R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
B. B RUNO
R. E. P HILLIPS
On multipliers of Heineken-Mohamed type groups
Rendiconti del Seminario Matematico della Università di Padova, tome 85 (1991), p. 133-146
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B. BRUNO - R. E.
PHILLIPS (*)
1. Introduction.
In this paper, a Heineken-Mohamed
type
group is anon-nilpotent
p- group G(p
aprime)
in which every propersubgroup
is both subnormal andnilpotent.
The firstexample
of such a group wasgiven
in[6].
Sub-sequently,
andusing
avariety
oftechniques,
several authors con-structed Heineken-Mohamed
type
groups(see [4], [5], [7]
and[9]).
Inparticular,
in[9]
Meldrum constructs an uncountable number of iso-morphism types
of Heineken-Mohamedtype
groups(see
also[7]).
Mel- drum’s groups arepresented
as direct limits of finite p-groups, a fact which enables that author to determine theisomorphism types
of thegroups constructed. All of the groups G constructed in the above refer-
ences are metabelian with
~(G) ~ ~.
Thequestion
of the existence of solvable Heineken-Mohamed groups of derivedlength greater
thantwo arises in the
study
of certainminimality
conditions(see [2;
p.50]).
In this paper, we will show that if G is any of the p-groups
(p
an oddprime)
constructedby Meldrum,
then the Schurmultiplier M(G)
is infinite.Further,
it will be shown that if H is any stem extension ofM(G) by G,
then H is a threestep
solvable(and non-metabelian)
group of Heineken-Mohamedtype. Evidently,
the(*) Indirizzo
degli
AA.: B. BRUNO:Dipartimento
di Matematica Pura edAp- plicata -
Universita diPadova,
Via Belzoni 7, 35131 Padova(Italy);
R. E. PHIL-LIPS :
Department
of MathematicsMichigan
StateUniversity
E.Lansing,
MI 48823, USA.The authors are indebted to Professor H. Heineken for several
suggestions
concerning
thesubject
of this paper.group H
has infinitecenter,
and it iseasily
seen that themultiplier M(H)
= 0.The existence of solvable Heineken-Mohamed groups of derived
length greater
than three remains an openquestion. Also,
it seemslikely
that any metabelian group of Heineken-Mohamedtype
has infi- nitemultiplier,
but we have not been able toverify
that this is so. Aswill be evident in the
sequel,
our methods are amenableonly
to thosegroups which are
presented
as certain direct limits of finite groups; it is for this reason that weprimarily
discuss the groups of Meldrum.2.
Description
of the groups and statement of the theorems.We
begin
with a briefdescription
of Meldrum’s construction. As much as ispossible,
we will use the sameterminology
as thatof[9].
Let x3 be a
positive integer satisfying
and
4}
a sequence ofpositive integers
withThe sequence
{xk ~k ~ 3}
is definedrecursively by (2.1)
andby
For the
positive integer k > 3,
letAk
be theelementary
Abelian p- group of rank xk, and fix a basis ofAk-
The group
is a
cyclic
group oforder pk
which acts onAk
viaand
Equivalently, [ac(i, k), bk ] = a(i -1,1~)
if1 i x,~,
and-1
It is shown in
[9;
p.438]
thatBk
actsfaithfully
onAk-
Now,
letthe
function 6k : Dk --~
definedby
is an
embedding
ofDk
intoDkll.
The direct limitis a group of Heineken-Mohamed
type,
andG({rk })
isisomorphic
toG({(r* )k })
if andonly
if the sequences and{(r* )k }
areequal [9;
pp.
440-444]. Thus,
the direct limitsG({rk }) provide
an uncountable number ofisomorphism types
of groups of Heineken-Mohamedtype.
These groups are
clearly
metabelian.Further,
Meldrum shows that G =G({rk })
has a non-trivial center if andonly
if there is ako
such thatfor all k >
ko,
rk = 1.Finally,
it is not difficult to show that for oddprimes,
the sequenceprovided by
yields
aG({rk })
which isisomorphic
to the group constructed in[6].
We now state our two
principal
results:THEOREM 1. For any odd
prime
p and any sequence as de- scribedabove,
the group G =G({rk })
hasinfinite multiplier M(G): fur- ther, M(G)
is anelementary
Abelian p-group.THEOREM 2. For any
of
the groups G in Theorem1,
let H be a stemextension
of M(G) by
G. Then H is athree-step
solvable and non-metabelian group
of
Heineken-Mohamedtype.
