I can't understand so much the second paragraph (page 160) of the proof of the lemma 31.17(1) (pages 160) in M. Aschbacher, Finite Group Theory about generalized fitting subgroup. Here I post the statement of the lemma and the proof given in the book.
(31.17) Let $O_{p'}(G)=1$ and $P$ a $p$-subgroup of $G$ where $p$ is a prime. Then
(1) $O_{p',E}(N_G(P))$ fixes each component of $G$.
Proof Let $H=N_G(P)$, $X=O_{p'}(H)$, $H^*=H/X$. Let $K\leq X$ or $X\leq H$ with $K^*\in Comp(H^*)$, and subject to these constraints pick $K$ minimal subject to moving a component of $G$. Let $P\leq P_0\in \mathrm{Syl}_p(C_G(K))$. As $K$ satisfies the same hypothesis with respect to $P_0$, we may take $P = P_0$. In particular, by 31.14, $O_p(G) \leq P$. Let $R\in \mathrm{Syl}_p(H \cap E(G))$.
Suppose first $K\leq X$. Then $K$ is a $q$-group for some prime $q$, and by coprime action, 18.7, there exists an $R$-invariant Sylow $q$-group $Q$ of $O_{p'}(H)$ containing $K$. By 24.4, $Q = [R, Q]C_Q(R)$. Now $[R, Q] \leq [E(G), Q] \leq E(G)$, so $[R,Q]$ fixes each component of $G$. Hence we may take $[K, R] = 1$. Thus, by choice of $P,~R\leq P$. So $P\cap E(H)\in \mathrm{Syl}_p(H)$ by Exercise 3.2. As $O_{p'}(G) = 1$, $p\in\pi(L)$ for each $L\in Comp(G)$, so $1\neq P \cap L\in \mathrm{Syl}_p(L)$ by 6.4. Then $P\cap L\nleq Z(L)$, so $L= [E(G), P\cap L]$ is $K$-invariant. $\cdots\cdots\cdots$
From my perspective, I guess $K$ is $R$-invariant. But why?
Why $P\cap E(H)\in \mathrm{Syl}_p(H)$?
Why pick such $R$?
Here some definitions.
The components of a group $G$ are its subnormal quasisimple subgroups. $Comp(G)$ is the set of components of $G$. And $E(G)=\langle Comp(G)\rangle$, $F^*(G)=F(G)E(G)$.
For a set of primes $\pi$. $O_{\pi}(G)$ is the maximal normal $\pi$-subgroup of $G$, $O^{\pi}(G)$ is the minimal normal subgroup of $G$ such that the quotient group is a $\pi$-group. And $O_{p',E}(G)/O_{p'}(G)=E(G/O_{p'}(G))$.
(31.14) Let $P$ be a $p$-subgroup of $G$. Then
(1) $O_{p',E}(G)\leq C_G(O_p(G))$, and
(2) If $P\leq O_p(G)$ then $O^p(F^*(N_G(P))) = O^p(F^*(G))$.
Exercise 3.2 Let $P$ be a $p$-subgroup of $G$. Then $P\in\mathrm{Syl}_p(G)$ or $P<P_0\in \mathrm{Syl}_p(N_G(P))$.
I agree that this is hard to read!
I am not sure why $K$ is $R$-invariant, but the $R$-invariant Sylow $q$-subgroup $Q$ contains a conjugate of $K$, so we can just replace $K$ by that conjugate to get $K \le Q$.
I can't make much sense of $P \cap E(H) \le {\rm Syl}_p(H)$, and I suspect that was not exactly what he meant, so let's ignore that and carry on reading.
Since $R \le P$, we have $R = P \cap E(G)$. Then we must have $P \in {\rm Syl}_p(PE(G)$, since otherwise $N_G(P) \cap E(G)$ would strictly contain $R$, contradicting $R \in {\rm Syl}_p(H \cap E(G))$. So $P \cap E(G) \in {\rm Syl}_p(E(G)$ and hence also $P \cap L \in {\rm Syl}_p(L)$. Note also that $P \cap L \le R$ is centralized by $K$, and so it is $K$-invariant.
I everything else OK for that case?