Mixed Groups with Decomposition Bases and Global k-Groups
Except where reference is made to the work of others, the work described in this thesis
is my own or was done in collaboration with my advisory committee. This thesis does
not include proprietary or classified information.
Chad Mathews
Certificate of Approval:
Overtoun Jenda
Professor
Mathematics and Statistics
William Ullery, Chair
Professor
Mathematics and Statistics
H. Pat Goeters
Professor
Mathematics and Statistics
Peter Nylen
Professor
Mathematics and Statistics
Stephen L. McFarland
Dean
Graduate School
Mixed Groups with Decomposition Bases and Global k-Groups
Chad Mathews
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
August 7, 2006
Mixed Groups with Decomposition Bases and Global k-Groups
Chad Mathews
Permission is granted to Auburn University to make copies of this thesis at its
discretion, upon the request of individuals or institutions and at their expense.
The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Michael Chad Mathews, son of Michael Brett and Tammy Elaine Mathews, was
born on October 27, 1981 in LaGrange, Georgia. He is a 2000 graduate of Heard County
Comprehensive High School in Franklin, Georgia. In the fall of that year, he entered the
University of West Georgia and received the degree of Bachelor of Science in Mathematics
on May 6, 2004. In August of 2004, he began his graduate study in the Department of
Mathematics and Statistics at Auburn University.
iv
Thesis Abstract
Mixed Groups with Decomposition Bases and Global k-Groups
Chad Mathews
Master of Science, August 7, 2006
(B.S., University of West Georgia, 2004)
38 Typed Pages
Directed by William Ullery
This thesis is devoted to proving assertions made without proof by Paul Hill and
Charles Megibben in their fundamental papers regarding knice subgroups and the Ax-
iom 3 characterization of global Warfield groups. The main theme throughout is the
relationship between the notions of a global k-group and a group with a decomposition
basis. Most of our results involve properties of the auxiliary notions of primitive element
and ?-valuated coproduct in both the mixed and torsion free settings.
v
Acknowledgments
The author would like to thank Dr. William Ullery for lending his algebraic knowl-
edge to this project and for his unyielding patience and understanding. The author also
wishes to thank his family for their love and support.
vi
Style manual or journal used Journal of Approximation Theory (together with the
style known as ?aums?). Bibliography follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (specifically
LATEX) together with the departmental style-file aums.sty.
vii
Table of Contents
1 Introduction 1
2 Preliminaries 2
3 Mixed Groups with Decomposition Bases 12
4 Torsion Free Groups with Decomposition Bases 20
5 Torsion Free Separable Groups 25
Bibliography 30
viii
Chapter 1
Introduction
Throughout this thesis, G will always denote an additively written abelian group
and we shall only consider such groups. We do not exclude the possibility that G is a
nonsplit mixed group. By this we mean that G may contain elements of both finite and
infinite order and the torsion subgroup of G is not necessarily a summand.
Our main goal in this thesis is to provide the justification for many of the basic
facts that were stated by P. Hill and C. Megibben in [7, 8] without proof. Indeed, unless
explicitly stated to the contrary, most of our results appear there.
We conclude this brief introduction with an outline of the remainder of the paper.
In chapter two, we state the definitions and provide the notation that will be used
throughout. Chapter three consists of various properties of decomposition bases leading
up to the proof that a group with a decomposition basis is a k-group. In chapter four,
we discuss torsion free groups with decomposition bases and we show that a torsion free
group is completely decomposable if and only if it has a decomposition basis. In chapter
five, we show that every torsion free separable group is a k-group, and we use this fact
to provide an example of a k-group without a decomposition basis.
1
Chapter 2
Preliminaries
This chapter is devoted to providing some basic results needed for the remaining
chapters. The terminology and notation used here is due to Hill and Megibben [7, 8].
