Suppose $G = \langle g \rangle$ is a cyclic group of order $n$, $k$ is an algebraically closed field, and $w \in k$ a primitive $n$th root of unity.
I want to show that in the group algebra $k[G]$, the elements
$$e_1 = \frac{1}{n}\sum_{k=1}^n g^k \quad \text{and} \quad e_j = \frac{1}{n}\sum_{k=1}^{n} (w^{j} g)^k \quad (j = 2, …, n)$$ satisfy $e_i \cdot e_j = \delta_{ij} e_i$ and $\sum_i e_i = \mathbb{1}_{k[G]}$ in $k[G]$.
The only thing I have been able to prove is $e_1^2 = e_1$, but I don't know what to do for the rest.
Best Answer
For notational convenience I will write $$ e_i = \sum_{r=0}^{n-1} (\omega^ig)^r $$ So that "my" $\{e_0,\ldots,e_{n-1}\}$ is "your" $\{e_1,\ldots,e_n\}$.
We can compute $e_ie_j$ by looking at the coefficient of each $g^k$, which is $$ n^{-2}\sum_{r+s=k} \omega^{ri+sj} = n^{-2}\sum_{r+s=k}\omega^{ri+si}\omega^{sj-si}=n^{-2}\omega^{ik}\sum_{s=0}^{n-1}(\omega^{j-i})^s$$ Since $\omega^{j-i}$ is an $n$th root of unity, this last sum is $0$ when $j\neq i$ and is $n$ when $j=i$.
Similarly we can compute $\sum e_i$. The coefficient of $g^k$ is then $$ n^{-1}\sum_{t=0}^{n-1} \omega^{tk} = n^{-1}\sum_{t=0}^{n-1}(\omega^k)^t $$ This last sum is $0$ when $k\neq0$ and $n$ when $k=0$.