TITLE: An exercise about p-solvable
QUESTION [6 upvotes]: I'm dealing with a problem about p-solvable in Isaac's finite group theory book. Question is the following: "Let $G$ be $p-$solvable and $P \in Sy{l_p}\left( G \right)$ and $K \le G$ such that $p$ doesn't divide $\left| K \right|$. Suppose that $P \le {N_G}\left( K \right)$. Show that $K \le {O_{p'}}\left( G \right)$." There is a hint.
Hint: Firstly, consider the case ${O_{p'}}\left( G \right) = 1$. For general case, consider the quotient group $G$/${O_{p'}}\left( G \right) $. I hope some of you can answer me.

REPLY [3 votes]: It's probably easiest to work by induction on $|G|$. As in the hint, if $O_{p'}(G) \ne 1$, then the result for $G$ follows immediately from the result for $G/O_{p'}(G)$, so we can assume that $O_{p'}(G)=1$.
So $O_p(G) \ne 1$, and then $O_p(G) \le P \le N_G(K)$ implies that $[K,O_{p
}(G)] \le K \cap O_{p}(G) = 1$, so $K \le C_G(O_{p}(G))$. Now, since $O_{p}(N) \le O_{p}(G)$ for any normal subgroup $N$ of $G$, we can assume that $G = C_G(O_{p}(G))$ - that is $O_{p}(G) \le Z(G)$.
We can clearly assume that $K$ is nontrivial, so $G$ is not a $p$-group, and hence $O_p(G) < O_{p,p'}(G)$, and $O_{p,p'}(G)$ must decompose as a direct product of $O_p(G)$ and a nontrivial $p'$-group, where the latter lies in $O_{p'}(G)$, contradicting $O_{p'}(G)=1$.
That seems a bit contorted, and there may well be a slicker way of doing it!