Let O? denote the class of ordinals with the symbol ? adjoined as a maximal
element with the convention that ? < ? for all ? ? O?. If x ? G, we write |x|p for the
height of x in G at the prime p. So |x|p = ? if x ? p?G and x /? p?+1G, while |x|p = ?
if x ? p?G for all ? ? O?. If P is the set of rational primes, a height matrix is a doubly
infinite P??0 matrix M = [mp,i] where mp,i ? O? and mp,i < mp,i+1 for all p ? P and
i < ?0. By a height sequence, we mean any sequence ? = {?i}i 1 where k = pn11 pn22 ???pnrr for
distinct primes pi and positive integers ni with i ? {1,2,...,r}. We proceed by induction
on r. If r = 1, then we are done by what we have shown above. So suppose r > 1. By
the induction hypothesis,
bardblpn22 ???pnrr xbardbl = pn22 ???pnrr bardblxbardbl.
Then again making use of the preceding paragraph, we have
bardblkxbardbl = pn11 bardblpn22 ???pnrr xbardbl = pn11 (pn22 ???pnrr )bardblxbardbl = kbardblxbardbl.
With each height matrix M and group G, we associate the fully invariant subgroups
G(M) = {x ? G : bardblxbardbl ? M} and G(M?) = ?x ? G(M) : bardblxbardbl notsimilar M?. (We make
3
the exception that if M ? ?, where ? is the height matrix with all entries ?, then
G(M?) = tG ? G(M). Here tG denotes the torsion subgroup of G.) For each prime
p and each height sequence ? = {?i}i 0. Moreover, we know that M ? bardblalbardbl and we obtain the contradiction that
bardblalbardbl ? M. Therefore, cl = 0 for all l, as claimed.
13
We now know thatsummationtextrl=1 cl = 0 and can conclude from condition (?) that
ssummationdisplay
j=1
dj = kn.
Let pe be the largest power of p that divides kn. Then, there is some dj that is not
divisible by pe+1. After reindexing if necessary, we may assume that d1 is not divisible
by pe+1. Since
piknb1 = d1pinx+d1,1pinx1 +???+d1,tpinxt
for all i < ?0, and since ?x???x1???????xt? is a valuated coproduct, we have that
|pe+ib1|p = |piknb1|p ? |d1pinx|p ? |pe+inx|p.
Because |pe+ib1|p ? mp,e+i for all i, and |pe+ib1|p negationslash= mp,e+i for infinitely many values
of i, we conclude that |pe+inx|p negationslash= mp,e+i for infinitely many values of i. Therefore,
|pinx|p negationslash= mp,i for infinitely many i, and the proof is complete.
Lemma 3.2. tG?G(M) ? G(M?) for every height matrix M.
Proof. We may assume that M notsimilar ?, since otherwise tG?G(M) = G(M?) by definition.
Now, if x ? tG?G(M), then x ? G(M) and there is a positive integer n such that nx = 0.
Note that bardblxbardbl notsimilar M. Indeed, if it were the case that bardblxbardbl ? M, we obtain
? = bardbl0bardbl = bardblnxbardbl ? bardblxbardbl ? M,
14
contrary to the assumption that M notsimilar ?. So, we have that x ? G(M) and bardblxbardbl notsimilar M.
Consequently, x ? G(M?).
Lemma 3.3. If x ? G(M) for some height matrix M and if n is a positive integer, then
the following conditions are satisfied.
(a) If nx ? nG(M?,p) for some prime p, then x ? G(M?,p).
(b) If nx ? nG(M?), then x ? G(M?).
Proof. To prove part (a), we have by hypothesis that nx = ny for some y ? G(M?,p).
Since both x and y are in G(M), x ? y ? G(M). Moreover, x ? y ? tG because
n(x?y) = 0. Therefore, by Lemma 3.2, x?y ? G(M?). Then,
x ? y +G(M?) ? G(M?,p)
because y ? G(M?,p) and G(M?) ? G(M?,p). The proof of part (b) is similar. For
again we have that x?y ? G(M?). But then
x ? y +G(M?) ? G(M?)
since y ? G(M?).
