Group algebra

Remarks

With this product $F[G]$ is an associative ring with identity $1_G$, and an algebra over $F$ of dimension $\abs{G}$. It is commutative if and only if $G$ is abelian. Example: for $G = \set{1, t}$ cyclic of order $2$ and $F = \Q$, the elements $e = \tfrac{1}{2}(1 + t)$ and $f = \tfrac{1}{2}(1 - t)$ satisfy $e^2 = e$, $f^2 = f$, $ef = 0$ and $e + f = 1$, since $t^2 = 1$. So $\Q[G] \cong \Q \times \Q$ as rings. Non-example: for infinite $G$, the set of all functions $G \to F$ is not the group algebra, since it fails the finiteness condition. For $G = \Z$ the product formula then breaks down, as the inner sum $\sum_{gh = k} a_g b_h$ can be an infinite sum. A representation of $G$ on $V$ is the same as a left $F[G]$-module structure on $V$. For a commutative ring $R$, the same definition gives the group ring $R[G]$.