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Higman's embedding theorem says that any finitely generated recursively presentable group is embeddable in a finitely presentable group. The converse is also true (a finitely generated subgroup of a finitely presentable group is recursively presentable) and, I think, rather straightforward.

I've been wondering if this is also true if we restrict to torsion-free groups? Of course the converse holds, but is it true that a torsion-free finitely generated recursively presentable group embeds in a torsion-free finitely presentable group?

What about torsion groups? Or $p$-groups?

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    $\begingroup$ Torsion-free: very probably this is known to hold, and maybe automatic from the original construction. Torsion: this is unknown. It's a famous open problem whether there's an infinite fp torsion group. $\endgroup$
    – YCor
    Commented Apr 30 at 21:38
  • $\begingroup$ @YCor: This was my intuition (that torsion-free is probably true, and torsion is probably false). $\endgroup$
    – tomasz
    Commented Apr 30 at 23:23
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    $\begingroup$ For torsion, I'm not saying it's probably false. I'm just saying we do not know and have no clue. $\endgroup$
    – YCor
    Commented May 1 at 9:07
  • $\begingroup$ @YCor: Well, okay, you're not saying that, but it seems dubious to me at a glance, and the fact that a positive answer would be a strengthening of a solution of a famous open problem reinforces that impression. :) $\endgroup$
    – tomasz
    Commented May 1 at 11:52

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Higman’s embedding theorem can indeed be made to preserve the property of being torsion-free. See, for instance, this paper of Chiodo—Vyas, who prove the stronger fact that torsion length can be preserved. (I guess the weaker fact you ask for was known much earlier.)

There are many other “relative” versions of the Higman embedding theorem. For instance, an old theorem of Clapham says that solvability of the word problem can be preserved.

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    $\begingroup$ There is also a nice paper of Mark Sapir, saying that asphericity can be preserved. $\endgroup$
    – ADL
    Commented May 1 at 8:27
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    $\begingroup$ @ADL: Yes, that's another great example. Although, when I tried to read that paper, I found the proof next-to-impossible to extract. It would be great if there were another account somewhere in the literature, since it's an important theorem. $\endgroup$
    – HJRW
    Commented May 1 at 8:53
  • $\begingroup$ I was actually looking at that that paper this morning and for a minute thought that it solves the problem, but it seemed to me that they only define torsion length starting at 1, not 0 (which is equivalent to being torsion-free). But I guess I misunderstood. Thanks! $\endgroup$
    – tomasz
    Commented May 1 at 11:50
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    $\begingroup$ The standard Higman embedding preserves being torsion-free. The finitely presented group in which one embeds in the Higman embedding is built from the given recursively presented group and the trivial group by a finite sequence of HNN-extensions and free products, and this operations preserve being torsion-free. $\endgroup$
    – Igor Belegradek
    Commented May 1 at 12:24
  • $\begingroup$ @IgorBelegradek: yes, I think any version of the construction should have this property. $\endgroup$
    – HJRW
    Commented May 1 at 13:53

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