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\begin{document}
\title{Projective Classes of Completely\linebreak Decomposable Abelian Groups}
\author{Carol L. Walker\thanks{Work on this paper was partially supported by NSF GP-21448.}}
\date{\ }
\maketitle
\paragraph{1. Introduction}
The relative homological algebras generated by taking classes of completely
decomposable groups as projectives have several interesting properties which
we investigate in this paper. A group\footnote{The word group will mean
\textit{Abelian group} throughout this paper.} is \textit{completely
decomposable} if it has a decomposition as a direct sum of rank one groups,
the rank one groups being the group $Q$ of rationals, groups of the type
$Z(p^{\infty})$ and subgroups of $Q$ and the $Z(p^{\infty})$'s.
If $\mathcal{C}$ is a class of groups, the relative homological algebra
generated by $\mathcal{C}$ consists of the class $\mathcal{E}=\mathcal{E}%
\left( \mathcal{C}\right) $ of all short exact sequences $A\rightarrowtail
B\twoheadrightarrow C$ of Abelian groups for which the sequence%
\begin{equation}
0\longrightarrow\operatorname*{Hom}(G,A)\longrightarrow\operatorname*{Hom}%
(G,B)\longrightarrow\operatorname*{Hom}(G,C)\longrightarrow0\tag{*}%
\label{homsequence}%
\end{equation}
is exact for every $G\in\mathcal{C}$. Associated with $\mathcal{E}$ are the
class $\overline{\mathcal{C}}$ of all groups $G$ for which (\ref{homsequence})
is exact for all sequences $A\rightarrowtail B\twoheadrightarrow
C\in\mathcal{E}$, and the class $\mathcal{I}=\mathcal{I}(\mathcal{E})$ of all
groups $G$ for which the sequence%
\[
0\longrightarrow\operatorname*{Hom}(C,G)\longrightarrow\operatorname*{Hom}%
(B,G)\longrightarrow\operatorname*{Hom}(A,G)\longrightarrow0
\]
is exact for every $A\rightarrowtail B\twoheadrightarrow C\in\mathcal{E}$. The
class $\mathcal{E}$ is the class of \textit{proper short exact sequences}. The
class $\overline{\mathcal{C}}$ is the class of \textit{projectives} and the
\textit{projective closure} of $\mathcal{C}$. The class $\mathcal{I}$ is the
class of \textit{injectives}.
If $\mathcal{C}$ is any set of rank one groups it follows immediately from
Theorem 1 in \cite{RWW} that there are enough projectives, i.e. for each group
$A$ there is a proper exact sequence%
\[
K\rightarrowtail C\twoheadrightarrow A
\]
with $C\in\overline{\mathcal{C}}$. It also follows from this theorem that
$\overline{\mathcal{C}}$ consists of all summands of direct sums of groups in
$\mathcal{C\cup}\{Z\}$, where $Z$ is the group of integers. It then follows
from theorems of Kulikov \cite{Kulikov}, Kaplansky \cite{Kaplansky}, and Baer
\cite{Baer}, and theorems about completely decomposable torsion groups that
$\overline{\mathcal{C}}$ consists of all direct sums of groups in
$\mathcal{C}\cup\{Z\}$. Kulikov proved that summands of countable completely
decomposable torsion free groups were again completely decomposable, and
Kaplansky reduced the general prohlem to the countable case. Baer proved that
any two decompositions of a completely decomposable torsion free group into a
direct sum of rank one groups are isomorphic, so $\overline{\mathcal{C}}$ does
not contain rank one torsion free groups outside of $\overline{\mathcal{C}%
}\cup\{Z\}$. Similar theorems for completely decomposable torsion groups are
well known.
Thus the classes of projectives in these relative homological algebras are
completely known. The aspects we will explore are descriptions of the proper
exact sequences, and determinations of the relative homological dimensions.
