C OMPOSITIO M ATHEMATICA
F RIEDRICH K NOP
Homogeneous varieties for semisimple groups of rank one
Compositio Mathematica, tome 98, n
o1 (1995), p. 77-89
<http://www.numdam.org/item?id=CM_1995__98_1_77_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
Homogeneous varieties for semisimple groups of
rank
oneFRIEDRICH KNOP
Department of Mathematics, Rutgers University, New Brunswick NJ 08903, USA Received 28 April 1994; accepted in final form 14 June 1994
© 1995 Kluwer Academic Publishers. Printed in the Netherlands.
1. Introduction
Finite
subgroups
ofSL2((C)
are animportant part
of classical mathematics.They
are linked to many different
topics
likeDynkin diagrams, singularities
ofsurfaces,
Platonic
solids,
andequations
of fifthdegree (see
Klein’s book with commentsby Slodowy [Kl]).
Another way to look at a finitesubgroup
H isby
its cosetspace
SL2(C)/H.
It is anhomogeneous algebraic variety.
All otherhomogeneous algebraic SL2(C)-varieties
are of lower dimension and areeasily
determined.In this paper we
study homogeneous SL2 (k) -varieties
X which are defined overground
fields k different from C and wegive
a classification in two situations:(a)
kis
algebraically
closed and X isarbitrary,
and(b)
k isarbitrary
but dimX 5
2.The main
intricacy
of thisproblem
appears when k is ofpositive
characteristic.First of
all,
there are many more finitesubgroups
than in characteristic zero, forexample,
the groupsSL2(Fq). Essentially,
thesesubgroups
have been classifiedby
Dickson
[Di].
Therefore,
the main focus of this paper is on the otherproblem
inpositive
characteristic: A
homogeneous variety
is nolonger
determinedby
theconjugacy
class of its
isotropy
groups.Instead,
one has to consider moregeneral objects,
so-called
subgroup schemes,
i.e.algebraic
groups whosering
of functions may havenilpotent
elements. The best knownexample
is the kemel of the Frobeniusmorphism, widely
used inrepresentation theory (see
J. C. Jantzen’s book[Ja]).
In this paper, we first define certain connected
subgroup
schemes ofSL2(k)
and
give
a concrete realization of their coset spaces. Most of them aretriangular (modulo
Frobeniuskemels),
i.e. contained in a Borelsubgroup. Only
in the caseof characteristic two we define two other series of
subgroup schemes,
namedB(q)
andC(q). Exactly
one of them is ofpositive dimension, namely B(~).
Thecorresponding homogeneous variety
is realized as the open orbit in theprojective
space
P(g),
where g is the Liealgebra
ofSL2(k).
We then show that our list contains all connected
subgroup
schemesup to
conjugation.
This is doneby
an induction.Combining
this with theabove-mentioned classification of
Dickson,
we obtain acomplete
list ofsubgroup
schemes ofSL2(k).
Actually,
my own interest forhomogeneous
varieties ofSL2(k)
comes fromthe fact that many
properties
of actions of anarbitrary
reductive group G can be reduced to the rank-one-case(see
e.g.[Kn]).
The results here will be used for that purpose in asubsequent
paper.NOTATION. All varieties and groups are defined over an
algebraically
closedfield k of
arbitrary
characteristic(except
in Section5).
Let p be its characteristicexponent,
i.e. p = char k if the characteristic ispositive
and p = 1 otherwise. In the whole paper q, ql, q2,q’
etc. denote either a power of p or 00. If twosubgroups
are
conjugated then
we denote thisby H1 ~ H2.
2. Construction of certain
homogeneous
varietiesFor
short,
asubgroup
of analgebraic
group G is in this paper what isusually
calleda
subgroup
scheme.Then, homogeneous
varieties for G are classifiedby conjugacy
classes of
subgroups
of G. We startby defining
somesubgroups
ofby equations,
where everyequation involving
an infinite power XOO has to be omitted.In
arbitrary
characteristic we defineIn particular, 1 =.4(1, 1), T := A(~, 1), U := A(1, ~) and B := A(~, oo )
is the trivial
subgroup,
a maximal torus, a maximalunipotent subgroup
and a Borelsubgroup, respectively.
