R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
Z VI A RAD
G IDEON E HRLICH
O TTO H. K EGEL J OHN C. L ENNOX
An application of Ramsey’s theory to partitions in groups - I
Rendiconti del Seminario Matematico della Università di Padova, tome 84 (1990), p. 143-157
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Application Ramsey’s Theory
to
Partitions in Groups - I.
ZVI ARAD - GIDEON EHRLICH - OTTO H. KEGEL - JOHN C. LENNOX
(*)
1. Introduction.
Ramsey theory
is a branch of combinatorics which deals with structures which arepreserved
underpartitions,
forbackground
de-tails see
[3].
The ideas andtechniques
ofRamsey theory
have foundbroad
applications
in settheory,
yprobability theory,
yanalysis,
andeven theoretical
computer
science. Our aim in this paper is to takeone of the earliest results of «
Ramsey type >>
andapply
it in grouptheory.
In
1916,
in connection with his work on the FermatConjecture,
I. Schur
[9] proved
thefollowing
result:THEOREM. In every finite
colouring
of thepositive integers
Nthere exists a monochrome solution to the
equation x -E-- y
= z.For interest’s sake we mention that 1927 van der Waerden
[11]
answered a
conjecture
of Schur[10] by proving
a related result onsubsets of N:
(*) Indirizzo
degli
AA.: Z. ARAD and G. EHRLICH:Department
ofMathematics and
Computer
Science, Bar-IlanUniversity,
Ramat-Gan, Israel;O. H. KEGEL: Mathematisches Institut,
Albert-Ludwigs-Universitat,
D-7800Freiburg
i. Br., WestGermany;
J. C. LENNOX:Department
of Pure Mathe- matics,University College
Cardiff, Cardiff CF 1 1XL,England.
The work on this paper was
supported by
agrant
of the German-Israeli Foundation.THEOREM. If N is
partitioned
intofinitely
manysubsets,
then atleast one of these subsets contains
arbitrarily long
arithmeticprogressions.
This,
of course, means that if N is2-coloured,
then there is amonochrome solution to the
equation x -f- y
= z. Both of these theorems have beengeneralised
in[3].
Our
objective
here is to prove a version of Schur’s Theorem forarbitrary
groups. We treat the case of infinite and finite groupsseparately.
THEOREM A. For any
n-colouring
of an infinite group there existsa monochrome solution to the
equation
xy = z, where x, y, and zare distinct
non-identity
elements.THEOREM B. For any
n-colouring
of a finite group of order atleast R(2, 8, (n2 - n)/2~ + 1
there exists a monochrome solution to theequation
xy = z, y, and z are distinctnon-identity
elements.
The numbers
b, c) appearing
here are theRamsey
numbers(see [3]
andbelow). They
areextremely large-in
factthey
havenot
yet
beencomputed. (However,
see[8]
for anotherapproach
tosuch
numbers.)
In the
special
cases n = 2 and n = 3 we use Theorem A to prove.THEOREM C.
a)
If G is a 2-coloured group of ordergreater
than 7 which is notelementary
abelian of order 9 then there is a monochrome solution of theequation
xy = z, where x, y, and z are distinct non-identity
elements.b )
If G is a 3-coloured group of order 17 orgreater
than 18 thenthere is a monochrome solution of the
equation
xy = z, where x, y, and z are distinctnon-identity
elements.The bounds on the group orders
appearing
in Theorems B and Cwill become smaller if we do not
require
the elements x, y, and z to be distinct.Indeed,
iff (t)
denotes thelargest n
such that[n] :
_: _
{lg 2, ... , nj
c N can be t-coloured in such a way that there is nomonochrome solution to the
equation x --~-- y
= z, thenaccording
to([5],
p.88)
one hasf (1 )
=1, f (2)
=4, f (3)
=13, f (4)
= 44. The eval-uation of
f(5)
appears to be a difficult combinatorialproblem [5].