Proposition 3.4. If G has a decomposition basis X, then circleplustextx?X?x? is a ?-valuated
coproduct.
15
Proof. Suppose that y ?circleplustextx?X?x? and write
y = c1x1 +c2x2 +???+ctxt,
where x1,x2,...,xt are distinct elements of X, and cj ? Z for j = 1,2,...,t. We need
to show that if y ? F, where F is one of the fully invariant subgroups of the form G(M),
G(M?), G(M?p,p) or G(M?,p), then each cjxj is in the same F. We consider, in turn,
each of the four natural cases.
Case 1. F = G(M). This case is clear since, by definition, circleplustextx?X?x? is a valuated
coproduct.
Case 2. F = G(M?). If M ? ?, then y ? G(M?) implies that y ? tG. Then y = 0
since each nonzero element of circleplustextx?X?x? has infinite order. It then follows that each
cjxj = 0 ? G(M?). Therefore, we may assume that M notsimilar ? and write
y = a1 +a2 +???+ar
where for i = 1,2,...,r, bardblaibardbl ? M and bardblaibardbl notsimilar M. Now select a positive integer k so
that kai ? ?X? for all i. Thus, for each i we have
kai = di,1x1 +di,2x2 +???+di,txt +dprimei,1z1 +???+dprimei,szs,
16
where x1,x2,...,xt are as above, x1,x2,...,xt,z1,...,zs are distinct elements of X, and
all di,j and dprimei,l are in Z (for j = 1,2,...,t and l = 1,2,...,s). Note that the inequalities
kM ? bardblkaibardbl ? bardbldi,jxjbardbl imply that, for all i and j, di,jxj ? G(kM) and bardbldi,jxjbardbl notsimilar kM.
Thus, each di,jxj is in G((kM)?) = kG(M?). Therefore, since bardblcjxjbardbl ? bardblybardbl ? M and
kcjxj =
rsummationdisplay
i=1
di,jxj ? kG(M?),
Lemma 3.3(b) implies that cjxj ? G(M?) for all j.
Case 3. F = G(M?p,p). In this case we have that
y = a1 +a2 +???+ar
where for i = 1,2,...,r, bardblaibardblp ? Mp and |peai|p negationslash= mp,e for infinitely many e < ?0.
Select a positive integer k such that kai ? ?X? for all i. We then have
kai = di,1x1 +di,2x2 +???+di,txt +dprimei,1z1 +???+dprimei,szs,
where the notation is the same as that in Case 2. For a given j, observe that
rsummationdisplay
i=1
di,j = kcj. (??)
Now temporarily fix j, and after reindexing if necessary, we may assume that j = 1.
Thus, the proof in this case will be complete once we have shown that c1x1 ? G(M?p,p).
17
Let pf be the largest power of p that divides kc1. Then condition (??) implies that
pf+1 does not divide di,1 for some i. For such an i,
|pf+eai|p = |pekc1ai|p ? |pec1di,1x1|p ? |pf+ec1x1|p
for all e < ?0. From this we conclude that |pec1x1|p negationslash= mp,e for infinitely many e.
Moreover, bardblc1x1bardblp ? bardblybardblp ? Mp. Hence, c1x1 ? G(M?p,p).
Case 4. F = G(M?,p). In this case we have that y = a1 + a2 where a1 ? G(M?) and
a2 ? G(M?p,p)?G(M). Select a positive integer k such that kai ? ?X? for i = 1,2. We
then have
kai = di,1x1 +di,2x2 +???+di,txt +dprimei,1z1 +???+dprimei,szs,
where the notation is the same as that in Cases 2 and 3. For i = 1, Case 2 says that
each d1,jxj ? G((kM)?). While for i = 2, Case 3 implies that each d2,jxj ? G((kM)?p,p).
