Does there exist an infinite group $G$ such that:

  1. There are no conjugacy classes containing infinitely many elements.
  2. For every $n \in \mathbb{N}$, there are only finitely many conjugacy classes containing exactly $n$ elements.

Some basic observations:

  • $G$ cannot be Abelian, otherwise it would have infinitely many conjugacy classes containing $1$ element.
  • $G$ must have infinitely many conjugacy classes.

A basic idea I had was to construct a group $$G := \bigoplus_{n \in \mathbb{N}} G_n,$$ where $G_n$ is a finite group with $2$ conjugacy classes: one containing the neutral element, of size $1$, and the other containing all other elements, of size $p_n$. If all $p_n$ are prime and $p_1 < p_2 < \cdots < p_n < \cdots$, I believe the conditions would be satisfied. However, I have no idea if there are infinitely many primes $p_n$ for which such groups $G_n$ exist...

  • 1
    $\begingroup$ For starters, a finite group with only $2$ conjugacy classes must be cyclic of order $2$. $\endgroup$ – Nicky Hekster Sep 6 '19 at 16:40
  • $\begingroup$ The group would be "very abelian", in that every centralizer will have finite index... $\endgroup$ – Arturo Magidin Sep 6 '19 at 20:36
  • $\begingroup$ Actually, finite groups with $\le m$ conjugacy classes have order $\le c_m$ for some $c_m$, see groupprops.subwiki.org/wiki/…. This is an elegant application of the fact that every positive rational has only finitely many representations as a sum of $k$ inverses of (possibly equal) positive integers for each given $k$ ("Egyptian fraction representations"). $\endgroup$ – YCor Sep 7 '19 at 8:48

Just take a direct sum $G=\bigoplus G_n$ where in $G_n$ the smallest size $c_n$ of a conjugacy class $\neq\{1\}$ satisfies $c_n\to\infty$. For instance $G_n$ the symmetric group works (in $S_n$ for $n\ge 3$ every conjugacy class has cardinal $\ge n$, and actually much more). Alternatively take prime powers $q_n\to\infty$ and the affine group $G_n=\mathbf{F}_{q_n}\rtimes \mathbf{F}_{q_n}^*$, in which every nontrivial conjugacy class has cardinal $\ge q_n-1$. This gives a solvable (metabelian) example.

Indeed if $g$ has a conjugacy class of size $c$ and $c_n>c$ for $n\ge n_0$, then $g\in\bigoplus_{n<n_0}G_n$, which leaves finitely many possibilities for $g$.

(Note: a group with only finite conjugacy classes is called an FC-group.)


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