If
f *(t)
is defined in the same way asf (t), except
that we insist thatthere be no monochrome solution to x
-~- y
= z fordistinct x, y, and z,
thenf *(1 )
=2, f*(2)
=8, f *(3)
= 23.PROOFS OF THEOREM A AND B.
Ramsey’s
Theorem states that for every infinite setS,
for everypair k,
r of naturalnumbers,
andfor every y
partition
p(colouring) X: g) x 7 Ik- - [r]
of the setL Ik
of all(unordered)
k-element subsets of S into r(disjoint)
subsetsCi:=
:=
X-1(i),
i =1,
..., r, there exists an infinite subset T of ~S such that all the k-element subsets of Tbelong
to the sameC;.
The
compactness
theoremyields
a remarkable finite version ofRamsey’s
THEOREM. There exists a
(smallest)
natural numberR(k, 1, r)
such that for every finite set S with at least
R(k, 1, r)
elements and any yppartition
:s
-+r
of the set of k-element subsets of S into rk []
subsets
i = 1, ..., r,
there exists a subset T of Scontaining
atleast I elements and such that
Ik - Ci
for some i E[r].
Let denote the commutative
partial semigroup
on the set[~;]:=={1~2y...~}c(Ny-t-)
definedby
the restriction of the usual addition of natural numbers to[n] :
The sum ac+
b is defined for a, bE Sn
if andonly
if a+
b E[n]. Essentially
Schur’sproof
nowyields
THEOREM 1. If the natural number n is at least as
large
asR(2, 4, r),
then for every
partition X:
p x[Sn] 1 - [r] []
ofS..
into r(disjoint) (
J)
sub-sets
Ci I i
=17
... , r, there exist two distinct elements ac, b ESn
suchthat a,
b,
and a-~- b belong
to some subsetCi
ofSn .
PROOF. Define a
partition z*
of2 s "
into r subsetsby x * ( x y)
=By
the finite version ofRamsey’s
Theorem and thechoice of
n ~ .R(2, 4, r)
there exists a 4-element subset T ={w7
x, y,~
w C x
y
z, ofSn
such thatX*(g, h)
= i for some i E[r]
and allg, h
E T h. But that meansx(x - w)
=x(y - x)
=x(y - w),
=
(x - ~,v) +
Putting a
=(x
-w)
and b =(y
-x)
we shall haveproved
ourassertion if b. But if
(x - w)
=(y - x), put a
=(x - w)
andb =
(z
-x)
to prove the theorem. 0REMARK. If the set
[n]
is embedded in such a way into a com- mutative(and additively written) partial semigroup
that the ad- dition of S induces on[n]
an addition that extends that of/S~
then-if
n>R(2, 4, r)-for
everypartition
of S into r subsets one has the statement of the theorem f or S(actually,
for the restriction of S to[n] ~ .
An
important example
of this sort ofembedding
will be the natural map of[n]
into thecyclic
groupZ/mZ,
Where m is any in-teger
m ~ n.The order of the element s of the
partial semigroup
~’ is the smal- lest natural number n such that either sn+l is not defined or = smfor some m with If no such n exists s has infinite order.
The
partial semigroup S
is calledperiodic
if every element of S is of finite order. With this definition one has thefollowing
consequence of the theorem and the remark:COROLLARY. If the
partial semigroup S
admits apartition
into rsubsets
Ci , z = 1, ... , r,
such that for none of these subsetsC,
the pro- duct of any two distinct elements ofCi
isagain
inCi,
then S isperiodic
and the order of every element of S is boundedby 4, r).
PROOF. For s E S assume that the elements
s, s2, ... , s~
are alldefined and
pairwise distinct,
then there is aninjection
of the par-tial
semigroup
into thepartial subsemigroup s~
of ~Sgenerated by s,
such that 1. --~ s. If now n >R(2, 4, r)
the remarkyields
a con-tradiction to the
assumption.
Hence one has ofnecessity n ~ .R(2, 4, r).