Further observe that cjxj ? G(M) for all j because bardblcjxjbardbl ? bardblybardbl ? M. Thus, for
j = 1,2,...,t,
kcjxj = d1,jxj +d2,jxj ? G(kM)?(G((kM)?) +G((kM)?p,p))
= G((kM)?,p) = kG(M?,p).
Therefore, Lemma 3.3(a) shows that cjxj ? G(M?,p) for all j.
Theorem 3.5. If G has a decomposition basis X, then G is a k-group.
18
Proof. We first note the fact that 0 is always a nice subgroup. Now, if S is a finite subset
of G, there is a positive integer k such that ks ?circleplustextx?X?x?. Then k?S? ?circleplustextx?X?x?. So
for all s ? S, we have that
ks ? ?x1???x2???????xm?
for some distinct x1,x2,...,xm ? X. Then, by Propositions 3.1 and 3.4,
k?S? ? ?x1???x2???????xm?
where the coproduct is a ?-valuated coproduct with each xi primitive.
19
Chapter 4
Torsion Free Groups with Decomposition Bases
In this chapter we show that a torsion free group has a decomposition basis if and
only if it is completely decomposable (Theorem 4.3). We also show that a k-group of
finite torsion free rank has a decomposition basis (Theorem 4.5). As a result, we are
able to give an example of a torsion free group that is not a k-group.
A torsion free group G is of rank 1 if G is isomorphic to an additive subgroup of
Q and has the property that if x,y ? G are nonzero, then mx = ny for some nonzero
m,n ? Z.
Definition 4.1. A torsion free group G is said to be completely decomposable if it is a
direct sum of rank 1 subgroups.
Lemma 4.2. If A is a subgroup of a group G and if p and q are relatively prime integers,
then pA?qA = (pq)A.
Proof. Clearly (pq)A ? pA ? qA. For the reverse inclusion, suppose that x ? pA ? qA.
Then, x = pa1 = qa2 where a1,a2 ? A. Since (p,q) = 1, rp + sq = 1 for some r,s ? Z,
which implies that
a1 = rpa1 +sqa1 = rqa2 +sqa1 = q(ra2 +sa1) ? qA.
20
But then,
x = pa1 ? p(qA) = (pq)A.
If N is a subgroup of a torsion free group G, define N? = {x ? G : nx ? N for some
nonzero integer n}. Observe that N? is a pure subgroup of G and is the smallest pure
subgroup of G that contains N.
Theorem 4.3. A torsion free group G has a decomposition basis X if and only if G is
completely decomposable.
Proof. Suppose that G is a torsion free abelian group and that X is a decomposition
basis for G. Observe that each ?x?? with x ? X has rank 1. For, suppose y,z ? ?x??.
Then my ? ?x? and nx ? ?x? for some nonzero integers m,n. So my = lx and nz = rx
for some nonzero integers l,r. But then
(rm)y = (rl)x = (ln)z.
Next we claim that the sum summationtextx?X?x?? is direct. Indeed, if for some x1 ? X and y ? G
we have that
y ? ?x1?? ?
summationdisplay
x?X\{x1}
?x??,
then there are a finite number of distinct elements x2,x3,...,xk ? X\{x1} such that
y ?summationtextki=2?xi??. Thus,
y = a1 =
ksummationdisplay
i=2
ai
21
where each ai ? ?xi??. Now select a positive integer n such that nai ? ?xi? for all i.
Then, sincecircleplustextx?X?x? is a direct sum,
ny ? ?x1??
ksummationdisplay
i=2
?xi? = 0.
Since G is torsion free and n negationslash= 0, y = 0. We conclude thatsummationtextx?X?x?? =circleplustextx?X?x??.
Since each ?x?? with x ? X is torsion free of rank 1, this part of the proof will be
complete once we have shown that G = circleplustextx?X?x??. For a given y ? G, the fact that
G/?X? is torsion implies there is a positive integer n, distinct x1,x2,...,xk ? X and
c1,c2,...,ck ? Z such that
ny = c1x1 +c2x2 +???+ckxk.