The group4(n, q)
is connected if andonly
if nequals
o0or is a power
of p.
For n oo, we set Mn :=4(n, 1).
Only
in characteristic two we defineand
It is
easily
verified that theseequations actually
definesubgroups
of G.Observe,
B(1) ~ A(4, 1) and C( 1 ) N A(2, 2).
For q =
pl
letFq :
G ~ G be the l th power of the Frobeniusmorphism
definedby (a, b,
c,d)
~( aq , bq, cq, dq).
For asubgroup H
of G and q oo setqH :
=F-1q(H) 9
G. Forexample, ,4 (n, q’)
is definedby
theequations
In
particular,
if the l th Frobenius kemel of analgebraic
group H is denotedby H(q) then G(q) = q1 = qA(1, 1).
Next,
1 want togive
a concrete construction of thecorresponding homogeneous
varieties. It suffices to do this for
subgroups
notcontaining
a Frobenius kemelG(q).
In
fact,
let X be aG-variety.
If we denote the samevariety
but with the twisted actionG
G ~ AutXby qX
thenG/qH
=g(G/F).
Consider H =
A(n, q) :
n = q = oo :
Then,
of courseG / H
=G / B
=P is
theprojective
line.n q = oo: Let
Ln - Pl
be the line bundleof degree n
andL’n
thecomplement
of the zero-section. Then .H
C
B realizesG/H
as bundle overG/B
=Pl
and infact
G / H = L’n.
q oo : Let B - be the group of lower
triangular
matrices. Thenq B -
is definedby
bq = 0 and arises as an
isotropy
group ofqPl.
Becausewe can realize
G/H
as an orbit inLn
xqPl (where
we setL’ 00 := P1).
Infact, it
is
exactly
thecomplement
of thegraph
of themorphism
Next,
consider char k = 2 and H =B(~).
Let g be theadjoint representation
ofG,
i.e. the space of 2 x 2-matrices of trace zero. Then G acts on theprojective
spaceP(g)
with three orbits: A curve definedby
det A =0,
a fixedpoint represented by
theidentity matrix,
and thecomplement Xo
of both. Then[ô 1]
EXo
and itsisotropy subgroup
isB(~). Therefore, G/B(~) ~ Xo.
For 1 q oo, we have
Hence,
the spaceG/B(q)
is realized as a dense open subset ofP(g)
xqP1.
To beprecise, let
be a
point
ofP(g)
xPl.
ThenG/B(q)
is thecomplement
of two divisors definedby
Finally,
consider H =G(q)
with 1 q oo. Consider therepresentation
Vo
= 9 EB qg and choose coordinates as followsLet W be the line
spanned by (12, 12)
and V :=Yo/W.
Then x : := xo -z i , yo, zo, y, Zl are coordinates on V. An easy calculation shows that
G/C(q)
isthe
generic
orbit in V describedby
theequations
More
precisely, C(q)
is theisotropy
group of any vector with yo =z1 ~
0 andzp = 0.
3. The normalizer
We want to show that this way be got all connected
subgroups
of G up toconjugacy.
For this we need their normalizers.
3.1. LEMMA. The normalizer
subgroup
Nof
asubgroup
Htogether
with theisomorphism
typeof N/H
isgiven by
the table below:The number in brackets
refers
to the character with whichGm
actsby conjuga-
tion.
Proof.
Withthe
equality XHX-1
= H leads for H =4(ql, q2)
tofollowing
system ofequalities:
The evaluation of these
equations
is easy and is leftmostly
to the reader. Let mework out
only
the case1
q12q2
oo,q1 ~ 2,
q2 > 1. Assume first ql = 1.Then a = 1 and
(3)
holds. Because q2 > 1 we haveb ~ 0. Hence, (1)
reducesto
x221
= 0 and(2)
to x11x21 = 0.Therefore,
x 21 = x 21 det X = 0 is theonly relation,
i.e. X E B.Consider now ql > 2. Then a -
a-’ 54
0 and(1) implies
x21x22 =x221
= 0. Asabove,
we get X21 = 0. Then(2)
is satisfied.Finally, (3)
follows from q12q2.
