# Universal group?

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.

Questions:

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

P.S.

• 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.
• 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 Mar 30, 2010 at 23:48
• 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... Mar 31, 2010 at 0:34
• 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". Mar 31, 2010 at 1:33
• 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.
– HJRW
Mar 31, 2010 at 1:41
• 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. Mar 31, 2010 at 3:07