Let n = pe11 pe22 ???pett be the prime factorization of n. Since circleplustextx?X?x? is a valuated
coproduct,
ej ? |pejj y|pj = |ny|pj ? |cixi|pj
for i = 1,2,...,k and j = 1,2,...,t. We then have that
cixi ? pejj G??xi? ? pejj G??xi?? = pejj ?xi??.
Therefore, for each i,
cixi ?
tintersectiondisplay
j=1
pejj ?xi??
22
so that cixi ? n?xi?? by repeated applications of Lemma 4.2. Hence,
ny = na1 +na2 +???+nak = n(a1 +a2 +???+ak)
with ai ? ?xi?? for i = 1,2,...,k. Since n negationslash= 0 and G is torsion free, it follows that
y = a1 +a2 +???+ak ?
circleplusdisplay
x?X
?x??
and we conclude that G =circleplustextx?X?x??.
Conversely, suppose that G is completely decomposable. Say G = circleplustexti?I Ai where
each Ai has rank 1. In each Ai, select a nonzero element xi. Now set X = {xi}i?I. We
claim that X is a decomposition basis for G. To see that G/?X? is torsion, suppose that
g ? G. Then there is a finite subset {i(1),i(2),...,i(n)} ? I with
g = ai(1) +ai(2) +???+ai(n)
and ai(j) ? Ai(j) for j = 1,2,...,n. For each j, there are nonzero integers kj,lj with
kjai(j) = ljxi(j) which implies that kjai(j) ? ?X?. Now, let k = lcm{k1,k2,...,kn}. Then
k has the property that kg ? ?X? and hence G/?X? is torsion. Finally, sincecircleplustexti?I Ai is
a valuated coproduct in G,circleplustexti?I?xi? is valuated.
Definition 4.4. The torsion free rank of a group G is the cardinality of a maximal
Z-independent subset of G consisting only of elements of infinite order.
23
Theorem 4.5. If G is a k-group of finite torsion free rank, then G has a decomposition
basis.
Proof. Let {a1,a2,...,ak} be a maximal Z-independent subset of G consisting of ele-
ments of infinite order. Since G is a k-group, there are primitive elements x1,x2,...,xn ?
G with N = ?x1???x2???????xn? a ?-valuated coproduct such that there is some nonzero
integer m with mai ? N for i = 1,2,...,k. Observe that if g is any element of G, there
is some positive integer l with lg ? ?a1???a2???????ak?. Hence, G/N is torsion. We
conclude that {x1,x2,...,xn} is a decomposition basis for G.
One consequence of Theorem 4.3 and the last result is that any torsion free group of
finite rank cannot be a k-group unless it is completely decomposable. For an example,
let p1,p2,p3 be distinct prime numbers and let
G = Z[1/p1]?Z[1/p2]?Z[1/p3]?(1,1,1)?
?
.
It is known that G is a torsion free group of rank 2 that is not completely decomposable.
For example, see [1]. Hence, G is not a k-group.
24
Chapter 5
Torsion Free Separable Groups
In this chapter we show that a torsion free separable group is a k-group (that does
not necessarily have a decomposition basis).
Definition 5.1. A torsion free group G is called separable if every finite subset of G is
contained in a completely decomposable direct summand of G.
Lemma 5.2. If G = A?B, then for every prime p and ordinal ?, p?G = p?A?p?B.
Proof. Clearly p?A ? p?B ? p?G. So it suffices to prove the reverse inclusion. We
proceed by transfinite induction on ?. If ? = 1, then x ? pG gives that x = py for some
y ? G. Now write y = a+b where a ? A and b ? B. Then
x = py = p(a+b) = pa+pb ? pA?pB ? pG.
Therefore, pG = pA?pB. We finish the proof by considering two cases.
