Deduce that, if kH is J-semisimple for any finitely generated subgroup H of G, then kG itself is J-semisimple.
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Expert's answer
2013-02-05T08:19:46-0500
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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