Corollary 1 (Maschke)

Suppose that the group G is finite and that K is a commutative field. The algebra K[G] is semisimple if and only if the element $|G|$ of the field K is not zero.

Suppose $|G| \not= 0$. By Theorem 1, every K[G]-submodule of $K[G]_s$ is a direct factor. Hence K[G] is semisimple.

Conversely, suppose that $|G|$ is zero, and denote by $\varepsilon$ the element $\sum_{g\in G}g$ of the center of K[G]. We have $\varepsilon \not= 0$ but $\varepsilon^2=|G|\varepsilon = 0$, so K[G] is not semisimple (VIII, p. 153, Proposition 3, a) and VIII, p. 157, Remark 1).