Next,
let H =C ( q )
with q > 1 and char k = 2. Thenimplies
Then
S e C(q)
isequivalent
toA2
=1, B2q
=0,
and C = ABq.Therefore,
weget
thefollowing equations
This
implies
Conversely,
the threeequations
are
equivalent
to(1)-(4) above,
i.e. N is thepreimage of A( q
+1, oo)
underF2
as claimed.
Finally,
consider H :=B (q).
Then a direct calculation as above becomesquite
involved.
Therefore,
weproceed differently.
Firstobserve,
thatB(1)
isconjugated
to
A(4, 1) ~ G(4) by
s =( i °). Moreover, multiplication gives
amorphism
which is an
isomorphism
of schemes but not ahomomorphism.
It is easy tocheck,
that U =
4(l, oo)
normalizesB(q).
Hence,B(oo)
=B(1)U
C N. To see theopposite inclusion,
consider the action of G onG/B(~) ~ Xo
CP(g).
Wecalculate the fixed
point
schemeXOH.
FirstXB(1)0
is reduced and consists of allpoints
of the formThen intersection with the fixed
point scheme of,A( 1, q) implies y
=0,i.e. XB(q)0 is
scheme-theoretically
thepoint
x0 definedby y
= 0.Hence, Nxo Ç XB(q)0 = {x0}
which
implies
N CGxo = B( (0).
The calculations for
N/H
are easy and left to the reader. For future reference letme just
mention thecomputation
for H =A(2, q),
1 q 00. In this caseN :=
Gm (G(2)a [- 1]
xGa[q])is isomorphictothe matrix group
Then t
= a2, x
=a-lc and y
=(ab) qdefines
ahomomorphism
of2B
onto Nwith kemel H. ~
4. The connected
subgroups
Now we are in the
position
to prove our main classification result:4.1. THEOREM.
Every
connectedsubgroup of
G =SL2(k)
isconjugated
toexactly
oneof
these:(a)
Inarbitrary
characteristic, G andq,4(ql, q2)
with1
q oo and1
q1, q2 s 00.
(b)
In characteristic two,additionally qB(q’)
with1
q oo, 1q’
oo andqC(q’)
with1
q oo, 1q’
oo.Proof.
IfH ~
G then there is a maximal Frobenius kemelG (q)
which is contained in H. Because the list above is closed undertaking preimages
of theFrobenius
morphism,
we mayreplace
Hby Fq(H).
Hence we may assume thatG(q)
c Himplies
q = 1 or what is the same LieH ~
Lie G.We consider first zero-dimensional connected
subgroups
H. Then there is a smallest power q =q(H)
of p such that H CG(q).
We prove, that our list above iscomplete by
induction onq(H). Obviously, q(H)
= 1implies
H = 1 =1A(1, 1).
Assume now
q(H)
> 1. Then H’ := H nG(q/p)
is propersubgroup
of Hwith
q(H’)
q =q(H), hence, by induction,
it is on our list. Because the Frobenius kemels are normalsubgroups,
H is contained in the normalizer N of H’.Furthermore, Q
:=H /H’ - G(q)/G(q/p) ~ G(p)
is a group schemeof height
one.
Hence,
it suffices tocheck,
that our list is closed under normal extensions ofheight
one. This is not very difficult because thesecorrespond bijectively
top-closed subalgebras
of LieN/ H’ (see [DG] II, 7.4).
Assume H’ is the second or third A-case in Lemma
3.1,
orB(q) or C(q):
In thiscase
dim Lie N/ H’
=1,
whichimplies
that H’ has aunique
normal extension ofheight
one which isA(pql, 1), A(pq1, q2), B(pq)
or2 A(1, q) respectively.
Allof these are on the list.
Assume H’ =
A(1, 1)
= 1: Then H is determinedby ap-closed
Liesubalgebra
of G. Hence, up to
conjugacy,
Hequals A(p, 1), 4(l, p), A(p, p)
orG(p) = pA(1, 1).
