Let be a conjugacy class in the finite group . Let be a ring with 1.

- Prove that the element is in the center of the group ring . [Hint: Check that for all .]
- Let be the conjugacy classes of , and for each , let be the sum of the elements in (as described in part 1). Prove that an element is in the center if and only if for some elements .

- Let . Note that conjugation by permutes the elements of , so that (as an element of ) we have . Thus for all . Then for all , we see the following.

= = = = = = = = Thus .

- First we show that . Let . Then we have the following.

= = = = = = = = = Thus .

Now let . First, let be arbitrary. By examining each coefficient of , we see that for all . Now recall that acts transitively (by conjugation) on each of its conjugacy classes. If is a conjugacy class of and , then we have for some . The coefficient of in is on one hand and on the other, so that in fact . In fact the coefficient of each is , and we have .

## Comments

In a group ring RG, R is a commutative ring so you need not show that r_i \in Z(R), since we know that r_i \in R = Z(R)

The ring doesn’t have to be commutative.

In the Dummit and Foote context, R was commutative, but in general this does not need to hold.