(q) x GF(q). and for some Q Â£ GF(q) we let T^(a) =
i +2), we can define the q2 linear functionals
T* : 11 h/Tl ^ GF(2) : [a. 3. 0] o 61 ^ Tc(a) + Ta{3) + i
The element [0. 0. 0] o <$> is in the kernel of T*. It follows that
= { M. i.ft] ::0T> '* J }
are q2 elation groups of H(3. q2) about p. There are exactly q2 elation groups
about p in lib. and each Eqm 0 E00 must be one of the exotic elation groups
about p.
As before, for a fixed (,cr E GF(q), we will use the notation
CCT
^c> ^ r^'T ^ ^
a = (r
and redefine the group E^a as
Â£<> = {[ 3 p\q.a : ap + 0a + p + p = 0}
with the following group operation.
[a. p. /./]C.CT [ah 10. =
Q
Q,a+J.d + /i'
/(if.3 /CI+/3 /.JAa+J
P + P + O{0)
a + 3
Q.(T
Jd) OA^ jA^
Put aa+) = a. na + J = aP, and an+J = a+T and define the map <Â£> so that
T : [o. 0. /i] i.o 1 [a + 3
We show that $ is a group isomorphism bewteen Â£T0 and Ti i. keeping in mind
that does not effect the required relationship on a.0.p.
(ja./I/./]i.o [o'. /. //]L0)
= ([a + o'5, i + /J'5 // + //5 + a(3'f + d(a')5]i.o)*
= [a + a'5 + 3 + {3's. 3 + f\ /i + ,,'5 + n( i' f + d(o')s]u
([ '? //]i.o) * (V ^.//']i,o)
= [ + /? 3, /'[ ;.i [o' + /?' *' //'[i.:
= [a + 3 + (o' + /? + p'a'+3r+3\ fx + +
+ (a + 3)(3'f^3' + 3(a + [3Y^%a
[a + a'5 + [3 + d'5.1 + /"h /./. + /h5 + o(W)S + ^(o'f]i.i
Next define the map $* so that
$* : [a. 3 /t]i.o 1 Id. o. //n 
We show that * is a group isomorphism bewteen Ei,0 and Eq,\ keeping in mind
that * does not affect the required relationship between o. i./i.
([<> >'//[ :.(j [o'. 3. //]l,0)
= ([a + o'5, 3 + l"\ // + //,s + o(d')5 + d(o,)s]I.o)
= [3+ 3"\ a + o'5. // + //'5 + r,. i'f' + d(b')5]o.i
44
(Jo,* ([a', i3, //]i.oj
; [3. ft, //]o.i l?ft7, //]o.i
= [3 + (3lS, a + a'5. // + //'" + a(3'f + 3(q')%a
Next, for S GF(q)* define the map <&,* so that
<&s [ft,/?,//]i.o 1 [h_1o. h_1d. S~2n]s o
We show that dq is a group isomorphism bewteen Ey,o arid E,){). keeping in mind
that dq does not affect the required relationship between n. .1. ft.
(^[q. d./t]i.o [fth /3.//,]i,o)
= ([ + a'a, (3 + /3's. n + //5 + a(3')s + tf(a')S]i.o)*'
 [Sl(a + a5), 6~l(3 + 3's), 6~2n + rVs +
+ 52a(3f + d23(3')%,0
(,<> M/q.o) (Vd'Vji.o)
= [<5_1a. S~li3, S~2ii]s.o [<5_1a', S~l3'. d'V]A.o
= [W1 (q + S~\3 + }. S~2fi + W2/,"^ +
+ s2q(0)^ +
= [r^a + ft/5), 6~\:3 + + ry5 +
+ r2a:(d'f + r2b(ft/)1
45
It is easy to see that is also an isomorphism between Â£0.1 and Â£0.<5 Next
define the map $^iCr such that
$C.
We show that ,. a is a group isomorphism, keeping in mind that does not
affect the required relationship between a. /.//.
The isomorphisms $. $*, show that all q2 1 exotic elation groups
are isomorphic. That is, there are exactly two elation groups of H(3.q2), up to
isomorphism.
Although we have shown isomorphism, to satisfy curiosity, we mention some
other isomorphisms between these exotic elation groups in the original represen
{[a + a'T B + Ts, n + n'a + q(B')s + pia'f] Li)^
[C> + a,S). + B'5). rV_1(/t + /''* + W>r + /3(a')a )]<.*
tat ion of the exotic elation groups.
Theorem 2.6.5 Let s.t be distinct nonzero elements ofGF(q). Then if t =
sd G GF(q). the two elation groups Et(^^ and E^iri^ are isomorphic under the
map
A :
a. 3. c
o 0
da.6 13.c
o
This gives us q 1 orbits of size q 1.
Proof: Suppose that t = sS G GF(q)* and consider the map:
a. ;3. c
o d)
6a. 6 L3.c
o((L
Then A is a group isomorphism of S?
Recall that
^CC
{ (cp, 02) bo), c o o' : + cio. b\ + bo. i) 0 j1
(fli.2). (61 b2),c
o 6l : 0 N /. \ (ai + a2b[ + bo. i) = 0
<>).<>10c) '
(i <^2) {bibo). c o o' : 0(s(.) ^_lt^ + 02] <5 + bo}. i) 0
47
We now apply A to an elation group EtQ.c.
AIÂ£<]
 S(ai,a2).S 1{bi,b2),c o o : + fl2^1 + b2,i) = o
=  S(ai,a2).S 1(bi,b2),c 00' : 0s(,s 1t( + 02]* $ 1 [Â£x H 62] /) = 0 j>
Oh, ;r2), (yi. y2). cod1: 0sC.(sit)f (xi + ;r2. y1 + y2, i) = o
= E,
sC.s^C
Hence EK< = EsC;Asit)f.
Next suppose that Es^_sitQ = Et$.p for any t G GF(q) and d. C G GF(q).
