PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 45, Number 1, July 1974
ALMOSTCOMPLETELY DECOMPOSABLE
TORSION FREE ABELIAN GROUPS
E. L. LADY
ABSTRACT.
A finite
pletely
decomposable
C with
finite
composable
index
mands.
G has,
is a prime
of G as a direct
If G and
finite
rank
subgroup
¿(G),
free
abelian
group
¿(G) = i(H)
in
group
pletely
unruly.
mands
of nonunique
This
paper
We show
is possible.
that
G has,
there
are only finitely
group
terizes
such
H.
groups
All groups
sion
in this
free abelian
Presented
groups.
to the Society,
AMS (MOS) subject
Key words
property,
unless
subgroup,
rank torsion
cancellation
February
Almost
there
2, 1973;
received
free
of direct
decomposable
exists
otherwise,
we will follow
need
is not com-
sumSum-
decomposable
We show
a finite
G © L % H © L. Theorem
indicated
In general,
classifications
and phrases.
regulating
groups
many summands.
H fot which
that
decom-
of all the most
the situation
completely
only finite
L such
paper,
decom-
We show tnat an almost completely
many groups
free abelian
that
situations
a maximal
up to isomorphism,
torsion
of such
of finite
however,
in certain
mand is unique up to isomorphism.
group
show,
We show that
a
to
C having finite index
is the source
decompositions
will
H contains
completely
subgroup
sum decompositions
examples
then
and prime
G is almost
decomposable
of groups
groups.
and
to a completely
class
abelian
groups,
G.
direct
familiar
sum-
are uniquely
\_H: G ] is finite
not be unique.
this
many
in any de-
groups
in G. It is well known that
In fact,
finitely
decomposable
L is isomorphic
free abelian
is a completely
(ii)
that
index
de-
completely
(i) G © L % H © L for some
L.
to G such
finite
rank torsion
if there
only
com-
subgroup
all completely
1 summands
completely
equivalent:
G is almost
An almost
sum of indecomposable
G © L ^ 77 © L where
with
by ¿(G).
the rank
are
isomorphic
subgroup
A finite
posable
torsion
then
are almost
statements
G
(iii)
posable
H
group
decomposable
of [G: C] over
up to isomorphism,
composition
following
abelian
a completely
minimum
power,
determined.
the
free
exists
C of G is denoted
group
If ¿(G)
torsion
in G. The
subgroups
decomposable
rank
if there
that
rank
11 charac-
are finite
rank tor-
the notation
and conven-
by the editors
June
18, 1973.
(1970). Primary 20K15.
completely
decomposable
group,
cancellation
/^-equivalence.
Copyright © 1974, American Mathematical Society
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
41
E. L. LADY
42
tions
of [2].
In particular,
if H is a subgroup
pure subgroup
of G containing
called
The type determined
a type.
The set of types
type
t(x)
a maximal
If r is a type,
by all
by writing
l((x)f).
1 groups.
of rank
summand
of
of G , so that
G(o) for a > r, and
G(S) to denote
the subgroup
of finite
G and
index
from
K-equivalent
then
G = 0LC
of those
.
x
of G generated
generated
by all
G(a)
S is a set of types.
free groups
monomorphisms
decomposable
G consisting
the pure subgroup
Torsion
of
decomposable,
G (t) to denote
to a subgroup
is
by t(A).
if Horn (A, B) f-
G is completely
If G is completely
G(z") is the subgroup
1 groups
G is homogeneous
l(x) > t. We will use
with a e S, where
the smallest
A is denoted
t(A) < t(B)
We say that
r-homogeneous
then
class
1 group
0 4- x e G, and that
sum of rank
G. will denote
by a rank
ordered
denotes
t if t(%) = t for all
if G is a direct
with
H. An isomorphism
is partially
0. If x e G , then
of G, H^ denotes
G into
H ate quasi-isomorphic
in H. This
is equivalent
H and from
H into
if G© L % H © L for some finite
if G is isomorphic
to the existence
G. We will
rank torsion
call
of
G and
H
free abelian
group
L.
1. Regulating
most completely
Definition.
denote
subgroups.
