rt.representation theory - Character table does not determine group Vs Tannaka duality

Edit: Somehow I totally misread the question. I talked about the group algebra $mathbb C[G]$, which is not at all the same as the character ring $R(G)$. Over $mathbb C$ (or any other field of characteristic 0), $R(G)$ is naturally a subalgebra of $(mathbb C[G])^*$, which is the algebra of functions on $G$ with pointwise multiplication, and now the comultiplication encodes the group structure. On the other hand, it is not a subbialgebra: the coproduct of a class function is not a class function.

Anyway, original post below, with the obviously wrong things struck out. So it's really an answer to Kevin, rather than anything else.

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Well, it depends on what you mean by "$R(G)$". I won't address TK duality, and most of what I'll say is essentially a follow-up to Kevin's answer, rather than an answer in its own right. Also, I'm only going to address finite groups and their finite-dimensional representations. Also, for me the word "ring" means (associative, unital, noncommutative) "$mathbb C$-algebra".

Recall that a complex representation of $G$ is the same as an algebra representation of $mathbb C[G]$. Let $R$ be a ring. As Kevin says, it's in general impossible to define an $R$-module structure on $Motimes N$ when $M,N$ are $R$-modules. (When $R$ is abelian, which is not the case here, one can define a tensor product $M otimes_R N$, but that's not the tensor product of representations anyway.) What would a tensor product of modules require? It would require a rule that assigns to each $rin R$ and each pair $M,N$ of $R$-modules an endomorphism of $Motimes N$, of course, and we should impose all sorts of axioms that force the tensor product to be well-behaved. Among other things, it's much easier if the endomorphism is an element of the tensor product $text{End}(M) otimes text{End}(N) subseteq text{End}(Motimes N)$. And we already have some distinguished elements of $text{End}(M)$ and $text{End}(N)$, namely the action of $R$.

So one way to try to construct a well-behaved tensor product on the category of $R$-modules is to find a nice map $Delta: R to Rotimes R$. Then the axioms for this map that assure that the tensor product is good are that $Delta$ be an algebra homomorphism, and that it be "coassociative": $(text{id}otimes Delta)circ Delta = (Delta otimes text{id})circ Delta)$. Let's suppose that there's also a distinguished "trivial" representation $epsilon: R to text{End}(mathbb C) = mathbb C$; if this is to be the monoidal unit, then we'd need $(text{id}otimes epsilon) circ Delta = text{id} = (epsilon otimes text{id})circ Delta$. The maps $Delta, epsilon$ satisfying these axioms define on $R$ the structure of a bialgebra.

By the way, the map is called "$Delta$" because if $G$ is a group (or monoid) and $R = mathbb C[G]$, then the map $R to Rotimes R$ given on the basis $G$ by the diagonal map $Delta: g mapsto gotimes g$ is such a structure.

Then here's a cool fact. Define an element $rin R$ to be grouplike if $Delta(r) = rotimes r$. Then the grouplike elements are a multiplicative submonoid of $R$. And when $R = C[G]$, the grouplike elements are precisely $G$.

So my answer to your question is that "the additional information contained in $R(G)$ as opposed to the character table" is its bialgebra structure.