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Suppose that $G$ is a cyclic group of order $n$. I want to show that $R[X]/(X^n-1)\cong R(G)$ where $R(G)$ is a group ring.

Could you give me an idea about it ? Thank you

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If $A$ is an $R$-algebra, we have natural bijections $$\hom_{\mathsf{Alg}(R)}(R[G],A) \cong \hom_{\mathsf{Grp}}(G,A^*) \cong \{a \in A : a^n=1\} \cong \hom_{\mathsf{Alg}(R)}(R[X]/(X^n-1),A),$$ hence, $R[G] \cong R[X]/(X^n-1)$ by Yoneda. If $G=\langle \sigma \rangle$, the isomorphism maps $\sum_i r_i \sigma^i$ to $\sum_i r_i X^i$. If you don't like universal properties and instead have fun with calculating things, you can also check directly that this defines an isomorphism.

The same proof shows: If $$G = \langle g_1,\dotsc,g_n : R_1(g_1,\dotsc,g_n)=\dotsc=R_p(g_1,\dotsc,g_n)=1 \rangle$$ is a group presentation, then we get an algebra presentation $$R[G] \cong R \langle X_1^{\pm 1},\dotsc,X_n^{\pm 1} : R_1(X_1,\dotsc,X_n)=\dotsc=R_p(X_1,\dotsc,X_n)=1 \rangle.$$

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