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Not long ago I asked a question about groups where some finite union of conjugacy classes is left-syndetic. Here's a stronger variant.

Suppose $G$ is a finitely-generated group such that there exists $n$ and $A \Subset G$ such that whenever $g_1, g_2, ..., g_n \in G$, we have $g_i^{-1} g_j \in A^G$ for some $i \neq j$. Does $G$ have necessarily finitely many conjugacy classes?

Here, $A \Subset G \iff A \subset G \wedge |A| < \infty$ and $A^G = \{gag^{-1} \;|\; a \in A, g \in G\}$.

Some observations:

  • If there are finitely many conjugacy classes, then the above property holds (with $n = 2$), by taking $A$ to be the representatives.
  • The above property implies that there is a finite union of conjugacy classes which is left-syndetic (otherwise any tuple $(g_1, g_2, ..., g_n)$ contradicting the above for $A \subset G$ is right-extendable).
  • The infinite dihedral group has a finite union of conjugacy classes which is left-syndetic (see the linked post), but does not have the above property, because it contains a copy of $\mathbb{Z}$ whose elements have conjugacy classes with cardinality at most two (for any $A, n$ you can pick the $g_i$ from that subgroup).
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  • $\begingroup$ As is standard to ask: why the downvote? $\endgroup$
    – Ville Salo
    Commented Nov 30, 2020 at 15:34
  • $\begingroup$ If $G$ is a group with only 2 conjugacy classes and s is an automorphism of order 2, I see no reason that $H=G\rtimes_s C_2$ to have finitely many conjugacy classes (i.e., whether $G$ has finitely many twisted conjugacy classes, i.e. orbits under $v\cdot u = vusv^{-1}s$). If there's an example, then $H$ would answer negatively the question. But I'm not sure at all it exists (and not sure there's no easier approach). $\endgroup$
    – YCor
    Commented Nov 30, 2020 at 18:10
  • $\begingroup$ Maybe if you have not two but finitely many conjugacy classes, you could use an inner automorphism, at least IIRC Osin's construction allows an arbitrary set of orders, and certainly the involutions can't be central. Then again maybe it's the part where $H$ has infinitely many conjugacy classes that is not clear. $\endgroup$
    – Ville Salo
    Commented Dec 1, 2020 at 7:43
  • $\begingroup$ To be more explicit: Let $G$ be the infinite dihedral group and apply Theorem 1.1 from [Osin 2010, "Small cancellations ..."], to obtain a $2$-generated group $C$ which contains $G$, has no elements of orders $n \notin \{2, \infty\}$, and has exactly three conjugacy classes. Let $g \in G$ have order $2$ and $s(h) = h^g$ the inner automorphism, which is nontrivial of order $2$ because $g$ is not central (since $G \leq C$ so there are at least two involutions in $C$). Define $H = C \rtimes_s C_2$ where $C_2$ acts by $s$. The property holds, but possibly $H$ has infinitely many conjugacy classes. $\endgroup$
    – Ville Salo
    Commented Dec 1, 2020 at 8:09

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