For discrete group $G$ and $H\leq G$. Show that $G$ also satisfies the Folner condition if $H$ satisfies it and $[G:H]<\infty$.

Question

A finitely generated group $G=\langle S \rangle$ is said to have the Folner condition if $\forall \varepsilon>0$, there exists a finite subset $F\subset G$ such that $$\#((S\cup S^{-1})F\setminus F)<\varepsilon\# F,$$ where $\#S$ is the cardinality of a set $S$. Or equivalently, $\forall\varepsilon>0$, for any finite $T\subset G$, there exists a finite $F\subset G$ such that $$\#(TF\setminus F)<\varepsilon\#F.$$

Let $G$ be a discrete group and let H be a subgroup of $G$. Suppose that the index $[G:H]$ is finite and that $H$ satisfies the Folner condition. Prove that $G$ also satisfies the Folner condition.

I have shown that if $T\subset H$, then it is trivial. So I am thinking of considering the case where $T\nsubseteq H$ and show the existence of $F\subset G$ satisfying the inequality. — Danilo Similatan, Feb 07 '21 at 11:16

score 2 · Answer 1 · answered Feb 07 '21 at 12:43

Here are some ideas. Let $S$ be a generating set of $G$, and let $\{t_i: 1 \le i \le n \}$ be a set of left coset representatives of $H$ in $G$. For $g \in G$, denote the unique $t_i$ with $gH = t_iH$ by $\bar{g}$.

Now let $T = \{\overline{st_i}^{-1}st_i : s \in S \cup S^{-1}, 1 \le i \le n \}$. Then $T$ generates $H$ by a result of Schreier.

Now, given $\epsilon > 0$, choose $F$ as a corresponding Fölner set for $H$ with respect to the generating set $T$. Then $\cup_{i=1}^n t_iF$ is a Fölner set for $G$ with respect to $S$.

For discrete group $G$ and $H\leq G$. Show that $G$ also satisfies the Folner condition if $H$ satisfies it and $[G:H]<\infty$.

1 Answers1