Demonstrate that $\mathbb{C}[\mathbb{Z}_{2} \times \mathbb{Z}_{3}] \simeq \mathbb{C}^{6}$. This is a review problem that I'm solving. I have been told that
$ \mathbb{C}[\mathbb{Z}_{2} \times \mathbb{Z}_{3}] $ is a $ \mathbb{C} $ algebra so a vector space where the field is $\mathbb{C}$ and the 'vectors' are $[\mathbb{Z}_{2} \times \mathbb{Z}_{3}]$. I'm quite confused as to what the 'action' is in this case, since $[i]_{2}$ does not make any sense, so I'm assuming its kind of like a group ring where you simply have $i(0,1)$ where $(0,1) \in \mathbb{Z}_{2} \times \mathbb{Z}_{3}$. So really, this question is asking me to demonstrate a ring isomorphism between these two algebraic objects. So we define a map, $\phi:\mathbb{C}[\mathbb{Z}_{2} \times \mathbb{Z}_{3}] \rightarrow \mathbb{C}^{6}$ where,
$$\phi(0,0) = (0,0,0,0,0,0), \phi(0,1) = (1,0,0,0,0,0), \phi(0,2) = (0,1,0,0,0,0), $$
but this idea does not work because $\phi(0,2) + \phi(0,1) \ne \phi(0,0)$. I know that $\mathbb{Z}_{2} \times \mathbb{Z}_{3} \simeq \mathbb{Z}_{6}$, but I still don't know how to 'encode' the vectors into $\mathbb{Z}_{6}$.
Any help would be much appreciated. Thanks!
 A: I'm not sure if you're allowed to use Maschke's theorem or not, but let's look at solutions that involve that.
Basically, if $G$ is an abelian group of order $n$, then $\mathbb C[G]$ is a commutative semisimple $\mathbb C$ algebra, so it has to be copies of $\mathbb C$, and we know there will be $n$ of them.  From that you could conclude that $\mathbb C[\mathbb Z_2\times \mathbb Z_3]\cong \mathbb C^6$ since the group has order $6$.
Actually, you can work out that $\mathbb C[G\times H]\cong\mathbb C[G]\otimes_\mathbb C\mathbb C[H]$.  For similar reasons the pieces of the product have dimension $2$ and $3$, and so the dimension of their tensor product will be $6$ in total.  Using this latter piece you can see how the two smaller groups are 'encoded' into the larger one, and their group rings are encoded into the full group ring.
A: If $R$ is a commutative ring, and $G$ is a group, the group ring $R[G]$ is an $R$-algebra, together with a map $i_G \colon G \to R[G]$, characterized by the following universal property:

For any $R$-algebra $S$, every group homomorphism $f$ from $G$ to $S^\times$ (the group of units of $S$) can be uniquely ‘extended’ to an $R$-algebra homomorphism $R[G] \to S$; that is, there exists a unique $R$-algebra homomorphism $\bar f \colon R[G] \to S$ such that $f \circ i_G$ is the same as $f$ followed by $S^\times \hookrightarrow S$.

Any other $R$-algebra $A$ with a map $G \to A$ satisfying the above property is isomorphic to $R[G]$.

So, what we need to do here is, first, give a map $i \colon \Bbb Z_2 \times \Bbb Z_3 \to \Bbb C^6$; and then prove that if $S$ is a $\Bbb C$-algebra and $f \colon \Bbb Z_2 \times \Bbb Z_3 \to S^\times$ a group homomorphism, then there exists a unique $\Bbb C$-algebra homomorphism $\bar f \colon \Bbb C^6 \to S$ such that $f \circ i$ is $f$ followed by $S^\times \hookrightarrow S$. (Hint: Try to define $i$ so that its image is the canonical basis for $\Bbb C^6$.)
