# 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 $G$.

## 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.