# Recognize this countably generated abelian group?

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 $1<q$.

• I think this is the subgroup of $p$-adic integers consisting of those $p$-adic integers with finite expansions. – Anthony Quas Oct 30 '15 at 19:07
• @AnthonyQuas: I thought something similar, but wouldn't $p$-adic integers with finite expansions (and a $+$ or $-$ sign) just be regular integers? – Aeryk Oct 30 '15 at 19:18
• If you believe my answer, then $G_{(p,q)}/\langle g_k\rangle \cong \mathbb{Z}[1/q]/\langle p^k/q^k \rangle$ is an infinite group, so cannot be $\mathbb{Z}/p^k\mathbb{Z}$. – Julian Rosen Oct 30 '15 at 19:57
• @JulianRosen: Egads, you're right! I was computing my factor group incorrectly. Instead of modding out by $<g_k>$, I was effectively modding out by $<g_k, g_{k+1}, \dots>$. I'll correct that above. Thanks! – Aeryk Oct 30 '15 at 20:31

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 $1<q<p$.
• Uh, if $q=1$, the group is no longer infinitely generated. I think the condition $1 \lt q \lt p$ shows where the interesting part is. I do like that your answer does cover the case $q=1$. Gerhard "For My Idea Of Interesting" Paseman, 2015.10.30 – Gerhard Paseman Oct 30 '15 at 20:59