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 ? 




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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 $G$ ? 




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**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". 



Nilpotency degree of the augmentation ideal 
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 
$$ \sum i (p-1) \dim\_{\mathbb{F}\_p} (\Gamma\_i/\Gamma\_{i+1}) $$


