I have to prove that $G=\mathbb{Z}/6\times\mathbb{Z}/2$ and $H=\mathbb{Z}/12$ are not isomorphic, but the group rings $\mathbb{C}[H]$ and $\mathbb{C}[G]$ are isomporphic. First part was quite easy. Second part is a bit trickier for me. I know that $\mathbb{C}[H]=\{\sum_{h\in H}Z_hh|Z_h\in\mathbb{C}\}$ and similarly $\mathbb{C}[G]=\{\sum_{g\in G}Z_gg|Z_g\in\mathbb{C}\}$. I tried defining a general isomorphism from $\phi:\mathbb{C}[G]\rightarrow\mathbb{C}^{|G|}$ by $\phi(\sum_{\in G}Z_gg)$ is the vector of coefficients, but I don't think it works. I thought that this has something to do with the fact that the field of complex numbers is algebraiclly closed, but I feel like I'm missing some basic properties of such group rings.
We did define the dual group $G^*$ the fourier transfrom of finite abellian groups, but I don't know if this has anything to do with the question itself. Any hint would help, even some formal properties that might help me find the solution myself.
By Maschke's Theorem both algebras are semisimple. Now Wedderburn's Theorem tells us that as $\mathbb{C}$ is algebraically closed any abelian group $G$ with $n$ elements must give $\mathbb{C}(G)\cong M_{n_1}(\mathbb{C})\times...\times M_{n_k}(\mathbb{C})\overset{G \text{ abelian}}{\cong}\underset{n \text{ times}}{\mathbb{C}\times...\times\mathbb{C}}$ as the $n_i$ represent the number of elements in each conjugacy class of $G$.