A long exact sequence%
\[
\ldots\rightarrow G_{n+1}\overset{f_{n}}{\longrightarrow}G_{n}\overset
{f_{n-1}}{\longrightarrow}G_{n-1}\rightarrow\ldots
\]
is a \textit{proper} exact sequence if each short exact sequence
\[
\operatorname{Im}f_{n}\rightarrowtail G_{n}\twoheadrightarrow\operatorname{Im}%
f_{n-1}%
\]
is a proper short exact sequence. A \textit{proper projective resolution} of a
group $G$ is a proper exact seqence%
\begin{equation}
\ldots\rightarrow P_{n+1}\rightarrow P_{n}\rightarrow P_{n-1}\rightarrow
\ldots\rightarrow P_{1}\rightarrow P_{0}\twoheadrightarrow G\tag{**}%
\label{resolution}%
\end{equation}
with each $P_{i}\in\overline{\mathcal{C}}$. It follows easily from the
existence of enough projectives that every group has a proper projective
resolution. If $G$ has a proper projective resolution (\ref{resolution}) with
$P_{m}=0$ for $m>n$, $G$ has $\mathcal{C}$\textit{-projective dimension} $\leq
n$. If $G$ has no such resolution for any integer $n$, $G$ has
\textit{infinite} $\mathcal{C}$\textit{-projective} \textit{dimension}. The
projective dimension of the relative homological algebra is the least upper
bound of the $\mathcal{C}$-projective dimensions of all groups. We can
ascertain dimensions only in a few special cases. Also we can determine the
class of injectives only in a few special cases.
\paragraph{2. Classes of divisible and free groups.}
Projective classes which contain divisible groups were investigated in
\cite{RWW}. It was proved there that the group $Q$ of rationals is in
$\overline{\mathcal{C}}$ if and only if for each short exact sequence
$A\rightarrowtail B\twoheadrightarrow C$ in $\mathcal{E}=\mathcal{E}\left(
\mathcal{C}\right) $, the sequence of subgroups%
\[
A\cap dB\rightarrowtail dB\twoheadrightarrow dC
\]
is exact (where $dG$ denotes the maximum divisible subgroup of $G$) and $A\cap
dB$ is the direct sum of a cotorsion group and a divisible group (i.e.
$\operatorname*{Ext}\left( Q,A\cap dB\right) =0$). The group $Z(p^{\infty})$
is in $\overline{\mathcal{C}}$ if and only if for each short exact sequence
$A\rightarrowtail B\twoheadrightarrow C$ in $\mathcal{E}=\mathcal{E}\left(
\mathcal{C}\right) $ the sequence of subgroups%
\[
dA_{p}\rightarrowtail dB_{p}\twoheadrightarrow dC_{p}%
\]
is exact (where $dG_{p}$ is the maximal divisible $p$-primary subgroup of
$G$). Also, $\overline{\mathcal{C}}$ contains all divisible groups if and only
if for each short exact sequence $A\rightarrowtail B\twoheadrightarrow C$ in
$\mathcal{E}=\mathcal{E}\left( \mathcal{C}\right) $ the sequence of
subgroups%
\[
dA\rightarrowtail dB\twoheadrightarrow dC
\]
is exact. These theorems characterize the proper exact sequences in all cases
where $\mathcal{C}$ is a set of divisible groups.
If $\mathcal{C}$ is any set of rank one torsion divisible groups or the set of
all rank one divisible groups then the class of divisible groups in
$\overline{\mathcal{C}}$ is closed under homomorphic images. Thus each group
$G$ has a unique maximum divisible subgroup $\mathcal{C}(G)$ belonging to
$\overline{\mathcal{C}}$, namely, $\operatorname{Im}\left( \sum_{C\in C}%
\sum_{\operatorname*{Hom}\left( C,G\right) }C\longrightarrow G\right) $.
Let $F\twoheadrightarrow G$ be a free group mapping onto $G$ and let
$\mathcal{C}(G)\rightarrow G$ be the inclusion map. It is easy to see that the
sum of these two maps%
\[
K\rightarrowtail F\oplus\mathcal{C}(G)\twoheadrightarrow G
\]
gives a proper projective resolution of length $\leq1$ The group $K$ must be a
free group, since projecting into $F$ is a monomorphism.
The only other possibilities are sets $\mathcal{C}$ of divisible rank one
groups with $Q\in\mathcal{C}$ and $Z(p^{\infty})\notin\mathcal{C}$ for some
prime $p$. We show that in this case $Z(p^{\infty})$ has $\mathcal{C}%
$-projective dimension 2, and that the relative homological algebra generated
by $\mathcal{C}$ has projective dimension 2.