Case 1. ? = ? + 1 for some ?. By induction, p?G = p?A ? p?B. The base case then
provides that p(p?G) = p(p?A)?p(p?B). That is, p?G = p?A?p?B.
Case 2. ? is a limit ordinal. Then p?G = p?A?p?B for all ? < ?. Now if x ? p?G for
each ? < ? (that is, if x ? intersectiontext? ? and |b1|p > ?, then ? < |(a1 + b1)|p = |x|p, a
contradiction. We conclude that |a1|p = ? or |b1|p = ?. Therefore,
|x|p = min{|a1|p,|b1|p} = |a1|p ?|b1|p.
Observe that Corollary 5.3 says that if G = A ? B, then G(M) = A(M) ? B(M)
for every height matrix M.
Proposition 5.4. If G = A?B, then A?B is a ?-valuated coproduct.
Proof. Suppose x ? F where F is one of the fully invariant subgroups G(M), G(M?),
G(M?p,p) or G(M?,p). We need to show that x ? (A ? F) ? (B ? F). We consider, in
26
turn, each of the four natural cases.
Case 1. x ? G(M). Corollary 5.3 provides that the coproduct is valuated.
Case 2. x ? G(M?). If M ? ?, then G(M?) = tG(M). So since G(M) = A(M)?B(M)
we have that tG(M) = tA(M)?tB(M). More precisely,
G(M?) = A(M?)?B(M?) ? (A?G(M?))?(B ?G(M?)).
If M notsimilar ?, then x = x1 + x2 + ??? + xn where bardblxibardbl ? M and bardblxibardbl notsimilar M. Also, for
each i, xi = ai +bi where ai ? A and bi ? B. We claim that bardblaibardbl notsimilar M for all i. Indeed,
if bardblaibardbl ? M, there are positive integers k,l such that M ? kbardblaibardbl and bardblaibardbl ? lM. But
then bardblxibardbl ? bardblaibardbl ? lM and bardblxibardbl ? M. That is, bardblxibardbl ? M, a contradiction. Therefore,
bardblaibardbl notsimilar M, and by symmetry, bardblbibardbl notsimilar M. We now obtain
x =
nsummationdisplay
i=1
(ai +bi) =
nsummationdisplay
i=1
ai +
nsummationdisplay
i=1
bi ? A(M?)?B(M?)
? (A?G(M?))?(B ?G(M?)),
as desired.
Case 3. x ? G(M?p,p). If x ? G(M?p,p), then x = x1 + x2 + ??? + xn where each xj
has the property that bardblxjbardblp ? Mp but |pixj|p negationslash= mp,i for infinitely many i. Now write
xj = aj +bj where aj ? A and bj ? B. Then
bardblajbardblp ?bardblbjbardblp = bardblxjbardblp ? Mp
27
gives that both bardblajbardblp ? Mp and bardblbjbardblp ? Mp. Hence, for all i < ?0,
|piaj|p ?|pibj|p = |pixj|p
gives that both |piaj|p negationslash= mp,i and |pibj|p negationslash= mp,i for infinitely many i. Therefore,
x ? A(M?p,p)?B(M?p,p) ? (A?G(M?p,p))?(B ?G(M?p,p)).
Case 4. x ? G(M?,p). In this case,
G(M?,p) = G(M)?(G(M?p,p) +G(M?))
= (A(M)?B(M))?[(A(M?p,p)?B(M?p,p)) + (A(M?)?B(M?))]
= (A(M)?B(M))?[(A(M?p,p) +A(M?))?(B(M?p,p) +B(M?))]
? (A(M)?(A(M?p,p) +A(M?)))?(B(M)?(B(M?p,p) +B(M?)))
= A(M?,p)?B(M?,p)
? (A?G(M?,p))?(B ?G(M?,p)).
Corollary 5.5. Let G = A?B with A torsion-free of rank 1. If 0 negationslash= a ? A, then ?a??B
is ?-valuated and a is primitive in G.
