Existence of proper I.C.C. subgroup

Question

A countable discrete group $G$ is called I.C.C.(infinite conjugacy class) if for any $e\neq g\in G$, $\#\{sgs^{-1}\mid s\in G\}=\infty$.

My question is:

Is it possible for a group $G$ to be I.C.C. but also contain no proper I.C.C. subgroup?

Answer 1

Yes. Tarski monster groups are infinite finitely generated groups all of whose proper, non-trivial subgroups are cyclic of order a fixed prime $p$. Tarski monsters are I.C.C. but as all proper subgroups are finite Tarski monster groups can contain no proper I.C.C.subgroup.

To see that a Tarski monster $T$ is I.C.C., suppose otherwise. Denote the fixed prime by $p$. Then there exists some non-trivial element $g\in T$ and $p$ non-trivial elements $h_1, h_2, \ldots, h_{p}$, each non-equal to $g$, such that $h_igh_i^{-1}=g$ but each $h_i\neq h_j$ (this is because $T$ is infinite but the conjugacy class of some element $k$ is finite, and then picking an appropriate conjugate of $k$). Hence, $g$ and $h_i$ commute for all $i$. As there are $p$ elements $h_i$, $g$ is contained in two separate cyclic subgroups, so $a^i=g=b^j$ with $\langle a, b\rangle=T$. However, because $a$, $b$ and $g$ all have order $p$, we can also write $a=g^{i_0}$ and $b=g^{j_0}$. Hence, $\langle a, b\rangle=\langle g\rangle$ is cyclic, a contradiction.