If $G$ is any group, there is a decomposition $G=D\oplus R$ with $D$ divisible
and $R$ reduced (i.e. $R$ contains no non-zero divisible subgroup). Then $D$
decomposes as $D=D_{0}\oplus D_{1}$ with $D_{0}\in\overline{\mathcal{C}}$ and
$D_{1}$ containing no non-zero subgroups in $\overline{\mathcal{C}}$. If we
have a proper projective resolution of $D_{1}$%
\[
\ldots\longrightarrow P_{3}\longrightarrow P_{2}\longrightarrow P_{1}%
\longrightarrow P_{0}\twoheadrightarrow D_{1}%
\]
we may assume each $P_{i}$ is torsion free, since, for example,
$\operatorname*{Hom}(Z\left( p^{\infty}\right) ,D_{1})=0$ for all $Z\left(
p^{\infty}\right) \in\mathcal{C}$. Let $F_{1}\rightarrowtail F_{0}%
\twoheadrightarrow R$ be a free resolution of $R$ and let $D_{0}\rightarrow
D_{0}$ be the identity map. It is easy to see that the sum of these sequences%
\[
\ldots\rightarrow P_{3}\rightarrow P_{2}\rightarrow P_{1}\oplus F_{1}%
\rightarrow D_{0}\oplus P_{0}\oplus F_{0}\twoheadrightarrow D_{0}\oplus
D_{1}\oplus R=G
\]
is a proper projective resolution of $G$. This reduces the problems of
relative homo\-logical dimension to determining what happens for divisible
groups $D$ which have no nonzero subgroups in $\mathcal{C}$, and this quickly
reduces to finding resolutions for groups $Z\left( p^{\infty}\right) $ which
do not belong to $\mathcal{C}$.
Suppose $Z\left( p^{\infty}\right) \notin\mathcal{C}$ and let $P$ be the
additive group of $p$-adic integers. Then $P\otimes Q$ is torsion-free
divisible and there is an exact sequence%
\[
P\rightarrowtail P\otimes Q\twoheadrightarrow Z\left( p^{\infty}\right)
\]
This sequence is proper since $P$ is cotorsion. Let $K\rightarrowtail
F\twoheadrightarrow P$ be a free resolution of $P$. This is a proper
projective resolution, since $P$ is reduced. Composing the two sequences gives
a proper projective resolution%
\[
K\rightarrowtail F\rightarrow P\otimes Q\twoheadrightarrow Z\left( p^{\infty
}\right)
\]
of $Z\left( p^{\infty}\right) $. Thus $Z\left( p^{\infty}\right) $ has a
$\mathcal{C}$-projective dimension at most 2. Let%
\[
F_{1}\oplus D_{1}\rightarrowtail F_{0}\oplus D_{0}\twoheadrightarrow Z\left(
p^{\infty}\right)
\]
be a short exact sequence with $F_{1},F_{0}$ free and $D_{1},D_{0}$ torsion
free divisible. If the sequence is proper, then so is%
\[
F_{1}\rightarrowtail F_{0}\oplus\left( D_{0}/D_{1}\right) \twoheadrightarrow
Z\left( p^{\infty}\right)
\]
so we may as well assume $D_{1}=0$. Then if the sequence is proper, $F_{1}\cap
D_{0}$ is both free and cotorsion plus divisible. This implies $F_{1}\cap
D_{0}=0$. Thus the map $D_{0}\rightarrow Z\left( p^{\infty}\right) $ is a
monomorphism. But this implies $D_{0}=0$ since $D_{0}$ is torsion free. But
clearly the sequence $F_{1}\rightarrowtail F_{0}\twoheadrightarrow Z\left(
p^{\infty}\right) $ is not proper. We conclude the original sequence is not
proper. Thus $Z\left( p^{\infty}\right) $ has dimension exactly 2 if
$\mathcal{C}$ is a set of rank one divisible groups with $Q\in\mathcal{C}$ and
$Z\left( p^{\infty}\right) \notin\mathcal{C}$, and the relative homological
algebra has dimension 2.
The relative $\operatorname*{Ext}$ functors can be computed in terms of
$\operatorname*{Hom}$ and $\operatorname*{Ext}$ for $\mathcal{C}$ any set of
divisible groups. If $Q\notin\overline{\mathcal{C}}$ then any group $G$ has a
maximum divisible subgroup $D$ belonging to $\overline{\mathcal{C}}$, and it
is easy to show that%
\[
\mathcal{E}^{1}\left( G,A\right) \cong\mathcal{E}^{1}\left( G/D,A\right)
=\operatorname*{Ext}\left( G/D,A\right)
\]
and $\mathcal{E}^{n}\left( G,A\right) =0$ for $n>1$. Now suppose
$Q\in\overline{\mathcal{C}}$ and let $I=\{p:Z\left( p^{\infty}\right)
\notin\overline{\mathcal{C}}\}$. A group $G$ has a decomposition
$G=D_{1}\oplus D_{2}\oplus R$ with $D_{1}\in\overline{\mathcal{C}}$, $D_{2}$ a
divisible subgroup of $G$ which has no non-zero subgroups in $\overline
{\mathcal{C}}$, and $R$ reduced. Let $r_{p}$ be the $p$-rank of $D_{2}$. Then
for any group $A$, we have $\mathcal{E}^{n}\left( G,A\right) \cong
\mathcal{E}^{n}\left( D_{2},A\right) \oplus\mathcal{E}^{n}\left(
R,A\right) $. Also, $\mathcal{E}^{1}\left( R,A\right) =\operatorname*{Ext}%
(R,A)\cong\operatorname*{Ext}(G/dG,A)$, and $\mathcal{E}^{n}\left(
R,A\right) =0$ for $n>1$. The problem comes down to computing $\mathcal{E}%
^{n}\left( Z\left( p^{\infty}\right) ,A\right) $ for $p\in I$, $n=1,2$.
