There is a natural notion of a presentations in the category of residually finite groups. Namely, if $X$ is set and $R$ is a set of words in the free group $FG(X)$ on $X$, then define $G=RF\langle X\mid R\rangle$ to be $FG(X)/N$ where $N$ is the intersection of all finite index normal subgroups of $FG(X)$ containing $R$. Equivalently, $N$ is the closure in the profinite topology of the normal closure of $R$. The group $G$ is residually finite and is the universal residually finite quotient of the group $H$ with presentation $\langle X\mid R\rangle$ (equivalently, it is the quotient of $H$ by the closure of $\{1\}$ in the profinite topology).

Every residually finite group has a presentation in this sense.

Having a finite presentation in the residually finite category does not (or at least should not) imply being finitely presented in the usual sense.

One can then ask about the uniform word problem for residually finite groups in this setting. That is, we can ask given a finite set $X$, a finite set of relations $R\subseteq FG(X)$ and a word $w\in FG(X)$, is $w=1$ in $G=RF\langle X\mid R\rangle$? Notice that the set of such $w$ is co-r.e. because we can enumerate all $X$-generated finite groups satisfying $R$ and determine if $w\neq 1$ in some such finite group. But there is no procedure to enumerate the words $w$ equal to $1$ in $G$. Indeed, Slobodoskoii proved the uniform word problem is undecidable for residually finite groups.

One can then ask the following restricted problem.

Question. Is the uniform word problem decidable for groups with a one-relator residual finite presentation? That is, given a finite set $X$ and a word $r\in FG(X)$, is there an algorithm to determine if a word $w\in FG(X)$ is trivial in $G=RF\langle X\mid R\rangle$?

This is basically asking for a Magnus theorem in the category of residually finite groups. Not all $1$-relator groups (in the usual sense) are residually finite so being $1$-relator as a residually finite group does not (a priori) mean being a $1$-relator group.

My question is equivalent to asking whether there is an algorithm to compute membership in the profinite closure of the trivial subgroup of a $1$-relator group.

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    $\begingroup$ You can add that there exists a number $N$ such that the uniform word problem for residually finite groups with at most $N$ relations is undecidable. It follows from Slobodskoy's proof. I would guess that his proof gives an upper estimate for number $N$ as about 10,000. $\endgroup$ – Mark Sapir Sep 13 '13 at 14:35
  • $\begingroup$ If the maximal residually finite quotient of a 1-relator group is linear, then the answer to your question is yes. Do you know of a 1-relator profinite group which is non-linear? $\endgroup$ – Ian Agol Sep 13 '13 at 18:12
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    $\begingroup$ @IanAgol: An example is in Druţu, Cornelia, Sapir, Mark Non-linear residually finite groups. J. Algebra 284 (2005), no. 1, 174–178. $\endgroup$ – Mark Sapir Sep 13 '13 at 19:35
  • $\begingroup$ Thanks Mark, I came across your paper before but forgot about it. $\endgroup$ – Ian Agol Sep 13 '13 at 20:23
  • $\begingroup$ I suppose there are also 1-relator examples which are not residually finite and whose profinite completion is non-linear. $\endgroup$ – Ian Agol Sep 13 '13 at 20:31

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