The concept
decomposable
group
If G is an almost
the minimum
of
decomposable
subgroups
decomposable
subgroup
[G: C]
of a regulating
subgroup
is the key to everything
completely
where
decomposable
C ranges
of G. A regulating
C with finite
of an alfollows.
group,
z'(G) will
over the set of all completely
subgroup
index
that
in G such
of G is a completely
that,
for each
type
r,
G(r) = CT © G*(r).
Notice
Cr is pure
lowing
that
if C is a regulating
in G. The existence
subgroup
of regulating
of G, then,
subgroups
for each
follows
type
r,
from the fol-
theorem:
Theorem
1. A completely
pletely
decomposable
= i(G).
If D is any completely
decomposable
group
subgroup
G is a regulating
decomposable
subgroup
subgroup
C of an almost
if and only
of G, then
com-
if [G: C]
\_G: D] is
a multiple of i(G).
Proof.
We show
[G : C] = [G: Di,
phic.
first
Notice
that
that
if C and
C and
D are regulating
Choose a minimal type r such that
Cs = ©2$
CŒ, Ds = ©2^.
by induction,
D . Then
subgroups
D ate quasi-isomorphic,
so that,
we may assume
Since
CT © GiS)^ = G(r) + G(S)„ we have
isomor-
CT4 0 and let S = ía|CCT^ Q, o f t\,
Cs and Ds are regulating
G(S)
of G, then
hence
that
[G(S)
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
subgroups
: Cs] = [G(S)^.: Ds].
of
ALMOST COMPLETELY DECOMPOSABLE TORSION FREE GROUPS
43
[G:C]= [G:Ct © G(5)J[Cf ©G(5)^:C]
= [G:G(r)+G(S)J[G(i)^:Cs]= [G:Dr © G(S)*][G(S)*:DS]= [G:Dl.
It now suffices
group of G and
pletely
to show
decomposable
of [G: C] (hence
computation
that
if D is a completely
D is not a regulating
subgroup
subgroup,
C of G so that
[G: £>] ^ z'(G)). Let
r and
decomposable
then there
a regulating
subgroup
a com-
[G: D] is a proper
S be chosen
multiple
as before.
above shows that [G: D] = [G: Dt © G(S)*][G(S)*:
by induction,
sub-
exists
Cç of GiS),,. and choose
The
D.J. Choose,
by [2, Proposia.
tion 86.5, p. 114], a subgroup
is not a regulating
is a proper
subgroup,
multiple
either
case,
then
Proof.
Let
D
[G: Dr © G(S)+]
or D~ is not a regulating
subgroup
2. // G a/za? ß are K-equivalent
able groups,
and
subgroup
[G(S)*: Ds] is a proper multiple of [G(S)*: Cs].
C. © C^ is the desired
Corollary
G(r) = Cr © G (r). Since
G(t) 4 DT © G^(r)
of [G: CT © G(S)*\
of GCS)^ and, by induction,
In either
CT of G so that
C.
almost
completely
decompos-
i(G) = z'(ß).
C and
D be regulating
subgroups
of G and
H, and let
r
and 5 be as in the proof of Theorem 1. If G © L % ß © L, it follows that
G/(G(t) + GiS)J % H/(H(t) + HiS)*). Also G(S)* © L(S)* % H(S)* © L(S)*,
hence
G(S)*
and H(S)*
are K-equivalent,
so by induction
[G(S)*:
Cs] =
[H(S)m: Ds]. Thus
[G : C] = [G: G(r) + G(S)J[G(S)* : Cs] = [ß : H(r) + H(S)J[H(S)* : D^] = [H : D].
Remark.
If C and
case
that
G/C
rank
1 groups
% G/D.
and let
D are regulating
For instance,
K be a rank
p be a prime
subgroups
let
1 group
such
of G, it need
A, B and
with
not be the
ß be noncomparable
t(K) < infi t(A),
t(B)î
and
t(K) ¿ t(ß).
Let
that none of A, B, H, K ate ¿/-divisible
and choose
a, b, h, k in A, B, H, K all with ¿»-height 0. Let G be genera-
ted by A, B, ß, K and (a + b)/p, (h + k)/p.
of G generated
B ® H ® K'
by k + (a + è)/p.
ate regulating
Let K
be the pure subgroup
Then C = A © B © H © K and D = /l©
subgroups
of G. But
G/C
** Z(/») © Z(p)
where-
as G/D *Z(p2).
