9
$\begingroup$

Define a group to be 2-locally finite if, for any two elements, the subgroup generated by them is finite.

Define a group to be locally finite if the subgroup generated by any finite subset is finite.

I want an example of a 2-locally finite group that is not locally finite.

This paper https://arxiv.org/pdf/1403.0331.pdf discusses a special case where 2-locally finite groups must be locally finite (namely, the case of groups with a planar lattice of subgroups).

More generally: for a nonnegative integer n, call a group n-locally finite if every subset of size at most n generates a finite group.

I'm interested more generally in examples of n-locally finite groups that are not (n + 1)-locally finite. Currently I know of solutions for:

  • n = 0: This is trivial; any group that has elements of infinite order will do.
  • n = 1: This means a periodic group (every element has finite order) where there is an infinite subgroup generated by two elements. Examples here include the Grigorchuk group and some negative solutions to the Burnside problem (specifically, things like Tarski monsters); I've compiled this info at https://groupprops.subwiki.org/wiki/Periodic_not_implies_locally_finite
$\endgroup$
  • $\begingroup$ I also tried to search whether there are known examples of group varieties in which 2-generated free groups are finite but not 3-generated free groups, but I didn't succeed. But I sort of suspect it's known. $\endgroup$ – YCor Apr 15 '17 at 22:44
  • $\begingroup$ If $\mathcal{P}$ is a group-theoretical property then a group is called weakly $\mathcal{P}$ if every 2-generator subgroup is a $\mathcal{P}$-group. See page 38 of [Finiteness Conditions and Generalized Soluble Groups, Part 2 By Derek J.S. Robinson, Springer ,1972] to find "weakly nilpotent". Groups which are infinite weakly finite share some properties as locally finite groups: e.g. they all have infinite abelian subgroups. $\endgroup$ – Alireza Abdollahi Apr 16 '17 at 11:19
10
$\begingroup$

Golod (MR link; Some problems of Burnside type. (Russian) 1968 Proc. Internat. Congr. Math. (Moscow, 1966) pp. 284–289 Izdat. "Mir'', Moscow; English translation Amer. Math. Soc. Transl. (2) 70 (1968), 49) produced infinite $n$-generator groups in which all $(n-1)$-generated subgroups are finite, for all $n$.

From the MR review: (...) The main theorem follows: For each field $k$ and each integer $d\ge 2$, there is a $d$-generator infinite-dimensional $k$-algebra in which every $(d−1)$-generator subalgebra is nilpotent and $\bigcap_{n=1}^\infty A^n=(0)$. Corollary 1 states that for each $d\ge 2$ and each prime $p$ there is a $d$-generator infinite residually finite $p$-group G whose $(d−1)$-generator subgroups are finite.

(I haven't checked but I expect that $A$ is meant to be associative but not not unital and that $A^n$ denotes the subspace generated by products of $n$ elements, which is a two-sided ideal.)

$\endgroup$
  • $\begingroup$ Thank you! And thanks as well for including the abstract; irritatingly I can't access the abstract when I click on the MR link because I'm not at a university or academic institution. So having it in the answer here is helpful. $\endgroup$ – Vipul Naik Apr 16 '17 at 0:39
  • $\begingroup$ Just to know, can you access it through this link? ams.org/mathscinet/search/… $\endgroup$ – YCor Apr 16 '17 at 1:22
  • $\begingroup$ No, it asks for a username and password in order to be able to access that page. $\endgroup$ – Vipul Naik Apr 16 '17 at 22:59
  • $\begingroup$ OK, sounds bad news :( Before, links to MathSciNet reviews used to be in open access (while the search system was in restricted use). $\endgroup$ – YCor Apr 16 '17 at 23:11
  • $\begingroup$ Further references can be found in Mark Sapir's answer here: mathoverflow.net/questions/137678/… $\endgroup$ – YCor Aug 10 '17 at 11:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.