A question about Burnside normal p-complement theorem

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".

Proof:

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?

Thanks!

$$\ker \tau$$ 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.
• @ 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? Nov 28, 2022 at 15:48
• 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 $p$. Nov 28, 2022 at 15:55
• Oh,maybe I misunderstand the definition of $p'$-group. I thought $p'$ was also a prime number. I know it .Thanks ! Nov 28, 2022 at 16:00