Assume H’ =
A(2, 1 ),
with p = 2. Then we have to determine thesubalgebras
of Lie
PGL2(k)
which are the 2 x 2-matrices modulo scalars. Weget,
up to G-conjugacy,
To these
subalgebras correspond
H =A(2, 2), A(4, 1), G(2)
=2A(1, 1), A(4, 2),
and2A(2, 1), respectively.
Assume H’ =
4(ql, q2)
with1
q12q2
oo andq1 ~ 2,
1 q2:Up
toB-conjugacy
there areonly
three non-trivialsubalgebras
of LieN/H’ leading
toA(pq1, q2), 4(ql, pq2) and A(pq1, pq2).
Assume H’ =
A(2, q)
with p = 2 and 1 q oo: This is the mostinteresting
case because here the groups
Ci(q)
andC(q)
appear.By
the calculation at the end of theproof
of Lemma3.1,
LieN/H’
has apresentation
asA short calculation shows that there are, up to
(N/H’)"-conjugacy, only
thefollowing subalgebras s
which are closed under A -A2:
This finishes the
proof for
finitesubgroups.
Let H be a
positive
dimensionalsubgroup
of G with LieH ~
Lie G. Assumefirst that H is smooth. Then
dim H
2implies
that H issolvable,
henceconjugate
to a
subgroup
of B.Hence,
either H - B =4(oo, oo) or
dimH = 1. In thelatter case we
get either
H c-’2Ga
or H ~Gm. Therefore, H -
U =4(l, oo)
orH N T =
4(oc, 1).
This settles inparticular
the case of chark = 0. From nowon we assume p 2. Then
Hq
= H nG(q)
is a finite normalsubgroup
of H withq(Hq)
= q.Let q
> 2.Assume that
Hq
=8(q).
ThenHred ç NG(B(q))red =
U andB(~) = 8(q)U
=NG(B(q))
we conclude H =B(~).
Next assume
Hq
=C(q/2).
Thenagain H-d =
U. ButG(2)
(2) CC(q/2)U
C Hin contradiction to the choice of H.
We are left with the case that all
Hq
areconjugated
toA(ql, q2)
where qi, q2depend
on q. Moreprecisely, max(ql, q2)
=q(Hq)
= q. ThenHq 9 Hpq
and thefact that
4(ql, q2)
isconjugated
to asubgroup
of4(q’, q2)
if andonly
ifq1 q’1
and q2
q2 imply
that three cases mayhappen:
ci is constant and q2 = q: Then for
2q
> ql there is aunique subgroup
oftype
A(q1, pq2) containing A(q1, q2).
Hence we may assume thatHq
notonly
isconjugated but equals A(q1, q) for every q. Therefore, H n G (q)
=.4(qi, 00 )nG(q)
for all q which
implies H
=A(ql, oo) (see [MO] 2.1).
q2 is constant and ql = q, or ql = q2 = q. As
above,
one concludes H -A(~, q2)
or H N4(oo, oo).
~5. The infinité
subgroups
From Theorem 4.1 it is not difficult to derive a
complete
classification of allsubgroups
H of G: It is determinedby
its connectedcomponent Ho,
known fromTheorem
4.1,
and a finite reducedsubgroup
ofNG(H0)/H0,
known from Lemma 3.1. Let us start with the case dim H > 0. Finitesubgroups
are handled in the nextsection.
5.1. THEOREM. Let H
be a positive dimensional subgroup of G
=SL2 (k).
ThenH is
conjugated
toqoH’
where H’ is oneof G, B,
03BCn U where1
noo, T x
U (q)
where1
q oo,NG(T),
orB(oo)
when p = 2.Now it is easy to
generalize
this to anarbitrary ground
fieldko.
LetSL2(ko)
acton the Lie
algebras s(2(ko), 012(ko)
andpg[2(ko) by conjugation.
For a,(3
Eko
we define the
quadrics Q(a, Q) :
={A
Eg(2ltr A
= a, detA =B}.
5.2. THEOREM. Let
ko
be anyfield
and X ahomogeneous SL2(ko)-variety
defined
overko.
