Let H be a normal subgroup of G. If H is finite, show that I is also nilpotent.
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Expert's answer
2013-02-05T08:13:23-0500
Any σ ∈ G defines a conjugation automorphism on thesubring kH ⊆ kG, and this automorphism must takeradkH to rad kH. Therefore, (rad kH)σ ⊆ σ · rad kH ⊆ I, which shows that I is anideal of kG. This method also shows that In= kG · (radkH)n for any n ≥ 1, so if H is finite,then I is nilpotent..
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