In the same vein as Theorem 1 we now prove THEOREM 2. Let G be any group with at least
elements. For every
partition
y: of GXX =GB~1~
into rpairwise disjoint
subsetsCi =
E[r],
there is at least one suchsubset
C$
of Gcontaining
two distinctelements a,
b as well as theirproduct
ab.PROOF.
Starting
from thepartition
X of Gxx define apartition X*
of T] [G]
intor2 - r 2
subsetsby
The
group G
is chosenlarge enough
toapply
the finite version ofRamsey’s
Theorem: There exists an 8-element subset T of GXX such thatx *
restricts to a constant functionon T 12
i.e.*
x associateswith every
pair
of distinct elements of T the samepair
of numbersfrom
[r], (1, l’),
say. Let v be any element ofT;
then there exist at least four furtherelements w, x, y, z
E T such thatX(vw-1)
=x(vx-1)
==
X(vy-1)
=x(vz-1) (= 1, say). Assume,
without loss ofgenerality,
that also =
1, ;
then one has =vx-1,
and the twofactors on the left have the desired
property,
unless vw-1= wx-1.Thus we may assume that among the
quotients gh-1
of distinct elements in T there are no twohaving
the desiredproperty:
For every suchtriple f, g, h
E T withx( f g-1 )
=x(gh-1)
theequation
Consider the case
X(wy-1)
= 1. Then(vw-1)(wy-1)
=vy-1,
so that=
wy-1.
But this leads to wx-1=wy-I
and x = y, a contra- diction. So one must haveX(yw-1)
=1’). By
the sameargument
we obtain
x(xy-1)
= 1 and alsox(wz-1)
= 1. But now theequations
together
with ourassumption
lead to x = z, a contradiction!Thus,
the
assumption
cannot be correct onT,
and apair
with the desiredproperties
must exist. 0This
completes
theproof
of Theorems A and B.2. Generalities.
Having
obtained that the class rP of groups (~admitting
apartition
intodisjoint
subsets{1}, C1,
...,C,
such that forevery i
E[r]
and every choice of distinct
elements x, y
ECi
theproduct xy 0 Ci
isa class of groups of bounded
order,
one would like to determine thefinitely
manyisomorphism types
of groups in this class rP.Exploit- ing
the group structure ofrP-groups
onehopes
toget
muchsharper
bounds on order and
exponent
ofrP-groups
than thosegiven by
theRamsey
numbers in Theorem B or theCorollary
to Theorem A.But,
as
long
as one hasonly
anessentially
combinatorialapproach,
suchan
enterprise
seems to be feasibleonly
for very small values of r. We shall treat hereonly
the cases r = 2 and r = 3.The case r = 2 may be
approached directly by
easy combinatorialtrickery.
PROOF OF THEOREM C.
a)
Let G E2P,
that is G ={11
U A u Bwith the
product
of any two distinct elements of A(resp. B) being
in B u
(resp.
A V~1~ ) .
We shall showSuppose
, and Then of the four distinct elementsaclai , i = 2, 3, 4, 5,
at least three lie inB ;
henceChoose
x1 E A.
There is an element X2 E so one has XlX2 c B.If the
IAI
elements(xlx2) a,
allbelonged
toA,
then onewould have = 1 since
(xlx2) b E A
u{11
for all b EThus there exists an element ~3 c A with
(XlX2)Xa
E B.There exists an element x4 E
An x31,
with E Band XlX2: Hence
(XlX2)(X3X.)
E A. Now there exists an elementX5 E
AB~x4, (We
owe this choice of X5 toa
suggestion
of M. J.Tomkinson, Glasgow.)
One has then x4x, EB,
or Hence we
get
IfI then contradiction. Thus we must have X5 = x1x2x3x4 and A =
{x4,x-14,x-14x1x2x3,(x1x2x3x4)-1, x5}.
Nowx5
= as x1x2xa=l=If then a contradic-
tion.