Assume1
dimX
2. Then X is a Frobenius twistof one of the
varieties in the
right-hand
column. Thefirst
column describes theisotropy
groupover an
algebraic
closureof ko.
B: The
projective
linePl .
pn x U:
L’n,
the line bundle overPl of degree
n 1 with the zero-section removed.
p2 x U,
P = 2:Q(0, 03B2)B[12] where 03B2
Eko is
nota square.B(~),
p = 2:{[A]
EP(sl2)|[A] ~ l121, det A 0 0}.
T U(q),
q > 1 :Pl q P1B0393q where 0393q is the graph of the Frobenius morphism Fq.
T:
Q(1, 03B2)
where,8
Eko, 403B2 ~
1.NG(T):
The open orbit inP(PO[2),
which is describedby
4 detA -(tr A)2 ~
0.REMARKS: 1. We denote the line
through
a matrix Aby [A].
2. The intersection
Q(0, ,8)
n[12]
consistsonly
of onepoint, namely 03B212.
Furthermore, Q(0, 03B21) ~ Q(0, 02) if and only if 03B21 - Q2
Ekô.
3.
If p ~ 2
thenQ(1, 03B2)
~Q(0, ,Q’)
withQ’
=f3 - ! =F
0.Furthermore,
Q(0, 01) Q(0, /?2)
if andonly
if03B2103B2-2 1 is
a square inko.
4.
If p = 2 then consider f : ko
~ko : t - t2 - t. Then Q ( 1, 03B21) ~ Q ( 1, 02)
if and
only if 01 - /?2
Ef (ko).
Proof.
Let k be analgebraic
closure ofko.
ThenXk ~ G/H
where H is one ofthe groups in Theorem 5.1. Let N be the normalizer of H in G. Because
AutG Xk - N/H,
the different forms ofXk
overko
are classifiedby
thecohomology
setH1(k/k0, N/H).
Moreprecisely,
an element of thiscohomology
set is aN/H-
torsor over
ko.
In our case, it is the fixedpoint
schemeX H.
Because
NG(qH)
=QNG(H)
we getAutGqXk
=AutG Xk. Therefore,
wemay assume that
Xk
is not a Frobenius twist.It is known that the
cohomology
set for any smooth connected solvable group is trivial([DG] IV,
Section4, 3.7a). By
thecomputations
in Lemma 3.1 this meansthat there is
only
oneform, namely
the onegiven above,
unless H is either T orp2 x U with p
= 2.For H = T we get
N/H
=Z/2Z
which means that anN/H-tensor
isjust
aGalois extension of
ko
ofdegree
two. Let x ij be coordinates of gl2. ThenQ ( 1, 03B2)T
is described by
theequations
x12 = X21 =0,
X22 =1- x11, and x11(1-x11) = 0.
Hence,
Q(1, 03B2)T ~ Spec k0[x]/(x2-x
+03B2). Every
Galois extension ofko
ofdegree
two is of this form.Moreover,
two of them areisomorphic
if andonly
if theconditions in Remark 3 and 4 are met.
Finally,
consider the case H =03BC2 U
in characteristic two. ThenN/H
=Gm Gr2),
henceHI(k/ko, N/H)
=HI(klko, G(2)a).
The last set classi-fies
inseparable
field extensions ofdegree
two. NowQ(0, 03B2)H
isgiven by
X21 =
0,
X22 = xm· andx211
=(3.
This showsagain
that wegot
allN/H-
torsors. D
REMARK: If H =
NG(T)
and the characteristic is two thenG/H
is thecomple-
ment of the
hyperplane
tr A = 0 inP( pg[2).
This meansG/H ~ A2.
Inparticular, SL2(k)
can acttransitively
on an affine space. Thisprovides
acounterexample
toTheorem
l.l(ii)
in[Bo]. (The
erroneous statement in[Bo]
occurs on p.77, 1.
10.The
argument
is valid if H is connected. Inparticular,
the main result of the paper, Theorem1.1(i),
iscorrect.)