Then s = t6Cl and
slK = (t[3cY1tC
= r1/T1C*C
= .r1c2
48
But this equals 3 if and only if C = 3 That is. under the map A. is not
the image of for any t G GF(q)* when 3 ^ Â£. This gives us q 1 orbits
for each t GF{q)*. each orbit containing q 1 groups for each <) G GF(q)*.
Theorem 2.6.6 Let Â£, <7 be distinct nonzero elements of GF{q). Then if Q =
ad G GF(q). Â£Â£ 0 Ea0 and Eop = Eqm under the map
A: ajd.e o
This gives us 2 orbits of size q 1.
Proof: Recall the following group isomorphism:
A: a,3.e o#h 5a. 6 i. ( of1
Then since Â£ = ad for some 5 G GF(cj) we get
T'o.c  (fli G2). (frii b'2). e o h : 0o.c(
= { (ay.a2).(bi,b2).c o
=  (rii. a2). (61, b2).c o cf : 0O(T(d(a 1 + a2). 5 l(b^ + 62)0 = o
, c op! :
Using the same argument as in the previous theorem we get
49
Since for all (.a Â£ GF(q)* there is some 5 such that Â£ = crh. we must have
Â£o.C = E0a for all 11011zero (.a Â£ GF(q). Similar computations show that
Eq q = Ea,o for all nonzero Â£, a G GF(q).
We have partitioned the q2 1 exotic elation groups into q + 1 orbits of
size q 1. We will list the orbits as [^iC.c] [E'i.o] [Eo,i]. where tt
ranges over GF(q)*. Next, consider the map
and Et^ = Et3_3 for all Â£,Â£? G GF(q). So A provides an isomorphism between
the q 1 Aorbits [Â£flC.c],....
We mention one other known group isomorphism. Consider the map Aa
where a = 2s. Then
A : [a. 3 c] o [ha, 5,3. h2c] o
Now consider the following elation group where Â£ = /i 5.
=  (ai. a2). (bi,b2).c o (pl : 0M. , + a.2), S(b1 + bo), i) = 0 j
A =  5(ai.a2).5(bi.b2).Sc o </>l : (),;(5{a: + a2),S(bi + b2),i) oj
50
A17 : [a. ;3. c] o#h [o'7. ;3a. caJ o
is also a group isomorphism for any automorphism a. To see the image of
such a map, let q = 2e and consider the following, where (a) = GF(q)* and
a. c G GF(q).
tr (c Aa(a)) = tr (caa))
= tr (a1 (cd)2 )
n=0
r n2+s
Â£ [a12"' (J)
n=Q
e1
=E
n=0
= tr (c2 a
where the second to last equality follows since raising to the 2s power is an iso
morphism and does not change the value of the absolute trace function.
2.7 Building Subgroups A{t) in the Exotic Elation Group E
Let q = 2e and denote a G GF{q2) by (ai.ao) where , G GF{q). Using the
absolute trace function tr : GF(q) h> GF{2) we define the map T : GF(q) x
GF(q) ^ GF{2) by
51
T(a) = tr(a.i) + tr(a2)
Similar to when we worked with the unitary representation of these groups,
we put. aa = aPr(cvF We now define the elation group Fffio as follows
ELo = {
with group product
Q, f3, c
: a. 3 Â£ GF(q) x GF(q), c. Â£
3 v /
Q. , 3. c * a , i. c
or', [3' c' = [or + o,Q, /i + 3,a. c + d + lo')0 / r\.
For the remainder of this paper we will assume that E = E\,0. We also note
that in some computations we will revert back to the old notation, using aP
instead of aa.
For each t Â£ GF(q), let d (f) be a function from GF(q) xGF(q) into GF(q) x
GF(q). and let gt be a map from GF(q) xGF(q) into GF(q). For each t Â£ GF{q)
we also have a subset A(t) of order q2 such that
.4(() = {
Q. Or
m
What are the necessary and sufficient conditions on d(t) and gt so that A(t) is
a group?
52
2.8 The Function S(t)
Consider the product
9 9
f
a.am,gt{a)
a + (a'f:am
a'. a,6{t), gt(a')
+ (a'm)a,gt(a) + gt(a') + (a6^)sPa'T
Hence am + {a,sw)R = (a + (a'f)S(t).
We show that S(t) is an additive function.
First let a = 0. Then (F'w + a/dW = o'Sit) and = 0. Now if T(a) = 0.
we get q5^ + a'* = (a + awhich holds regardless of the value of T(a').
We next show that a6^P = (aP)6^ for all a G GF{q) x GF(q). We first
suppose that T(a) = 1. Then asw + a1 Sit)P = (a + a'P)sw. and if we put
a = aP. then
am + {aP)mP = (a + aPP)m = 0 => amP = {aP)m
Next, we show that F r P = (dP)d(<) when T(3) = 0. If we have some 3
such that T(3) = 0 we know that we can choose two elements a = (cti.O) and
a' = (q^.0) such that T(a) = T(tx') = 1 and o +a'P = p. Letting 3 = cx + aP
it is clear that that 8{t) satisfies (aP + cy')6w = [(o + a'P)P]6^.
53
From above we know S(t) satisfies odil> + o' = ( + a'P)S{1). Using the
fact that T(a') = T(a'P) = 1 we get
am + a'm P = [a + a'P)m
am + a'Pm = (a+ oc'P)m
(.am + q'P *lt)) P = (q + a P)m P
amP + a'PmP = (a + a'P)6it) P
aPm + a'PP6w = (a + a'P)m P
aPm + am = (a + a'P)m P
(aP + a') m = (a + a'P)m P
jpd(t) __ p
So when T(j3) = 0 we get = i3Ps^K This completes the argument that
Q/d(t)p Qtp6(t) for ap a ^ GF(q) x GF(q). Xow recall that when T(a) = 1.
S(t) must satisfy cUw + a'^P = (a + a.'P)6^. Using the above computations
we see that when T{o) = 1 we get a6w Pa'PS{t) (o + a'P)sw. It follows that
S(t) is an additive function on all of GF(q) x GF(q). Fortunately, there is an
easy classification of additive functions from GF(q) into GF(q).
