# Existence of proper I.C.C. subgroup

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?

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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.
I do not understand why "g is contained in two separate cyclic subgroups". I think the following argument also works. Clearly, we can pick some $h_i$ with $h_i\not\in \langle g\rangle$ then $\langle h_i, g\rangle=T$ so $T$ is abelian, a contradition. – ougao Aug 11 '14 at 12:28
@ougao Yeah, that idea works. I was trying to say that if $g$ and $h_i$ commute then they are contained in a common cyclic subgroup of order $p$, denoted $H_i$, but as there are $p$ such $h_i$s then two of these $H_i$s must be non-equal. – user1729 Aug 11 '14 at 13:10
my fault, I should mention that $G$ is inner amenable group if there exists a finite additive probability measure on the power set of $G$ which is invariant under conjugate action of $G$.(Recall for amenable group, we require that measure is invariant under left action of $G$). – ougao Aug 11 '14 at 17:08