R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
N OBORU I TO
On permutation groups of prime degree p which contain (at least) two classes of conjugate subgroups of index p
Rendiconti del Seminario Matematico della Università di Padova, tome 38 (1967), p. 287-292
<http://www.numdam.org/item?id=RSMUP_1967__38__287_0>
© Rendiconti del Seminario Matematico della Università di Padova, 1967, tous droits réservés.
L’accès aux archives de la revue « Rendiconti del Seminario Matematico della Università di Padova » (http://rendiconti.math.unipd.it/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/conditions).
Toute utilisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit conte- nir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
DEGREE p WHICH CONTAIN (AT LEAST) TWO
CLASSES OF CONJUGATE SUBGROUPS
OF INDEX p
NOBORU ITO
*)
Let p
be aprime
and let F(p)
be the fieldof p elements,
cal-led
points. Let 0
be a transitivepermutation
group on such that(1) G
contains asubgroup 18
ofindex ?
which is not thestabilizer of a
point.
.V
has twopoint orbits,
say D D(cf. [3]).
Let kbe the number of
points
in D. Then 1~ k p -
1. Furthermore D = D( p, k, A)
can be considered as a difference setmodulo p
suchthat the
automorphism group A (D)
of Dcontains G
as asubgroup (cf. [5]).
Replacing
Dby F (p) D,
if needbe,
wealways
can assumethat
k 1 (p - 1).
Now the
only
known transitivepermutation groups G
of de-gree p
satisfyng
the condition(I)
are thefollowing
groups :(i) Let F (q)
be the fieldof p
elements. .LF(r, q)
and
SF (r, q)
be the r-dimensional vector space, the r-dimensionalprojective special
linear and semilinear groups overF (q) respectively
*) This research was partially supported by National Science Foundation Grant GP-6539.
Indirizzo dell’A. : Depart, of Mathematics, University of Illinois at Chicago Circle, Box 4348, Chicago, Ill. 60680 USA.
288
f20131
where r 2
3and p = qr - 1
1. Let II be the set of one dimensionalq - 1
subspaces
of V(r, q). q)
can be considered as apermutation
group on II.
Identify
II withF (p).
Then anysubgroup G
ofS.F (r, q) containing
.LF(r, q)
satisfies(I)
withparameters
and(ii) G
=LF (2, 11),
where p = 5 and A = 2.Now among the groups mentioned above
only
LF(2, 11)
sati-sfies the
following
condition:(II)
the restriction of36
to D is faithful(cf. [5]).
Thus it is natural to ask whether this is the
only
group sati-sfying (I)
and(II).
The purpose of this note is to make a firststep
towards the solution. We prove the
following
theorem.Let 0
be a groupsatisfying (I)
and(II).
If k is aprime,
thenLF (2, 11).
PROOF. First of
all,
we recall thefollowing
fundamentalequality
for the difference setSince k is a
prime by assumption,
from(1)
we see that divi-des p -
1. Putwhich
implies by (1)
that(b)
Let0
be aSylow p-subgroup of G
and letNs1P
be thenormalizer of
1P
inG.
Thensince G = PB, Nsf)
= withQ ==== J6
nNs1P. Q
iscyclic
of order q,where q
is a divisorof p -
1.Clearly Q
leaves D fixed. Alsoclearly Q
leavesonly
onepoint
fixed. Thus either k - 1
(mod q)
or k - 0(mod q).
In the former case,by (2)
1) For the theory of difference sets see
[7].
In the latter case, since is
prime,
(c)
The restrictionof ?
to D isdoubly
transitive.Otherwise, by assumption (II)
andby
a theorem of BurnsideJ8
ismetacyclic
of orderk~, where ~
is a proper divisor of lc - 1.Hence the
order g of G
isequal
to On the otherband, by Sylow’s Theorem, g pq (1 -~- np),
where n ispositive, since (5
isclearly
nonsolvable. ThusIf k = q, then from
(6)
1+ p
~ 1+
np =~.
This is a contradiction.Thus
1 -f - np - 1 -f - n - 0 (mod k).
Put n = 1. Then from(2)
and
(6)
we obtainSince
N ~
1 and k> 1
s Nk ~ N+
k. Thus from7)
kC ~ .
This isa contradiction.
(d)
Let1k
be aSylow k-subgroup of G
contained in38. By assumption (II)
the restriction of1k
to D is faithful. ThusR
is of order k. IfIt
leaves fixed at least twopoints,
thensince G
isdoubly
transitive on
I’ ( p),
the index of1k in G
is divisibleby p -
1. Thiscontradicts
(2).
Thus K leaves fixedexactly
onepoint,
say ~. Then ibelongs
toI’ (~) D.
Let be the normalizer of1k
inG.