For each prime $p\in I$, there is a proper exact sequence%
\[
P(p)\rightarrowtail D\twoheadrightarrow Z\left( p^{\infty}\right)
\]
with $P(p)$ the additive group of $p$-adic integers and $D$ torsion free
divisible. This leads to the exact sequence%
\[%
\begin{tabular}
[c]{l}%
$\operatorname*{Hom}(D,A/dA)=0\rightarrow\operatorname*{Hom}%
(P(p),A/dA)\rightarrow\mathcal{E}^{1}(Z\left( p^{\infty}\right)
,A/dA)\rightarrow$\\
$\ \rightarrow\mathcal{E}^{1}(D,A/dA)=0\rightarrow\mathcal{E}^{1}%
(P(p),A/dA)\rightarrow\mathcal{E}^{2}(Z\left( p^{\infty}\right)
,A/dA)\rightarrow$\\
$\ \rightarrow\mathcal{E}^{2}(D,A/dA)=0$%
\end{tabular}
\]
and we have isomorphisms%
\[
\operatorname*{Hom}(P(p),A/dA)\cong\mathcal{E}^{1}(Z\left( p^{\infty}\right)
,A/dA)\cong\mathcal{E}^{1}(Z\left( p^{\infty}\right) ,A)
\]
and%
\[
\operatorname*{Ext}(P(p),A)\cong\mathcal{E}^{1}(P(p),A/dA)\cong\mathcal{E}%
^{2}(Z\left( p^{\infty}\right) ,A/dA)\cong\mathcal{E}^{2}(Z\left(
p^{\infty}\right) ,A).
\]
Combining these results, we have%
\[
\mathcal{E}^{1}(G,A)\cong\operatorname*{Ext}(G/dG,A)\oplus\prod_{p\in I}%
\prod_{r_{p}}\operatorname*{Hom}(P(p),A/dA)
\]
and%
\[
\mathcal{E}^{2}(G,A)\cong\prod_{p\in I}\prod_{r_{p}}\operatorname*{Ext}%
(P(p),A)
\]
whenever $Q\in\overline{\mathcal{C}}$.
The other aspect of these relative homological algebras is the class of
injectives. In this aspect, the relative homological algebras are deficient.
If $G$ is any reduced group there is a prime $p$ for which $pG\neq G$. For
such a prime, there is a non-splitting extension $G\rightarrowtail
H\twoheadrightarrow Z(p)$. But such an extension is proper for any class
$\mathcal{C}$ of rank one divisible groups. Thus there are no reduced
injectives for any of these relative homological algebras.
\paragraph{3. Completely decomposable groups.}
We turn now to the set $\mathcal{C}$ of all rank one groups. As mentioned in
the introduction, $\overline{\mathcal{C}}$ is the class of all completely
decomposable groups. A couple of the properties of the proper exact sequences
in $\mathcal{E}=\mathcal{E}(\mathcal{C})$ are easily observed. First
$\mathcal{C}$ contains all primary cyclic groups, so the sequences in
$\mathcal{E}$ are pure, i.e. if $A\subset B$ then $A\rightarrowtail
B\twoheadrightarrow B/A\in\mathcal{E}$ implies $A\cap nB=nA$ for all positive
integers $n$. Second, $\mathcal{C}$ contains all rank one divisible groups so
the sequcnee $dA\rightarrowtail dB\twoheadrightarrow dC$ of maximum divisible
subgroups is exact whenever $A\rightarrowtail B\twoheadrightarrow C$ belongs
to $\mathcal{E}$. In the case both $A$ and $C$ are torsion, these two
conditions are sufficient to imply an extension $A\rightarrowtail
B\twoheadrightarrow C$ is proper.
\begin{theorem}
If $A$, $C$ are torsion, then a short exact sequence $A\rightarrowtail
B\twoheadrightarrow C$ is proper with respect to the projective class of
completely decomposable groups if and only if the sequence is pure exact and
the sequence $dA\rightarrowtail dB\twoheadrightarrow dC$ of maximum divisible
subgroups is exact.
