Abelization of a group is infinite cyclic Suppose i have a group $G$ with the following presentation:
$$G= \langle a,b:a^p=b^q \rangle$$ with $p$ and $q$ coprime. I want to conclude that the abelization $G_{ab}$ infinite cyclic is. I have done the following step (i try to write out $G_{ab}$ and want to use Tietze transformations):
$$G_{ab}= \langle a,b:a^p=b^q \rangle_{ab}= \langle a,b:a^p=b^q,ab=ba\rangle$$
But then?? i don't know how to go further. Can someone help me with this question? Thank you.
 A: By Bezout's identity there exist integers $u$ and $v$ such that $pu+qv=1$. This means that
$$
b=b^1=b^{pu}b^{qv}=b^{pu}a^{pv},
$$
so in the abelianization we have $(b^ua^v)^p=b$. On the other hand we also have
$$
a=a^1=a^{pu}a^{qv}=b^{qu}a^{qv},
$$
so in the abelianization we have $(b^ua^v)^q=a$. Therefore the abelianization is generated by (the coset of) $c:=b^ua^v$. Leaving it to you to check that $c$ has infinite order As per the comment by Derek Holt (and another by Andreas Caranti) it is clear that the abelianization is infinite, so $c$ has infinite order proving the claim.
A: First, if $p=0$ and $q \neq 0$ then $G_{ab} \simeq \mathbb{Z} \times \mathbb{Z}_q$, if $p \neq 0$ and $q=0$ then $G_{ab} \simeq \mathbb{Z} \times \mathbb{Z}_p$.
Suppose that $p,q \neq 0$. According to von Dyck's theorem, there exists a morphism $\rho : G_{ab} \to \mathbb{Z}$ such that $\rho(a)=q$ and $\rho(b)=p$; because any nontrivial subgroup of $\mathbb{Z}$ is infinite cyclic, we can suppose that $\rho$ is onto. Let $g \in \text{ker}(\rho)$. You can write $g=a^ib^j$ with $0 \leq j \leq q-1$, so $0=\rho(g)=iq+jp$. Thus, $q$ divides $j$ and $j \geq q$ if $j \neq 0$, a contradiction. Therefore, $i=j=0$: $\rho$ is an isomorphism.
