R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
L. F UCHS F. L OONSTRA
On direct decompositions of torsion-free abelian groups of finite rank
Rendiconti del Seminario Matematico della Università di Padova, tome 44 (1970), p. 175-183
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ABELIAN GROUPS OF FINITE RANK L. FUCHS and F. LOONSTRA
*)
1. In his book
[5],
I .Kaplansky asked: if
G fl3 G and H fl3 Hare
isomorphic,
are G and Hisomorphic?
B.J6nsson [4]
hasgiven
anexample
of two torsion-freeindecomposable
abelian groups1)
G and Hof rank 2 which are not
isomorphic,
but their direct squares are isomor-phic.
Puzzledby
thisexample,
we haveinvestigated
the direct powers of torsion-freeindecomposable
groups, and have found that to everyinteger m ? 2,
there exist twoindecomposable
groups, G andH,
of rank 2, such that the direct sum of ncopies
of G isisomorphic
to the directsum of n
copies
of Hexactly
if m divides n(Theorem 1). Moreover,
we can choose G such that the direct sum of m
copies
of Gdecomposes
into the direct sum of m
pairwise non-isomorphic
groups(Theorem 2) 1).
Any
two directdecompositions
of a torsion-free group of rank~ 2 are
isomorphic:
infact,
the group is eitherindecomposable
orcompletely decomposable ~). However,
it is an openproblem
whether ornot a torsion-free group of rankm ? 3 can have
infinitely
many,pairwise non-isomorphic
directdecompositions.
We make astep
towards the so- lution in our Theorem 3 which asserts that to every m ? 3 and every*) Indirizzo degli AA.: L. Fuchs: Tulane University, Dept. of Mathematics, New Orleans, Louisiana 70118, U.S.A.
F. Loonstra: Technical University, Delft, The Netherlands.
1) All the groups in this note are additively written abelian groups. For the definition of rank and related concepts see e.g. Fuchs [2].
2) For another type of pathological direct decomposition of torsion-free groups of finite rank, see Corner [1].
3) See e.g. Fuchs [2].
176
large t
there is a torsion-free group of rank m which has at least tnon-isomorphic
directdecompositions.
Thisyields
anegative
solutionto Problem 15 in Fuchs
[ 3 ] 4).
2. We start off with the construction of groups needed in the
proofs.
Let
Ai(i=1,
...,m)
beisomorphic
groups of rank1,
and=1, ...,
m) isomorphic
groups of rank 1 such thatand for their
endomorphism rings 5)
’holds. For
instance,
we can choose twodisjoint,
infinite sets ofprimes,
say PI and
P~ ,
and define Ai as asubgroup
of the rational vector space with basis ai ,generated by
for allpiePi :
.and similarly,
for
f==l,
.., m. In view of(2),
we can choose elements aiEAi and such that ai , ..., am andbi ,
...,bm correspond
to each other underisomorphisms
of the A1 and therespectively,
andp .~
bi for someprime
p(which
will besuitably
chosen lateron).
Then the groupsare
isomorphic
amongthemselves,
and as standard argumentsshow, they
areindecomposable.
4) This problem asks for the existence of an integer k(n) such that every torsion-free group of rank n has at most k(n) non-isomorphic direct decompositions
into indecomposable summands.
5) Z denotes the ring of integers.
Let
CZ = Ai(i =1,
...,m)
andDI - Bi(i =1,
...,m)
be other sets of groups, and letdiEDi correspond
to ai , bi under some isomor-phisms.
Then for any choice ofki =1,
...,p -1,
we can form thegroups
These are
again indecomposable.
AllXt,
Yi are of rank 2.Assume there is an
isomorphism
From
(1)
we conclude that ... + Am and B=Bifli ... arefully
invariantsubgroups
ofX*,
thus cp must map A onto C=C1 EÐ ...and B onto
where
owing
to(2)
the oci; , areintegers
such that the matricesII II Oij 11
areinvertible,
i.e. the determinantsNotice
that,
forevery i,
An
isomorphism
must map an element divisibleby
p upon an element of the same sort, thus from theindependence
of thedj
we obtainfor all i and
j.
From(6)
and(8)
it follows thatLet us observe that
(6)
and(8) together
guarantee that(5)
and(7) yield
anisomorphism
between X* andY*,
since(6) implies
that(5)
induces an
isomorphism
between A+B andCOD,
while(8) implies,
because of
(7)
and(6),
that all ai+ hi are divisibleby
p if andonly
178
if all the are divisible
by
p.3. Now we are
ready
to prove:THEOREM 1. Given any
integer m ? 2,
there exist twotorsion-free indecomposable
groups Xand Y, of
rank 2 such thatif
andonly if
We choose X as in
(3)
and Y as in(4)
withki =
... = km = k in order to haveisomorphic
groups YI. Ourgoal
is to assure that thedirect sums of m
copies
of X andY, respectively,
areisomorphic,
butthe same fails to hold for a smaller number of
copies.
In view of(9),
this meansbut
(We dropped
thepossibility km---1,
because this can occur in the pre-sence of
(11) only
for oddm).
