Nilpotency and the Augmentation Ideal (7.4.3)

Dummit and Foote Abstract Algebra, section 7.4, exercise 3:

(a) Let $G$ be an abelian group of order $p^n$. Prove $Aug(\mathbb{F}_pG)=\mathfrak{N}(\mathbb{F}_pG)$.
(b) Let $G=\{g_1,...,g_n\}$ be a group and assume $R$ is commutative. Prove $r∈Aug(RG)⇒r(g_1+...+g_n)=0$.

Proof: (a) ($\subseteq$) By the preceding exercise, $Aug(\mathbb{F}_pG)$ is generated by $\{g-1~|~g∈G\}$, so for arbitrary $x∈Aug(\mathbb{F}_pG)$ write $x=\sum_{g∈G}a_g(g-1)$ for some $a_g∈\mathbb{F}_pG$ for all $g∈G$. Prove $(g-1)^{p^n}=0$ and is thus nilpotent for any $g∈G$, so that by the idealness of $\mathfrak{N}(\mathbb{F}_pG)$ we have $x$ is nilpotent. By the binomial theorem for commutative rings, since $p^n$ divides $p^n \choose k$$ = \dfrac{p^n!}{k!(n-k)!}$ when $0 < k < p^n$, we can observe $(g-1)^{p^n}=g^{p^n}-1=1-1=0$. ($\supseteq$) Since $\mathbb{F}_pG/Aug(\mathbb{F}_pG)≅\mathbb{F}_p$ is a field, we have $Aug(\mathbb{F}_pG)$ is a maximal ideal, so that $\mathfrak{N}(\mathbb{F}_pG)∈\{Aug(\mathbb{F}_pG),\mathbb{F}_pG\}$, and is evidently not the latter as $1∈\mathbb{F}_pG$ is not nilpotent.$~\square$

(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 $$r(g_1+...+g_n)=r'(g-1)(g_1+...+g_n)=r'(g(g_1+...+g_n)-(g_1+...+g_n))=$$$$r'((g_1+...+g_n)-(g_1+...+g_n))=r'0=0~\square$$