Corollary
3. // G and
H are almost
completely
decomposable
groups,
then KG © H) = i(G)i(H).
Proof.
If C and
is a regulating
D are regulating
subgroups
of G and
C ® D
subgroup of G ® H, and [G © «: C ©D] = [G: C] [ß: D].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
Corollary
ß, then
4.
An almost
completely
decomposable
group
G has,
up to
•E. L. LADY
44
isomorphism,
groups
only finitely
H which
Proof.
are
Let
K-equivalent
Then
there
pletely
decomposable
groups
subgroup
er hand,
to a summand
2. Completely
groups
shows,
however,
special
case
G and
do not have
5. Let
decomposable
B and
C have
Outline
com-
D is a regulating
is a finite
of D containing
nD.
D is isomorphic
to a regu-
z'(ß) is a divisor
subgroup
of ß.
Now
On the oth-
of z'(G) and
D is
So in either
case,
In general,
property.
is sometimes
completely
decom-
The following
theorem
possible.
This
theorem
is a
in [l].
summands
on ß
ß(r)
z'(ß) = n and
the cancellation
= B®C
group and
where
A is a r-homogeneous
H has no quasi-summands
B . and
C.
so that
com-
of type
t. Then
G = B, © C. © H. In particular,
then H ^ H ..
of proof.
hypothesis
decomposable
many almost
summands.
G = A(BH
if also G ** A ©ßj,
many
H.
cancellation
of a result
Theorem
D a completely
ß are quasi-isomorphic).
of a regulating
decomposable
pletely
that
of G, then
many such
that
finitely
only finitely
z'(G) = z'(ß) and
of G (because
posable
and
many subgroups
to G, then
are only finitely
Hence
ß such
if ß is a summand
isomorphic
there
are only
ß % nH and nD CnH C D. But D/nD
are only finitely
if ß is K-equivalent
lating
integer
are, up to isomorphism,
subgroup of H. In fact,
so there
There
to G.
72 be a positive
group.
group,
many summands.
Let
implies
a and
ß be the projections
that the restriction
is invariant
under
ß
so that
onto
A and
to H(r) of aß
there
exist
must
summands
B. The
be zero
B,
and
C.
of B(t) and C(t) such that G(r) = B t© Ct© ß(r). It follows that G = Bj © Cx © H.
It is well
no rank
known
that
1 summands
For almost
completely
ing weaker
theorem
Theorem
6.
of type
Proof.
groups.
Note
decomposable
// G z's an almost
G). Thus
fails
instance,
groups,
that
summand
completely
A, B are r-homogeneous
t , then
By [2, Lemma
decomposable
theorem
r. (For
if we merely
cf.
suppose
[2, Theorem
however,
that
90.4,
we do have
ß has
p. 138].)
the follow-
:
A © ß = B © K where
summands
this
of type
A and
86.8,
decomposable
and
group
and
G —
H, K have no rank 1
A «¿ B.
B are necessarily
p. 115]
decomposable
H n B is a r-homomoneous
of B and hence
ß O B = 0 by the hypothesis
completely
of ß (because
on H. Thus
the projection
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
completely
ß is a summand
onto
of
A maps
ALMOST COMPLETELY DECOMPOSABLE TORSION FREE GROUPS
B monomorphically
into
A, so rank
B < rank
A. By symmetry,
rank
45
A =
rank B, so A % B.
Corollary
maximal
7. // G zs a« almost
completely
It does
completely
decomposable
not follow
summand
from Theorem
6 that
sum of indecomposable
groups
summands
of a given
r. This
is because,
even
when neither
may have
a rank
cannot
occur
lowing
technical
Lemma
power.
then a
will involve
the same
as is well
Gj
The proof
of G in-
nor
number
known,
G2 do.
of this
of
Gj © G2
However,
depends
this
on the fol-
lemma:
8.
a regulating
1 summand
if z'(G) is a prime
group,
up to isomorphism.
any two decompositions
to a direct
type
decomposable
of G is unique
Let
G be an almost
subgroup,
and suppose
completely
decomposable
i(G) = p
for some prime
group
and
C
p. For some
type T, let C, be a rank 1 summand of C and let S = \o\o~ f t\. Then C,
z's a summand
of G if and only
if C. ffi G(S)*
is a pure
subgroup
of G. Other-
wise (Cj © G(S))*/(C j © G(S)*) is a cyclic p-group.
