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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:

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

P.S.

The group which I construct is in fact hyperbolic.
My 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 "all-inclusive" actions now.

edited Mar 22 2012 at 19:03

asked Mar 30 2010 at 23:30

Anton Petrunin
16.5k●41●110

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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 – Gerhard Paseman Mar 30 2010 at 23:48

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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... – Anton Petrunin Mar 31 2010 at 0:34

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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". – Agol Mar 31 2010 at 1:33

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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. – HW Mar 31 2010 at 1:41

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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. – Igor Belegradek Mar 31 2010 at 3:07

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Anton Petrunin · Accepted Answer · 2011-10-09 02:48:47Z UTC

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I was asked write in an answer to move the question to answered status.

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

answered Oct 9 2011 at 2:48

Anton Petrunin

community wiki

I think if you accept your own answer, not only will members not think less of you, the question will not reappear because the MathOverflow user will not care anymore about your answer not being voted up. Gerhard "Ask Me About System Design" Paseman, 2011.10.08 – Gerhard Paseman Oct 9 2011 at 4:49

Universal group?

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