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Let $k$ be a field of characteristic $p$ and $G$ a finite group. How do you prove that if $kG$ is local then $G$ is a $p$-group? (I know how to prove the converse but not this implication).

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A theorem of Brauer says that the number of simple $kG$-modules (up to isomorphism) is equal to the number of conjugacy classes of $p$-regular elements of $G$ ($g \in G$ being $p$-regular means the order of $g$ is not divisible by $p$).

If $G$ is not a $p$-group, there are at least two such conjugacy classes. Thus there are at least two simple $kG$-modules, hence at least two indecomposable projective $kG$-modules. This in turn means that the regular $kG$-module $_{kG} kG$ is not indecomposable, hence its endomorphism algebra, which is ${kG}^\text{op} \cong kG$ is not local.

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  • $\begingroup$ The hint on the notes I'm reading says to use the fact that the nilradical is the same as the jacobson radical, is there a simpler proof using this hint? $\endgroup$ – user343345 May 30 '16 at 12:44
  • $\begingroup$ tinyurl.com/z8sccj2 has an argument along those lines. $\endgroup$ – Cihan May 30 '16 at 13:51

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