## Exhibit an element in the center of a group ring

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 $R[G]$.

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.

 $NM$ = $\left( \displaystyle\sum_{i=1}^n g_i \right) \left( \displaystyle\sum_{j=1}^n r_jg_j \right)$ = $\displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_jg_ig_j$ = $\displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_j g_j g_j^{-1} g_ig_j$ = $\displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_j^{-1}g_ig_j \right)$ = $\displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_i \right)$ = $\left( \displaystyle\sum_{j=1}^n r_jg_j \right) \left( \displaystyle\sum_{i=1}^n g_i \right)$ = $MN$.

Thus $N \in Z(R[G])$.

• Bobby Brown  On November 5, 2010 at 3:45 pm

indices i and j are mixed up, and got quite confusing, but I think I sorted it out. Nice proof.

• Will  On November 15, 2011 at 2:36 am

Hi,

I might be totally lost, but I think that the g_i at the very end of the third to last line should be deleted, no?

Definitely a nice proof.

• nbloomf  On November 15, 2011 at 11:32 am

You’re right; I’m not sure how it got there. Thanks!

• Ali  On May 10, 2012 at 8:06 am

In the step where you go from (the summation of) gj^(-1)gigj to just gi, don’t you have to assume the elements gi,gj, etc. of G are commutative? And we don’t know that the group is commutative, so can we make that step?

• nbloomf  On August 17, 2012 at 11:45 pm

This step is tricky. Note that the element $\sum_{i=1}^n g_j^{-1}g_ig_j$ is the sum over all the elements $g_i$ in the group after they have been conjugated by $g_j$. But conjugation by $g_j$ merely permutes the elements of the group, so this sum is precisely $\sum_{i=1}^n g_i$. No commutativity is required.