So x35
= 1.From 5=
(
it follows thatIGI = 9, 10,
or 11. But thegroup G has order divisible
by
3. Hence[G[ 9.
If
[G[
= 8 or G ~09
then one checks thatG 0
2P. On the otherhand,
it is not difficult to choose subsets ~. and B in G ~C3
XC.
showing
that G E 2P. ClFor r > 2 we shall need some
helpful organisation.
3. The
computer
prograxnme CP.In order to decide
constructively
whether a set of n elements hasan
r-partition,
i.e. apartition
into rdisjoint subsets,
that fulfills someproperty P,
one cangenerate
all suchpartitions
and check whether there is one with thisperoperty
P. Analgorithm
for this isgiven
in[1];
it
generates
each newr-partition
from itspredecessor by moving
asingle
element from one subset to another. The number ofr-partitions
of a set of 11; elements is known as the
Stirling
number of the second kindSerb, r),
it isgiven by
the recursive formula(see [1]) :
Solving this,
one obtainsOne can see that
This relation shows that
generating
allpartitions
ispractical only
for small n and r. The sizes of some of the groups we have to check
are
larger
than40,
and thepresent
supercomputers
need more thana life time to deal with them! We can
generate
allt-partitions,
r, of n elements as follows:
0)
Start with thepartition
of theempty
set.1)
Once we havea t-partition p
of the first k nelements,
weget
from it min(t + 1, r) -partitions p’
of the first k-f-
1 elementsby adding
the(k + l)-st
element to each of the subsets of p or, if t r,as a set
by
itself.Let T be a directed
graph
the nodes of which are all thet-parti-
tions of the first k-elements of a
given
set of nelements, tr,
the
edge p ~ p’
exists in T if andonly
ifp’
isgen erated
from pby
adding
asingle
element to a subset of p.T is a tree with root the
empty partition;
its leaves are all thet-partitions
of the n elements.A backtrack
algorithm
is ageneral
method ofgenerating
all thenodes of similar
trees,
where the nodes are combinatorial items.We start
by generating
the root and then wegenerate
all the nodes of subtrees-branches-rooted on the sons of theroot, by
the samemethod,
subtree after subtree.Examples
of backtrackalgorithms,
both recursive and
iterative,
aregiven
in[6].
The value of backtrack
algorithms
in our context lies in the fact thatthey
facilitatepruning
a tree. Once we know Lhat a subtree does not have a«good
»leaf,
we prune the wholesubtree;
after gener-ating
the root of such asubtree,
we do notgenerate
any of its sons, but passimmediately
to its nextbrother,
if it has any. We do notdispose
of anymagic
rule to decide whether a subtree has agood
leaf.Otherwise we could
apply
thatmagic
rule to the root and would knowimmediately
whether agiven
group is in rP.We prune the tree in two cases:
1)
if the(k + l)-st
element violates theproperty P,
i.e. in ourcase if in the subset to which it was added there now is a
triple
ofelements one of which is the
product
of the others.2)
If among the elements not in any subsetof p
there is anelement that cannot be added to p
withoug violating
theproperty
P.4. The
algorithm.
In order to use one
algorithm
for all cases each group is trans- formed to thefollowing
standardrepresentation:
The n elements are the
integers 0, ... ,
n - 1 where 0represents
the unit element.
The group
operation
is an n X ninteger
matrix MULT:MULT is used
just
forbuilding
another n X n setmatrix UNFIT:
A
partition
of{1,
... ,kl
iskept
in two ways : as an arrayshowing
foreach j,
to which of the subsets1,
... , r itbelongs,
y and as anarray of r stacks each
holding
one of the subsets.Once an element k is added to a subset t we build a set
(The
stacks are not treated as purestacks.)
Using
CAN(1~) helps
usperform prunings
of the first kind. Thistype
ofprunine along
was found to bepowerful enough; just
one toten minutes were needed for
checking
each of the groups on a P.C.using
Turbo-Pascal.For the second
pruning
we check whetherThe
input
to thealgorithm
is: a list ofgenerators
and relations and.a list of group elements.
