What is this group? (Recognising a group from a presentation).

Question

I am trying to find out what the following group is:

$$G = \langle a, b \mid ab^2 = b^2a,\ a^4 = b^3\rangle.$$

Due to the isomorphism problem for groups, there is no algorithmic way to approach questions like this in general. The only technique I know of is to consider the abelianisation of $G$, which, if I'm not mistaken, is the group given by the same generators and relations, together with the additional relation $ab = ba$. In this case, aside from the fact that you can remove the first relation, it doesn't seem be any simpler to determine.

So my questions are as follows:

1. What is the group $G$?
2. Without using the answer to the first question (i.e. using only the presentation), what is the abelianisation of $G$?

In addition, any descriptions of other techniques that one can use to get a better understanding of a group from a presentation would be very much appreciated.

Answer 1

Firstly, $a$ commutes with $b^2$ by the first relation, while clearly $a$ commutes with $a^4$ so $a$ commutes with $b^3$ by the second relation. This means that the subgroup $H=\langle a, b^2, b^3\rangle$ is abelian, as the generators pairwise commute. However, $b^3b^{-2}=b\in H$ and so $G=H$. Thus, $G$ is an abelian group. As has already been noted, the abelianisation of $G$ is infinite cyclic, and thus we conclude that $G$ is infinite cyclic.

Answer 2

The abelianization is generated by $c=a^{-1}b$ because $c^{3}=a^{-3}b^3=a$ and $c^4=ac=b$. Hence all relations become redundant and we end up with $\mathbb Z$.

Answer 3

$b^2$ commutes with $a,$ (by the first relation), so every element in the group can be written as $$(b^2)^k \prod (a^{i_j}b^3)^l a^m=b^{2k} a^n,$$ (by the second relation). Given the abelianization, the group should be $\mathbb{Z}.$