COROLLARY. For each odd
prime
p, there are an uncountable num-ber
of isomorphism types of
p-groups G which have eachof
thefollow- ing properties.
(i)
G isthree-step solvable;
(ii)
G isnon-metabelian;
(iii)
G isof
Heineken-Mohamedtype;
(iv)
G hasinfinite
center.Theorems 1 and 2 will be
proved
in thefollowing
Section3;
theCorollary
is an obvious consequence of Theorems 1 and 2. The results embodied in Theorems 1 and 2 remain open for theprime p
= 2.3. Additional notation and
proofs.
The essential
ingredient
of theproof
is toapply
the direct limitproperty
ofmultipliers; i. e. ,
themultiplier
of a direct limit is a direct limit of themultipliers [1:
p.57].
In Section3.1,
wedevelop
the nota-tion necessary to
apply
thistecnique:
insubsequent subsections,
weuse the methods
of § 3.1
in theproofs
of the theorems.3.1. Notation. - The
spirit
of what follows is as in[8;
pp.296-300].
We assume the notation of Section 2: in
particular,
we assume that thesequences and
satisfy
the conditions(2.1), (2.2),
and(2.3)
andthat
Throughout,
we will use thegenerators
and relations
as a
presentation
ofDk-
Thus,
ifFk
is free on thegenerators
and the homomor-
phism
definedby
then
Further,
there is anembedding
definedby
and it is
easily
checked that thefollowing diagram
is commu-tative
Denote the Schur
multiplier
ofDk by
the abovediagram implies (see [1;
Section1.3])
that thehomomorphism
induced
by 6k (see 2.7)
isgiven by
Finally,
= themultiplier
ofG,
isgiven by
(see [I ;
pp.56-57]).
3.2. The
multipliers of
thefinite
groupsDk. -
In the free groupFk,
letand
Then
Uk/Sk
isisomorphic
to theelementary
Abelian groupFurther,
is
isomorphic
to the Schurmultiplier
ofEk: also,
withthe elements
{y2, ~ (k)~ 1 i j xk -1}
form a basis for the groupWk.
’
The
automorphism bk
ofEk
induces anendomorphism Bk
ofUk:
here
and
The
subgroups U’k
and[Sk, Uk ]
are bothBk-invariant, and Ak
inducesan
automorphism (also
called on thequotient Wk.
Inparticular,
wehave
(in
additivenotation)
The
equations (3.2.3)
can be stated in thefollowing,
sometimesmore
convenient,
form:It has been shown
by
Evens[3;
pp.170-171] that,
for oddprimes
p, the group isisomorphic
to asubgroup
ofM(Dk) =
= Rk specifically
if weput
the
mapping
defined
by
is an
embedding (this
is theonly point
in all of ourproofs
where we usethe fact
that p
isodd).
We are able to determine a basis and to derive.
other
specific
informationregarding
the vector spaceThis,
inconjunction
with(3.2.6),
willgive
considerable information about themultiplier M(D~ )
and thehomomorphism trk
of(3.1.3).
Wecollect various
properties
of the vector space in thefollowing
(iii)
Consider the element Wr,r+k where 1 ~ k n - r, r > 1. Thenthere are elements
{ei(r, k)}
inGF(p),
thefield of
pelements,
suchthat:
where 8k = 2 if k is
even, andif
k is odd.(iv)
Let{Yl , ... , yt I
and{ ~l , ... ,
be setsof positive integers
sat-isfying
thefollowing
three conditions:Then i ~
tj
is anindependent
set.Proof. Throughout
thisproof
wereplace by
yi,j; recall also that yi,j is defined for 1 ~ ij ~
n. Theequations (3.2.4)
can be trans-lated to read
Part
(i)
of thelemma
followseasily
from(a), (A)
and(~).
We now moveto the
proof
ofpart (iii).
Notice first that for allrelevant r,
= -- Wr,r+l; thus the
equations (3.2.7)
hold for k = 1 and k = 2(for
allr).