Theorem 2.8.1 If f : GF(q) i> GF(q) is an additive function, then
f(x) = a0x + aprp + a2xp~ + + ae_ pU'
where a, E GF{q).
Proof: See Theorem A.G in Appendix A.
54
Suppose that S(t) is the additive function that is used in the definition of
the subgroup A(t). Given the above theorem, for all a Â£ GF(q) x GF(q) we
must have
Unfortunately, we do not have a complete answer about which additive func
tions will suffice. However, we do have the following conjecture about S(t).
be a subset of Ei. .4 necessary condition that A(t) be a subgroup is that S(t) be
additive function of the form
.Td'w = ax
for some a Â£ GF (q).
2.9 The Function gt
We need to determine properties of gt so that A(t) is a subgroup. Multiplying
two elements in A(t) we easily get the following condition on gt.
[am]P
(qqCl)P + (cpcffi) P + ^a.20f ^ P + + ^(l2elMP
= a0(aP) + ai{aP)p + a2{aP)p~ Hh a2e^i{aP)r''
= (aP)m
Conjecture 2.8.2 Let S(t) : GF (q2) i> GF (q2). Then let
A(t) = {LSWj((q) : q Â£ GF(q) X GF(q), c Â£ GF(q)~j
9t(a) + gt(a') + gt(a + (af) = (a,)sP(at>)T
(2.1)
(2.2)
9t(a) + gtW) + gt(aa' + ft') = a'P(a'm)T
We choose the additive function aS{f'1 = fa. for t G GF(q). Then gt must
satisfy
gt{a' + ft') + gt((a')s + ft) = t. a?'Pa'T + a*'P{a')r
Note that if we let q = ft' we get <7t(0) = 0.
(2.3)
Next suppose that a (qi.q2) and ft' = (a), a)), and we consider the
following cases.
(i) T(a) = T(a') = 1: Then from equation 2.3 we get
g{ot i + a2 ^2 + ft)) = g{ct2 + ftq. O' i + ft))
(ii) T{a) ^ T(a'): Then from equation 2.3 we get
<7(oX + ft'), ft2 + ft)) + (/(ftq + ft) ftl + ft)) : t ftftft) + ft2ft) T ftlft) + ft2ft)
(iii) Let T(a) = 0: Then from equation 2.1 we get
9t{o) + 9t(c*') + 9t(a + a) = ct
H
W
ft)
i\T
Before we determine acceptable functions gt. we characterize all functions
from GF(q) x GF(q) to GF(q) as polynomials.
Theorem 2.9.1 Let gt(a) be a function from GF(q) xGF(q) into GF(q). Then
there is a distinct polynomial associated with gt(a) of the form
fft(a) = 22 i2 where c.j.k e GF(q)
0
Proof: See Theorem A. 7 in Appendix A.
Thinking of gt(ct) as a polynomial, if we evaluate the function gt{ct) only
on elements from GF(q) = {a = (qi.O) : Qi G GF(q)}, because the function is
additive on these elements, it must be of the form
gt(a) = oicv'i + ci2af Hb ae_iof 1 + 22 c'LJa\Qi
0
1<3
where a0 = 0 since gt{0) = 0. Then recalling that gt{o) = gt(aP) we have the
following restriction for gt.
9t{a)
+ Q'i) + + 22 cjk(ail2 A aia2) F
i
22 di{aia2y
l
Using the fact that gt(a + o') + f/f(cw) + gt{o') = f [opo^ + cua)] we compute the
following:
gt{a + a) + gt(a) + gt{ o')
Qi(ai+Q'2) ++ d^ayUj)1 + Cj^,{Qiai + a{a2) +
1<1
ci\{oi^ + a'^) + + 'y ^ d,{a[a'2y + y ^ T o^\Oi'2) T
l
Qi(o!i + aj + cb + o^) d~ y di((\i + o,1)(q*2 + ft2)1" "h
l<(<2el
cj.lt([Ql + a',l]A;[a2 + a'2^ + [ftl + ttlF[Q2 + Â£*2]*)
1 < j < A; < 2e 1
= 'y2 cj.k y*iai + 1Q2 + aiQ2 + o/{a2 + [1 + a'J^cnj + +
l
[1 + Oi],7[2 + + y ^ di
l
(OlO'2)i + (o^O^)* + [al + Oi]?[a2 + Ct'iY
\
t 0
Q
iT
= fjonO^ + OjOj]
We now make the substitution a' = aP and get
y ^ di(tt 1 + O2)* =
l
for all cv 1, OL'2 G GF{q). That is.
y ^ did' = t[3 =t (f + d\)f3 + dod^ + + dqid(/ 1 = 0
l
58
for all [3 Â£ GF(q). But this is a polynomial of degree q 1. and if every element
of GF(q) is a root it must be the zero polynomial. It follows that di = t and
di = 0 when i ^ 1. We have shown that necessarily.
Qt{Ot) flq (c*i + G2)2 + WClel (W +09)'^ 1+toiQ9+ y ' +
Lets make the assumption a\ = on and a2 ^ 0 = a'2. Then
(ot\
tct\ CV2
Q
iT
\t0/
= gt{a + a) + gt(a) + gt{a)
= l0 [0:2 + y cLfr(alQ2 + kl2)
l
So we have
cjk (QiQ^ + Qia2 j = 0.
l
Since for a fixed q2. this is a polynomial of degree less than or equal to 2e 1
that is zero at all cq e GF(q). we must have Cj= 0 for all 1 < j. k < 2t 1.
We have shown the following.
