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Let $p$ be a prime number, $G=:G_0$ a (non-Abelian) finite p-Group and $K$ a finite field with $\operatorname{char}(K)=p$. It is well-known that the group $G_1:=1+\operatorname{rad}(KG)$ is a p-group containing $G$. We can construct now $G_2:=1+\operatorname{rad}(KG_1)$ and so on. By this way can construct a sequence of p-groups with growing quantities.

My question is, whether the exponent of these groups resp. of the centers of these groups is bounded. As theses sequences are growing the bounding is equivalent to the fact that these exponents are getting constant after finite many steps.

I have no conjecture because I was not able to control the exponents at least for ($p=2$ and $G=Q_8$ or $G=D_8$) or (a p-group of order $p^3$).

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  • $\begingroup$ If I'm not mistaken this is the same question as bounding the nilpotency index of the augmentation ideal, right ? $\endgroup$ – Max Apr 17 at 20:59
  • $\begingroup$ No, if you consider the sequence of growing p-groups I want to know whether the exponent of these groups resp. the exponent of the centers of these groups is bounded. $\endgroup$ – Sven Wirsing Apr 18 at 15:25
  • $\begingroup$ Yes of course, but the exponent of $G_2$ for instance is the same as the nilpotency index of the augmentation ideal of $KG_1$ $\endgroup$ – Max Apr 18 at 15:38
  • $\begingroup$ Hi Max, can you quote this result, please or present the argumenation for this? KR Sven $\endgroup$ – Sven Wirsing Apr 18 at 19:16
  • $\begingroup$ Well $ I(G) \subset KG $ is nilpotent (with nilpotency index a power of $ p $ ) when the characteristic is $ p $ and $ G $ is a $ p $ - group. If $ x \in I(G) $ and $ x^{p^k}=0 $, then $ (1+x)^{p^l} = 1+x^{p^l}$ so it's $ 1 $ if and only if $ k \leq l $ $\endgroup$ – Max Apr 18 at 19:21

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