Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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".

share|improve this question
2  
I believe Loewy length is the name. –  Benjamin Steinberg Apr 26 '12 at 20:06
    
Yes, the definition of "Loewy length" agrees with the one above. Thanks. –  Ralph Apr 26 '12 at 20:30
add comment

1 Answer

up vote 4 down vote accepted

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 $$ \sum i (p-1) \dim_{\mathbb{F}_p} (\Gamma_i/\Gamma_{i+1}) $$

share|improve this answer
    
Many thanks for advising to Jennings' theorem. –  Ralph Apr 26 '12 at 21:05
    
This is (essentially) theorem 3.6 in Passman's group ring book (just for a "paper" reference). –  Steve D Apr 26 '12 at 23:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.