Let $G$ be an amenable (countable, discrete) group and let $F\_1,F\_2,...,F\_n,...$ and $G\_1,G\_2,...,G\_n,...$ be two Følner sequences. Is the product sequence (i.e. the sequence $(H\_n)$ where $H\_n$ is all elements of the form $f\_ng\_n$ for $f\in F\_n, g\in G\_n$) necessarily also a Følner sequence? If not, is this at least true for some nice class of groups (say, locally virtually nilpotent groups, or solvable groups or when the $F\_i=G\_i$ for all $i$? 



Is a product of Følner sets Følner?
**some details added January 7th**


It seems that the answer is **"no" even for virtually abelian groups**, and even
if $G\_i$ is chosen by your arch enemy. The sequence $(F\_i)\_i$ can also be chosen rather freely in the proof. This answers some subquestions of the question, but not all of them. In particular do not know what happens if $F\_i = G\_i$ is required or if there are strong constraints on how fast the sets become invariant.



> 
> Suppose $G$ is a countable amenable group where some conjugacy class is infinite. Then for every Følner sequence $(G\_i)\_i$ there exists a Følner sequence
>  $(F\_i)\_i$ such that $(F\_iG\_i)\_i$ is not a Følner sequence.
> 
> 
> 


Proof. Let $h$ have infinite conjugacy class, and $(G\_i)\_i$ be a Følner sequence. Since $h$ has infinite conjugacy class, for each $i \in \mathbb{N}$
there exists infinitely many $f\_i \in G$ such that $f\_i^{-1}hf\_i \notin G\_iG\_i^{-1}$. Then $hf\_iG\_i \cap f\_iG\_i = \emptyset$, since otherwise $hf\_ig\_1 = f\_ig\_2$ for some
$g\_1, g\_2 \in G\_i$ and thus $f\_i^{-1}hf\_i = g\_2g\_1^{-1} \in G\_iG\_i^{-1}$. Since there are infinitely many ways to choose $f\_i$, we can also ensure $f\_iG\_i \cap G\_i = \emptyset$.


Let now $(F\_i')\_i$ be any Følner sequence for $G$. For each $i \in \mathbb{N}$ pick maximal $\alpha(i) \leq i$ so that, writing $\chi\_A$ for the characteristic function of a
set $A \subset G$, and $| \cdot |$ for the $\ell^1$-norm, we have
$$ |\chi\_{F\_{\alpha(i)}' G\_i} - \chi\_{G\_i}| < 1/(\alpha(i) + 1) |G\_i|. $$
Since $(G\_i)\_i$ is Følner, $\alpha(i)$ tends to infinity with $i$, so $(F\_{\alpha(i)}')\_i$ is a Følner sequence (using the definition of the asker).


Define $F\_i = F\_{\alpha(i)}' \cup \{f\_i\}$. Obviously $(F\_i)\_i$ is a Følner sequence since the new element can contribute at most $2$ to the symmetric
differences $|\chi\_{gF\_i} - \chi\_{F\_i}|$. With the shorthand $o(|G\_i|)$ for $\ell^1$ functions $x$ with $|x/|G\_i|| \rightarrow 0$ with $i$, and also for
its usual meaning for reals, we have (by the properties of $f\_i$ chosen in the first paragraph)
$$ \chi\_{F\_iG\_i} = \chi\_{f\_iG\_i} + \chi\_{F\_{\alpha(i)}' G\_i} - \chi\_{f\_iG\_i \cap F\_{\alpha(i)}' G\_i} = \chi\_{f\_i G\_i} + \chi\_{G\_i} + o(|G\_i|) $$
By $f\_iG\_i \cap G\_i = \emptyset$, we then have $|\chi\_{F\_iG\_i}| = 2|G\_i| + o(|G\_i|)$.


We similarly have
$$ \chi\_{hF\_iG\_i} = \chi\_{h f\_i G\_i} + \chi\_{h G\_i} + o(|G\_i|). $$
Using these equalities, the fact $G\_i$ is a Følner sequence, and the inverse triangle inequality, we get
$$ |\chi\_{hF\_iG\_i} - \chi\_{F\_iG\_i}| = |\chi\_{h f\_i G\_i} + \chi\_{hG\_i} + o(|G\_i|) - \chi\_{f\_i G\_i} - \chi\_{G\_i} - o(|G\_i|)|
\geq |\chi\_{h f\_i G\_i} - \chi\_{f\_i G\_i}| - |\chi\_{h G\_i} - \chi\_{G\_i}| - o(|G\_i|) = 2|G\_i| - o(|G\_i|). $$


This gives $\frac{|\chi\_{hF\_iG\_i} - \chi\_{F\_iG\_i}|}{|\chi\_{F\_iG\_i}|} = \frac{2|G\_i| - o(|G\_i|)}{2|G\_i| + o(|G\_i|)} \rightarrow 1$,
and thus $(F\_iG\_i)\_i$ is not a Følner sequence.



> 
> The infinite dihedral group admits Følner sequences $(F\_i)\_i$ and $(G\_i)\_i$ such that $(F\_iG\_i)\_i$ is not a Følner sequence.
> 
> 
> 


Proof. Every element of finite order has an infinite conjugacy class.


**original**


Not a solution but too long for a comment.


I think it's not necessarily Følner, as long as you can find a Følner sequence $G\_i$ such that for some fixed $\epsilon > 0$ and $h \in G$,
for all $i$ there exist arbitrarily large $g$ such that $|hgG\_i \setminus gG\_i|/|gG\_i| > \epsilon$. I imagine can happen in the non-abelian case, and it happens on the Heisenberg group.


Here's the idea of the construction. Pick $G\_i$ such a Følner sequence. For each i construct $F\_i$ like so: Pick a tiny set $F\_i'$ around the
identity, containing the identity, in such a way that $F'\_iG\_i = G\_i \cup A\_i$ where $|A\_i|$ is much smaller than $G\_i$ (using that $G\_i$ is Følner) and so that $F\_i'$ itself is a Følner sequence.


Now, recall our assumption was that $gG\_i$ is not a very left Følner-ish set for infinitely many g and a fixed translation $h$. So pick a couple of such $g$ and add them to $F\_i'$ to get $F\_i$. As long as we pick
much fewer than the cardinality of $F'\_i$, $F\_i$ will also be a Følner sequence.


But if we pick at least $\ell$ such $g$, then
if we pick them very separated from each other, $F\_iG\_i$ will actually consist of $F'\_iG\_i \approx G\_i$ plus some disjoint
$gG\_i$'s. If you pick a random element of this $F\_iG\_i$, it's going to be one of the $gG\_i$ with probability roughly $(\ell-1)/\ell$ once $i$ is large (and $\ell$ can grow to infinity with $i$), and then with probability at least $\epsilon > 0$ the $h$-translate gets outside $gG\_i$ (and by picking $g$ separated enough, we can make sure they don't hit any of the other $g'G\_i$ either). So we don't even get a left Følner sequence for $h$-translations.


On the Heisenberg group, $G\_i$ can be any Følner sequence:
Consider Heisenberg with generators $x$, $y$ and $z = [x, y]$, and any Følner sequence $G\_i$. Consider $g = x^j$ and $h = y$. Since $h g G\_i = g h z^j G\_i$
we have $h g G\_i \cap g G\_i \neq \emptyset \implies h z^j G\_i \cap G\_i \neq \emptyset$ which means
$hz^j \in G\_iG\_i^{-1}$, which happens only for finitely many $j$.
So for the Heisenberg group, the answer is that the product is not necessarily Følner.