\end{theorem}
\begin{proof}
We have already observed the necessity of the conditions. Suppose
$A\rightarrowtail B\twoheadrightarrow C$ is pure exact and the sequence of
maximum divisible subgroups is exact. Then $\operatorname*{Hom}%
(G,B)\rightarrow\operatorname*{Hom}(G,C)$ is an epimorphism for all groups
$G\in\mathcal{C}$ which are either torsion or divisible. If $Z\subseteq
G\subseteq Q$ and $f:G\rightarrow C$ then the image of $f$ is torsion, hence
isomorphic to a subgroup of $Q/Z$. Thus $\operatorname{Im}f\in\overline
{\mathcal{C}}$, in fact $\operatorname{Im}f=\sum T_{p}$ where $T_{p}\cong
Z(p^{n})$ for some $0\leq n\leq\infty$. The inclusion $\operatorname{Im}%
f\subseteq C$, can thus be factored through $B\rightarrow C$. But this factors
$f$ as well, so we have $\operatorname*{Hom}(G,B)\rightarrow
\operatorname*{Hom}(G,C)$ an epimorphism for all $G\in\mathcal{C}$, and the
sequence is proper.\medskip
\end{proof}
In the case both $A$ and $C$ are torsion free, the proper sequences have been
characterized by Lyapin \cite{Lyapin}. Lyapin's theorem and other relevant
information are in \cite{Fuchs}, \S46. We need a couple of definitions before
stating the theorem. If $G$ is a torsion free group and $x\in G$, the
\textit{height of} $x$ \textit{in} $G$ is the sequence $H_{G}\left( x\right)
=\left\langle k_{1},k_{2},\ldots\right\rangle $ where, if $p_{1},p_{2},\ldots$
is a listing of the primes in their natural order, $k_{i}=k$ if $x\in
p_{i}^{k}G$ and $x\notin p_{i}^{k+1}G$, and $k_{i}=\infty$ if $x\in p_{i}%
^{k}G$ for all positive integers $k$. If $G$ and $G/H$ are both torsion free
then a coset $x\in G/H$ is \textit{regular} if $x$ contains an element $a$
such that $H\left( b\right) \leq H\left( a\right) $ for all $b\in x$.\medskip
\noindent\textbf{Theorem} (Lyapin). \textit{Let the factor group }%
$G/H$\textit{\ of the torsion free group }$G$\textit{\ be a completely
decomposable torsion free group. Then }$G$\textit{\ is the direct sum of }%
$H$\textit{\ and a completely decomposable group if and only if every coset of
}$G/H$\textit{\ is regular.}\medskip
Our second theorem follows as an immediate corollary.
\begin{theorem}
If $A$, $C$ are torsion free, then a short exact sequence $A\rightarrowtail
B\twoheadrightarrow C$ is proper with respect to the projective class of
completely decomposable groups if and only if every coset of $B$ mod $A$ is
regular (i.e. contains an element of maximal height).
