# Answer to Question #23477 in Abstract Algebra for Tsit Lam

Question #23477

Deduce that, if kH is J-semisimple for any finitely generated subgroup H of G, then kG itself is J-semisimple.

Expert's answer

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.Need a fast expert's response?

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