For a discrete group $G$, let $C^*_r(G)$ be its reduced group C$^*$-algebra.
Question: Do there exist discrete, torsion-free non-isomorphic groups $G,H$ such that $C^*_r(G)$ and $C^*_r(H)$ are stably isomorphic? Stable isomorphism means $C^*_r(G)\otimes K\cong C^*_r(H)\otimes K$ where $K$ are the compact operators on a separable, infinite-dimensional Hilbert space.
Motivation: It is a well known(?) open question whether or not there exist non-isomorphic, discrete torsion-free groups $G,H$ such that $C^*_r(G)\cong C^*_r(H).$ If you are like me and your intuition suggests there should exist such a pair of non-isomorphic groups, then you are naturally led to this "easier" version of the rigidity question. (There are a few classes of groups (e.g. abelian, finitely generated 2-step nilpotent) for which it is known that $G\cong H$ implies $C^*_r(G)\cong C^*_r(H).$)
For years I assumed one could follow this procedure to positively answer my question:
Write down two (smartly chosen) groups arising as (non-split) extensions $0\to \mathbb{Z}^d\to G_i\to F\to 0$ where $F$ is finite and the actions of $F$ on $\mathbb{Z}^d$ are the same but the groups are non-isomorphic. Then use a Packer-Raeburn type trick to untwist the actions and get matrix algebras over $C^*_r(G_1)$ and $C^*_r(G_2)$ isomorphic to each other. I recently decided to actually sit down and do this, and it doesn't work (at least in my examples...). So here I am.
Notice that for abelian C$^*$-algebras stably isomorphic implies isomorphic, so stably isomorphic group C$^*$-algebras of torsion-free abelian groups implies isomorphic groups.
