(b) Let G={g1,...,gn} be a group and assume R is commutative. Prove r∈Aug(RG)⇒r(g1+...+gn)=0.
Proof: (a) (⊆) By the preceding exercise, Aug(FpG) is generated by {g−1 | g∈G}, so for arbitrary x∈Aug(FpG) write x=∑g∈Gag(g−1) for some ag∈FpG for all g∈G. Prove (g−1)pn=0 and is thus nilpotent for any g∈G, so that by the idealness of N(FpG) we have x is nilpotent. By the binomial theorem for commutative rings, since pn divides (pnk)=pn!k!(n−k)! when 0<k<pn, we can observe (g−1)pn=gpn−1=1−1=0. (⊇) Since FpG/Aug(FpG)≅Fp is a field, we have Aug(FpG) is a maximal ideal, so that N(FpG)∈{Aug(FpG),FpG}, and is evidently not the latter as 1∈FpG is not nilpotent. ◻
(b) Once again write r=r′(g−1) for some r′∈RG, and recalling that the action of multiplication on a group by one of its elements is a permutation of that group's elements, we observe r(g1+...+gn)=r′(g−1)(g1+...+gn)=r′(g(g1+...+gn)−(g1+...+gn))=r′((g1+...+gn)−(g1+...+gn))=r′0=0 ◻
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