Suppose (3.2.7)
true for 2 ~ h k(with
h inplace
of k in(3.2.7));
wecomplete
the inductiveproof by showing
that(3.2.7)
holds for 1~.By (a) above,
= r +k,
the inductionhypothesis yields
Now,
if k is even, we have r + Sk-1 = r + 1 + = r + sk,while,
if 1~ isodd, r
+ s-i r + 1 + Sk-2 = r + sk. It is then clear that the above sumis of the desired
type
and thiscompletes
theproof
ofpart (iii).
Observe that
part (i), together
with theequations (3.2.7),
show thatthe set of vectors in
(ii) (of cardinality m)
forms aspanning
set ofIf we can show that contains a set of m inde-
pendent
vectors(n
= 2m or n = 2m +1),
theproof
ofpart (ii)
would becomplete. Producing
such a set of vectors in appears diffi-cult,
and we instead show that the null space has dimension at least m. SinceWk/[Wk ,
and have the samedimension we conclude from this that = m, and
part (ii)
will now follow.To show that
NS((PK - 1))
has dimension at least m, note that the el-ements Yr
ofWk,
1 r m, definedby
are in The
proof
of this fact isroutine,
and we will not pre- sent it here. The essential observation now isthat yr
contains the term±y,,,,,
and all other terms in yr are of the form where i r andj >
r + 1. From thisobservation,
it is routine to show that the set{ yl , ... , y~ }
is anindependent
subsetFurther,
the vec-tors yr are defined in both of the cases n = 2m and n = 2m + 1. We con- clude that has dimension at least m, and in view of the re-
marks
above,
this concludes theproof
ofpart (ii).
For the
proof
ofpart (iv),
observe that the formulas(3.2.7) together
with the conditions on -(i
and ri imply
that theequation (3.2.7)
for thevector involves the term while this vector does not occur
in the
expression (3.2.7)
of any otherz , j ~ i.
Theindependence
ofthe set
-- tj
followseasily,
and thiscompletes
theproof
ofthe lemma.
LEMMA 2.
Let p
be an oddprime
and use the notationo, f
3.2.5and 3.2.6.
Then
(i)
uk is anembedding
andis a basis
of
Im
(uk ),
whileif
1basis
of
Im(uk ).
It
follows
that Im is anelementary
Abelian p-groupof
orderthus, from (i), M(Dk)
is anelementary
Abelian p-groupof
orderp m+ 1.
.(Vi) If
the setsof positive integers
andsatisfy
thehy- potheses of (iv) of
Lemmac1,
then i ~t}
is anindependent
subset
of M(Dk ).
" ’
Proof.
The first statement of(i)
is a consequence of a theorem of Evens[3:
p.1671,
as it has been noticed before. Moreover it is not hard to prove is asubgroup of M(Dk )
of or-der p and is
isomorphic
toHI«bk),Ek); further,
~ [s(a(1,1~)),
is not contained in Im(uk ).
From these con-siderations and
again
from the work of Evens[3;
Theorem2.1], (i)
fol- lows. Part(ii)
is a consequence of(i)
and ofpart (ii)
of Lemma1,
whilepart (iii)
follows from(i)
and from(iv)
of Lemma1;
thiscompletes
theproof
of Lemma 2.3.3. The
proof of
Theorem 1. - Recall that G is the direct limit of the finite groupsDk
with theembeddings ek (see (2.6)
and(2.7))
and thatM(G)
is the direct limit of the groupsM(Dk )
withhomomorphisms nk (as
in(3.1.3)). Keep
in mind also thatDk
is definedonly
for 1~ > 3.The fact that
M(G)
is anelementary
p-group followsdirectly
frompart (ii)
of Lemma 2. Weproceed
to show thatM(G)
is infinite.Let k ~
3,
nk= xk -1,
and be theidentity
map onM(Dk );
for thepositive integer
s > 1define nk, s by
To avoid excessive
subscripting
in thisproof,
we writeFor
any i
with 1 ~i ~ nk -1,
and s >1,
writeIt follows
easily
from(2.7), (3.1.2), (3.1.3)
and(3.2.5)
thatFurther,
it follows from(2.3)
that the dimensionsnk = xk - 1
satis-fy
The essential features of the
proof
of Theorem 1 are embodiedin
For the
proof
ofparts (a)
and(b)
of(3.3.5),
use(3.3.3)
and an obvi-ous induction on s. We prove
part (c)
alsoby
induction on s. If s =1,
wehave y1
(i)
=~(i -1)
+ rk+1 and ~nk +rk,l - 1 (see (2.3)).