Proof. Let cf>:(Cj © G(S))* —*Cj be defined by multiplication
followed
by the projection
induces
a monomorphism
of (C,
© G(S))*/(C,
is a cyclic /»-group. Now if G = Cj©ß,
pure.
Conversely,
C = Cj©
suppose
G = C.(B
G(S)*.
mand of (C, © C,)*
Then
since
G/(Cj
© C2 © C,*)
Otherwise,
Then, for some
c2 have
many exceptions
powers
is trivial
a generator
ate pure
and
or acyclic
p-height
0 and
Cj ^0.
into
C./p
C,,
which
is
We may suppose
that
< r and C(S) C C,, since
on the rank
of G, C.
is a sum-
so that (C, © C,)* = C, © C,*.
C2 has rank
//-group.
s < k, psg = c: + c2 + c
1, it follows
In the former
for a generator
£ Cy © C2 © C,*
Now since
that
case
we are
of this
group.
where both
t(C2) < t(C,),
Cj +
we have
for a = p, whenever ht Ac f) = », and with finitely
otherwise.
Choose
of the exceptional
m so that
primes.
g + mc:
Then
£(C'2
1 + mps
is divisible
by
C, © C, C C, © Cl where
ffi C,)+
and
C' =
g + t/zCj represents
of G/(Cj ffi C2 © C3Hî). Hence G = Cj © (C2 © Cj^.
Theorem
that
by induction
g £ G be a representative
(c2 + (I + mp )cj ) *. Then
such
is pure.
may be chosen
Cj © C,*
let
that ht (cf) < ht(cf)
large
C. © G(S)*
Then
and C
C2 and
done.
c3 and
that
© G(S)*)
then G(S)* C H, so Cl(BG(S)*
C2 © C, where C2 has rank 1 and t(CJ
otherwise
by p
of Cj ®G(S)* onto C,. Since Cj and GTS)* are pure, <f
i(G)
9.
Lei
and
G area" ß
i(H)
H has a rank 1 summand.
èe almost
are powers
Then
completely
of the same
G ©ß
prime
decomposable
Aas no rank 1 summand.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
groups
p, and neither
G nor
Any two
E. L. LADY
46
decompositions
the same
of G as a direct
number
Proof.
C and
of rank
sum of indecomposable
1 summands
Suppose that
D be regulating
of each
type
of A in C
will
involve
t.
G®H = A © K where rank A = 1 and t(A) = r. Let
subgroups
of G and
ß, and let
onto A. Then A Ç CT © D? ©G**(r) ffi ß*(r).
nents
groups
and Df and let
a be the projection
Let Cj and Dj be the compo-
S = !ct|ct ^l t\.
It follows
from Lemma
8 that
a((Cj ffi GÍS^/aíCj
ffi G(S)*) is trivial or a cyclic ¿»-group. In the latter
case, since a(G(S)*) = 0, a(C f) Ç pa(G) = pA 4 A. Likewise either a(D A
ÇpA 4 A or a(Dx) = a((D x ffi H(S))*). But a(G^(r)) = a(ß#(r)) = 0, so A =
a(Cj)
+ a(D,).
Hence we may assume,
say, thata((C[
ffi G(S))*) = a.(Cf).
This
means that (C: ffi G(S))* Ç Cj ffi K. Since G(S)* C K, this means that Cj ffi
GTS)* is pure,
so, by Lemma
8, C,
last
of the theorem
now follows
statement
decomposable
summands
in the following
Lemma
morphic
Let
applies
G and
Proof.
that
G 777 is rpure,
'
hence
n is a multiple
G
for almost
part of the proof,
given
finite
rank torsion
H contains
free abe-
a subgroup
G
iso-
to n.
ß are reduced.
For each
positive
subgroup
m\G mm= G . Each
e
r G m of G so that
large
m,1 we have G m = G 777+. ,1 = • • • ,' hence
I
O
«=0. Without
of ml. Hence
Now suppose
then
G and
The
to the in-
rank groups.
is prime
that
is a largest
o
so for sufficiently
is divisible,
integer,
part
7.
