每日定理4

Isaacs, $\textit{Character Theory of Finite Groups}$, Problems(1.9)

Let $G$ be a group and $F$ a field of characteristic $p$. Suppose $p\mid|G|$, then $F[G]$ is not semisimple.

Pf: Consider regular module $F[G]^{\circ}$ and prove by contradiction.

• $0\neq v=\sum_{g\in G}g\in Z(F[G])$
• $F[G]^{\circ}=\langle v\rangle\oplus W$, $\langle v\rangle=\{va|a\in F[G]\}$
• $v=1v=(va+w)v=vav+wv=v^2a+wv=wv\in W$

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来源： https://www.cnblogs.com/zhengtao1992/p/10739828.html