Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$ and let $I(G)=(g-1\mid g \in G)$ be the augmentation ideal of the group ring $k[G]$. It's known that $I(G)$ is nilpotent, i.e. there is $n> 0$ such that $I(G)^n=0$. Call the least such $n$ the nilpotency degree of $I(G)$ and denote it by $\operatorname{nildeg}I(G)$.

Question 1: Are there known upper and lower bounds for $\operatorname{nildeg}I(G)$ ?

Question 2: Does the invariant $\operatorname{nildeg}I(G)$ have a particular name in the literature ?

An inspection of the proof that $I(G)$ is nilpotent can be used to detemine upper bounds for $\operatorname{nildeg}I(G)$: Let $C \le G$ be central. Then $$\frac{k[G]}{I(C)k[G]} \cong k[G/C]\;,\qquad \frac{I(G)}{I(C)k[G]} \cong I(G/C).$$ By taking $C=\mathbb{Z}/p$ and iterating, one obtains $$\operatorname{nildeg}I(G)\le |G|.$$ By taking $C=Z(G)$ this can be futher refined: If $1 = Z_0 \le Z_1 \le ... \le Z_c = G$ is the upper central series of $G$ and $Z_i/Z_{i-1}=\prod_{j=1}^{r_i}\mathbb{Z}/p^{e_{ij}}$ then $$\operatorname{nildeg}I(G) \le \prod_{i=1}^c\;\big((e_{i,1}-1) + \cdots + (e_{i,r_i}-1)+1\big)$$ Since $(g-1)^{\operatorname{nildeg}I(G)}=0$ for each $g \in G$, a trivial lower bound is $$\operatorname{nildeg}I(G) \ge \operatorname{exp}(G)/p$$ Hence a more acurate formulation for quest 1 is:

Question 3: Are there better bounds than these or bounds that use other invariants of $G$ ?

Edit: Apart from the exact formula given by Jennings' theorem as described in mt's answer, I found the lower bound
$$\operatorname{nildeg}I(G) \ge m(p-1)+1$$ if $|G|=p^m$ in the book "Karpilovsky: The Jacobson Radical of Group Algebras".

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    $\begingroup$ I believe Loewy length is the name. $\endgroup$ – Benjamin Steinberg Apr 26 '12 at 20:06
  • $\begingroup$ Yes, the definition of "Loewy length" agrees with the one above. Thanks. $\endgroup$ – Ralph Apr 26 '12 at 20:30

Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$, and define a series $\Gamma_i$ of subgroups of $G$ by letting $\Gamma_1 = G$ and $$ \Gamma_{i+1} = \langle [ \Gamma_i,G ], \Gamma ^p _{\lceil (i+1)/p \rceil} \rangle .$$ Then $\Gamma_i / \Gamma_{i+1}$ is elementary abelian, so we can fix elements $f_{i1}, \ldots, f_{id_i}$ of $G$ whose images in $\Gamma_i/\Gamma_{i+1}$ form a basis. Consider all products of the form

$$ \prod_{i,j} (f_{ij}-1)^{\alpha_{ij}} \in kG \qquad (1) $$ where the product is taken in lexicographic order and $0 \leqslant \alpha_{ij} \leqslant p-1$. Define the weight of such a product to be $\sum_{i,j} i\alpha_{ij}$.

Jennings' Theorem says that if $J=I(G)=\operatorname{rad}(kG)$ then the set of products (1) of weight at least $s$ form a basis of $J^s$, and a basis for $J^s/J^{s+1}$ is given by the images of the products of weight exactly $s$.

In particular, the largest non-zero power of the radical is $$ \sum i (p-1) \dim_{\mathbb{F}_p} (\Gamma_i/\Gamma_{i+1}) $$

  • $\begingroup$ Many thanks for advising to Jennings' theorem. $\endgroup$ – Ralph Apr 26 '12 at 21:05
  • $\begingroup$ This is (essentially) theorem 3.6 in Passman's group ring book (just for a "paper" reference). $\endgroup$ – Steve D Apr 26 '12 at 23:28

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