Let $\Delta$ = {$ \delta_1,\dots, \delta_m$} be a conjugacy class in a finite group $G$. Prove that the element $D = \delta_1 + \cdots + \delta_m$ is in the center of the group ring $RG$ .
Hint: Check that $g^{-1}Dg = D$ for all $g \in G$.
Attempt proof: Let $X = \Sigma_{j = 1}^n r_jg_j$ and $D =\Sigma_{i = 1}^m \delta_i$.
$g^{-1}Dg = g^{-1}(\Sigma_{i = 1}^m \delta_i)g = \Sigma_{i = 1}^m(g^{-1}\delta_i g) = \Sigma_{i = 1}^m \delta_i = D$.
Then suppose $X = \Sigma_{j = 1}^n r_jg_j$ is an element of $Z(RG)$. Then we need to show $DX = XD$.
Then for any $X \in RG$, we have
$$DX = (\Sigma_{i = 1}^m \delta_i) (\Sigma_{j = 1}^n r_jg_j) = \Sigma_{i = 1}^m \Sigma_{j = 1}^n \delta_ir_jg_j = \Sigma_{j = 1}^n \Sigma_{i = 1}^m \delta_ir_jg_j = \Sigma_{j = 1}^nr_j(\Sigma_{i = 1}^m \delta_i)g_j = \Sigma_{j = 1}^nr_jDg_j = \Sigma_{j = 1}^n (r_jg_j)D = (\Sigma_{j = 1}^n r_jg_j)D = XD,$$
so $D \in Z(RG)$.
Can someone please verify this? Any suggestions or better approach will be appreciated it. Thanks!
Yes that's fine. You can unclutter it by removing indices for one:
$$g Dg^{-1}=g\left(\sum_{\delta\in\Delta}\delta\right)g^{-1}=\sum_{\delta\in\Delta} g\delta g^{-1}=\sum_{\delta\in g\Delta g^{-1}}\delta=\sum_{\delta\in\Delta}\delta=D. $$
Also when going on to prove $D$ is central, there's no need to write out $D$:
$$DX=D\left(\sum_{g\in G}x_gg\right)=\sum_{g\in G}x_g(Dg)=\sum_{g\in G}x_g(gD)=\left(\sum_{g\in G}x_gg\right)D=XD. $$