I want to prove $\mathbb{Z}(G\times C_2)\cong (\mathbb{Z}G)C_2$, where $C_2=<x| x^2=1>$, where $\mathbb{Z}(G\times C_2)$ and $(\mathbb{Z}G)C_2$ are integral group rings and I am looking for ring isomorphism.
Now I am not sure what map should I consider to make it an isomorphism. I tried $\phi$ which takes $\alpha=\sum_g a_g.(g,x) \to (\sum_g a_g g+(\sum_g a_gg)x)$, and $\beta=\sum_h b_h.(h,x) \to (\sum_h b_h h+(\sum_h b_hh)x)$,
But this is not working. How can prove them isomorphic.
you have a correct argument, I 'm just going to improve the notation of the elements in each group algebra, you observe that every element in $\Bbb Z[G\times C_2]$ can be write as $$\sum_{(g,y)\in G\times C_2} a_{(g,y)}(g,y)=\sum_{(g,1)\in G\times C_2}a_{(g,1)}(g,1)+\sum_{(g,x)\in G\times C_2} a_{(g,x)}(g,x).$$ Then the map \begin{align*} \varphi:\Bbb Z[G\times C_2]&\rightarrow \Bbb ZG[C_2]\\ \sum_{(g,1)\in G\times C_2}a_{(g,1)}(g,1)+\sum_{(g,x)\in G\times C_2}a_{(g,x)}(g,x)&\mapsto \left(\sum_{g\in G} b_g g \right)1+\left(\sum_{g\in G} c_g g \right)x \end{align*} where $b_g=a_{(g,1)}$ and $c_g=a_{(g,x)}$ for all $g$ in $G$. it's well defined and that it's a isomorphism.