6. The finite
subgroups
We start with a
simple
observation which followsimmediately
from Lemma 3.1.6.1. LEMMA. Let H C G be a
subgroup
withHo = 4(qi, q2)
and q2 > 1. Then H ~ B.For this reason we start
by looking
forsubgroups
in B or moregenerally
inGm Ga[d],
i.e. the group definedby
Note,
that for d = 2 we get B via theisomorphism
Recall,
that ap-polynomial
is apolynomial f
Ek[t]
which is of the form03A3iaitpi.
It isadditive,
i.e. satisfiesf (x
+y)
=f (x)
+f (y).
Let n, 1 beintegers.
Then
f
=03A3iaiti
Ek[t]
is calledn-homogeneous of degree
l ifai 0
0implies
i ~ 1 mod n. In this case,
f (at)
=alf(t)
when an = 1.6.2. LEMMA.
Every subgroup of Gm x Ga[d]
isconjugated
to onedefined by
the
equations
xn =1, f(y)
= 0 where n 0 andf
is an (n,d)-homogeneous
p-polynomial.
Proof. By [Wa]
Thm.4,
H is a semidirectproduct
of asubgroup
S ofGm,
and a
subgroup
N ofGa.
Then S is definedby
xn = 1 and Nby f (y)
= 0for some
integer
n, andp-polynomial f.
The fact that S normalizes N meansf(xdy)
=xlf(y)
for some1, i.e., dq -
1 mod n for every tqoccurring
inf.
Thismeans that
f
isn (n,d) -homogeneous.
~The case d = 2
implies
that everysubgroup
of B isconjugated
to one of thefollowing
form:DEFINITION.
Let n j
0 be aninteger
andf
ann (n,2) -homogeneous p-polynomial.
Then define a
subgroup
ofSL2(k) by
theequations
In
particular,
we getA(n, q) = A(n, tq), 4(oo, q) = A(0, tq), 4(n, ~) =
A(n, 0), .4(oxo, oo) = A(0, 0)
for1 n
oo, q oo.Next,
we considersubgroups
H withHo = B(q), 1 q ~, p = 2.
Theny =
(ab)g
identifiesCi(oo)/,Ci(q)
withGa. Hence, H
is definedby f-«ab)q)
= 0where
f
is a2-polynomial
withf’(0) ~
0. Asf
and q vary,f(t)
=f(tq)
runsthrough
the set of all2-polynomials
withf’
= 0.Therefore, H
is one of these:DEFINITION. For p = 2 let
f
be a2-polynomial
withf’ -
0. Then define asubgroup
ofSL2(k) by
theequations
In
particular,
weget B(q)
=B(tq)
for 1 q oo andB(oo) = B(0).
Now consider the case
H0
=C (q),
1 q oo, p = 2. Thenis a
surjective homomorphism
with kemelC (q). Hence,
H is describedby equations a2n = 1, f(a-1c + (ab)q)
= 0 wheren|q
+ 1 and1
is ann-homogeneous
2-polynomial
withf’(0)
=03BB ~
0. Thenc2
= 0implies i(a- 1 c) = Àa -1 e,
hencec =
a03BB-1f((ab)q).
Because the linear term off
does notvanish,
thedegree
off
is
actually
one.Hence, f(t)
=f(tq) is
ann-homogeneous 2-polynomial
ofdegree -1 ~ q
mod n.Therefore,
H is one of these:DEFINITION. For p = 2
let n )
1 be odd andf
ann-homogeneous 2-polynomial
of
degree -1
withf’ -
0. Then define asubgroup
ofSL2(k) by
theequations
Note that
only
thoseintegers
n occur which divide a number of the form 2m + 1. Asspecial
cases, weget C(q)
=C ( 1, tq),
1 q ooand,A(2n, oo)
=C(n, 0).
It remains the classification of
ordinary,
i.e.reduced,
finitesubgroups
ofSL2(k)
which has been
done by
Dickson. Then weget:
6.3. THEOREM. Let H be a
finite subgroup of
G =SL2(k)
with LieH ~
Lie G.Then H is
conjugated
to oneof
thefollowing:
1.
