This [question](https://math.stackexchange.com/questions/547087/finitely-generated-group-which-is-not-finitely-presented/547144#comment4975221_547144) was answered by @Jim Belk And he defined $G\_n$ as follows:
$$
G\_n \;=\; \langle a,b \mid [a^{-1}ba,b] = \cdots = [a^{-n}ba^n,b]=1\rangle
$$
My question is:  

Are $G\_n$'s mutually non-isomorphic? (i.e., for every two distinct natural numbers $n$ and $m$, $G\_n$ and $G\_m$ are not isomorphic). 


Could you please help me to find my answer.


Thanks,



$G_n$ 's mutually non-isomorphic
Yes. For a group $G$, define $a(G)$ as the greatest $k$ such that $\mathbf{Z}^k$ embeds as a subgroup of $G$.


Then $a(G\_n)=n+1$ for all $n\ge 0$. Thus the $G\_n$, for $n\ge 0$, are pairwise non-isomorphic.


We have $a(G\_n)\ge n+1$ since $s\_k=a^{-k}ba^k$, $0\le k\le n$ generate a free abelian subgroup on $n+1$ generators. Let us show the other inequality.


We can add the $s\_k$ among the generators, so a presentation is given by $$\langle a,s\_0,\dots,s\_n\mid a^{-1}s\_ia=s\_{i+1}:0\le i\le n-1,\;[s\_i,s\_j]=1,0\le i,j\le n\rangle.$$
This is an HNN presentation, namely the HNN extension of $\mathbf{Z}^{n+1}=\langle s\_0,\dots,s\_n\rangle$ for the isomorphism between two copies of $\mathbf{Z}^n$ given by $s\_i\mapsto s\_{i+1}$.


Let $A$ be an abelian subgroup in $G\_n$ isomorphic to $\mathbf{Z}^d$. If $A$ has a loxodromic element (for the action on the Bass-Serre tree), then its axis is $A$-invariant and hence a copy of $\mathbf{Z}^{d-1}$ fixes this axis pointwise. In particular, this $\mathbf{Z}^{d-1}$ fixes an edge and hence is conjugate into the edge group, namely $\mathbf{Z}^n$. Hence $d-1\le n$, that is, $d\le n+1$. Otherwise, every element of $A$ fixes a vertex, and since $A$ is finitely generated, this implies that $A$ fixes a vertex. Hence $A$ is conjugate into a vertex group which is isomorphic to $\mathbf{Z}^{n+1}$, hence $d\le n+1$ again.


PS $G\_\infty$ is the wreath product $\mathbf{Z}\wr\mathbf{Z}$: this is a motivation for studying these "approximating" groups. In the original MathSE question (linked by the OP), it is checked that the canonical quotient map (the homomorphism given by $a\mapsto a$, $b\mapsto b$) from $G\_m\to G\_n$ when $m<n$ is not an isomorphism.