Then(see (3.3.4)),
and this verifies the s = 1 case.Suppose
now that(by (3.3.4))
and thiscompletes
theproof
ofpart (c)
of(3.3.5).
Now that
(3.3.5)
isestablished,
3 and consider the set of vectorsIt is routine to
verify
thatin all cases. We will now prove
is an
independent
subset ofM(Dk+8)
andFor the
proof
of(3.3.6)
letil, z2
=il
+1, i3 = ~
+ be all theintegers
between nk + 1and nk -1 (including endpoints
where appro-priate).
Consider now the sets ofintegers
=1, 2,..., t},
and=
1, 2, ... , t} :
we will prove thatLs
islinearly independent by using (iii)
of Lemma2;
inparticular
we will prove that the two setsabove
satisfy,
forall s,
conditions(a) (b)
and(c)
of(iv)
of Lemma 1. Theindependence
ofLs easily implies
thatcardinality
assertion.For the verification of the
hypotheses
of Part(iv)
of Lemma 1 note that(c)
of(3.3.5) gives
+ 1 and from this it is easy to de- duce that(a)
of Lemma 1(iv)
holds. Notice nowthat,
from our defini-tion,
thusby (a)
of(3.3.5)
we have ys- =
ps = rs(ij) - ys(ij), j =1, ..., t,
and so(b)
of Lemma 1(iv)
alsoholds;
since(c)
of Lemma 1(iv)
istrivially verified,
theproof
of(3.3.6)
is
complete.
It now follows that in the direct limit
M(G)
=~k },
theequivalence
classes~
form an
independent
set.Thus,
the rankof M(G)
must begreater
thanor
equal
tonk/2 -
2 for all k > 3. Since the sequence tends to infin-ity, M(G)
is anelementary
Abelian group(as
a direct limit of element- ary Abeliangroups)
of infiniterank,
and this concludes theproof
ofTheorem 1.
3.4. The
proof of
Theorem 2. - Let G be any of the groups in Theo-rem 1 and M =
M(G)
be themultiplier
of G. Also let H be any central«stem» extension
of M by G (for
existence see[1;
p.92]).
We view M asa
subgroup
of H and so M c H’ n~(H).
Thegroup H
isobviously
threestep solvable;
we show that H" ~ 1. As we have seen in Theorem1,
M is an infiniteelementary
abelian p-group. Let S be a finitesubgroup
ofM of rank > 2. There is then a finite
subgroup
K of H such that(1)
for some1~ ~ 3, KM/M = K/K n M = Dk
and(2) S ~ K’ n M.
Now,
since K n M is anelementary
Abelian p-group, K’ n M is a di- rect factor of K nM;
thus there exists X c K n M such that K n M ==
(K’
nM) O
X. Now W =K/X
is a central «stem» extension of K nM/X by (K/X)/(KnM/X)
Moreover n M and S c K’ n M. Thus
K n M/X
hasrank > 2. We will show that W" #
1,
which in turnimplies
that1.
Using
the notationof [1; Prop.
3.4 pp.92-93],
there is a normal sub-group
Tk
ofFk satisfying:
(iv) Rk/Tk = K
nM/X
has rank at least 2 and is ahomomorphic image
ofM(Dk ).
From Lemma 2
(i)
we see that there must be an i such thatwe see that
[s(a(i,1~)), s(a(i
+1, k))]Tk
E Thus W =Fk/Tk
isnot metabelian.
To
complete
theproof
of Theorem 2 we must prove that H is a group of Heineken-Mohamedtype.
To thisend,
let N be a propersubgroup
ofH;
then NM is a propersubgroup of H,
for otherwise H’ = N’ cN,
andsince M c H’ c
N,
we have N =H,
a contradiction. ThusNM/M
is aproper
subgroup of H/M
= G and so is subnormal inH/M
andnilpotent.
This
implies
that NM is subnormal in Hand,
since M is central inH,
that N is
nilpotent
and subnormal. Thiscompletes
theproof
of Theo-rem 2.
REFERENCES
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and the SchurMultiplier,
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BRUNO,
On p-groups with«nilpotent-by-finite»
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EVENS,
The Schurmultiplier of
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Manoscritto pervenuto in redazione 19