When I read the proof of Burnside normal p-complement theorem, I have a question.

The theorem is :

If for some prime $p$ a Sylow $p$-subgroup $P$ of a finite group $G$ lies in the centre of its normalizer, then $G$ is p-nilpotent.

The following proof comes from "A course in the theory of groups".


By hypothesis $P$ is abelian and $P = C_P(N_G(P)$, We deduce at once from $10.1.6$ that $P ~\cap $ $Ker ~\tau$= $e$ where of course
$\tau$: $G \rightarrow P$ is the transfer. This means that Ker $\tau$ is a $p'$-group, which in turn implies that $G$ is $p$-nilpotent since $G/\mathrm{ker} \tau \cong \mathrm{Im} \tau $, a $p$-group.

The theorem $10.1.6$ is :

Let the finite group $G$ have an abelian Sylow $p$-subgroup $P$ and let $N$ denote $N_G(P)$. Then $P = C_P(N) \times [P, N]$. Moreover, if $\tau $: $G \rightarrow P$ is the transfer, Im $ \tau $ = $C_P(N)$ and $P ~\cap $ Ker $\tau$ = [P,N].

My question is how to get ker $\tau$ is a $p'$-group?



1 Answer 1


$\ker \tau$ is a normal subgroup of $G$, so $P \cap \ker \tau$ is a Sylow $p$-subgroup of $\ker \tau$.

But $P \cap \ker \tau = [P,N] = 1$, so $\ker \tau$ is a $p'$-group.

  • $\begingroup$ @ Derek Holt If I suppose $|G|=p^{l}p_1^{l_1}\cdots p_k^{l_k}$,I can get |ker $\tau$|=$p_1^{l_1}\cdots p_k^{l_k}$.I wonder why $k=1$?Am I thinking wrong? $\endgroup$
    – fusheng
    Nov 28, 2022 at 15:48
  • $\begingroup$ I don't understand your question. We don't necessarily have $k=1$, and nobody claimed that. A (finite) $p'$-group is defined to be a group whose order is not divisible by $p$. $\endgroup$
    – Derek Holt
    Nov 28, 2022 at 15:55
  • $\begingroup$ Oh,maybe I misunderstand the definition of $p'$-group. I thought $p'$ was also a prime number. I know it .Thanks ! $\endgroup$
    – fusheng
    Nov 28, 2022 at 16:00

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