59
Theorem 2.9.2 Let gt(a) : GF(q) x GF{q) ^ GF(q). Then let A{t) =
 a.ta.gt(a)
he a subset of the elation group E. A necessary condition that
A(t) be a subgroup is that gt(a) be a function of the form
(
gt(a) = a
m t
6 1
Q
V
+ (1 + ftj)'
i=l
o f(t)
We are now ready to talk about 4gonal families when
defined above. Define the subgroup .4(oc) as
.4(oc) =  0, q. 0 
and form a set of 1 + q subgroups of order q2 as
F = {.4(f) : t E GF(q) U oc}
and then let .4(0* equal .4(f) Z{E) which gives
4(?)* = {
Q.fQ.C
: Q e
GF{q) X GF{q), c GF(q)}
It is clear that if we form ^4*(f) in this manner, then only subgroups A(t) that
trivially intersect the center will allow F and F* to satisfy K2.
We want to know which of the families built from F and F* satisfy Kl, as
we already know they satisfy K2. Essentially, we are looking at relationships
between values on the diagonal entries of a g(t) and g(s) that allow the sets to
form an EGQ. Once this is decided will will decide which GQ are isomorphic to
the classical H(3.q2).
60
Recall that AT says that A(t)A(s) f] A(k) = {id} when s.t. and k are
distinct. The following result will be helpful.
Theorem 2.9.3 (Payne) Let A{t), A(s), and A(k) be subgroups of order q1.
Then A(t,)A(s) f)A(k) = {id} if and only if A(tn) A(s7T)f}A(k~) = {id} for
every permutation ~ of s, t. k.
We will check A(t)A(s) f) A(k) = {id,} when k = oc and s. t. k ^ oc. Choose
g = n
a, ta. gt(a)
E A(t) and also choose g' ~ ah sex', gs(a') E T(.s), with
t 7^ s and gt(a) defined as
gt{a) = a
^/(O t \ ti r , >2'
\ a + V, K(qi + av
\ 0 f{t) J L
Then for the product g g' we get
/  ' m t N T / / / ^ f(s) S tT
li a.to.a a a.sa a a
^ 0 m/ v o f(s)J
= a + o'3, ta + so'3. *
where
= /(0[Q1 + Ql\ + /(s)[(ai)2 + (q2)2] + + saia'2 +
e 1
+ Â£b(
i=i
/ V2'
Ol\ + of,)
G1
For any a G GF(q2). g g' is in A(k). where k = oc. if and only if the
following conditions hold.
(i) q = o' which means that aPc/r = 0.
(ii) fit) + f(s) (a'i + a) +
t + s
a]a2 = 0 for any id a G GF(q2).
If we were to have f(t) = f(s) for distinct s and t, then choose a with
Qi = 0 and q2 7^ 0 to violate A'I. It follows that f(t) must be a bijection on
GF{q). Furthermore, property (ii) is equivalent to
7 {t) + f{3y
tr
t + s
 1.
Next suppose that k G GF(q). Then we violate AT provided we satisfy the
following two conditions.
(i) {a, a'} is a GA(g)dependent set with
a =
k +1,
k + s
a
a, =
k +1
k T s
a. and of, =
k +1
k + s
09
f (k) + f {k) [7,^ +/7) + /(7
k + s
k + k
k +1
k + s
k + s
+ t + s
(ay + a.]) +
k + t
k + s
OiQ2 = 0
for all nonidentity a = (au. a2) This condition, when simplifed. says that
we satisfy AT if and only if for all distinct s. t. k G GF(q) we satisfy
ff{k)[s2 + t2} + f(t)[k2 + s2] + /(s)[A~2 + t2}\ _
V k[.s2 + t2} + t[k2 + .s2] + .s[A2 + t2} ) ~
62
Since all cases for A'l will be some permutation of the last two cases we
know that we have found necessary conditions for F = {A(t) : t G GF(q) U oc}
to satisfy Al.
From here we would like to see if WLOG we can assume that /(0) = 0.
If a = (011.02). consider the map
A :
a. p. c
a, 3. c + S(q( + o.j)
where S e GF(q). Clearly. A : id t> id. Next see that
A(W) = a(
Q. P. C
/ of f
a . J C
= a (jo (u'f.p + {.i' f.c + F + ia'rr.i1
a + (a'f. 0 + (p'f. c + d + [a'fPP +
<5( [Q1 + (<^l )Ct]2 + [
o. 3. c + 5 (oj + a*)
= A(g) Mg)
= 7T
O (pr^a'> 7T
a
+ + a?)
o
T(a')
and A is an automorphism of E. It is easy to see that when f(t) is a Injection
on GF{q), then
A : Af(t) Ah(t)
G3
where
A,(t) = j 1 T + E ai(i + aiY
1 1 /, i=1 L
A,(t) = 1 a, tea, ca t \ T + E Q;(l + a^Y
1 V 0 M0 i j=i _
where h(t) = f(t) + /(0).
Furthermore. A fixes every point A*(t). Since A is an automorphism of E
and its coset geometries we can WLOG assume that
A(t) = < < q, ta, a f m t ' 1 oY + N Gj(cVi + O9)2
< y 0 fit) j * ^ L J t=l J
where f(t) is a bijection such that /(0) = 0. In this setting we satisfy Kl if and
only if the following conditions hold.
(0
tr
7(0 +/(g)
t + s
= 1. t^s
11
tr
7(0"
= 1. t 0
in
tr,w +/(.)>, = 1
ts2 + St2
64
These conditions are certainly true if we choose the function f(t) = 3t.
where tr(p) = 1. We aim to show the following theorem.
Theorem 2.9.4 Consider the set
F = ^A(t) =  a,ta,gt(a) J : t G GF(c/)J (J .4(oc) = { [O.o.O]} 
r i j i t i
y i(a i +02)'
where gt{ot) = a
m t
\
0 m
Z=1
Then if we let F* = jyT(f) = A(t) Z(Â£') U j^T(oc) = A(oc) Z{E) j, a
necessary condition that F be a 4gonal family is that f(t) = ;3t for some fixed
3 G GF(q) such that tr(3) = 1.
Proof: Using condition (i) above, the result follows from Theorem 2.9.5
due to Pent ilia and O'Keefe, letting 7 = 1.