The
output
of thealgorithm
is: allr-partitions
of theelements #
1.as groups in rP or a note that this group does not
belong
to rP.We shall refer to these
computer
programmes as CP.PROOF OF THEOREM C.
b)
The situation for 3Pbeing
much morecomplicated
than for2P,
we resort to the intensive use of our CP.LEMMA 1. For the
cyclic
groupCn
of order n one hasCn
E 3Pif and
only
if4 ~~~15.
PROOF. It is easy to check that if
4 n 15. By
our CPwe
proved
b)
thepartial semigroup S24 C{N, +1
may not be 3-coloured;such that the sum of any two elements of the same colour has a dif- ferent colour.
Therefore for
n > 16.
REMARK. The
partial semigroup 823 ç; ~~T, +}
hasexactly
the fol-lowing
threepartitions
such that in thecorresponding colourings
thesum of any two distinct elements of the same colour has a different
colour:
LEMMA 2. Abelian groups of orders
25, 3 ~, 5 2, 7 2, 11 ~, 132, 2 2 ~ ~, 2 2 ~ 7, 3~ ~ ~, 2 ~ ~ 3,
and 22-32duo notbelong
to 3P. If the abeliangroup A
has order 16 and
belongs
to 3P then it iselementary
abelian or adirect
product
of twocyclic
groups of order 4.PROOF. This was checked
by
our CP. Of thegiven
orders > 16 there exist 31isomorphism types
of abelian groups none of whichbelongs
to 3P. The directproduct
of thecyclic
group of order 8 with the group of order 2 does notbelong
to 3P.LEMMA 3. For the abelian group one has
IA I 18
andIA I =F
17.PROOF.
By
Lemma 1, the set ofprime
divisors ofIAI
is con-tained in the set
{2, 3, 5, 7, 11, 13}. Thus,
if18,
the group Awould have a
subgroup Ao
of one of the orders > 16 discussed in Lemma 2. But thecolouring
of A E 3P would induce such a colour-ing
ofAo . However,
Lemma 2gives
3P. Thusc
18.Checking through
the well-known groups of order 16 our CPprovides
us with the usefulLEMMA 4. There are
precisely
two non-abelian groups of order 16 thatbelong
to 3P:In both these groups the
subgroup x, z, u~~
is a characteristic sub- group of order 8.LEMMA 5. The soluble groups of orders
2 ~ 11, 2 ~ 13, 3.7, 3 ~ 13,
PROOF. The
following
non-abelian groups were checkedby
ourCP;
none of them
belongs
to 3P. We arrange them in a table:a )
The soluble non-abelian group of orders 2 3 ~ 5 and 22.3.5 are not in 3P:By Sylow’s
theorem and itsgeneralisation by
P. Hall any such group G contains asubgroup H
of order 22.5.By
Lemma 3 andthe result
reported
in thepreceding table, H 0
3P. But then G cannotbelong
to3P,
for otherwise any3P-colouring
of G would induce sucha
colouring
in H.b)
A group of order 2$ ~ 7 does notbelong
to 3P: A group G of this order has a normalsubgroup
.g ofprime
index p. If p =2,
then= 22.7.
By
Lemma 3 and the above tableH w 3P.
Hence wemay assume p = 7 and .H~ is an
elementary
abelian2-subgroup
oforder
8;
G is a Frobenius group of order 2 a ~ 7.Using
our CP we checkedthat this Frobenius group does not
belong
to 3P.c)
A group of order 23.32 does notbelong
to Agroup G
of this order contains a normal
subgroup
H ofprime
index p.If p
=2,
then H has order 22.32. Lemma 3 and the table
yield
that 3P.Hence we may assume that p = 3 and
[H[
= 2 3 ~ 3.Again
Lemma 3and the table
yield H 0
3P.Thus G 0
3P.d)
A group of order24 ~ 3, 24 ~ ~,
or 24 ~ 7 does notbelong
to 3P:In each case, such a group contains a normal
subgroup
H ofprime
index p.
