The Free Burnside group $G=B(2,665)=\langle a,b|g^{665} \rangle$ is infinite, by the work of Adyan and Novikov. Furthermore, the centralizer of any nonidentity element in $G$ is finite cyclic, and so the group is an i.c.c. group and the associated left group von Neumann algebra $LG$ is a type $II\_{1}$ factor. It is a fact, due to of Adyan, that this group is not amenable, so the group von Neumann algebra is not injective. 


A type $II\_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset $\{ x\_{1}, x\_{2},..., x\_{n} \} \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux\_{j}-x\_{j}u||\_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||\_2=(\tau(T^{\*}T))^{1/2}$ for $T\in M$.) 


I should mention that if a group is **not** inner amenable in the sense described in 


[Is there an i.c.c. nonamenable simple group that is inner amenable?](https://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable) 


then its left group von Neumann algebra does **not** have property $\Gamma$. (There exist i.c.c. inner amenable groups whose group von Neumann algebras don't have $\Gamma$, as recently shown by Stefaan Vaes: <http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.1485v1.pdf>.)


My question is:



> 
> Does the group von Neumann algebra $LG$
>  have Property
>  $\Gamma$?
> 
> 
> 



Do Burnside Group Factors have Gamma?
The answer is no, the group von Neumann algebra of the Burnside group $B(2,665)$ does not have property Gamma. In fact, $B(2,665)$ is not inner amenable. 


To see this, first note that, as pointed out in the question, $B(2,665)$ is nonamenable, and the centralizer of every non-identity element of $B(2,665)$ is finite. However, every nonamenable inner amenable group $G$ contains a non-identity element whose centralizer is nonamenable. This is a consequence of applying the following well-known folklore lemma to the conjugation action of $G$ on $G-1$: 


**Lemma:** if an action of a nonamenable group $G$ on a set $X$ admits an invariant finitely additive probability measure $\mu$, then there is a point $x\in X$ whose stabilizer subgroup $G\_x$ is nonamenable.


There are a variety of proofs of this Lemma. Here is a standard averaging argument: Suppose that every $G\_x$ is amenable. Let $X\_0\subset X$ contain one point from each $G$-orbit and for $x\in X\_0$ let $\nu \_x$ be a finitely additive probability measure on $G$ which is invariant under left translation by $G\_x$. Then $x\mapsto \nu \_x$ extends to a well-defined assignment on all of $X$ by taking $\nu \_{g\cdot x} = g\cdot \nu \_x$ for $x\in X\_0$, $g\in G$. The barycenter $\nu = \int \_X \nu \_x \, d\mu (x)$ is then invariant under left translation by $G$, so $G$ is amenable, contrary to the hypothesis of the lemma.