Theorem 2.9.5 Let f : GF[q) ^ GF(q) be a function satisfying /(0) = 0 and
let 7 G Aut GF{q). The equation
, \fir) + fid) 1 1
' L J
for all x.y G GF(q) with x y if and only if /(:r) = 3x~ for some 3 G
GF{q) with tr(3) = 1.
65
Proof: We rely on the proof in [16].
Let rn be an integer satisfying 1 < rn < q 1. (rn, q 1) = 1 and
/(?') + fiy)
{x + y)m
for all x, y G GF(q) with x ^ y. It follows immediately that F is a permutation
of GF(q), for otherwise there exist x,y G GF(q) with f(x) = f(y) and
Let 7Z be the minimal set of residues modulo q 1 such that for any z G
{0___,q 2} there exists a unique y G 1Z and i G {0,..../; 1} such that
z = 2'y(mod q 1). We remark that 0 G 1Z. For p G 7Z. let lp = \{p2l : i =
0_______ h 1} (that is. lp is the length of the orbit of p under Aut GF(q)). Let
/() = Zto QF'1 and let
where p2k + m is the unique integer such that 0 < p2k + m < q 1 and
p2k + rn = p2k + rn(mod q 1). It was shown in Glynn [11] that
for all ^ G GF(q) and for all p G 1Z \ {0}. Further. tr(am) = 1.
Suppose that m = q G AutGF(q). Since m is a power of 2. the arguments
in Glynns Lemma 4.10 [11] show that the only nonzero apk is o00 = am (the
66
coefficient of ;rm in /), hence f(x) = ax7 for some a G GF{q) with tr(a) = 1.
The converse is immediate, since
ax~ + ay7
(x + y)y
= tr(a) = 1
2.10 Property(G)
In the remaining sections we show that when we consider the function
QS(t) jie oniy EGQ which arise are classical. We start with the fol
lowing definition for Property (G).
Definition 2.10.1 A GQ S with parameters (q2.q) has Property (G) at the
point, p provided the following holds. Let L\ arid Mi be distinct lines incident with
the point p. Let M2, M3, il/4 be distinct lines and L1.L2.L3.L4 be distinct
lines for which Lt ~ Mj whenever i + j < 7. Then L4 ~ i\/4.
Figure 2.1 may be helpful.
To help see the motivation behind this definition we restate the following
theorem.
Theorem 2.10.2 (R.C. Bose) [3] Let S beaGQ with parameters (s,t). Then
the following statements are equivalent:
(0 t = s2
(ii) For some pair {.r. y) of noncoUmear points, each triad (x.y.z) has a con
stant number of centers, in which case this constant is 1 + s.
67
Figure 2.1: Propertv(G)
(Hi) Every triad of points has a constant number of centers, in which case this
constant is 1 + s.
This theorem tells us that every triad of points in a GQ with parameters
(q. q2) has exactly 1 + q centers. Dually, it the follows that every triad of lines
in a GQ with parameters (q2. q) has exactly q + 1 transversals. This result is
known as the theorem of Bose. Now let Li.L2.Li be distinct skew lines and
Mi. M2. Ms AL\ be transversals of Li.L2.La such that M\ and Li are both in
cident with the point p. We would like to know if each of the transversals
of Mi. M2, Ma also intersect the line A/4. If so. we are guaranteed to have a
(q + 1) x (q + 1) grid about the point p.
Results by both J. Tlias and Matt Brown have shown that any GQ with
68
parameters (q2. q) and having property (G) at a point must be a flockGQ. We
wish to show that all EGQ arising from these 4gonal families in exotic elation
groups of H(3. q2) are flockGQ.
We start with the following facts about incidence in these exotic GQ. where
the notation f e F, suggests that two points are collinear via a line of type
A(t) g. Also, for the remainder of this section, disregard the notation a = PT^a)
and simply consider a as an element of GF(q2).
i
[
= {[a, 13' + 3. c' + c + aP1+3,T] : (3' G F*. c' eF,}
= {[a./?.c]:/3eF2,ceF9}
= {[0,/Ac] : 3 G F^. c G Fg [a. 0. 0]
= A*{oo) [a. 0.0]
(ii)
[a. f3, c] ~ A(oc) [o, 3 c]
= {[0.^.0] :,r Â£F,2}[q,;J.c]
= {[a..3' + ;i.c + oP.rT]:/)'eF'}
69
(iii)
[a, p. c] ~ A*(t) [a, 1. e]
=  [a7, to', c] : a e F'\ c Â£ F9 [a. 3, c]
=  a' + aPTin').ta' + pPTia'\P + c + aPl+3rf }
= { a' + aPT[a'Kt (o' + aPT(Q,)) + taPT[Q'] + pPT(a'\c* J
= { a' + aPT(a'\t (a' + aPT(ft,)) + (to + 3) PT(a'\c
= A*(t) [0, (to + 3)PT{a\ 0]
[a, P, c] ~ .4(f) [a, p, c]
= { [a'J.tx\gt(a')\ } [a,p,c]
=  a' + aPT{Q'].ta' + pPTin').c + gt{a,) + ta,P1+T{a')aT j
= { a' + aPT{a'Kt (o' + o:PT(a,)) +
gt(aPT{a']) + +gt{a' + aP^)] }
= A(t) [0, (ta + 3)P,c + gt(a)]
To see this, we use gt(a + a') + gpo) + gt{ot') = a'PaT and gt{a) = gt(aP).
In Figure 2.2 we have constructed a 3 x 3 grid at the point (oc). with lines
[.4(oc)] and [.4(f)], and containing the fixed point [a. /hr].
The coordinates of each point and line are given in the following list:
70
Figure 2.2: 3x3 Grid.
A(x)l0 = A{x)[a,P
A(oc)Ll = A(oc) [a*, 0, gt(a*) + gt(a) + c]
A(t)L 0 = A(t)\0.(ta + 3)PTia\gt(a) + c]
A{t)Ll = A(t)\0.{ta + 3 + 3r)P.c + gt(a)+aP3'T}
P0 = [a*.ta* + (ta + 3)PT(a+0,'),c + gt(a)+gt(a*)]
Pi = [a. 3' + /i,c + aP3ri]
f r
a. ta + (to + 3 + 3/)F,T(a+"). (//(q) + gt(a) + c + to.Pa.
