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Let $n$ be a natural number, $p$ a prime number, $G$ a finite $p$-group and $K$ a finite field with $p^n$ elements. We focus on the group $1+J(KG)$, where $J(KG)$ is the Jacobson radical of $KG$, which is in this case the augmentation ideal of $KG$.

My question is: Is there a connection between the number of conjugacy classes of $G$ (denoted by $c(G)$) and of $J:=1+J(KG)$ (denoted by $c(J)$)?

It is well-known that each conjugacy class of $G$ can be extended uniquely to a conjugacy class of $J:=1+J(KG)$: If $g_1^G,...,g_c^G$ are the conjugacy classes of $G$, then $g_1^J,...,g_c^J$ are distinct conjugacy classes of $J$.

It is also possible to prove that the number of conjugacy classes of $J$ and of $G$ are not identical: $c(G)<c(J)$.

Background of the question: I want to know how the dimension of the center of $KG$ -- which is $c(G)$ -- is related to the dimension of the center of $KJ$ -. which is $c(J)$ --.

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  • $\begingroup$ What is meant by "class number" of a group ($p$-group?) $G$? nilpotency class? $p$-nilpotency class? $\endgroup$
    – YCor
    Commented Sep 15, 2019 at 17:08
  • $\begingroup$ @YCor: I think "number of conjugacy classes" is what is intended here by "class number". $\endgroup$
    – Geoff Robinson
    Commented Sep 15, 2019 at 17:28
  • $\begingroup$ class number = number of conjugacy classes $\endgroup$
    – Sven Wirsing
    Commented Sep 15, 2019 at 17:54
  • $\begingroup$ we should close the topic. $\endgroup$
    – Sven Wirsing
    Commented Dec 16, 2020 at 20:02

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