Proof. Observe that {a} is a decomposition basis for A. Then by Proposition 3.2, a
is primitive in A. So if na ? G(M?,p), it must be that either M notsimilar bardblabardblA = bardblabardblG or
28
mp,i negationslash= |pina|Ap = |pina|Gp for infinitely many i. Thus, a is primitive in G. Finally, note
that since A?B is ?-valuated, ?a??B must be as well.
Theorem 5.6. If G is a torsion free separable group, then G is a k-group.
Proof. Suppose S = {x1,x2,...,xn}is a finite subset of G. Then S ? C where G = C?B
for some B and completely decomposable C of finite rank. Write C = A1?A2?????Am
where each Ai is torsion free of rank 1. For each i, select a nonzero ai ? Ai. Then there
is a positive integer k such that kxi ? ?a1???a2???????am?. Observe that repeated
applications of Corollary 5.5 then gives that each ai is primitive and that the coproduct
is ?-valuated.
Example 5.7. We claim that G = producttext?0 Z is a k-group that does not have a decom-
position basis. We note that G is indeed a k-group since by Theorem 139 of [4], G is
separable, and by Theorem 5.6, torsion free separable groups are k-groups. Now, if G
had a decomposition basis, it would be a direct sum of rank 1 groups by Theorem 4.3.
Then Proposition 96.2 of [3] (due to Mishina [15]) provides that each rank 1 summand
of G is isomorphic to Z. This would mean that G = producttext?0 Z is free, a contradiction in
light of Corollary 52 of [4] which states that producttext? Z is not free for any cardinal ? ? ?0.
Hence, G does not have a decomposition basis.
We conclude by noting that Example 3.1 of [6] provides an example of a torsion free
k-group that is not separable.
29
Bibliography
[1] D. Arnold, Finite rank torsion free abelian groups and rings, Lecture Notes in Math.,
vol. 931, Springer-Verlag, New York, 1982.
[2] L. Fuchs, Infinite abelian groups, vol. I, Academic Press, New York, 1970.
[3] L. Fuchs, Infinite abelian groups, vol. II, Academic Press, New York, 1973.
[4] P. Griffith, Infinite abelian group theory, Univ. of Chicago Press, Chicago, Ill., 1970.
[5] P. Hill, On the classification of abelian groups, photocopied manuscript, 1967.
[6] P. Hill and C. Megibben, Torsion free groups, Trans. Amer. Math. Soc. 295 (1986),
735-751.
[7] P. Hill and C. Megibben, Knice subgroups of mixed groups, Abelian Group Theory,
Gordon-Breach, New York, 1987, pp. 89-109.
[8] P. Hill and C. Megibben, Mixed groups, Trans. Amer. Math. Soc. 334 (1992), 121-
142.
[9] R. Hunter and F. Richman, Global Warfield groups, Trans. Amer. Math. Soc. 266
(1981), 555-572.
[10] R. Hunter, F. Richman, and E. Walker, Warfield modules, Lecture Notes in Math.,
vol. 616, Springer-Verlag, New York, 1977, pp. 87-123.
[11] R. Hunter, F. Richman, and E. Walker, Existence theorems for Warfield groups,
Trans. Amer. Math. Soc. 235 (1978), 345-362.
[12] C. Megibben and W. Ullery, Isotype Warfield subgroups of global Warfield groups,
Rocky Mountain J. Math. 32 (2002), 1523-1542.
[13] C. Megibben and W. Ullery, The sequentially pure projective dimension of global
groups with decomposition bases, J. Pure and Applied Algebra 187 (2004), 183-205.
[14] C. Megibben and W. Ullery, On global abelian k-groups, Houston J. Math. 31 (2005),
675-692.
[15] A. Mishina, On the direct summands of complete direct sums of torsion free abelian
groups of rank 1, Sibirsk. Mat. ?Z. 3 (1962), 242-249.
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