Since
clearly
D is theonly
block left fixedby
is contained in:13. By assumption (II) ’[~
coincides with its own centralizer.Thus the order of
equals kC, where C
is a divisor of k - 1.(e) Let H (i)
be the stabilizer of i inG.
IfG EZ
LF(27 11), 2)
then the restriction of
is
nH (i)
to D isdoubly
transitive.Otherwise, by assumption (II)
andby
a theorem of BurnsideR (i)
is contained inNs1k.
Since leaves ifixed,
_R (i). Thus C
is a proper divisor of k-1.Since :JB : :13 n :a (i) = - p k,
the orderof 36
isequal
to( p - k) Ir~.
Now let be a minimal normal
subgronp
of)8.
Then23’
isa direct
product
ofmutually isomorphic simple
groups. Since the restrictionof 23
to D isdoubly transitive,
the restriction of to2) Read: 4D is not isomorphic to ...
290
D is transitive. Since
1k
hasorder k, 16’
issimple. By Sylow’s Theorem B = B4 (NsR).
Thus3B’
has order(p
-k) where C’
isa divisor of k - 1.
Now
by (2) P - k = (N - 1) k -+-
1. Ifthen by a theorem
of
Ostrom-Wagner ([6]) (5 does
notsatisfy
theassumption (II).
Hencek - 1 k - 3
by (3)
N - 1 =k - 1 -
K201320132013 .
* Thereforeby
a theorem of Brauer([1],
Theorem10)
either(a) ~=2,B’==Lj~(2~)
or(~)
N -
2 -
1:rB’ LF(21
k -1),
k - 1 2-.By
aprevious
result([3]) G
cannot betriply
transitive onF(p).
lf
(a)
occurs and if p> 11,
thenby
aprevious
result([4]) 0
isquadruply
transitive on.F’ ( p). Thus p
= 11. Then it is easy to checkthat G = LF(2, 11).
Suppose
that occurs. Thenby (3)
2=2. Now from g = pq(1 -t- np)
= p( p - k) k~
it follows thatBy (2) k2
- 2(mod p).
ThusSince we obtain from
(8)
which
implies
thatThen
by (4)
and( 5 ) q
=1~. Nowagain
fromit follows that which
implies
thatand ~ = 2 it follows that
Now let
(5’
be a minimal normalsubgroup
of0.
ThenC3’
has order pq(1 + n’p)
with n’ Henceagain by
a theorem of Brauer([1],
Theorem10)
n’ = 1 andG’ 2i L.F (2, p).
Then k = q=p - 1*
2 .Thus
k=5,p= 11,
andG = G4 = LF(2, 11).
( f )
The linethrough
two distinctpoints i and j
is the inter- section of all the blokscontaining
both iand j (cf. [2]). Since G
isdoubly
transitive onF((),
every line contains the same number ofpoints.
Let s be the number ofpoints
on a line. ThenIn
particular,
if N ~4,
thenIn
fact,
the number of lines isequal
toSince p
and k areprimes
and since~, ~ s,
we obtain(12).
(g)
Assumethat G N LF (2, 11).
Let 0 and 1 be two distinctpoints
of D.Let B (0) abd B (1 )
be the stabilizers of 0 and 1 in~ respectively.
Thenby (e)
we see at once thatThus the orbit
of zt (0)
nit (1) containing
i contains F( p)
- D.Clearly
this is thecase for every block
containing
both 0 and 1. Thus the orbit ofH (0)
nR (1) containing i
coincides with the line determinedby
0and 1. Now
considering
the indexof A (0) n Zt (1)
nA (i)
inH(0)nH(l)
we obtain292
where t is the index of From
(14)
we obtainSince
by (13)
ispositive.
From(2)
and(15)
itfollows that which
implies
thatFrom
(2), (15), (16)
we obtainwhich
implies
thatBut
by (3)
and(13)
This contradiction establishes
BIBLIOGRAPHY
[1]
R. BRAUER, On permutation groups of prime degree and related classesof groups,
Ann. of Math. (2) 44 (1943), 57-79.
[2]
P. DEMBOWSKI - A. WAGNER, Some characterizations of finite projective spaces, Arch. Math. 11 (1960), 465-469.[3]
N. ITO,Über
die GruppenPSLn
(q), die eine Untergruppe von Primzahlindex en-thalten, Acta Sci. Math. Szeged 21 (1960), 206-217.
[4]
N. ITO, Transitive permutation groups of degree p = 2q + 1, p and qbeing
primenumbers III, Trans. Amer. Math. Soc. 116 (1965), 151-166.
[5]
N. ITO, On a class of doubly, but not triply transitive permutation groups, to appear in Arch. Math.[6]
T. G. OSTROM - A. WAGNER, On projective and affine planes with transitive colli- neation groups, Math. Zeitschr. 7 (1959), 186-199.[7]
H. J. RYSER, Combinatorial mathematics, MAA (1963).Manoscritto pervennto in redazione il 17 marzo 1967