# Solvable word problems and quasi-isometries

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?

• 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 – Paul Plummer Mar 26 '17 at 23:55
• 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. – Derek Holt Mar 27 '17 at 8:47
• It is unclear (to me) if quasiisometries preserve recursive presentability. – Moishe Kohan Mar 28 '17 at 3:13
• 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). – YCor Apr 17 '17 at 20:48