K-equivalence
H be K-equivalent
[H : G ]
We may suppose
m,' there
the first
The most difficult
to all finite
If n is a positive
to G such
integer
o
groups.
lemma,
10.
lian groups.
Corollary
We will now characterize
decomposable
of G, a contradiction.
by applying
of G and using
3. K-equivalence.
completely
is a summand
loss
of generality,
nA 4 A fot every
G ® L ?*sH (B L where
L has
we may suppose
nonzero
finite
subgroup
rank.
G 771
that
A of G.
Let
E(G)= End(G)///End(G)
and for any group
A let
H(A)= Hom(G, A)/72Hom(G, A).
Then
H(A)
is a finite
length
right
E(G)-module
and
H is an additive
functor.
Hence H(G) ffi H(L)% H(ß) ffi H(L) and so H(G) % H(ß) by the Krull-Schmidt
theorem.
erator
In particular,
for H(ß).
F(ß)
is cyclic.
Now for any group
erated by (/(G) + nA)/nA
tor, so that
H(ß)
Let
A, let
t/> €Hom(G,
F(A)
ß)
represent
be the subgroup
of A/nA gen-
fot all / e Hom(G, A). Then F is an additive
% F(G) = G/nG
by the Krull-Schmidt
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
theorem,
a gen-
func-
and we con-
ALMOST COMPLETELY DECOMPOSABLE TORSION FREE GROUPS
elude that F(ß) = H/nH, since
H/nH % G/nG and H/nH is finite.
clearly
we see that
G/nG
that
F(ß)
= (d>(G) + nH)/nH,
onto
H/nH.
From the fact that this
d> is monic,
elements
and second
with order
Remarks.
lent and
that
dividing
G
from Lemma
pG = G for all but finitely
= 0(G)
10 that
many primes
(2) If G and ß are K-equivalent,
then
groups G .^ G 2^i G of ß so that [ß: G.]
It follows
(as in the proof
of G ® G.
In particular,
summands,
then there
Theorem
Then
(i)
(ii)
G and
G and
p, then
ß
has no
subgroup.
are K-equiva-
G *fcH.
H is isomorphic
up to isomorphism,
many such
H be almost
statements
H/dAG)
is the desired
if G and
from
we infer first
by Lemma 10 there exist
and [ß: G2] are relatively
11) that
are only finitely
11. Let
the following
of Theorem
if G © G has,
Since
an isomorphism
map is monic
<f(G) n nH = ncpAG), so that
n. Hence
(1) It follows
<f induces
induced
47
subprime.
to a summand
only finitely
many
H.
completely
decomposable
groups.
are equivalent:
H are K-equivalent.
z'(G) = z'(ß) and
H contains
a subgroup
G
with
G
^ G and[H:G']
finite and prime to i(G).
(iii)
G ® L ^ H ® L where
L is isomorphic
to a regulating
subgroup
of
G.
Proof, (i) ■"*(ii). By Corollary 2 and Lemma 10.
(ii) =» (iii). Let n = i(G), m = [H: G ' ] and choose integers
r and s so
that rm + sn = 1. Let D be a regulating
subgroup of ß and map ß into G ©
D by h\-*(rmh,
snh)
and map
snh = h, H has been
tary summand,
decomposable.
then
embedded
L and
G
® D onto
ß by (x, y) h->x + y. Since
as a summand
in G © D. If L is a complemen-
D are quasi-isomorphic,
By Corollary
rmh +
so
3, z'(L) = i(D) = 1, so
L is almost completely
L is completely
decompos-
able. Thus L%D,
(iii) => (i). Obvious.
Remark. It can in fact be shown that if G and H ate K-equivalent
most completely
decomposable
groups of rank r, and if L is isomorphic
rank r - 1 summand of a regulating
proof is tedious and computational
subgroup
of G, then
and will not be given
alto a
G ® L <%sH © L. The
here.
REFERENCES
1. D. M. Arnold
sion
free
abelian
2. L. Fuchs,
and E. L. Lady,
groups,
Infinite
Trans.
abelian
Amer.
Endomorphism
Math.
Soc.
rings
and direct
sums
of tor-
(to appear).
groups. Vol. I, Pure and Appl. Math., vol. 36,
Academic Press, New York, 1970 and 1973. MR 41 #333.
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF KANSAS, LAWRENCE, KANSAS
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
66044