Group Ring Isomorphism Suppose I have a cyclic group $G$ with order $n$. Then I will need to show $\mathbb{C}[G]\cong\mathbb{C}\times\cdots\times\mathbb{C}$ $n$ times.
I was thinking using the map
$\sum_{i=1}^{n}a_{g_i}g_i\rightarrow(a_{g_1},\cdots,a_{g_n})$ and it is clear that it is bijective. But I do not know how to exhibit the homomorphism. The addition part is easy, but how about the multiplication part?
 A: The map you propose is not multiplicative. Let's introduce some notation. Let us write $G= \langle g \rangle$. Thus $G= \{g^i: i= 0, \dots, n-1\}$ and an element in $\mathbb{C}[G]$ is of the form
$$\sum_{i=0}^{n-1} \lambda_i g^i.$$
Now, your map is
$$\Psi: \mathbb{C}[G]\to \bigoplus_{i=0}^{n-1} \mathbb{C}: \sum_i \lambda_i g^{i} \mapsto (\lambda_0, \dots, \lambda_{n-1}).$$
This is not a multiplicative map (check for instance that $\Psi(g^2) \ne \Psi(g)\Psi(g)$). Another way to see that this is not a good map is to note that $\Psi$ is not unital.

The isomorphism you are looking for follows by the Wedderburn-Artin theorem. Indeed, consider the Wedderburn Artin decomposition
$$\mathbb{C}[G]\cong M_{n_1}(\mathbb{C}) \oplus \dots \oplus M_{n_k}(\mathbb{C}).$$
Since $\mathbb{C}[G]$ is commutative, we must have $n_1 = \dots = n_k = 1$ and thus
$$\mathbb{C}[G]\cong \mathbb{C}\oplus \dots \oplus \mathbb{C}$$
(compare dimensions).
A: I'm assuming this is a homework problem, so I don't want to give a full solution. Here's a sketch:

*

*Show that $\mathbb{C}[G] \cong \mathbb{C}[x]/(x^n-1)$ (idea: these rings have the same representations)

*Show that $\mathbb{C}[x]/(x^n-1) \cong \mathbb{C}^n$ (idea: factor $x^n-1$)

*Profit!

