Answer to Question #23713 in Abstract Algebra for jeremy

Question

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

Answer 1

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.