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

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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:

Statement for local powering-invariant subgroups

It is possible to have the following:

Related facts

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 \ge 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 \ge 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 \ge 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.