\end{theorem}
In order to state a general theorem we need to modify the definition of
regular coset. If $x\in G$, $o(x)$ denotes the smallest positive integer $n$
such that $nx=0$, if such exists. Otherwise $o(x)=\infty$. If $\alpha$ is an
ordinal, $p^{\alpha}G$ is defined inductively as follows. If $\alpha$ is a
limit ordinal and $p^{\beta}G$ is defined for $\beta<\alpha$, then $p^{\alpha
}G=\cap_{\beta<\alpha}p^{\beta}G$. If $\alpha=\beta+1$ and $p^{\beta}G$ is
defined, then $p^{\alpha}G=p(p^{\beta}G)$. Now if $x\in G$, $H_{G}\left(
x\right) =\left\langle k_{1},k_{2},\ldots\right\rangle $ where $k_{i}$ is the
ordinal such that at $x\in p_{i}^{k}G$ and $x\notin p_{i}^{k+1}G$, if such
exists, and $k_{i}=\infty$ if $x\in p_{i}^{k}G$ for all ordinals $k$. We will
encounter only heights with the $k_{i}$'s either integers or $\infty$. Note
that to say $x$ has height $\infty$ at the prime $p$ means $x$ belongs to the
maximum $p$-divisible subgroup of $G$. If $x$ also has order a power of the
prime $p$, this means $x$ belongs to the $p $-primary divisible subgroup $dG$
of $G$.\medskip
\noindent\textbf{Definition}. Let $H$ be a subgroup of a group $G$. A coset
$x\in G/H$ is \textit{regular} if for each rank one subgroup $A/H$ of $G/H$
containing $x$, there is an element $a\in x$ such that $o(a)=o(x)$ and
$H_{A}(a)=H_{A/H}(x)$. The subgroup $H$ is a \textit{regular subgroup} of $G$
if every coset of $G/H$ is regular. A short exact sequence $H\rightarrowtail
G\twoheadrightarrow K$ is \textit{regular} if the image of $H$ is a regular
subgroup of $G$.\medskip
We need to verify that this definition is equivalent to Lyapin's in the event
both $G$ and $G/H$ are torsion free. It is easy to see that if a coset is
regular in the sense just defined, then every coset has an element of maximal
order. If $x\in G/H$ and both $G$ and $G/H$ are torsion free, then there is a
pure rank one subgroup $P/H$ of $G/H$ containing $x$. Then use the equalities
$H_{P/H}(x)=H_{G/H}(x)$ and for $a\in x$, $H_{P}(a)=H_{G}(a)$, to get an
element $a\in x$ of maximal height. For the converse, we need to apply
Lyapin's theorem. If $G$ and $G/H$ are both torsion free and every coset of
$G/H$ contains an element of maximal height, let $x\in A/H\subseteq G/H$ with
$A/H$ rank one. There is a pure rank one subgroup $P/H$ of $G/H$ with
$A/H\subseteq P/H$. Applying Lyapin's theorem to $P/H$ we get $P=H\oplus R$
for some $R$, and consequently $A=H\oplus\left( A\cap R\right) $. Now pick
$a\in x$ such that $a\in A\cap R$ and we get $H_{A/H}(x)=H_{A}(a) $.
Our aim is to prove that the regular exact sequences are precisely the proper
short exact sequences in the relative homological algebra generated by the
projective class of all completely decomposable groups. Most of the work lies
in proving the following\medskip
\noindent\textbf{Lemma}. \textit{If }$H$\textit{\ is a regular subgroup of
}$G$\textit{\ and }$G/H$\textit{\ is a reduced rank one torsion free group,
then }$H$\textit{\ is a direct summand of }$G$\textit{.\medskip}
\begin{proof}
Let $0\neq a+H$ $\in G/H$, with $H_{G}(a)=H_{G/H}(a+H)=\left\langle
k_{1},k_{2},...\right\rangle $. Let $F=\left\{ n:k_{n}\text{ is
finite}\right\} $, $I=\left\{ n:k_{n}=\infty\right\} $. For $i\in F$, let
$a=p_{i}^{k_{i}}b_{i}$. For $i\in I$, we know $a\in p_{i}^{\infty}G$, so we
can find $b_{i,1}\in p_{i}^{\infty}G$ with $p_{i}b_{i,1}=a$. Pick
$b_{i,j+1}\in p_{i}^{\infty}G$ with $pb_{i,j+1}=b_{i,j}$, $j=1,2$. Let $A$ be
the subgroup of $G$ generated by the sets $\left\{ b_{i}\right\} _{i\in F} $
and $\left\{ b_{i,j}\right\} _{i\in I,j\in Z^{+}}$. We first show $A$ is
torsion free. If $x\in A$, $x=\sum_{i\in J}n_{i}b_{i}$ for $J$ some finite set
of positive integers, and where for $i\in J\cap I$, $b_{i}=b_{i,j_{i}}$.
Suppose $mx=0$ for some $m\neq0$. Let $t_{i}$ be a non-negative integer for
which $p_{i}^{t_{i}}n_{i}b_{i}$ is a multple of $a$. Let $q=\prod_{i\in
J}p_{i}^{t_{i}}$ and $q_{i}=q/p_{i}^{t_{i}}$. Then we have $0=qmx=m\sum_{i\in
J}qn_{i}b_{i}=m\sum_{i\in J}q_{i}m_{i}a$, where $p_{i}^{t_{i}}n_{i}b_{i}%
=m_{i}a$. Then $\sum_{i\in J}q_{i}m_{i}=0$ since $a$ has infinite order. We
infer that $p_{i}^{t_{i}}|\,m_{i}$ for $i\in J$. This means $p_{i}^{t_{i}%
}n_{i}b_{i}=p_{i}^{t_{i}}m_{i}^{\prime}a$ for some $m_{i}^{\prime}$. Now
$a=p_{i}^{h_{i}}b_{i}$ for some $h_{i}>0$. This gives $p_{i}^{t_{i}}n_{i}%
b_{i}=p_{i}^{t_{i}+h_{i}}m_{i}^{\prime}b_{i}$. Since $b_{i}$ has infinite
order, we infer that $n_{i}=p_{i}^{h_{i}}m_{i}^{\prime}$. Thus $\sum
n_{i}b_{i}=\sum m_{i}^{\prime}a$. This element is either $0$ or of infinite
order. We conclude that $A$ is torsion free. Now if $h\in H\cap A$, $h=\sum
n_{i}b_{i}$ as before, and for some integers $m\neq0,r$ we have $mh=ra$. But
$o(a+H)=\infty$ implies $r=0$. Since $A$ is torsion free, this implies $h=0$.
