Pure subgroup of torsion-free abelian group not implies direct factor

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., pure subgroup of torsion-free abelian group) need not satisfy the second subgroup property (i.e., direct factor)
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Statement

Statement for pure subgroups

It is possible to have the following:

• $G$ is a torsion-free abelian group
• $H$ is a pure subgroup of $G$ (this means that for any $h\in H$ and $n\in \mathbb {N}$ such that $nx=h$ has a solution for $x\in G$, there is a solution for $x\in H$).
• $H$ is not a direct factor of $G$.

Statement for local powering-invariant subgroups

It is possible to have the following:

• $G$ is a torsion-free abelian group
• $H$ is a local powering-invariant subgroup of $G$ (this means that for any $h\in H$ and $n\in \mathbb {N}$ such that $nx=h$ has a unique solution for $x\in G$, that unique solution is in $H$).
• $H$ is not a direct factor of $G$.

Related facts

• Pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes

Proof

Example where $H=\mathbb {Z}$, $G/H=\mathbb {Z} [p^{-1}]$ for some prime number $p$:

Define $G$ as the subgroup of $\mathbb {Z} [1/p]\times \mathbb {Z} [1/p]$ with the following countable generating set:

$g_{n}=\left\{\left({\frac {1}{p^{n}}},{\frac {1+p+p^{2}+\dots +p^{n-1}}{p^{n}}}\right)\mid n\in \mathbb {N} _{0}\right\}$

Explicitly, the generators are:

$(1,0),\left({\frac {1}{p}},{\frac {1}{p}}\right),\left({\frac {1}{p^{2}}},{\frac {1+p}{p^{2}}}\right),\left({\frac {1}{p^{3}}},{\frac {1+p+p^{2}}{p^{3}}}\right),\dots$

We note that this generating set has the property that, for all $n\geq 1$:

$pg_{n-1}-g_{n}=(0,1)$

Say $h=(0,1)$ and take $H$ as the subgroup $\langle h\rangle =\{(0,a)\mid a\in \mathbb {Z} \}$.

We therefore get, for all $n\geq 1$:

$\langle g_{0},g_{1},\dots ,g_{n}\rangle =\langle h,g_{n}\rangle$

$G$ is a torsion-free abelian group

This is obvious from it being a subgroup of $\mathbb {Q} \times \mathbb {Q}$.

The element $(0,1/p)$ is not in $G$

If $(0,1/p)\in G$, it must be in the subgroup generated by finitely many generators:

$\langle g_{0},g_{1},\dots ,g_{n}\rangle$

As shown above, therefore, it must be in the subgroup:

$\langle h,g_{n}\rangle$

The image modulo $\mathbb {Z} \times \mathbb {Z}$ is a cyclic group of order $p^{n}$. It's easy to see that there is no element in this image that equals the image of $(0,1/p)$; in fact, forcing the first coordinate to map to zero in the quotient also forces the second coordinate to map to zero.

$H$ is a pure subgroup

For this, it suffices to show that any element of the form $\left(0,{\frac {r}{s}}\right)$ that is in $G$ is also in $H$. First, if $r/s$ is reduced, $s$ must be a power of $p$, say $p^{k}$, and $r$ relatively prime to $p$. Next, the existence of $(0,r/s)=(0,r/p^{k})\in G$ would force the existence of $(0,1/s)=(0,1/p^{k})\in G$. If $k\geq 1$, this forces $1/p\in G$, which we saw is false from above. Thus, $k=0$, so $s=1$, and therefore the element $(0,r/s)\in H$.

There is no complement to $H$ in $G$, so $H$ is not a direct factor

The quotient group $G/H$ is isomorphic to the first coordinate projection, which is $\mathbb {Z} [1/p]$, but there is no element in $G$ that has arbitrarily large $p^{th}$ power roots.