**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.2 Exercise 7.2.12**

Let $R$ be a ring with $1 \neq 0$, and let $G = \{g_1, \ldots, g_n \}$ be a finite group. Prove that the element $N = \sum_{i=1}^n g_i$ is in the center of the group ring $R[G]$.

Solution: Let $M = \sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.\begin{align*}NM =&\ \left( \displaystyle\sum_{i=1}^n g_i \right) \left( \displaystyle\sum_{j=1}^n r_jg_j \right)\\

=&\ \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_jg_ig_j\\

=&\ \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_j g_j g_j^{-1} g_ig_j\\

=&\ \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_j^{-1}g_ig_j \right)\\

=&\ \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_i \right)\\

=&\ \left( \displaystyle\sum_{j=1}^n r_jg_j \right) \left( \displaystyle\sum_{i=1}^n g_i \right)\\

=&\ MN.\end{align*}Thus $N \in Z(R[G])$.