In order to get a solution for(10)
and( 11 ),
let us choose theprime
p so as tosatisfy
by
Dirichlet’s theorem onprimes
in arithmeticprogressions,
such a p p-ldoes exist. If I is a
primitive
rootmod p,
then setk = l 2~2 ;
it iseasily
seen that both
(10)
and(11)
hold.Next we show ai; and
Oij
can be chosen so as tosatisfy (6)
and(8)
with as chosen above. Let xsatisfy
kx ---1(mod p)
andchoose,
forinstance,
the lowertriangular
m X m-matrixwhere in the main
diagonal
we have 1’s except for the last entryand,
below thediagonal,
we havealternately x
and 0 in the firstcolumn,
and x and 1 in the other columns. Then det(ai;) _ -1.
In accordancewith
(8),
wemultiply [[ I by k
and from each entry we subtracta
multiple of p
to get amatrix [[ I
with determinant -f-1:which is obtained
from [[ I by replacing
1by k
and xby
1through-
out and
by putting y = ( -1 )m + 1(km -f-1 )
in the upperright
corner. Con-sequently,
we can choose rLij, and k such that det(ocl;)= -1,
det and
(mod p).
We conclude that the direct sumof m
copies
of X isisomorphic
to the direct sum of mcopies
of Y, while(11)
ensures that the same fails to hold for ncopies
if n issmaller than m, and hence if it is not divisible
by
m. Thiscompletes
the
proof
of the theorem.It is worth while
noticing
that our methodyields
a somewhat180
better result for odd
integres
m.Namely,
to any odd m and anyinteger
t, there exist t groups of rank 2 such that any two of them behave like X and Y in Theorem 1
(in
this case wereplace
pby
aproduct
ofprimes).
For evenintegers m
we could construct tnon-isomorphic
groups such that the direct sums of m
copies
areisomorphic (but m
is not
necessarily
the smallest number with thisproperty).
4. Let us
point
out that for the direct sum of mcopies
of X in the last theorem wemight
have acompletely
different sort of decom-position :
THEOREM 2. Given an
integer m ? 2,
there existpairwise
non-isomorphic torsion-free indecomposable
groupsX, Yi ,
..., Ymof
rank 2such that
but ... EÐ X
(n summands)
is notisomorphic
to the direct sumof
n groups
Choose X=Xi in
(3)
and the Yi as in(4).
In this case we chooseincongruent ki
so as tosatisfy
but
For
instance,
if we let ..., for a ksatisfying
such that no power of k with a smaller exponent is --- ± 1
(mod p),
then
(12)
and(13)
will be fulfilled. As in theproof
of Theorem 1,we can find a
prime
p and a k with thisproperty.
Wechoose 11
as
earlier,
but the x’s in thejth
column will bereplaced by x3i-l ,
whilein the earlier form
of [[
the k’s in thejth
columnought
to bereplaced by
andy = ( -1 )"z T 1(k2 ~3m-1~ -f-1 ).
Thiscompletes
theproof
of Theorem 2.REMARK 1. It is
straightforward
to show that our choice of kiimplies
thatwith
can hold
only
ifREMARK 2. The fact that the summands in Theorem 1 and 2 are
of rank 2 is not essential. In
fact,
Xi in(3)
can bereplaced by
and Yi in
(4) by
where every
pair
amongAi , Bi ,
..., Ei satisfies(1)
and(2).
5. We
proceed
to theproblem
ofconstructing
a group of rank m? 3 which has many directdecompositions.
It is obvious that it suf- fices to consider the rank 3 caseonly.
THEOREM 3. To any
given integer
t, there exists atorsion-free
group
of
rank 3 which has at least tnon-isomorphic
directdecompo-
sitions.
Let and C be rank 1 groups whose
endomorphism rings
are
isomorphic
to Z andHom (Ai , C)=0= Hom (C, A1).
We choose elements and CEC which are not divisibleby
agiven prime
p
(to
bespecified later)
such that al H bi under someisomorphism.
182
Put
where Hi is
indecomposable
of rank 2.If ai , y; ,
Bi
areintegers
such thata;6;- oiyi= 1,
and if weput
then where
Ai, Bi
are the puresubgroups generated by b; ,
andbi H bi
under suitableisomorphisms
We wish to choose the
A 1 ,
Bi sothat,
for someinteger
we have
with
indecomposable
Hi . Thenmust be divisible
by
p, thusIt is
easily
seen thatconversely, (15) implies
that(14)
holds. If to anywe choose Yi==P, and
Si
so as tosatisfy
=1, then
(15)
will be satisfied. Notice that anisomorphism
cpl; : Hi ~Hi
must
map Bi
uponBi
and C upon itself such that bi and cConsequently,
(pij maps upon Sincedivisibility by
p is
preserved
underisomorphism
andp .~ b; ,
we conclude that gii can existonly
if(mod p).
Therefore,
if we choosethen
k1 ~ ± k; (mod p)
fori 0 j,
and thus the groupsHi ,
..., Ht arepairwise non-isomorphic
andindecomposable, establishing
the existence of at least tnon-isomorphic decompositions
for G.REFERENCES
[1] CORNER, A. L. S.: A note on rank and direct decompositions of torsion-free abelian groups, Proc. Cambr. Phil. Soc. 57 (1961), 230-233.
[2] FUCHS, L.: Abelian Groups, Budapest, 1958.
[3] FUCHS, L.: Recent results and problems on abelian groups, Topics in Abelian Groups, Chicago, 1963, 9-40.
[4] JÓNSSON, B.: On direct decompositions of torsion-free abelian groups, Math.
Scand. 5 (1957), 230-235.
[5] KAPLANSKY, I.: Infinite Abelian Groups, Ann Arbor, 1954.
Manoscritto pervenuto in redazione ii 13 maggio 1970.