Suppose that $C^\*\_r(G)\cong C^\*\_r(H)$, can we conclude that $G\cong H$?



Two isomorphic reduced group $C^*$-algebras
The simplest counterexample is $G = \mathbb{Z}\_2\times \mathbb{Z}\_2$ and $H = \mathbb{Z}\_4$. These groups are not isomorphic but $C^\*(G) \cong \mathbb{C}^4 \cong C^\*(H)$.


