Orthonormal basis in group ring necessarily characters?

Question

Let $G$ be a finite abelian group and $C(G,\mathbb{C})$ the group ring on $G$ with multiplication being the convolution $\ast$ and unit $\delta_0$. Let $B \subseteq C(G,\mathbb{C})$ be a $|G|$-element subset of orthogonal idempotents with respect to $\ast$, s.t. $\sum_{b \in B} b =\delta_0 $. Is $B = \widehat{G}$, necessarily?

Answer 1

You know that $C(G,\mathbb C)$ is isomorphic to $\mathbb C\times\cdots\times\mathbb C$ as an algebra, so you can translate your question to:

is a complete set of orthogonal idempotents in $\mathbb C\times\cdots\times\mathbb C$ necesarily the standard basis?

Is it?

Answer 2

The answer is yes. As Mariano Suárez-Álvarez pointed out, it suffices to show that the standard basis is the unique set of orthogonal idempotents with sum equal to $1$ in the algebra $\mathbb{C} \times ... \times \mathbb{C}$. Let $f_i:=Xe_i$ be another basis. Then the orthogonal idempotent condition translates to $X_{ki}X_{kj}=\delta_{ij}X_{ki}$, thus the entries of $X$ are either $0$ or $1$. The sum equal to $1$ condition translates to $\sum_k\,X_{ki}=1$, which means there is exactly one $1$-entry in every column. Since $X$ must be invertible, there is exactly one $1$-entry in every row. Hence, the $f_i$'s must be a permutation of the $e_i$'s.