It remains to show $G$ is generated by the two subgroups $A$ and $H$. Since
$G/H$ has rank one, $G/\left( A\oplus H\right) $ is torsion, so it will
suffice to show this quotient has no elements of order $p$ for any prime $p$.
Suppose $px\in A\oplus H$. Then $px=\sum n_{i}b_{i}+h$, and for some $0\neq
m$, $mpx=ra+mh$. Let $m$ be the smallest such integer. Now if $p|\,r$, we have
$mh\in pH$, and $mh=ph^{\prime}$ for some $h^{\prime}\in H$, since $H$ is
pure. This means $mx-r^{\prime}a-h^{\prime}$ has finite order. But $H$
contains all elements of finite order, since $G/H$ is torsion free. Thus we
have $mx\in A\oplus H$, and $mx=r^{\prime}a+h^{\prime\prime}$ with
$h^{\prime\prime}\in H$. Now $p\nmid m$, otherwise $(m/p)px=r^{\prime
}a+h^{\prime\prime}$ contradicts the choice of $m$. Thus there are integers
$s,t$ with $sp+tm=1$. Then $x=spx+tmx\in A\oplus H$. We conclude that
$G=A\oplus H$, and the lemma is proved.
\end{proof}
\begin{theorem}
Let $H$ be a subgroup of a group $G$. The short exact sequence
$H\rightarrowtail G\twoheadrightarrow G/H$ is proper with respect to the
projective class of completely decomposable groups if and only if $H$ is a
regular subgroup of $G$.\footnote{Another characterization came to light after
publication thanks to a suggestion of E. A. Walker. \textit{Theorem.} A short
exact sequence $A\rightarrowtail B\twoheadrightarrow C$ belongs to
$\mathcal{E}(\mathcal{C})$ if and only if for every sequence $\left\{
n_{p}\right\} $ of non-negative integers and symbols $\infty$, indexed by the
set of primes, the sequence $\cap_{p}p^{n_{p}}A\rightarrowtail\cap_{p}%
p^{n_{p}}B\twoheadrightarrow\cap_{p}p^{n_{p}}C$ is exact. By restricting to
the relevant set of sequences $\left\{ n_{p}\right\} $, this easily
generalizes to characterizations of the exact sequences that arise for
different classes $\mathcal{C}$ of rank one groups.}
\end{theorem}
\begin{proof}
Suppose the sequence is proper. Let $x\in A/H\subseteq G/H$ with $A/H$ rank
one. Then $A=H\oplus B$ for some subgroup $B$ of $G$, and $x=b+H$ for some
$b\in B$. Clearly $H_{A}(b)=H_{A/H}(x)$ and $o(b)=o(x)$. Now suppose the
sequence is regular. Let $C$ be any rank one group and $f:C\rightarrow G/H$ a
homomorphism. Let $A/H$ be the image of $f$. We will show $H$ is a direct
summand of $A$. Then writing $A=H\oplus B$, $C\rightarrow
A/H\twoheadrightarrow B\rightarrowtail G\twoheadrightarrow G/H$ factors $f$
through $G\twoheadrightarrow G/H$, showing the sequence is proper. If $A/H$ is
torsion free, the lemma says $A$ is a summand of $H$. Otherwise, $A/H=\sum
_{p}A^{p}/H$ with $A^{p}/H\cong Z(p^{n})$ for some $0\leq n\leq\infty$. If
$n<\infty$, every coset of $A^{p}/H$ having a representative of the same order
implies $H$ is a summand of $A^{p}$. If $n=\infty$, let $a+H\in A^{p}/H$ have
order $p$. Then we may pick $a$ with $o(a)=p$ and $p$-height $\infty$. This
implies $a\in D^{p}\subseteq A^{p}$ with $D^{p}\cong Z(p^{\infty})$. Since
$D^{p}$ and $H$ have no elements of order $p$ in common we conclude $D^{p}\cap
H=0$. Then since every proper subgroup of $Z(p^{\infty})$ is finite, and
$(D^{p}\oplus H)/H$ is infinite, we have $D^{p}\oplus H=A^{p}$. Thus for every
prime $p$ we have a decomposition $A^{p}=D^{p}\oplus H$ (with $D^{p}$ either
isomorphic to $Z(p^{\infty})$ or cyclic). It follows easily that
$A=H\oplus\sum_{p}D^{p}$.\medskip
\end{proof}
We do not know the dimension of this relative homological algebra. However, we
can determine the dimension of certain groups. We need only consider reduced
groups, since all divisible groups are projective.
