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

2$\begingroup$ Such a group has to be torsion. $\endgroup$ – The Masked Avenger Jul 5 '14 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 '14 at 23:49

$\begingroup$ I am not sure of the equivalence. I think closure under inverse is important (consider riffs on N under min). $\endgroup$ – The Masked Avenger Jul 6 '14 at 0:34

$\begingroup$ Clearly all $A_n$ contain $1$ and $\bigcap _{n}A_n$ is a subgroup. $\endgroup$ – H. Khas Jul 6 '14 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 nontopologizable groups. $\endgroup$ – YCor Jul 6 '14 at 8:25
Here an extended comment. Call Property (P) the assumption of nonexistence 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 ("quasifinite").
Indeed, a group with Property (P) satisfies the following 2 properties:
 it admits no nondiscrete Hausdorff topology "nontopologizable"
 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 quasifinite subgroup: indeed if $G=G_0$ is not quasifinite, 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 quasifinite subgroup (hence with (P)).
Now a quasifinite group is either finitely generated, or is an abelian Prüfer group $\mathbf{Z}[1/p]/\mathbf{Z}$ (HallKulatilaka), but the latter is topologizable as we see by embedding it densely in the Lie group $\mathbf{R}/\mathbf{Z}$. Hence the resulting quasifinite groups with (P) have to be finitely generated.
To conclude, the question boils down to whether there exists an infinite, quasifinite, finitely generated group with (P).

$\begingroup$ I think
it admits no nondiscrete Hausdorff topology
is equivalent toit 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 '14 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 '14 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 '14 at 21:37

$\begingroup$ OK, but once we are reduced to finitely generated quasifinite groups, this discussion is of little relevance. $\endgroup$ – YCor Jul 6 '14 at 21:42
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 intersectiontriviality 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 nontopologizable groups do exist.