A finitely presented, countable discrete group $G$ is amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gX)=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$ 


A countable discrete group $G$ is inner amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gXg^{-1})=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$


The growth $b:\mathbb{N} \rightarrow \mathbb{N}$ of $G$ (with respect to a given word length metric on $G$) is defined as the number of elements $b(n)$ in $G$ lying inside the ball of radius $n$ around $e$. 


It is possible to detect the amenability of $G$ in terms of the growth of G (c.f. R. I. Grigorchuk, “Symmetric random walks on discrete groups”, UMN, 32:6(198) (1977), 217–218).


Can the growth of G detect inner amenability?


I'd like to know if there is an i.c.c. discrete nonamenable simple group that is inner amenable? 


On a related note, what about an answer to Owen's question below?



Is there an i.c.c. nonamenable simple group that is inner amenable?
Here is a construction of a countable i.c.c. nonamenable simple group that is inner amenable. First consider the following condition:


(\*) For every finite subset $S\subseteq G$, there exists $g\in G\setminus \{ 1\} $ such that $[g,s]=1$ for every $s\in S$.


Using paradoxical decomposition for non-inner amenable groups, it is not hard to show that (*) implies inner amenability. Indeed let
$$
G\setminus \{ 1\} =A\_1\sqcup \ldots \sqcup A\_k\sqcup B\_1\sqcup \ldots \sqcup B\_m
$$
and let $x\_1, \ldots x\_k, y\_1, \ldots y\_m$ be elements of $G$ such that 
$$
G\setminus \{ 1\} =(A\_1)^{x\_1}\sqcup \ldots \sqcup (A\_k)^{x\_k}=(B\_1)^{y\_1}\sqcup \ldots \sqcup (B\_m)^{y\_m}
$$
By (*) there exists $g\ne 1$ that commutes with all $x\_i$ and $y\_j$. Let $A\_i^\prime =A\_i\cap \langle g\rangle$, $B\_i^\prime =B\_i\cap \langle g\rangle$. Intersecting $\langle g\rangle$ with the above decompositions of $G$ and noting that $(A\_i)^{x\_i}\cap \langle g\rangle=A\_i^\prime$ and similarly for $B\_i$'s, we obtain
$$
\langle g\rangle\setminus \{ 1\} = A\_1^\prime \sqcup \ldots \sqcup A\_k^\prime \sqcup B\_1^\prime \sqcup \ldots \sqcup B\_m^\prime = A\_1^\prime \sqcup \ldots \sqcup A\_k^\prime = B\_1^\prime \sqcup \ldots \sqcup B\_m^\prime.
$$
This is impossible for nontrivial $g$.


Now let us construct a group $G$ by induction. Let $G\_0=F\_2$, the free group of rank $2$. For $n> 0$ let $G\_n$ be a countable simple group that contains $G\_{n-1}\times \mathbb Z$ (every countable group embeds in a countable simple group). Let $G$ be the union of the chain $G\_0\subset G\_1\subset \ldots $. Clearly $G$ is simple being a union of simple groups and satisfies (\*) by construction. Hence $G$ is inner amenable. As $G$ is simple and infinite, it is i.c.c. Finally $G$ is non-amenable as it contains $F\_2$. 


Modifying the above argument one can also construct an i.c.c. simple inner amenable non-amenable group without nontrivial free subgroups.


