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The famous Higman embedding theorem says that every recursively presented group embeds in a finitely presented group. This is a convenient tool to construct finitely presented groups with bizarre properties from recursively presented ones, which are usually easier to construct.

One cannot hope for an exact analogue of Higman's theorem in the setting of residually finite groups because finitely presented residually finite groups have solvable word problem and hence their finitely generated subgroups do as well. But Meskin constructed finitely generated recursively presented groups with undecidable word problem. I know of no other obstruction to embedding a finitely generated residually finite group into a finitely presented one, so I ask the following question.

Does there exist a finitely generated residually finite group with decidable word problem that cannot be embedded in a finitely presented residually finite group?

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    $\begingroup$ Same question seems to be also unclear in the category of f.g./f.p. linear groups. $\endgroup$
    – Misha
    Commented Jan 31, 2014 at 5:40
  • $\begingroup$ As far as I know it is not known whether the Grigorchuk group embeds into a f.p. residually finite group. $\endgroup$
    – Mustafa Gokhan Benli
    Commented Jan 31, 2014 at 6:11
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    $\begingroup$ I am less optimistic about linear groups. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 31, 2014 at 16:04
  • $\begingroup$ @Misha, Benjamin: Benjamin you're right: a countable infinite simple linear group such as $PSL_2(\mathbf{Q})$ can't be embedded into a f.g. linear group, because the former is not residually finite. $\endgroup$
    – YCor
    Commented Feb 10, 2014 at 10:38
  • $\begingroup$ @YvesCornulier: True, but I was thinking about finitely-generated groups (embedding f.g linear groups in f.p. linear groups). (I guess, I was not very clear about it.) I do not see any obstructions for this. $\endgroup$
    – Misha
    Commented Feb 10, 2014 at 13:50

2 Answers 2

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A preprint by E. Rauzy appeared today on the arXiv, and gives a negative answer to this question. In other words (if the proof is correct), there exists a f.g. residually finite group with decidable word problem which does not embed in a f.p. residually finite group.

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In paragraph 1.1.7 of of a recent preprint of Kharlampovich--Myasnikov--Sapir, the authors write:

We expect the approach used in this paper to be useful in solving other problems that are still open. For example, the residually finite version of the Higman embedding theorem would be very desirable.

In particular, the problem seems to be open.

I'm not sure how the approach of that paper might apply to the problem at hand.

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  • $\begingroup$ Doesn't that paper just give solvable groups? $\endgroup$
    – Benjamin Steinberg
    Commented Jan 31, 2014 at 15:26
  • $\begingroup$ @BenjaminSteinberg - I believe so. Hence my last sentence. $\endgroup$
    – HJRW
    Commented Jan 31, 2014 at 18:21
  • $\begingroup$ my hope is such a construction would give an alternate proof of their result and answer your recent question. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 31, 2014 at 19:16

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