(Cross-post from [math.stackexchange](https://math.stackexchange.com/questions/3920898/if-the-union-of-finitely-many-conjugacy-classes-is-syndetic-are-there-finitely).)


Let $G$ be a finitely-generated group. Write $A^G = \{g^{-1} a g \;|\; a \in A, g \in G\}$, and $A \Subset G \iff A \subset G \wedge |A| < \infty$. Is the following true:
$$ \exists A \Subset G: AA^G = G \implies \exists B \Subset G: B^G = G? $$
In words, if the union of finitely many conjugacy classes is left syndetic, are there finitely many conjugacy classes?


This reminds me a bit of Neumann's trick <https://math.stackexchange.com/questions/536479/group-covered-by-finitely-many-cosets> but if it can be used I don't see how. I have no other ideas.



If the union of finitely many conjugacy classes is syndetic, are there finitely many conjugacy classes?
No. The dihedral group $D\_\infty$ has two conjugacy classes of elements of order 2, and their union is the nontrivial coset of an infinite cyclic subgroup of index 2, in which $D\_\infty$ conjugacy classes consist of opposite pairs, so there are $\infty$ many. So the conclusion fails with $A=\{1,s,t\}$, $s,t$ being non-conjugate elements of order 2.