If p
=2,
then thepossible
orders for H are2 3 ~ 3,
2~-5 or23 ~ 7, respectively. By
thetable,
Lemma3,
and the result instep b)
we obtain
H 0
3P. Thus we may assumethat p
is odd and H is anormal
subgroup
of order 16. If .H isabelian,
Lemma 2yields
that His either
elementary
abelian or a directproduct
of twocyclic
groups of order 4. AssumeH ~ C4 X C4 ,
then H contains a characteristicelementary
abeliansubgroup
V of order 4. Thus G contains a sub- group g on order4 ~p.
If p = 5 or 7 then Lemma 3 and the tableimply
3P. If p =3,
and aSylow 3-subgroup
T of G norma-lises a
subgroup
.E of order8,
then Lemma 3 and the tableyield
that .g =
TE 0
3P.Up
toisomorphism
there isonly
one group G of order 24 ~ 3 with normalsubgroup C4 X C4
and such that nosubgroup
of order 8 is normal in G. Our CP checked that this group is not in 3P.Hence we may assume that H G and .H is
elementary
abelian.If
p = 7,
then G is not a Frobenius group since7 ,~ (24 - 1).
Thus
14 ~ C~(~) ~, where
8 is aSylow 7-subgroup
of G. Inparticular,
~’ normalises a
subgroup E
of order 8 in .H. But thesubgroup K
=3P
by step b ) .
If p
= 5 then G must be a Frobenius group, since a group of order 22.5 or 2 3 ~ ~ cannot be in 3P. Our CP checked that this Fro- benius groups does notbelong
to 3P.If p =
3,
then either G is a Frobenius group, or theSylow 3-subgroup S
of G hasfixed-points #
1 in H. In the latter case S normalises asubgroup
E or order 8 in .H. As thesubgroup
.K ==
ES 0
3Pby
thetable, G 0
3P. The case that G is a Frobeniusgroup is ruled out
by
our CP.If the normal
subgroup
I~ of order 16 is non-abelian and in3P,
then
by
Lemma 4 there is a characteristic normalsubgroup
.E oforder 8 in ..g.
Thus,
if S is aSylow p-subgroup
of G thesubgroup
.g’ =
3P, by
theprevious
results.To
complete
theproof
of Lemma5,
weonly
have to consider groups of order 2 5.e)
A group of order 25 does notbelong
to 3P: If G is such agroup in 3P then
by
Lemma 3 G is non-abelian.By
Lemmas 2 and 4we know the
possible
structure of maximalsubgroups
of G.Case 1. All maximal
subgroups
of G are abelian.Suppose
G con-tains maximal
subgroups O2
XO2
XC,
XC. and S ~ C,
XC4.
ThenIR
J =8,
this intersection iselementary
abelian assubgroup
of I~and of
exponent
4 as maximalsubgroup
ofS,
a contradiction. Thus maximalsubgroups
of G must be of the sametype.
If all the maximal
subgroups
of G were ofelementary abelian,
thenevery non-trivial element of G would be an
involution,
and hence G would beabelian,
a contradiction. Thus assume that every maximalsubgroup
of G is oftype C4
XC4.
Let B and S be two distinct maximalsubgroups
ofG,
thenIR
r1S ~
= 8 and .1~ isisomorphic
toC4
XC2 . Clearly
.R =Z(G),
since G is non-abelian. For any x ES)
the
subgroup (I~
nS) x~
is abelian and thus proper. Thus this sub- group is oftype
and the element x has order 4. Therefore G containsexactly
3 involutions which form asubgroup T ~ C2
XO2
inside
Z(G).
For everyelement y
E G of order4,
onehas y2
E T. Henceis
elementary
abelian. Inparticular,
if y and z are elements of order 4 inG,
thesubgroup (y, z~
is a propersubgroup
of G contained in asubgroup isomorphic
toC4 X C4.
