Whether such an algebra has to be the Group algebra

Let $\mathbb C$ be the field of the complex numbers, $\mathbb Q$ the field of the rational numbers.

Let $G$ be an additive subgroup of $\mathbb Q$.

$R$ is an commutative algebra over $\mathbb C$, which is a domain and $G$-graded with all the $G$-graded spaces are 1-dimensional.

Is it true that we have $R \cong {\mathbb C}[G]$(the group algebra over $G$)?

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@BB, @Ralph, Thanks. I get the main point from the detailed proof of Ralph. So it can be proved as following directly Let $(\C[G],\circ)$ be any such algebra first find H_1\le H_2\le\cdots H_n\cdots with $H_i\cong \Z$, \cup H_i=G Then inductively we may find isomorphism $\a_i:(\C[H_i],\circ)\rightarrow (\C[H_i])$, with $\a_i|_{H_{i-1}}=\a_{i-1}$. Now we get the isomorphism $\a$ as the direct limit of $\a_i$. Thanks again. – r_l Jun 21 '12 at 0:42

Yes, it's isomorphic to the group algebra. We can do even a bit better:

Let $G$ be a torsion-free abelian group those finitely generated subgroups have rank one and let $k$ be an algebraically closed field. If $R$ is an associative zero-divisor-free $G$-graded $k$-algebra with $\dim_k R_g=1$ for all $g \in G$, then $R \cong k[G]$.

The proof is along Bugs' lines. Denote the product in $R$ by $\circ$ and write $g \circ h = a(g,h)gh$, $a(g,h) \in k$. Since $R$ is zero-divisor-free this defines a map $a: G \times G \to k^\times$. By associativity, $a$ satisfies for all $f,g,h \in G$: $$a(f,g)\cdot a(fg,h)=a(g,h)\cdot a(f,gh).$$ Futhermore, since the neutral element $e \in G$ is also the identity of $R$, we have $$a(f,e) = 1 = a(e,f).$$ These two equations just say that $a$ is a normalized cocycle, if we consider $k^\times$ as a trivial $G$-module (cf. (3.4),(3.9) in chap. IV of Brown: Cohomology of Groups). So $a$ represents an element in $H^2(G,k^\times)$.

Assume $H^2(G,k^\times)=0$. Then $a$ is a coboundary, i.e. there is a map $b: G \to k^\times$ such that $$a(g,h)=b(g)b(h)b(gh)^{-1}.$$ Define $\varphi: k[G] \to R,\; g \mapsto b(g)^{-1}g$. Obviously, $\varphi$ is an isomorphism of $k$-vector spaces and because of $$\varphi(g) \circ \varphi(h)=b(g)^{-1}b(h)^{-1}(g \circ h) = b(g)^{-1}b(h)^{-1}a(g,h)gh=b(gh)^{-1}gh=\varphi(gh)$$ it's an isomorphism of graded $k$-algebras.

Hence it suffices to show $H^2(G,k^\times)=0$. By universal coefficients (Hilton-Stammbach: A course in homological algebra, Theorem 15.1) there is a short exact sequence $$0 \to Ext_{\mathbb Z}^1(H_1(G),k^\times) \to H^2(G,k^\times) \to Hom(H_2(G),k^\times) \to 0$$ where $H_n(G) := H_n(G,\mathbb Z)$. As an abelian group, $G$ is the direct limit of its f.g. subgroups $H$ and because homology commutes with direct limits, $H_2(G)$ is the direct limit of $H_2(H)$. But by assumption $H \cong \mathbb Z$, whence $H_2(H)=0$.

Since $k$ is algebraically closed, the multiplicative group $k^\times$ is a divisible $\mathbb Z$-module and thus injective (Brown, III, 4.2). Hence $Ext_{\mathbb Z}^1(H_1(G),k^\times)=0$ and the short exact sequence implies $H^2(G,k^\times)=0$. q.e.d.

Remark: As the proof shows, the condition that $R$ is zero-divisor-free can be lowered to the assumption $g\circ h \neq 0$.

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Yes, such associative, not necessarily commutative algebras are classified by $H^2(G, {\mathbb C}^\ast)$. Indeed, try to define new multiplication on the group algebra by $g\cdot h = \alpha (g,h) gh$. Then $\alpha$ must be a cocyle, and isomorphic algebras correspond to cohomologous cocycles.

Finally, $H^2(G, {\mathbb C}^\ast)= Hom(\Lambda^2 G, {\mathbb C}^\ast)= 0$ since any 2 elements are linearly dependent. See here for the calculation of $H^2$.

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Does the condition $G$ is an additive subgroup of $\Q$ really make sense? It is easy to see that if $G$ is $\Z$, then the result hold. However, i have no idea on the general case, for example $G=\Q$. I don't know why you say it is huge? can you give one exact example to show such an algebra may not isomorphic to $\C[G]$? – r_l Jun 20 '12 at 9:23
or, Is there reference for H^2(\Q,\C^*)? Thanks for your help. – r_l Jun 20 '12 at 9:30
@BB : the OP asked for a commutative algebra, so I doubt the whole (or any ?) of $H^2$ will show up. – BS. Jun 20 '12 at 9:53
@ r_l Just click on HUGE @ BS Good Point. He needs symmetric cocycles... I will think about it... – Bugs Bunny Jun 20 '12 at 11:06
@BB: I would like a more direct proof or a direct calculation of $H^2(G,\C^*)$. Thanks. – r_l Jun 20 '12 at 11:58