P2 =L
a*. 3 + 3* + (ta + 3)PT{o+a].gt(a*) + gt(a) + c.+ a*P(3T
We must have a* = a, so we rewrite the coordinate for P2.
71
a*, to* + (to + 3 + 3>)PT{a+a">. gt{a*) + gt(a) + c. + taPa*T
a*. 3 + ta* + (to + 3)PT{a+a*\, gt{a*) + gt{a) + c + a*P[3T
The point P2 exists if and only if the following conditions hold:
ta* + 3'PT(a+an = 3 + ta *
taPa*T = a* P3t
The theorem of Bose guarantees that a solution will exist. We now consider
the case where Li = [.4(f)] and Mi = [/l(s)]. t ^ s.
In Figure 2.3 we have constructed a 3 x 3 grid at the point (oc). having lines
[.4(f)] and [.4(s)]. and the fixed point [a,f3,c\.
72
Figure 2.3: 3x3 Grid.
A(s)Lo = A(s) [0, (.so + i3)PT^,ga(a) + c]
A(t)Lo = A(t) [0. (to + ;3)Pt^. gt(a) + c]
A(t.)Ll = A(t) 0. ([t + s]as + (so + f3)PT^+n^pT^\
SV(to) + <7s(os) + <7.s (o) 4 c
Pi = [os. sos + (.so + 3)PT(a+as\ g(as) + <7.5(0) + c]
P0 = [af. tat + (to + 3)PT^\gt(,) + <7,(0) + c]
a*, ta* + (([f + s]a.5 4 (.so + ,5)Pr(n+ft*^ pn.)J
p _t ^ 9t(u*) + <7<(s) + gs(as) + (7.,(a) + c
a'. Sa' + ( ([^ + s]o4 + (to + jjjpTia+at)^ pPap'j pT(a)'
9s{oi') + gs(at) + (]t(at) + gt(a) + c
73
We must have a* = a', and so we rewrite the coordinates for P>.
to, to* +(([* + s]os + (sa + 3)PT{a+a*^ PT{a^ PrW).
9t{<**) + gt(c*s) + 9s(as) + 9s (a) + c
P2 =1
to. sa* +(([# + s]at + (to + 3)PT{a+nt)^j Pr(a())
9s{<**) + 9s(at) + 9t(at) + 9t(a) + c
For notational simplicity", given fixed pairs of elements a. 3 6 and
t Â£ Fg, define the following functions:
/a(7) = (\t + s], + (sq + 3)PT(Q+~i)^jPT{~)
/f(7) = {[t + + (to + 3)pna+3^pT( 7)
hs (7) = 9th) + 9shf) + 9 s(a)
^h) 9th) + 9sh') + 9t(a)
So we have a 3 x 3 grid if and only if the following conditions hold:
to* + fs(as)PTla] = sa* + /^(f)Pr(0*,
gt(a*) + hs(as) = ga(a*) + ht(at)
The theorem of Bose guarantees that such a 3tuple a*.at.as must exist.
In Figure 2.4 we have constructed a near 4x4 grid containing the 3x3 grid
with point \aj3. c].
M^)l2
Figure 2.4: Near 4x4 Grid.
The computations used to determine the previous 3x3 grid give us the
following information for the points and lines of the near 4x4 grid:
P* = \a.3.c\
Mt)Lo = M*) [0, (ta + ;l)l,ha]. gt{a) + c]
A{t)Ll = A(t) [0,/'fn,). hs(as) + c)
A(t)L.2 = A(t) [0, fs(as). hs(as) + c]
A{s)Lo = A(t) [0. (sa + f3)P.gs{a) + c]
A(s)Ll = A(t) [0, fiat), /i'(a,) + c]
A{s)l2 = A(t) [0, /(at), + c]
76
Po [at'tat + (to + j3)PT^a+at\ + gt(cy) + c]
Pi = sots + (so + p)PT(a+a*).gs(as) + gs{a) + c]
a\ta* + fa(a3)PT(am].gt(ci*) + hs{as) + c
p2 =i
a*..sa* + f s(c\t)PT{a'\ gs{a*) + ht(Qt) +c
a, to + f*(as)PT{a'1, gt(a) + hs(as) + c
P* =
d, S'Q + ft(at)PT{aK ga(6t) + /d(at) + c
<
q, td + fs(as)PJ (a\ (a) + hs(as) + c
f
S
d,sd + /f(at)PT(rt),
P5 = [a8, sas + (sa + ;3)PT(a*\ g8(aa) + r/8(a) + c]
P6 = [dt, to* + (fa + 3)PTiM}. gt(dt) + gt(a) + c]
Given this near 4x4 grid, the points P>. P$. and P4 tell ns that we satisfy
the following conditions:
77
to* + fs(a8)PT(a'] = sa* + f{at)PTia')
to + fs(as)PT{&) = sa + ft(at)PTlA)
to + fs{as)PT[6) = sa + /t(to)JPT()
Multiplying each equation by the appropriate power of P and adding equa
tions we see that if we put
Q = Q*pMa+o) + dpT(a*+n) + &pT^+&) (2.4)
then we also satisfy the following condition:
to + f8(as)PTm = sa + f\at)PnTS) (2.5)
Given this near 4x4 grid we also satisfy another set of conditions:
gt{a*) + hs(as) = gs(a*) + //(Of)
gt{a) + hs(as) = gs(a) + Jif{at)
gt(a) + hs(as) = gs(d) + /?(ot)
Adding these equations together we get the equivalent condition:
9t{cP) + gt(&) + gt{a) + h*{as) = g3(a*) + g,{d) + gs(d) + /if(t) (2.6)
78
We now ask if there is some point C which completes the 4x4 grid to give
us property (G). This point would have to have the following coordinates:
a, ta + fs(as)PTi&).gt{a) + hs(as) + c
C[=<
q, sdi + ft(at)PT{a'). gs{d) + /d(af) + c
We note that if there is a solution it will be a unique solution.
