I was trying to suppose for a contradiction that it is abelian, then:
$$ba^3 = a^2b = ba^2\implies a = 1$$
Then we have $G = \langle a,b\mid a = 1\rangle$, which I believe is isomorphic to $\mathbb{Z}$, but I cannot get a contradiction from here. Could someone give me some hints on where to start this problem, I would really appreciate that!
Thank you 🙂
Best Answer
The Baumslag-Solitar group $B(m,n)$ is the group given by the presentation $$ BS(m, n) = \langle a, b \mid b^{-1}a^m b = a^n \rangle $$ where $m,n \in \mathbb{Z} \setminus \{0\}$.
For your group $G$ we have $G=B(2,3)$ by definition, i.e., $G$ is the Baumslag-Solitar group with $m=2$ and $n=3$. This group is not even virtually abelian. Actually, $B(m,n)$ is abelian if and only if $mn=1$. A proof has been given, among other posts, here:
Condition for Baumslag-Solitar group to be virtually abelian