I can construct a finitely presented group $G$ with the following property (which I use to construct something else).

Given a finitely preseted group $\Gamma$, there is a subgroup $G'\le G$ of finite index such that $$\Gamma=G'/\langle\mathrm{Tor}\, G'\rangle ,$$ where $\mathrm{Tor}\, G'\subset G'$ is the set of all elements of finite order.

I think to call such group $G$ universal.


  • Was it already constructed?
  • Does it already has a name? Is there any closely related terminology?


  • The group which I construct is in fact hyperbolic.
  • The construction is simple, but it takes 2--3 pages. Let me know if you see a short way to do it.
  • Here, the term "universal group" was used in very similar context (thanks to D. Panov for the reference).
  • Thanks to all your comments, we call them "telescopic" actions now.
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    $\begingroup$ I advise against the word "universal", without more context at least. Call it Anton-universal or the Petrunin-Swiss-Army Group, or some useful modification of some synonym for "universal". Gerhard "Ask Me About System Design" Paseman, 2010.03.30 $\endgroup$ Mar 30, 2010 at 23:48
  • 3
    $\begingroup$ Well, Swiss-Army Group is a nice name. But why not universal? --- after quick search I did not see that term "universal group" is used... $\endgroup$ Mar 31, 2010 at 0:34
  • 7
    $\begingroup$ The closest condition I've heard of is "SQ-universal": en.wikipedia.org/wiki/SQ_universal_group Your group satisfies a very strong form of "SQ-universal in the class of finitely presented groups". $\endgroup$
    – Ian Agol
    Mar 31, 2010 at 1:33
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    $\begingroup$ This property seems far too specific to be called simply "universal". I'd go with something like "TQ-universal". If you want to know whether someone else has done this, I'd try looking at the work of Olshanskii and his students. $\endgroup$
    – HJRW
    Mar 31, 2010 at 1:41
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    $\begingroup$ Anton, by a "universal finitely presented group" one usually means a finitely presented group that contains each finitely presented group as a subgroup. Such groups can be constructed via Higman's embedding theorem. If $Q$ is such a group, it is possible to cook up a hyperbolic group $G$ such that $Q$ is a quotient of $G$, and the kernel is normally generated by elements of finite order. This is of course not the same as what you do. $\endgroup$ Mar 31, 2010 at 3:07

1 Answer 1


Аnswered to move the question to answered status.

We decided to use the term telescopic action.

Thank you all for your comments they were helpful for me and Dima.


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