Recovering a group from its C*-algebras and group algebra

Question

Let $G$ and $H$ be locally compact groups. Does anyone know the answers to these questions? Is it true that:

1. if $C^*(G)$ and $C^*(H)$ are $*$-isomorphic, then $G\cong H$?
2. if $C_r^*(G)$ and $C_r^*(H)$ are $*$-isomorphic, then $G\cong H$?
3. if $L^1(G)$ and $L^1(H)$ are $*$-isomorphic, then $G\cong H$?

Does anyone know if any of these questions have positive answers in general? Is it possible to recover a group from any of the algebras above?

Thanks.

Answer 1

Even for finite groups, one cannot expect to recover to be able to recover the group from the $C^*$-algebra. For example, there are two nonabelian groups of order 8. They both have group $C^*$-algebra $$ \mathbb C \oplus \mathbb C \oplus \mathbb C \oplus \mathbb C \oplus M_2(\mathbb C), $$ even though the groups are not isomorphic. Extra information (for example, a comultiplication) is needed before one can recover the original group.

In the commutative case, it is even worse. Any two finite nonisomorphic abelian groups of order $n$ will have group $C^*$-algebra $\mathbb C^n$.

My initial remark about group algebras was incorrect. There are non-isomorphic groups $G$ and $H$ such that $L_1(G)$ and $L_1(H)$ are isomorphic. However as Martin points out below, the existence of an isometric isomorphism $L_1(G) \simeq L_1(H)$ does imply $G \simeq H$.

Answer 2

It is possible to recover the group $G$ from the Banach algebra $L^1(G)$.

Wendel proved in 1950 that there exists an isometric isomorphism between the Banach algebras $L^1(G)$ and $L^1(H)$ if and only if $G$ and $H$ are isomorphic as topological groups.

Wendel's first proof relied on a theorem on positive isomorphisms due to Kawada. A few years later Wendel discovered that the left centralizers $L \colon L^1(G) \to L^1(G)$, i.e., the linear maps $L$ satisfying $L(f \ast g) = L(f) \ast g$, are given by convolutions by a Radon measure on $G$: $L(f) = \mu \ast f$. He then proved that such a convolution operator is an isometric isomorphism if and only if $\mu = e^{it} \delta_g$. From there it is only a small step to show that an isometric isomorphism $L^1(G) \to L^1(H)$ must be given by an isomorphism $\varphi \colon G \to H$ twisted by a character $\chi \colon G \to S^1$.