Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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 at 12:28
    
Thanks for your quick answer, do you know any I.C.C. non amenable inner amenable groups satisfy this property? –  ougao Aug 11 at 12:31
    
@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 at 13:10
    
Sorry - forgot to reply to your amenability question! Short answer is: I do not know. But partly, this is because I do not know what an "inner amenable group" is! However, amenable groups are suitably "small", so I would not be surprised if such a group existed. –  user1729 Aug 11 at 14:58
    
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 at 17:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.