Computing $\mathbb{Z} \ast_{\mathbb{Z}} \mathbb{Z}$

Question

I want to compute $\mathbb{Z} \ast_{\mathbb{Z}} \mathbb{Z}$ with respect to homomorphisms:

1. $\varphi_1:\mathbb{Z} \ni n \longmapsto an \in \mathbb{Z}$

2. $\varphi_2:\mathbb{Z} \ni n \longmapsto bn \in \mathbb{Z}$

for given $a,b \in \mathbb{Z}$.

My intuition says me that in this case $\mathbb{Z} \ast_{\mathbb{Z}} \mathbb{Z} = \mathbb{Z}_{a+1} \ast \mathbb{Z}_{b+1}$.

$\mathbb{Z} \ast \mathbb{Z}$ is a free group on 2 generators, consisting of words $c^{n_1}d^{k_1} \dots c^{n_l}d^{k_l}$. $\mathbb{Z} \ast_{\mathbb{Z}} \mathbb{Z}$ is by definition $(\mathbb{Z} \ast \mathbb{Z})/\langle \varphi_1(n)\varphi_2^{-1}(n)\rangle_{n \in \mathbb{Z}}$, i.e. $c^{an}$ is identified with $d^{bn}$.

I'm thinking of applying "descent" (divide with remainder powers of $c$ greater than $a$ and the same for the powers of $d$). This gives us a word with powers of $c$ lying in the set $\{0,\dots,a\}$ and powers of $d$ lying in $\{0,\dots, b\}$. But I have no clue how to write this down in a rigorous way.

Thanks in advance.

Answer 1

$\langle x, y | x^a = y^b\rangle$ has no elements of finite order unless $a$ or $b$ is $0$ (in which case — let's assume that $b$ is zero — it's actually a free product $\Bbb Z \star \Bbb Z/a$)

Flaw in your reasoning is when you "cancel out" occurences of $x^a$ and $y^b$; part of word can be discarded if it has form $x^{ka}y^{-kb}$ — but $x^{ka}y^{kb}$ is nontrivial! (and equal to $x^{2ka}$ or, equivalently, $y^{-2kb}$)

So, you can rewrite any word $x^{e_1} y^{d_1} \dots x^{e_n} y^{d_n}$ in such form that all $e$'s are less than $a$, and $d$'s less than $b$ except for the last one (or vice versa), by inserting trivial words $x^{ka}y^{-kb}$.