It is known that a finitely presented group has a solvable word problem if and only if it satisfies a recursive isoperimetric inequality (ie., its Dehn function is bounded above by some recursive function), so that having a solvable word problem turns out to be a quasi-isometric invariant among finitely presented groups. So my question is:

Let $G,H$ be two quasi-isometric (finitely generated) groups which are not finitely presented. Suppose that $G$ has a solvable word problem. Has $H$ a solvable word problem as well?

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    $\begingroup$ It seems quite believable that there some QI classes could have uncountable many groups, in which case there would be groups which are not recursively presentable, so you definitely could not have an algorithm solve the word problem for those groups $\endgroup$ – Paul Plummer Mar 26 '17 at 23:55
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    $\begingroup$ So you could ask whether the statement is true for recursively presentable groups. I am finding this question confusing because I am not sure to what extent we can apply the quasi-isometries algorithmically. $\endgroup$ – Derek Holt Mar 27 '17 at 8:47
  • $\begingroup$ It is unclear (to me) if quasiisometries preserve recursive presentability. $\endgroup$ – Moishe Kohan Mar 28 '17 at 3:13
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    $\begingroup$ As mentioned by Paul, I answered negatively this in MathOF here: mathoverflow.net/questions/160161/…: both having solvable word problem and being recursively presented are not QI-invariant, nor even stable by extensions with finite kernel. Derek's question whether the word problem is QI-invariant among recursively presented groups is reasonable, I'll think about it (I expect a negative answer). $\endgroup$ – YCor Apr 17 '17 at 20:48

In general there are quasi-isometry classes with uncountably many groups, so the must be groups without recursive presentations in that class, and you can have such classes contain groups with solvable word problem.

The Grigorchuk group is an example of a group which has uncountably many groups quasi-isometric to it, but has solvable word problem. In fact the proofs in this answer, or a paper by Anna Erschler: Not residually finite groups of intermediate growth, commensurability and non-geometricity show that there are uncountably many such groups commensurable up to finite kernel. The basic idea is that the groups has many extensions by a finite group. This means you even get commensurable groups up to finite kernel fail to have "word problem rigidity".

In Erschler's paper, she also proves that if you restrict to looking at recursively presented groups then you do get "word problem rigidity" among groups commensurable up to finite kernel. She asks if "work problem ridgity" holds for quasi-isometry classes restricted to recursively presented groups. I couldn't find an answer to that question (and there was no obvious argument I could think of) This suggests your question in this setting could be open.


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