I am trying to solve the problem $3E.1$ from Martin Isaac's book "Finite group theory".
Let $A$ act on $G$ via automorphism, and assume that at least one of $A$ or $G$ solvable, but do not assume that $A$ and $G$ have coprime orders. If $G$ nontrivial, prove that $G$ contains a nontrivial $A$-invariant $p$-subgroup for some prime $p$.
It is proposed to start with the proof of this statement:
If $P$ $\triangleleft$ $A$ is a $p$-subgroup and $C_{G}(P) = 1$, show that $p$ does not divide $|G|$. Show in this situation that for each prime $q$, there is an $A$-invariant Sylow $q$-subgroup in $G$.
I have an idea how to start to prove this exercise:
$\textbf{3.23 Theorem}.$ Let $A$ act via automorphism on $G$, where $A$ and $G$ are finite groups, and suppose $(|A|, |G|) = 1$. Assume also that at least one of $A$ or $G$ is solvable. Then for each prime $p$ there is exists an $A$-invariant Sylow $p$-subgroup of $G$. Also if $S$ and $T$ are $A$-invariant Sylow $p$-subgroups of $G$, then $S^{c} = T$ for some element $c \in C_{G}(A)$.
If we assume previous statement to be true, then by Theorem 3.23, we can understand that for any prime $q$ there is exists the unique $P$-invariant Sylow $q$-subgroup in $G$.
However, I don't understand why then there would be an A-invariant Sylow subgroup in this case.
In addition, I don't understand why it's necessary to consider this particular case to prove this exercise.
Use the following result, which is probably in Isaacs' book as a lemma or an exercise.
So you have seen that if $P \trianglelefteq A$ is a $p$-subgroup with $C_G(P) = 1$, then there is a unique $P$-invariant $q$-Sylow in $G$ for any prime $q \neq p$. Hence by (*), there is an $A$-invariant $q$-Sylow.
The other possibility is that $C_G(P) \neq 1$. By (*) the group $A$ acts on $C_G(P)$, so you are done by induction if $C_G(P) \neq G$.
That leaves you with the case where for any normal $p$-subgroup $P \trianglelefteq A$, we have $C_G(P) = G$. I think you should be able to finish it from here.