I wonder whether the underlying complex group algebra of a group $C^*$-algebra is unique? I.e. if $G$ and $H$ are discrete groups such that $C^*(G) \cong C^*(H)$ (or $C^*_r(G) \cong C^*_r(H)$) as $C^*$-algebras, does it follow that $\mathbb{C}G \cong \mathbb{C}H$ as complex $*$-algebras?
I know that this does not hold on the level of groups. Since $\mathbb{Z}_4 \ncong \mathbb{Z}_2 \times \mathbb{Z}_2$, while their group $C^*$-algebras are both $\mathbb{C}^4$.
The reason I am interested in the discrete case is that then we have an underlying algebraic object, $\mathbb{C}G$, which is dense when endowed with the norm. I am interested in the uniqueness of such "dense underlying algebraic objects" in $C^*$-algebraic constructions.
Thanks!
Suppose $G$ and $H$ are abelian. Then their group C*-algebras are isomorphic iff their Pontryagin duals are homeomorphic. So take, for example, $G$ to be the Prufer 2-group and $H$ to be the Prufer 3-group. Their Pontryagin duals are the 2-adic and 3-adic integers, which are homeomorphic but not isomorphic even as abstract groups. I'm not sure off the top of my head if the group algebras are isomorphic though. But I suspect they can be distinguished by their self-adjoint subalgebras.