**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.29**

Solution:

We begin with some lemmas.

Lemma 1: Let $\pi : G \rightarrow H$ be a group homomorphism, and let $R$ be a ring. Then the mapping $\varphi : R[G] \rightarrow R[H]$ given by $\varphi(\sum r_i g_i) = \sum r_i \pi(g_i)$ is a ring homomorphism. Moreover, if $\pi$ is surjective, $\varphi$ is surjective. Finally, $$\mathsf{ker}\ \varphi = \mathsf{Aug}(R[H]) R[G] = R[G] \mathsf{Aug}(R[H]).$$ Proof: It is clear that if $\pi$ is surjective then $\varphi$ is surjective. Now If $a = \sum r_i g_i$ and $b = \sum s_i g_i$ are in $R[G]$, we have \begin{align*}\varphi(a+b) =&\ \varphi((\sum r_ig_i)+(\sum s_ig_i)) \\=&\ \varphi(\sum (r_i+s_i)g_i) \\=&\ \sum (r_i+s_i)\pi(g_i) \\=&\ (\sum r_i \pi(g_i)) + (\sum s_i \pi(g_i)) \\=&\ \varphi(a) + \varphi(b).\end{align*} Similarly, \begin{align*}\varphi(ab) =&\ \varphi((\sum r_i g_i)(\sum s_i g_i)) \\=&\ \varphi(\sum_i \sum_j r_is_j g_ig_j) \\=&\ \sum_i \sum_j r_is_j \pi(g_ig_j) \\=&\ \sum_i \sum_j r_is_j \pi(g_i)\pi(g_j) \\=&\ (\sum r_i \pi(g_i))(\sum s_i\pi(g_i)) \\=&\ \varphi(a)\varphi(b).\end{align*} Thus $\varphi$ is a ring homomorphism. Now we show that $\mathsf{ker}\ \varphi = \mathsf{Aug}(R[H]) R[G]$; the other equality is similar since $H$ is normal.

($\subseteq$) Note that we can write every element of $G$ in the form $g = hk$ where $h \in H$; write $H = \{h_i\}$ and let $K = \{k_j\}$ be a set of coset representatives of $G/H$, and let $g_{i,j} = h_ik_j$. Suppose $\sum r_{i,j}g_{i,j} \in \mathsf{ker}\ \varphi$; then we have $\sum_{j} (\sum_{i} r_{i,j}) k_jH = 0$; comparing coefficients, we see that $\sum_{i} r_{i,j} = 0$ for each $j$. Then $$\sum_{i,j} r_{i,j} g_{i,j} = \sum r_{i,j} h_i k_j \sum_j (\sum_i r_{i,j} h_i) k_j \in \mathsf{Aug}(R[H]) R[G].$$ ($\supseteq$) Note that if $\sum r_i h_i \in \mathsf{Aug}(R[H])$, then $$\varphi(\sum r_i h_i) = \sum r_i \cdot 1 = (\sum r_i) \cdot 1 = 0.$$ Clearly then this ideal is contained in the kernel. $\square$

Lemma 2: Let $\varphi : R \rightarrow S$ be a surjective ring homomorphism and let $A \subseteq R$. Then $\varphi[(A)] \subseteq (\varphi[A])$.

Proof: Elements of $\varphi[(A)]$ have the form $$\varphi(\sum r_ia_is_i) = \sum \varphi(r_i)\varphi(a_i)\varphi(s_i)$$ where $r_i,s_i \in R$ and $a_i \in A$. $\square$

Lemma 3: Let $R$ be a ring and let $I,J \subseteq R$ be ideals. If $I \subseteq Z(R)$, then $IJ = JI$ and $(IJ)^n = I^nJ^n$.

Proof: Arbitrary elements of $IJ$ have the form $\sum a_ib_i$ where $a_i \in I$ and $b_i \in J$; since $I \subseteq Z(R)$, we have $$\sum a_ib_i = \sum b_ia_i \in JI.$$ Similarly, $JI \subseteq IJ$. Thus $IJ = JI$. The second conclusion then follows. $\square$

Now we move to the main result.

Since $G$ is a finite p-group, $|G| = p^n$ for some $n \geq 1$. We proceed by induction on $n$. We will also provide a bound on the exponent $m$ such that $\mathsf{Aug}(R)^m = 0$; namely, $p^n$.

For the base case, let $|G| = p$. Then $G = \langle \alpha \rangle \cong Z_p$ is cyclic of order $p$. Moreover, $R$ has characteristic $p$ and is commutative. By Exercise 7.4.2, $\mathsf{Aug}(R)$ is generated by $\alpha - 1$. By Exercise 7.3.26, we have $$(\alpha - 1)^p = \alpha^p - 1^p = 1-1 = 0.$$ Thus $\mathsf{Aug}(R)$ is nilpotent.

For the inductive step, suppose that for some $m \geq 1$, for all groups $H$ of order $p^m$, the augmentation ideal of $(\mathbb{Z}/(p))[H]$ is a nilpotent ideal. Let $G$ be a group of order $p^{m+1}$. Since $G$ is a $p$-group, it has a nontrivial center. By Cauchy’s Theorem, let $x \in Z(G)$ have order $p$. Now $H = \langle x \rangle$ is normal in $G$. As in the lemma above, let $\varphi : (\mathbb{Z}/(p))[G] \rightarrow (\mathbb{Z}/(p))[G/H]$ be the ring homomorphism obtained by letting $\pi$ be the natural projection $G \rightarrow G/H$.

Note that because the augmentation ideal of a group ring is generated by $\{g_i-1\}$ and $\pi$ is surjective, we have $$\varphi[\mathsf{Aug}((\mathbb{Z}/(p))[G])] = \mathsf{Aug}((\mathbb{Z}/(p))[G/H])$$ By the induction hypothesis, the augmentation ideal of $R[G/H]$ is nilpotent of exponent at most $p^m$. Thus we have $$\mathsf{Aug}((\mathbb{Z}/(p))[G])^{p^n} \subseteq \mathsf{ker}\ \varphi = \mathsf{Aug}((\mathbb{Z}/(p))[H]) (\mathbb{Z}/(p))[R],$$ so that by the base case, $$\mathsf{Aug}((\mathbb{Z}/(p))[G])^{p^{n+1}} \subseteq (\mathsf{Aug}((\mathbb{Z}/(p))[H]) (\mathbb{Z}/(p))[R])^p = \mathsf{Aug}((\mathbb{Z}/(p))[H])^p (\mathbb{Z}/(p))[R]^p = 0.$$ Thus the augmentation ideal of $R[G]$ is nilpotent with exponent at most $p+1$.