Answer to Question #23480 in Abstract Algebra for sanches
Let H be a normal subgroup of G. If rad kH is nilpotent, show that I is also nilpotent.
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 rad kH isnilpotent, then I is also nilpotent.
No comments. Be first!