Answer to Question #23477 in Abstract Algebra for Tsit Lam
Deduce that, if kH is J-semisimple for any finitely generated subgroup H of G, then kG itself is J-semisimple.
Assume rad kH = 0 for anyfinitely generated subgroup H ⊆ G. For any α∈ rad kG, we have α ∈ kH for somesuch H, so α ∈ kH ∩ rad kG ⊆ rad kH = 0. Thisimplies that rad kG = 0, as desired.
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