Answer to Question #23713 in Abstract Algebra for jeremy
For any field k and for any normal subgroup H of a group G, show that kH ∩ rad kG = rad kH.
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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