11
$\begingroup$

Is there an infinite group $G$ such that there is not any sequence $(A_n)$ of its subsets such that always $$A_n=A_n^{-1}, \quad A_{n+1}A_{n+1}\subsetneqq A_n$$ ?

link

$\endgroup$
8
  • 2
    $\begingroup$ Such a group has to be torsion. $\endgroup$ Jul 5, 2014 at 22:59
  • $\begingroup$ It's equivalent to say that $A_{n + 1} \subsetneq A_n, A_{n + 1} A_{n +1} \subseteq A_n$, by going through $A_{2n}$. $\endgroup$
    – user44191
    Jul 5, 2014 at 23:49
  • $\begingroup$ I am not sure of the equivalence. I think closure under inverse is important (consider riffs on N under min). $\endgroup$ Jul 6, 2014 at 0:34
  • $\begingroup$ Clearly all $A_n$ contain $1$ and $\bigcap _{n}A_n$ is a subgroup. $\endgroup$
    – H. Khas
    Jul 6, 2014 at 0:45
  • 1
    $\begingroup$ @Misha I'm not sure of any link with approximate groups, which involve a parameter and finite subsets. On the other hand there should be a link with non-topologizable groups. $\endgroup$
    – YCor
    Jul 6, 2014 at 8:25

2 Answers 2

5
$\begingroup$

Here an extended comment. Call Property (P) the assumption of non-existence of such a sequence, so that the question is whether there exists an infinite group with (P)

My comment is the observation that if such a group exists, then it can be chosen to be finitely generated with all its proper subgroups finite ("quasi-finite").

Indeed, a group with Property (P) satisfies the following 2 properties:

  • it admits no nondiscrete Hausdorff topology "non-topologizable"
  • it's artinian (or min): it admits no properly descending chain of subgroups

Moreover Property (P) passes to subgroups.

Now I claim that if $G$ satisfies (P), then it contains an infinite quasi-finite subgroup: indeed if $G=G_0$ is not quasi-finite, then it admits a proper infinite subgroup $G_1$, and by induction we define a properly descending chain of subgroups, which has to stop, i.e. we eventually get a quasi-finite subgroup (hence with (P)).

Now a quasi-finite group is either finitely generated, or is an abelian Prüfer group $\mathbf{Z}[1/p]/\mathbf{Z}$ (Hall-Kulatilaka), but the latter is topologizable as we see by embedding it densely in the Lie group $\mathbf{R}/\mathbf{Z}$. Hence the resulting quasi-finite groups with (P) have to be finitely generated.

To conclude, the question boils down to whether there exists an infinite, quasi-finite, finitely generated group with (P).

$\endgroup$
4
  • $\begingroup$ I think it admits no nondiscrete Hausdorff topology is equivalent to it admits no nondiscrete metric. It seems such a group cannot have an infinite chain of (normal) subgroups. It suggests to me (somehow) that Hausdorffness may have a better substitution. Btw, I hope I can find a relation to Banach measure. $\endgroup$
    – H. Khas
    Jul 6, 2014 at 21:28
  • $\begingroup$ I'm not sure what you mean by "no nondiscrete metric". What do you require about the metric? $\endgroup$
    – YCor
    Jul 6, 2014 at 21:33
  • $\begingroup$ I'm not sure but if there is a sequence of neighborhoods of $1$ which make a base around $1$ in a group topology the topology must be pseudometrizible. If Hausdorff, then metrizable. $\endgroup$
    – H. Khas
    Jul 6, 2014 at 21:37
  • $\begingroup$ OK, but once we are reduced to finitely generated quasi-finite groups, this discussion is of little relevance. $\endgroup$
    – YCor
    Jul 6, 2014 at 21:42
2
$\begingroup$

This is a partial answer. There exists an infinite group $G$ having no sequence of subsets $\{A_i\}$ such that $$ A_{n+1}A_{n+1}\subsetneq A_n, \quad\hbox{and}\quad \bigcap A_i=\{1\}. $$ (We omit the inverse condition but add the the intersection-triviality condition.)

Indeed, such a sequence defines a Hausdorff topology on the group where $A_i$ are neigbourhoods of $1$ and all other points are isolated. The multiplication is continuous at $(1,1)\in G\times G$ with respect to this topology. It remains to note that infinite locally non-topologizable groups do exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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