Does there exist an infinite group $G$ such that:
- There are no conjugacy classes containing infinitely many elements.
- 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…
Best Answer
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.)