# Answer to Question #23713 in Abstract Algebra for jeremy

Question #23713

For any field k and for any normal subgroup H of a group G, show that kH ∩ rad kG = rad kH.

Expert's answer

Even without

*H*being normalin*G*, we have in general*kH ∩*rad*kG**⊆**rad**kH*. Therefore, it suffices to prove that, if*H*isnormal in*G*, then rad*kH**⊆**rad**kG*. Consider any simpleleft*kG*-module*V*. By Clifford’s Theorem,*kHV*is asemisimple*kH*-module. Therefore, (rad*kH*)*V*= 0. Thisimplies that rad*kH**⊆**rad**kG*.
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