Condition for Baumslag-Solitar group to be virtually abelian The Baumslag-Solitar group is the group given by the presentation
$$
BS(p, q) = \langle a, b \mid a^{-1}b^p a = b^q \rangle
$$
where $p,q \in \mathbb{Z} \setminus \{0\}$.
I have read that this group is virtually abelian (i.e. contains a finite index abelian subgroup) if and only if $|p| = |q|$, but this is given without citation. I have also been unable to find this (or an equivalent statement) in any other papers. However I have found the result that $BS(p,q)$ is residually finite if and only if $|p| = |q|$ or at least one of $|p|$ and $|q|$ is $1$.
I have been trying to extend this condition to show that in the first case $BS(p, q)$ is virtually abelian (and the converse), but have so far been unsuccessful.
I've also been trying that if $|p| \neq |q|$ but at least one is $1$, then $BS(p,q)$ is not virtually abelian. It is clear that in this case you get a semidirect product of two copies of $\mathbb{Z}$, but I've been unable to show no finite index abelian subgroups exist.
Thanks for any hints/help!
 A: I'll expand my comment into an answer. We can assume that $1 \le p\le |q|$.
If $p \ge 2$, then the free reduction of a word in the generators $a$ and $c := bab^{-1}$ does not involve any power $b^k$ of $b$ with $|k| \ge 2$, so by Britton's lemma on  HNN-extensions, the subgroup $\langle a,c \rangle$ is freely generated by $a$ and $c$, and is not virtually abelian.
When $p=1$ and $|q| \ge 2$, the normal closure of $\langle b \rangle$ in $G$ is abelian but not finitely generated. Since any subgroup of a finitely generated virtually abelian group is itself finitely generated, $G$ cannot be virtually abelian in this case.
When $p=|q|=1$ it is easy to see that $G$ is either free abelian of rank $2$ or it has a subgroup of index $2$ (i.e.$\langle a^2,b \rangle$) that is free abelian, so $G$ is virtually abelian in the case.
When $|p|=|q|$ we can describe the structure of the group very precisely, but it is still not virtually abelian when $|p|>1$. For example, if $p=q>1$, then the normal closure of $\langle a \rangle$ in $G$ is a free group of rank $q-1$, with a free basis cycled by conjugation by $b$.
A: If $|m|\neq |n|$ then it has the solvable quotient $\mathbf{Z}\ltimes_{m/n}\mathbf{Z}[1/mn]$, which is not virtually abelian.
If $|m|=|n|\ge 2$, then killing $b^n$ yields the free product $\mathbf{Z}\ast(\mathbf{Z}/n\mathbf{Z})$, which is not virtually abelian.
In both cases, $\mathrm{BS}(m,n)$ having such a quotient is therefore not virtually abelian (actually not virtually nilpotent and even has exponential growth: it has a free semigroup on 2 generators).
In the remaining case, $|m|=|n|=1$, and then $\mathrm{BS}(m,n)$ is virtually abelian (abelian if $mn=1$, has an abelian subgroup of index 2 if $mn=-1$).