A(n, f),
where n 1 andf fl
0.2.
n
:= 03BCn Us. J-ln where s =(0-1 10),
n3,
and n is evenif p ~ 2;
3. When
p ~
2: Thebinary
tetrahedral groupÂ4,
thebinary
octahedral groupS4,
and thebinary
isosahedral groupAs;
4.
SL2(JFq)
where q >1;
5.
PGL2(Fq),
q > 1: The(schematic) preimage of PGL2 (IFq)
CPGL2(k)
inG;
6. When p = 2 :
B(f)
withf ~ 0;
All the redundancies
of this
list arethe following:
REMARKS: 1. For p >
2,
the normalizer ofSL2(Fg)
isPGL2(Fq)
in whichit has index two. An element of the non-trivial coset is
(03B10 0 03B1-1)
where a EFq2 - Fg, a2
EIFq. Forp
= 2 we getsimply PGL2(JFq) ==
M2 xSL2 (IF.).
2. In characteristic zero, one
gets
the well-known classification: Thecyclic
groups Ecn
(case 1),
and thebinary polyhedral
groups2n(n 2), Â,, 4,
andÂ5 (case
2 and3).
3. In characteristic two, G contains a
subgroup isomorphic
toA5, namely SL2(F4).
4.
For p
> 1 we have theequality B(IF.) = A(q - 1, t - tq).
Proof of the
Theorem: After the discussionabove,
theonly remaining
cases tocheck are
H0 = A(q1, 1), 1
ql oo. If ql > 2 then H CNG(T)
= T U 8. T.Then H n T = pn for some n 1.
Therefore,
either H = pn =A(n, 1)
orH ~ n. If p ~ 2 then 82 = -12
E pn whichimplies n
= 0 mod 2.Now assume ql =
1, i.e.,
H isreduced,
or p = ql = 2. In the latter caseH = p2 x
Hred.
But the list of the theorem is closed formultiplication
with03BC2 :
For n
1odd,
we get 03BC2 xA(n, f )
=A(2n, f),
p2 xn
=2n
andp2 x
SL2(IFg)
=PGL2(IFq). Therefore,
it suffices to consider reducedsubgroups
H of G.
We may assume that the
representation
of H on V= k2 is irreducible,
because otherwise H is contained in a Borelsubgroup
and thereforeconjugated
to someA(n, f).
Because a finite group hasonly
a finite number of irreducible represen- tations some Frobenius twistqV
isisomorphic
toV, i.e.,
V is defined over thefield
1Fq. Therefore,
we may assume that H is contained inSL2(JFq).
Now we useDickson’s classification of
subgroups
inPSL2(IFq) (see [Di]
260 or[Hu]
II.8.27).
It follows that every such
subgroup
which has no fixedpoint
inPl(k)
has evenorder.
Hence,
the order of H is even, too.If p ~ 2 then -12
is theonly
element oforder two in G. This
implies -12 E H, i.e.,
H is the fullpreimage
of asubgroup
ofPSL2(Fq ) .
This holds alsoforp
=2,
because theprojection SL2(JFq)
~PSL2(JFq)
is
bijective.
Now the theorem followseasily
from the table in[Di], [Hu].
The discussion in[Di]
also shows that for each type there isonly
oneconjugacy
classin G. This finishes the classification.
The redundancies are easy except
maybe
those for C.First, conjugation
witha
diagonal
matrix showsC(n, f(t)) ~ C(n, 03B1f(03B1t)). Conversely,
assumeHl
=C(ni, fl) N H2
=C(n2, f2).
ThenH0i
=C(l, 03BBitqi)
where03BBitqi
is the lowestdegree
term offi.
Afterconjugation
with elements of T we may assumeAI
=À2
= 1, henceHO
:=H01 = H02
=C(l, q)
where q = ql =q2. In particular, if Hl
=gH2g-1
then g ENa(HO)red ==
03BCq+1 U.Therefore, Hl
fl U =g(H2
~U)g-l.
This shows
fi(t)
=af2(at)
with03B1q+1 =
1.Finally, Hi fl B/Hi
~ U = 03BC2niwhich shows ni = n2. D
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