Thus yz = zy, and G isabelian,
a contradiction.
Case 2. G contains a non-abelian maximal
subgroup
H.By
Lem-ma 4 the
subgroup
.H isisomorphic
to one of the two groupsgiven
there. Let x E
GBH, G
=J?(.r).
Now we checkedby
our CP thatthe
partial semigroup T c ~G, ~ -17
T =~H~ : =
U U of31 elements of G does not admit a
3-colouring. Thus G E
3P. DLEMMA 6. The non-abelian
simple
groupsA, Åa, Å7,
7PSL(2, 7 ),
7and
PS.L(2, 8)
do notbelong
to 3P.PROOF. Our CP showed that
A5 ~ 3P.
thegroups
Åa
andA7
cannotbelong
to 3P. The groupsPSL(2,7)
and8)
contain solublesubgroups
of orders 21 and23.7, respectively.
By
Lemma 5 thesesubgroups
cannotbelong
to3P,
neither can thecontaining
group.We now come to the final
step
in theproof
of TheoremC b) :
LEMMA 7. If
G E 3P,
thenPROOF. Assume this statement is
false,
and let G be a minimal(finite!) counter-example. Thus,
propersubgroups
of G will be in 3P and hence be of orderc 18. Further, by
Lemma1,
the seta(G)
ofprime
divisors ofG ~
is contained in{2, 3, 5, 7, 11, 13}.
Assume His a
subgroup
ofprime
order p = 11 or 13 of G. Then H cannot be containedproperly
in any propersubgroup
ofG;
inparticular,
eitherH = or H is normal in G of
prime
index. In the first caseH =
Na(.H), by
Burnside’s theorem H has a normalcomplement
Kin
G,
which is a minimal normalsubgroup
of G. Since~g~ c 18
thesubgroup
.g issoluble,
henceelementary
abelian.By
Lemmas 3 and 5the
subgroup K
cannot be ofprime
order. So theonly possibilities
for are
2i, 2 c i c 4,
and 33. As H =NG(H),
one has ==
IG: NG(H) I
the number ofSylow p-subgroups
of G.By Sylow’s
theorem this number is
congruent
to 1mod p.
But this isclearly
not the case
for p
= 11 or 13 with =4, 8, 16,
or 9. So this casecannot arise.
If H is normal in
G,
thenIG: HI
is aprime.
But then Lemmas 3 and 5 show thatG 0 3P,
a contradiction. Therefore we may assume that neither 11 nor 13 divide y i.e.~G~
E{33 , 25,
divisor of24.32.5.7}.
.If G is
soluble,
thent5,
and3~ ~ ~ ,f’
since G cannotcontain an abelian
subgroup
of order > 18. For G one therefore ob- tains one of thefollowing
orders:2-3-5, 2 ~ ~ 3 ~ 5, 22.32, 22.5, 2 2 . 7, 23.37 23.5, 257 33
723.7, 23-327 24-3, 24-57 24-7, 3~’5,
or aproduct
oftwo
primes.
Butby
Lemmas 3 and 5 groups of these orders do notbelong
to 3P.There remains the case that G is non-soluble. A non-abelian com-
position
factor of G must-for order reasons-be one of thesimple
groups considered in Lemma 6. If a minimal normal
subgroup
of Gis
non-soluble,
it would be a directproduct
ofisomorphic copies
ofthe
simple
groups considered there. But these groups do notbelong
to
3P,
so G could notbelong
to 3P. This contradiction shows thatS(G)
~1>,
whereS(G)
denotes thelargest
normal solublesubgroup
of G. A minimal normal
subgroup
ofG/8(G)
is a directproduct
of thesimple
groups considered in Lemma 6. As each of thesesimple
groups contains a solublesubgroup
oforder >10,
the non-soluble group G contains a solublesubgroup
.K of orderK>20.
This contradicts theminimality
of G. DThis
completes
theproof
of Theorem Cb).
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