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(Cross-post from math.stackexchange.)

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.

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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.

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  • $\begingroup$ Hah! I am glad I decided against adding "obviously any counterexample would be some sort of monster group" in this post. $\endgroup$
    – Ville Salo
    Commented Nov 30, 2020 at 13:07
  • $\begingroup$ Or maybe the dihedral group is a monster group, this is not the first time it is a counterexample to a property I figured all natural groups should have (e.g. splendidness from my other MO post). $\endgroup$
    – Ville Salo
    Commented Nov 30, 2020 at 13:08
  • $\begingroup$ The fact this is virtually cyclic is extra interesting for my application, so double thanks. $\endgroup$
    – Ville Salo
    Commented Nov 30, 2020 at 13:27
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    $\begingroup$ By the way a single conjugacy class is syndetic in this example: $AB^G=G$ where $B=\{s\}$ and $A=\{1,st,s,t\}$. $\endgroup$
    – YCor
    Commented Nov 30, 2020 at 13:30

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