\begin{theorem}
If every rank one torsion free subgroup of a reduced group $G$ is free then
$G$ has projective dimension $0$ or $1$ with respect to the projective class
of all completely decomposable groups.
\end{theorem}
\begin{proof}
For such a group $G$, every rank one subgroup is cyclic. Let $P$ be tbe
(external) direct sum of all cyclic subgroups of $G$, and $P\twoheadrightarrow
G$ the map induced by inclusions. Clearly $\operatorname*{Hom}(C,P)\rightarrow
\operatorname*{Hom}(C,G)$ is an epimorphism for all rank one groups $C$. Then
$K\rightarrowtail P\twoheadrightarrow$ $G$ is a proper projective resolution,
since subgroups of direct sums of cyclic groups are again direct sums of
cyclic groups.\medskip
\end{proof}
This theorem says, in particular, that all torsion groups have projective
dimension $\leq1$. We get a theorem for certain torsion free groups by
applying a theorem of Baer \cite{Baer}. By a \textit{homogeneous} group we
mean a torsion free group all of whose non-zero elements have the same type.
\begin{theorem}
If $G$ is a homogeneous group then $G$ has projective dimension $0$ or $1$
with respect to the projective class of all completety decomposable groups.
\end{theorem}
\begin{proof}
Let $P$ be the external direct sum of all pure rank one subgroups of $G$. Then
$K\rightarrowtail P\twoheadrightarrow G$ is a proper exact sequence. Since $P$
is a direct sum of torsion free groups of rank one and of the same type, then
every pure subgronp of $P$ is again completely decomposable \cite{Baer}. Thus
$K$ is completely decomposable and $K\rightarrowtail P\twoheadrightarrow G$ is
a proper piojective resolution of $G$.\medskip
\end{proof}
It remains an open question whether a regular subgroup of a completely
decomposable group is again completely decomposable.
Since regular sequences are, in particular, pure exact sequences, the
injective class contains all algebraically compact groups. It is not known
whether the injective class coutains groups which are not algebraically compact.
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\bibitem {Baer}R. B\textsc{aer}, Abelian groups without elements of finite
order. Duke Math. J. \textbf{3}, 68-122 (1937).
\bibitem {Fuchs}L. F\textsc{uchs}, Abelian Groups. Budapest 1958.
\bibitem {Kaplansky}I. K\textsc{aplansky}, Projective modules. Ann. of Math.
\textbf{68}, 371-377 (1958).
\bibitem {Kulikov}L. Y\textsc{a}. K\textsc{ulikov}, On direct decompositions
of groups [Russian]. Ukrain. Math. \u{Z}. \textbf{4}, 230-275 and 347-372 (1952)
\bibitem {Lyapin}E. C. L\textsc{yapin}, On the decomposition of abelian groups
into direct sums of rank one groups [Russian] Mat. Sb. \textbf{8}, 205-237 (1940).
\bibitem {RWW}F. R\textsc{ichman} C. W\textsc{alker} and E. A. W\textsc{alker}%
, Projective classes of abelian groups. in: Studies on Abelian Groups,
pp.335-343. Paris 1967.
\end{thebibliography}
\bigskip
\begin{center}
Eingegangen am 5. 10. 1971
\end{center}
\noindent Anschrift des Autors:\medskip
\noindent Carol L. Walker
\noindent Department of Mathematics
\noindent New Mexico State University
\noindent Las Cruces, New Mexico, USA
\vfill
\noindent\hrulefill
\begin{center}
Sonderabdruck aus
ARCHIV DER MATHEMATIK
Vol. XXIII, 1972\hfill\textsc{birkh\"{a}user verlag, basel und stuttgart}%
\hfill Fasc. 6
\end{center}
\noindent\hrulefill
\end{document}