Recognize this countably generated abelian group?

Question

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 $1<q$.

Answer 1

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 $1<q<p$.