# When can we establish an isomorphism between two not-finitely presented groups?

Assume that finitely generated groups $G$ and $H$, are not finitely presented. Fix a generating set $\mathfrak g:=\{g_1,\dotsc,g_n\}$ of $G$. Let $\mathfrak R:=\{R_1,R_2,\dots\}$ be the set of all relations in $G$. For all non-empty finite subsets $\mathfrak S$ of $\mathfrak R$, there exists a generating set of $H$ whith same size as $\mathfrak g$, which its corresponded set of relations, contain $\mathfrak S$. What we can say about $G$ and $H$. Are they isomorphic? Are they quotient of each other?

• In what set does $\mathfrak R$ live? Is it another name for $\ker(\operatorname{Free}(\mathfrak g) \to G)$? – LSpice Jan 20 '16 at 18:32
• @LSpice: Yes, it is exactly all of $\ker(\text{Free}(\mathfrak g)\rightarrow G)$. – Meisam Soleimani Malekan Jan 20 '16 at 19:06
• You may like to look up the Gromov--Grigorchuk space of marked groups. If I've understood you correctly, your condition means that $G$ is in the closure of the subspace of groups isomorphic to $H$. – HJRW Jan 20 '16 at 22:44
• @YCor, it seems to me that the OP would benefit from an example of a group G which is a limit of groups H, but such that H is not a quotient of G. I'm under the impression that such a (necessarily infinitely presented) G should exist, but don't know a construction. Do you know one? – HJRW Jan 21 '16 at 12:06
• @YCor, I agree that what you describe is a contradiction to the question as literally read. But I presume the OP is looking for some non-trivial phenomenon in this vicinity, and so I thought an example of that form might be helpful to him. It was just a suggestion. – HJRW Jan 21 '16 at 19:58