This point is on both A{t)i, and A(s)i2 if and only if
ta + fs(as)PTiA) = sa + f(dt)PT{6) (2.7)
gt(d) + hs(ds) = gs(a) + /?f(at) (2.8)
If we let a = a as defined in equation 2.4 we have satisfied equation 2.7.
Then using equation 2.6 and the fact that
gt{a + a') = gt{a) + gt(a') + a
we see that satisfying equation 2.8 depends on satisfying
t (a*Pa + a*Pa + a Pa) = s {a*Pd + a*Pd + dPd).
yfOy
a
>T
79
Equivalently, since t s, we must satisfy
a* Pa + a* Pa + aPa = 0
The point is that a solution to equations 2.7 and 2.8 does not depend on
the choice of the function gt. Then recalling that at least one choice gt produces
H(3. q2), which S.E.Payne has shown, see [8]. has property(G) at the point (oc).
we see that we will satisfy equations 2.7 and 2.8 for any choice of gt.
Therefore, when the near 4x4 grid contains lines [/1(f)] and [,4(s)] incident
with (og), we have satisfied property(G) at the point (oc). Similar computations
show this is also true when the near 4x4 grid contains the lines [,4(oc)] and
[.4(f)], although we omit the details here.
2.11 A Theorem of Matt Brown
We now state a powerful result due to Matt Brown, see [5].
Theorem 2.11.1 Let S = (V.B.I) be a GQ{q.q2). q > 1. and assume that S
satisfies property(G) at some line /. Then S is the dual of a flockGQ.
It follows that each EGQ constructed from the exotic elation groups of
H(3.q2) are flockGQ.
80
2.12 The Final Push
Consider the following theorem from [28].
Theorem 2.12.1 Let S be a GQ and let H be a group of whorls about the point
x acting transitively on the set X = P \ {x}1. The set of elations in H does not
form a group if and only if (at least) one of the following conditions is satisfied:
(1) There is a j > 2 for which \fix(a)\ = j for some a H.
(2) There is a proper thick subGQ of S containing x (and all the lines through
x) fixed pointwise by a nonidentity element of H.
Theorem 2.12.2 Let S(tF) be a nonclassical flock generalized quadrangle of
order (q2,q), q > 1. qeven. Then the set of all elations about (oc) does form a
group.
Recall that 6 is a whorl about (oc) that fixes more that one point in
P \ {(oc)}. It follows that every "exotic EGQ that we have constructed is
classical, and hence isomorphic to the Hermitian surface H(3.q2).
2.13 Suggested Problems
In this section we suggest how to further this research. First, and possibly
the most obvious question:
81
Question (1): Are there any new examples of EGQ that may be constructed
from these new groups?
We might also ask the following question:
Question (2): Are there conditions that guarantee an EGQ have non
isomorphic elation groups?
The following conjecture was recently made by K. Tlias.
Conjecture 2.13.1 (K. Thas) If Property (F) does not. hold for {S^KG). then
S has nonisomorphic elation groups.
Property (F) depends on whether each A* E F* is normal in the elation
group.
82
Appendix A. Theorems
d
Lemma A.l d'^^t2 >
i=i
Proof: Taken from
&
[!]
9
d
Set dm = ti. Then after some straightforward manipulations we see that
i=i
d
0 < ^(m ti)2
i=1
The result easily follows.
d
i = 1
Wo will use this lemma to help prove the followng inequality due to Higman.
Theorem A.l Let S = (P.B.I) be a GQ of order (s. t). Then .s < f2 and
dually t < s2. Furthermore, t = s2 if and only if for some pair (x.y) of non
collinear points every triad (x.y.w) has exactly s +1 centers if and only if every
triad of points has exactly 1 + s centers.
Proof: Taken from [22],
Now we let x. y be fixed noncollinear points with perp given by
{.r.i/}1 = {^o....t}
83
and let V = {ui.....uy} be the set of all points such that {x.y.w} is a triad.
It follows that d s2t st s + t. Now for 1 < / < d. let L be the number of Zj
such that zj E {x. y. tLj}1. And for 0 < i < t + 1 let be the umber of triads
{x. y..} with exactly i centers. Counting the total number of triads we get
f+i
77; = .S2t st S + t
1=0
Xow counting the total number of ordered pairs (u\ z) such that 3 e {x, y. a'}1
we get
<+i
= Y1ini = (t+i)(f i)s = (t2 !)s
i=i
i=0
This follows since for each of the t + 1 points in {x. y}1 there are s points on
each of the t 1 lines not through x or y.
Now counting the number of ordered triples (w,z,z') such that w E
{x. y, z.z'}1 we get
cl t+1 , I \
= l 2 ] (t 1) 2 = (A2 l)t
i=1 i=0 ^ '
This follows since for each of the (,^1) ways to pick x, z' from {x. y}1, there are
exactly t 1 points (not in {x. y}1) which are the intersection of lines through
x. z'. But the triples are ordered and so it is obvious that
i = 1
d
If we put dm = ^ ti it follows from Lemma 1 that
;=i
84
d
0 <  uf = t(s l)(.s2 t)
i=1
It is now an easy conclusion that t < s2. Furthermore, if t = s2 we we must
have m = t; for each 1 < i < d ancl every triad has the same number of centers.
In this case there are s4 s3 + s2 s total triads each having the same number
k of centers. It follows that
d
A'(s4 sA + s2 s) = ^ ti = (s4 l).s = s3 s
1
and we must have k = s + 1.
Theorem A.2 Let G be a group of order s2t and let F = {A0. ..., At} be a
family of t + 1 subgroups, each with order s. arid let F* = {Ag. A*..... A*t] be
another family of t + 1 subgroups, each having order st where A, < A* for each
0 < i < r. Then if we build the coset geometry iS(oc) as prescribed above, S(oc)
is a. GQ, having order (s,t). if and if properties K1 and. K2 hold, where
Kl : AjAi n A/,. = {id} for all distinct i.j. k.
