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This question 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:
Why $G_n$'s are mutually non-isomorphic? (i.e., $G_n$ and $G_m$ are not isomorphic for all distinct natural numbers).

Could you please help me to find my answer.

Thanks,

Mojtaba
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  • @Jim Belk : Please help me to understand more. – Mojtaba Aug 29 '17 at 19:52
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    What Jim Belk said was that it is enough to prove that the natural quotient maps $G_n \to G_{n+1}$ are not isomorphisms in order to answer the previous question. It is not clear that proving this would prove that the $G_n$ are all mutually non-isomorphic. – Derek Holt Aug 29 '17 at 20:57
  • @Derek Holt: Oh sorry I got it. – Mojtaba Aug 30 '17 at 14:27
  • If it could be shown that all elements of $G_n$ with noncyclic centralizer lie in the normal closure $\langle b^G \rangle$ of $b$, then it would follow that any isomorphism from $G_n$ to $G_m$ would preserve the subgroup $\langle b^G \rangle$, which is not possible by Jim Belk's answer to the previous question. The claim about noncyclic centralizers seems plausible. You could show that by showing that, for any $1 \ne g \in \langle b^G \rangle$, $g$ and $g^{a^k}$ are not conjugate in $\langle b^G \rangle$ for any $k>0$, which seems likely, but I can't quite see how to prove it. – Derek Holt Sep 01 '17 at 13:07
  • Cross-posted to MathOverflow: https://mathoverflow.net/questions/280176/ – YCor Sep 02 '17 at 10:29
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    A remark: consider $p,q\ge 2$, and the group $H=(\mathbf{Z}[1/p]\times\mathbf{Z}[1/q])\rtimes\mathbf{Z}$, where the action is by multiplication by $(p,1/q)$. One can write $H=H_\infty$, where $H_n=\langle t,x,y|txt^{-1}=x^p,t^{-1}yt=y^q,[x,t^kyt^{-k}]=1,0\le k\le n\rangle$. Then the obvious surjective homomorphism $H_n\to H_{n+1}$ is not an isomorphism (so $H_\infty$ is not finitely presented). Still the $H_n, n\ge 0$ are pairwise isomorphic (because in $H_n$, the relators $[x,t^kyt^{-k}]$ for $k<n$ are redundant and one gets an isomorphism $H_0\to H_n$ by $(t,x,y)\mapsto(t,x,t^nyt^{-n})$. – YCor Sep 02 '17 at 11:41

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