Let $S$ be a symmetric subset of a group $G$ containing the identity, and let $S^n$ be the set of all products of $n$ elements of $S$. If $S^3\subset gS$ for some translate $gS$ of $S$ then it follows that $S^2=S^3$. My question is, if $S^2\subset gS$ does it follow that $S=S^2$?

My question is about the growth of a group in geometric group theory. For a finitely generated group $G$ the growth $G$ is the function that assigns to each natural number $n$ the cardinality $|S^n|$ for some finite generating set $S$. Up to an equivalence of growth functions this does not depend on the generating set chosen. Further, the function $(S^n:S)$ that assigns to each $n$ the smallest number of translates of $S$ that cover $S^n$ is in the equivalence class of $|S^n|$. So my question is asking in a general group $G$ does $(S^2:S)=1$ imply $S=S^2$. For $G$ finitely generated by $S$ the answer is yes but trivially because $gS$ and $S$ have the same cardinality.

In general I'm interested in the situation where $G$ is a topological group equipped with a coarse structure and $S$ is a small subset that generates $G$. So for example, it is relevant to assume $G$ is connected and Haar measurable and $S$ is a precompact open set.

Thank you for considering my question.

  • $\begingroup$ You have trivially $|S^2|\ge |S|$, so $S^2\subset gS$ implies $S^2=gS$. But $S^2=\bigcup_{h\in S}hS$. Since the cardinality of this is the same as the cardinality of $S$, you can deduce that $hS=h'S$ for each $h,h'\in S$. Since $e\in S$, you deduce that $hS=S$ for each $h\in S$. Now you deduce $S=S^2$. $\endgroup$ – Anthony Quas Aug 30 '15 at 23:57
  • $\begingroup$ Hello. I don't understand how $|\cup_{h\in S}hS|=|S|$ implies $hS=h'S$ for any $h,h'\in S$. I very much appreciate any additional details. $\endgroup$ – Jake Herndon Aug 31 '15 at 0:32
  • $\begingroup$ If the union of two sets of size 17 has size 17, then they are equal. $|A\cup B|=|A|+|B|-|A\cap B|$. $\endgroup$ – Anthony Quas Aug 31 '15 at 0:33
  • $\begingroup$ @AnthonyQuas Probably $S$ may not be finite...? $\endgroup$ – Dongryul Kim Aug 31 '15 at 0:40
  • $\begingroup$ I think you are assuming that $S$ is finite in the question. I already know this is true when $S$ is finite, sorry if that was not clear. $\endgroup$ – Jake Herndon Aug 31 '15 at 0:46

The answer is yes. In fact, it can be proved in a totally elementary way.

Because $1 \in S^2 \subset gS$, we have $g^{-1} \in S$ and thus $g \in S$. Then since $S^2 \subset gS \subset S^2$, it follows that $$S^2 = gS.$$ By symmetry of $S$, we obtain $S^2 = (S^2)^{-1} = (gS)^{-1} = S g^{-1}$. Hence $S \subset S^2 = gS = Sg^{-1}$, and thus $$g^{-1} S, Sg \subset S.$$ $ $

Let $x \in S$ be any element. By symmetry, $x^{-1} \in S$. Because $g^{-1} S, Sg \subset S$, both $$x^{-1} g, \; xg^2 \in S.$$ Since $S^2 = gS$, it follows that $g^{-1} S^2 = S$ and that $$g^{-1} (x g^2) (x) = g^{-1} x g^2 x \in S.$$ Also, $$g^{-1} (x^{-1} g) (g^{-1} x g^2 x) = gx \in S.$$ Thus for every $x \in S$, we have $gx \in S$. This means that $gS \subset S$, and as we have shown $S^2 = gS$, we get $$S \subset S^2 = gS \subset S. $$ Therefore $S^2 = S.$

  • $\begingroup$ Four years later I still find this answer surprising! Why is it so much more complicated than I think it should be? Anyways, a very late thank you. $\endgroup$ – Jake Herndon Sep 21 '19 at 0:54

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