Checking the Folner condition only on a set of generators

Question

Let $G$ be a countable, discrete group with a set of generators $S$. Let $F_n\subset G$ be a sequence of finite sets that satisfies $$\frac{|sF_n\triangle F_n|}{|F_n|}\to0$$ for all $s\in S$. Is it true then that $F_n$ is a Folner sequence? In other words, is it true that $$\frac{|gF_n\triangle F_n|}{|F_n|}\to0$$ for all $g\in G$? In particular, does it suffice to check the Folner condition on the generators instead of the entire group? Let's set $\alpha_n(g)=\frac{|gF_n\triangle F_n|}{|F_n|}$. I believe that this question boils down to showing that (1) if $\alpha_n(g)\to0$ then $\alpha_n(g^{-1})\to0$ and that (2) if $\alpha_n(g)\to0$ and $\alpha_n(h)\to0$ then $\alpha_n(gh)\to0$.

For (2) I observed that $$\alpha_n(gh)=\frac{|ghF_n\triangle F_n|}{|F_n|}=\frac{|(ghF_n\triangle hF_n)\triangle(hF_n\triangle F_n)|}{|F_n|}\leq\frac{|ghF_n\triangle hF_n|}{|F_n|}+\alpha_n(h)=$$ $$=\frac{|g(hF_n)\triangle hF_n|}{|hF_n|}+\alpha_n(h)$$ and I would be done if I could get some estimate of the form $|ghF_n\triangle hF_n|\leq C|gF_n\triangle F_n|$, but this seems impossible to me. Such an estimate would also be enough to show (1).

Any hint or reference is appreciated.

Answer 1

Apparently I "interpolated" the wrong thing. The problem is solved if one interpolates the set $gF_n$ instead of $hF_n$. To be more precise:

$$|ghF_n\triangle F_n|\leq|ghF_n\triangle gF_n|+|gF_n\triangle F_n|$$ and $ghF_n\triangle gF_n\subset g(hF_n\triangle F_n)$, because $gA\setminus gB\subset g(A\setminus B)$ for all $A,B$. therefore the above estimate goes on as $$\leq |g(hF_n\triangle F_n)|+|gF_n\triangle F_n|=|hF_n\triangle F_n|+|gF_n\triangle F_n|$$ and dividing everything with $|F_n|$ will yield $\alpha_n(gh)\leq\alpha_n(g)+\alpha_n(h)$.

A similar argument works for (1): we have that $$|g^{-1}F_n\triangle F_n|=|g^{-1}F_n\triangle g^{-1}gF_n|\leq|g^{-1}(gF_n\triangle F_n)|=|gF_n\triangle F_n|$$ for all $g\in G$, thus $\alpha_n(g^{-1})\leq\alpha_n(g)$. Actually by symmetry one concludes that $\alpha_n(g)=\alpha_n(g^{-1})$.

It was very elementary after all. Still, I think it makes a good exercise for beginners!