L(G) is a factor $\implies$ G is i.c.c.

Question

I have easily shown that if $G$ is an i.c.c. (infinite conjugacy class) group, then its group von Neumann algebra, $L(G) := \mathbb{C}[\lambda(G)]'' = (\text{span} \lambda(G))''$ is a factor (its center is $\mathbb{C} 1$, with no nontrivial projections). However, I can't see why it is guaranteed the other direction, that $L(G)$ being a factor would force $G$ to be i.c.c. I can't find anywhere on this website a full iff proof, just the above direction. But it appears that an iff statement is true. Can anyone point me to a source? I have tried assuming $G$ NOT i.c.c. and looking for a nontrivial projection in the center of $L(G)$ to prove it myself, but haven't gotten anywhere.

Answer 1

This one is proposition 7.9 in V.7 of Takesaki's Theory of Operator Algebras I. Essentially the argument is as follows.

The standard way is what you proposed: suppose that $G$ is not ICC and show its not a factor. If $G$ is not ICC, it has some finite conjugacy class $C(g)$ for $g \neq e$. Then $x = \sum_{h \in C(g)} u_h$ is nonscalar and central. It is central because $$ u_sxu_s^* = \sum_{h \in C(g)} u_{shs^{-1}} = \sum_{h \in C(g)} u_h = x$$ as conjugation by any $s$ gives a bijection $C(g) \to C(g)$, and so $x$ commutes with the generating unitaries. So the center is nontrivial.