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Let $p$ be a prime and let $G$ be a finite group of order a power of $p$ (i.e., a $p$-group).

Prove that the augmentation ideal in the group ring $\mathbb{Z}/p\mathbb{Z}G$ (to be read as $\left( \mathbb{Z}/p\mathbb{Z} \right) G$) is a nilpotent ideal. (Note that this ring may be noncommutative.)

Let $I_G$ be the augmentation ideal of the group ring $\mathbb{Z}/p\mathbb{Z}G$, i.e. $I_G$ consists of formal linear combinations $\sum_i n_i g_i$ ($n_i\in \mathbb{Z}/p\mathbb{Z}$, $g_i\in G$) such that $\sum_i n_i=0$. I cannot show that there is $m \in \mathbb{Z}_+$ such that $I_G^m=0$.

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  • $\begingroup$ Maybe, it helps that the group $G$ is nilpotent. $\endgroup$ – Peter Dec 9 '15 at 10:19
  • $\begingroup$ Just answered your question $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Dec 16 '15 at 16:09
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Lemma 1. Let $G$ be a group, $H$ be a normal subgroup, $\pi : G\to G/H$ the canonical morphism, $R$ be a commutative ring with unit, and $\varphi : R[G] \to R[G/H]$ the induced morphism of (perhaps non-commutative) rings induced by $\pi$. The $\varphi$ is surjective and $$\operatorname{Ker}(\varphi)=\operatorname{Aug}(R[H])R[G] = R[G] \operatorname{Aug}(R[H]).$$ Here, $\operatorname{Aug}(R[K])$ denotes the augmentation ideal of a group ring $R[K]$.

Proof of lemma 1. The surjectivity of $\varphi$ is obvious, due to the definition of $\varphi$ and the surjectivity of $\pi$. The second equality is obvious as $H$ is a normal subgroup of $G$. We show now the first equality. Write $H = \{h_i\}$ for distinct $h_i$'s, and let $K = \{k_j\}$ (with distinct $k_j$'s) be a set of coset representatives of $G/H$. Set $g_{i,j} := h_i k_j$. Obviously $G = \{g_{i,j}\}$. Now take a $\zeta\in\operatorname{Ker}(\varphi)$ and write, by what we just said, $\zeta = \sum r_{i,j} g_{i,j}$ with $r_{i,j} \in R$. Then $\sum_j \left(\sum_i r_{i,j}\right)k_j H = \varphi(\zeta) = 0$ and comparing coefficients shows that $\sum_i r_{i,j} = 0$ for all $j$. Therefore $\zeta = \sum r_{i,j} g_{i,j} = \sum_j \left(\sum_i r_{i,j}\right)k_j$ is in $\operatorname{Aug}(R[H])R[G]$. The reverse inclusion is obvious, as you remark that if $\sum_i r_i h_i \in\operatorname{Aug}(R[H])$ then it is mapped to $0$ by $\varphi$. $\square$

Lemma 2. If $\varphi : R\to S$ is a surjective morphism of rings and if $A\subseteq R$ then $\varphi[(A)] \subseteq (\varphi(A))$. Here, $(A)$ denotes the ideal of $R$ generated by $A$.

Proof of lemma 2. Obvious. $\square$

Lemma 3. Let $R$ be a ring, $I,J$ be additive subgroups of $R$, and suppose $I$ to be central, that is, included in $R$'s center. Then $IJ=JI$ and $(IJ)^n = I^n J^n$.

Proof of lemma 3. Obvious. $\square$

Theorem. Let $p$ be a prime number and $G$ be a finite $p$-group. Let $R$ be a commutative ring with unit such that $p \cdot 1_R = 0_R$. The augmentation ideal of the group ring $R[G]$ is nilpotent.

Proof of the theorem. As $G$ is a $p$-group we have $|G| = p^n$ for some $n\in\mathbf{N}$. We proceed by induction on $n$.

The case $n=0$ is obvious. Let's check at case $n=1$. Then $G$ is cyclic of order $p$, let's say generated by some $x$, and thus $R[G]$ is commutative of characteristic $p$, and you can see that $\operatorname{Aug}(R[G])$ is generated by $x-1$. As $(x-1)^p = x^p - 1^p = x^p - 1 = 1 - 1 = 0$, we see that $\operatorname{Aug}(R[G])$ is nilpotent indeed. Now we proceed to the inductive step. Suppose that for $n\geq 1$, for all groups $H$ of order $p^n$, the augmentation ideal of $R[H]$ is nilpotent of exponent $\leq p^n$, and let $G$ be a $p$-group of order $p^{n+1}$. Since $G$ is a non trivial $p$-group, it has a non-trivial center $Z$ (classic result on $p$-groups...). By Cauchy's theorem we can find an $x\in Z$ of order $p$. Let $H$ be the subgroup of $G$ generated by $x$. As $x\in Z$ the subgroup $H$ is normal. As in lemma 1 let $\varphi : R[G] \to R[G/H]$ be the ring morphism induced by the projection map $\pi : G\to G/H$. Now as the augmentation ideal of a group ring is generated by the $g-1$'s and as $\pi$ is surjective, we have $\varphi\left(\operatorname{Aug}\left(R[G]\right)\right) = \operatorname{Aug}\left(R[G/H]\right)$. By the induction hypothesis the ideal $\operatorname{Aug}\left(R[G/H]\right)$ is nilpotent of exponent $\leq p^n$, so that its $p^n$-th power is zero, so that $\varphi\left(\operatorname{Aug}\left(R[G]\right)\right)^{p^n} = 0$. As $\varphi$ is a ring morphism, this entails $\varphi\left(\operatorname{Aug}\left(R[G]\right)^{p^n}\right) = 0$, that is, $\operatorname{Aug}\left(R[G]\right)^{p^n}\subseteq \operatorname{Ker}(\varphi)$. But the ideal $\operatorname{Ker}(\varphi)$ is $\operatorname{Aug}(R[H]) R[G]$, and $\operatorname{Aug}(R[H])$ is central. By the base case and by Lemma 3, we thus have $\left(\operatorname{Ker}(\varphi)\right)^p = \left(\operatorname{Aug}(R[H])\right)^p \left( R[G]\right)^p = 0 \left(R[G]\right)^p =0$. We find finally that $\operatorname{Aug}\left(R[G]\right)^{p^{n+1}} = 0$. $\square$

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  • $\begingroup$ Nice proof! I have taken the liberty to generalize your theorem so that the $\mathbf{Z} / p\mathbf{Z}$'s get shortened to $R$'s. $\endgroup$ – darij grinberg Dec 16 '15 at 23:03
  • $\begingroup$ @darijgrinberg fine by me ;-) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Dec 17 '15 at 18:36
  • $\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj please help me to solve this problem math.stackexchange.com/questions/3025927/… ...thank you very much $\endgroup$ – neelkanth Dec 4 '18 at 18:30

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