Every subgroup of $\mathbf{Q}$ is a direct limit of $\mathbf{Z}\rightarrow\mathbf{Z}\rightarrow\cdots$

Question

In Hatcher's algebraic topology textbook there is an example in an appendix to chapter 3, stating that every subgroup of $\mathbf{Q}$ is the direct limit of a sequence of the form $\mathbf{Z}\rightarrow\mathbf{Z}\rightarrow\cdots$, "with an appropriate choice of maps". Can we prove this by simply combining the answer of anon, and the classification of subgroups of $\mathbf{Q}$?

Answer 1

It's very easy. Let $G \leq \mathbb{Q}$. Choose an enumeration $G = \langle g_1,g_2,\dotsc\rangle$, wlog $g_1 \neq 0$. Then $G$ is the colimit of the subgroups $\langle g_1,g_2,\dotsc,g_n \rangle$. These are finitely generated torsion-free abelian groups, hence $\cong \mathbb{Z}^d$ for some $d$. But $\mathbb{Q}$ has rank $1$, hence $d=1$.

Answer 2

Yes. A subgroup $G$ of $\mathbf{Q}$ is characterized by a generalized integer $\prod p_i^{k_i}$ where $k_i\in \mathbf{N}\cup\{\infty\}$ and an integer $n$ relatively prime to those $p_i$ such that $k_i>0$. So, enumerating a cofinal set of integer divisors of $\prod p_i^{k_i}$ by $(a_m)$ (for instance something like $p_1,p_1^2,p_1^2p_3,p_1^2p_3p_4,...$) we see $G$ is the union of $\frac{n}{a_i}\mathbf{Z}$, equivalently, a direct limit of $\mathbf{Z}$s with multiplication maps $a_{i+1}/a_i$.

Answer 3

If your subgroup is finitely generated then you can choose the sequence of maps $$\Bbb Z\xrightarrow{1} \Bbb Z\xrightarrow{1} \Bbb Z\xrightarrow{1}\cdots$$ This is because every finitely generated subgroup of $\Bbb Q$ is cyclic. The morphisms from each copy of $\Bbb Z$ to this subgroup is any (one) isomorphism.

The more interesting case is for non-finitely generated subgroups. Martin Brandenburg's answer gets you started in creating the morphisms necessary, and the approach parallels anon's answer in the first link. First instead of enumerating all of $G$ just enumerate a countable subset of $G$ chosen so that $g_{n+1}$ is not in $\langle g_1,\ldots,g_n\rangle$. We still have that $G$ is the direct limit of sequence of subgroups where the intermediate morphisms are inclusion. Set $G_n=\langle g_1,\ldots,g_n\rangle$. Now are system looks like $$G_1\hookrightarrow G_2\hookrightarrow G_3\hookrightarrow\cdots$$ For each $G_n$ there is a smallest positive rational number $r_n$. So we are actually working with the system $$r_1\Bbb Z\hookrightarrow r_2\Bbb Z\hookrightarrow r_3\Bbb Z\hookrightarrow\cdots$$ Now you should see that the most important thing about these numbers are their denominators when considered in their reduced form. We can always move over to an isomorphic system such that we only invert one integer at a time. More rigorously let $r_n = t_n/b_n$ (in reduced form). Then the isomorphism $(x\mapsto 1/t_n x)$ between $r_n\Bbb Z$ and $1/b_n\Bbb Z$ allows us to consider the isomorphic system $$1/b_1\Bbb Z\hookrightarrow 1/b_2\Bbb Z\hookrightarrow 1/b_3\Bbb Z\hookrightarrow\cdots$$ and this is isomorphic to the system $$\Bbb Z\xrightarrow{1}\Bbb Z\xrightarrow{b_2/b_1} \Bbb Z\xrightarrow{b_3/b_2} \Bbb Z\xrightarrow{b_4/b_3} \cdots$$ parallel to anon's answer.