I recently came across a construction that, in abstraction, leads to the following family of abelian groups: Fix $1<q<p$ with $q$ and $p$ relatively prime. The group $G\_{(p,q)}$ is given by the presentation
$$<g\_0, g\_1, g\_2, \dots \mid g\_i^p=g\_{i+1}^q,\ g\_ig\_j=g\_jg\_i>.$$
In retrospect, I quickly realized that none of my training in group theory covered infinitely-generated abelian groups. So for now I've picked up Fuch's "Infinite Abelian Groups" but I was hoping that perhaps $G\_{(p,q)}$ was already a known group with some literature specific to it. Does anyone recognize this group? 


If it helps, one of the main properties of the groups is that $\require{enclose}
 \enclose{horizontalstrike}{G\_{(p,q)}/<\!g\_k\!> \cong \mathbb{Z}/p^k\mathbb{Z}}$ $G\_{(p,q)}/<\!g\_k, g\_{k+1},\dots\!> \cong \mathbb{Z}/p^k\mathbb{Z}$ (Edit: This holds because of the relatively prime condition.)


EDIT: Other properties to consider: 


$G\_{(p,1)} \cong \mathbb{Z}$ under the map $g\_i \mapsto p^i$ and $G\_{(p,0)} \cong \bigoplus\_{i=0}^\infty \mathbb{Z}/p\mathbb{Z}$. Hence the restriction $1<q$. 



Recognize this countably generated abelian group?
There is an isomorphism
$$
\begin{align\*}
\varphi:G\_{(p,q)}&\to\mathbb{Z}[1/q],\\
g\_i&\mapsto\frac{p^i}{q^i}.
\end{align\*}
$$


To check surjectivity: for any $\frac{a}{q^n}\in\mathbb{Z}[1/q]$, we can find integers $b,c$ such that $b p^n+c q^n = a$ because $p^n$ and $q^n$ are relatively prime, and we will have $\varphi(g\_0^cg\_n^b)=\frac{a}{q^n}$.


To check injectivity: suppose some word $w:=\prod\_{i=0}^k g\_i^{a\_i}$ maps to $0$ by $\varphi$. For each prime power $\ell^n|q$, examining the $\ell$-adic valuation of $\varphi(g)$ shows that $\ell^n|a\_k$, so we conclude $q|a\_k$. Modulo the relation $g\_{k-1}^p=g\_k^q$, we can reduce $w$ to $$\prod\_{i=0}^{k-1}g\_i^{a\_i}\cdot g\_{k-1}^{p\frac{a\_k}{q}}.$$
Repeating the process, we can reduce $w$ to the empty word.


This works for arbitrary coprime integers $p$, $q$, even without the condition $1<q<p$.


