If [H,N]≠1 then H∩N≠1 and N≤H . Now NOp′(H) ,so that N∩Op′(H)=1. Hence N is a p-group. Now can we say CH(N)≥Op′(H) and H/CH(N)is p-group?

Question

If $G$ is a group. Let $H$ be a normal $p$-nilpotent subgroup of $G$ and let $N$ be a minimal normal subgroup of $G$ whose order is divisible by $p$. If $[H,N]\neq 1$ then $H \cap N \neq 1$ and $ N \leq H $ . Now $ N \nleqslant O_{p'}(H)$ ,so that $ N \cap O_{p'}(H)=1$. Hence $N$ is a $p$-group. Now can we say $C_{H}(N) \geq O_{p'}(H)$ and $ H/C_{H}(N) $is $p$-group? Why ?

Answer 1

Since $N,O_{p'}(H) \unlhd H$, $[N,O_{p'}(H)] \le N \cap O_{p'}(H)=1$, so $O_{p'}(H) \le C_H(N)$.

You are assuming that $H/O_{p'}(H)$ is a $p$-group and hence so is $H/C_H(N)$.