K2 : A* A, = {id} for all i ^ j.
Proof: We follow the proof in [20].
Two lines must be incident with at most one common point. We observe
that two distinct lines of type (ii) are incident only at the point (oc). Then for
85
each 0 < j < r. the cosets {Ajg : g G G} partition the group elements, from
which it follows that two lines A3g and Ajh may only be incident at a point A*g.
Next suppose that two points are incident with two lines Ajg and Ajh. i ^ j.
Since any element in a coset Ajg or AJi can be chosen as a coset representative,
we can WLOG assume that the points are group elements g and h. But then
g.h G Ajg D Ajh. or equivalently id. gh 1 G Aj fl Aj. We have shown that no
two points are incident with two different lines if and only if A, fl Aj = {id} for
each i ^ j. From now, we assume that the following property holds:
PI : Ai fl Aj = {id} for all i ^ j.
Next we show that there are no triangles in the geometry. First, there can
be no triangle with vertex (oc) as no two points A*g and A*h. i ^ j. are on a
common line, and no element g is incident with the point (oc). Furthermore,
since the cosets Ajg partition the group (and so intersect only on the line [A,]),
there can be no triangle with vertices A*g.A*h. and g, where A*g ^ A*h. We
have two additional cases two consider.
First, suppose that A*g.g. and h are the vertices of a traingle. Then there
are cosets A3g and Ajh which are contained in A*g. and the elements g.h are
in some coset Ajg. i ^ j. See Figure A.l.
But this is if and only if g. h G A*gDAjg. or equivalently id. hg~l G A*nAj.
It follows that no triangles of this type occur if and only if the following property
holds.
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Figure A.l: P2 : A* fl Ai = {id} for all i ^ j.
P2 : A* n Ai = {id} for all i j.
From now assume that property P2 holds. We now have to show that there
are no triangles in the geometry that have g, h. and u as its vertices as in Figure
A.2.
[^]
Figure A.2: P3 : AjA, fl Ak = {id} for all distinct i.j. k.
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Assume that such a triangle exists. Then the following coset relationships
would hold.
Akh = Aku
inr1 E Ak
Aj(J Ajh hg 1 e Aj
But then {hg*1) (gir1) = hu*1 implies that AjAt fl Ak\ > 2. Next assume
that AjAi fl Ak\ > 2. Then if id ^ g G AjAi fl Ak we see that Ak = Akg and
g = ttjtti, giving us Aja; = Atg and since g ^ at (which follows since g Ak
and property P> holds) the line Ajg has the points g and a,;. But then the line
Ai intersects Ajg at the point al and the line Ak at the point id. giving us a
triangle with vertices g,a,i. and id. See Figure A.3.
Figure A.3: P3 : AJAl fl Ak = {?'d} for all distinct i.j. k.
We have shown that no triangles exist, having vertices g.h and v. if and
only if the following property holds.
Ajai Ajg Akg
/
\ /
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P3 : A} A, D Af, = {id} for all distinct /, j. k.
From now on we assume that property P3 also holds. Our final step is to
show that given a line / and a point P not on L there is a unique pointline pair
(Q. m) such that PImIQIl.
First, if (oc) is not on a line Atg then there is unique line [Aj] and point A*h
such that (oc)I[Ai\IA*gIAjg. Next suppose that the point A*g is not incident
with a line Adi. Then then there is unique line [Aj\ and point A*h such that
A^gllA^IA^hLAjh. We have to consider two additional cases.
First, suppose that the point A*g is not incident with a line Ajh, i ^ j
We need to find, as is pictured in Figure A.4. a point x G A*g and a line Atg
(uniqueness follows from property P2) such that :r G Axg Pi Ajh.
Figure A.4: P4 : A*Aj = G for all i ^ j
So Atg = A,x. Ajh = AjX. A*g = A*x. and A*h = A*x. giving us x G
A*g fl Ajh. But. the cosets A*g and Ajh were chosen arbitrarily, and so for
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each g. h E G there would have to be elements a* E .4* and cij E Aj such that
a*g = a,jh, or gh~l = (a*)~1nJ E A*Aj. We have shown that this case holds if
and only if the following property holds.
P4 : A*Aj G for all i 7^ j.
Obviously, for each g E G. there is a unique line containing g and incident
with a point on each line [Ai). So finally, suppose that the point g is not incident
with a line AJi. that is g $ AJi. First define
Q = U{A, : 0 < r < r}
Since T\ and P> hold, each coset of At different from Az but contained in A*
is disjoint from Ar and we have the following
A* C At U {Atg : Atg O fi = 0}
Since g $ AJi. we can WLOG assume that the point is id Â£ A\g. First
consider the case g E A* \ Ah Here A*g = A* is a point on the lines Atg and
*4j. the second of which is collinear with the point id.
So we may assume that g Â£ A*. We need to find a line Aj such that
 Aj D A,g\ = 1 So there must be an i A j such that Aj D A,g / 0.
So if Atg is a coset of At disjoint from Â£2. it must be that Ajg C A*. But we
have already shown that A* C A, U {A,g : A,g fl Q = 0}. So assuming that I\
through P4 holds, this case holds if and only if the following property holds.
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Figure A.5: P5 : A* = At U {Aig : Atg fl Cl = 0}
[Aj]
Figure A.6: P5 : A* = At U {Atg : Atg fl Cl = 0}
P.5 : A* = Ai U {A,g : A,# fl Q = 0}
Looking back at the properties PI.......P5 it is easy to see that P2 implies
PI. and since G is finite, P2 also implies P4. Hence we need only assume that
properties P2. P3. and P5 hold. It turns out that P5 holds if and only if r = t.
We show this, and that the coset geometry is a GQ(s,t) if and only if r = t
and properties P2 and P3 hold. Assuming that P'2 and P3 hold we need the
