The following question/problem has been bugging me on and off for some time now: so I thought it might be worth broaching here on MO, as a case of "ask the experts".

Let $G$ be a discrete group. Kaplansky observed that since the group von Neumann algebra $VN(G)$ is a finite von Neumann algebra, each left-invertible element in $VN(G)$ is invertible. A proof is outlined in
> M.S. Montgomery, Left and right inverses in group algebras, Bull. AMS 75 (1969)

(Well, she actually states a weaker result, but inspection shows that her argument extends to give what we claim. See also my remarks on this [previous MO answer](https://mathoverflow.net/questions/18508/is-a-left-invertible-element-of-a-group-ring-also-right-invertible/18509#18509).)
The basic idea is to exploit the **faithful** trace $T\mapsto \langle T\delta_e,\delta_e\rangle$ and how it behaves on idempotents: for if $ab=I$, then $ba$ is an idempotent.
 
In particular, each left-invertible element of the reduced group $C^*$-algebra is invertible.

**Question.** What can we say for the **full** group $C^*$-algebra? Is every left-invertible element in $C^*(G)$ automatically invertible?

Some basic observations:

 - The case where $G$ is the free group on two generators follows from a result of M-D Choi [no relation] who showed that $C^*({\mathbb F}_2)$ embeds into a direct product of matrix algebras.

 - More generally, if $C^*(G)$ has a faithful trace then one can use the same argument as for the reduced $C^*$-algebra to get a positive answer.

 - If $C^*(G)$ has no non-trivial projections then $ab=I$ implies $ba=I$. (I think this was known to be true for $G={\mathbb F}_2$ but I've forgotten the reference at present.)

 - There are examples of $G$ where $C^*(G)$ has no faithful trace; these can be found in work of Bekka and Louvet, and come from exploiting Property (T).

> Bekka, M. B.(F-METZ-MM); Louvet, N.(CH-NCH)
Some properties of $C^*$-algebras associated to discrete linear groups. $C^*-algebras (Münster, 1999), 1–22, Springer, Berlin, 2000.