A group $A$ acts on another group $G$ if every $a \in A$ corresponds to some automorphism of $G$, i.e., we have $$ g^1 = g, \qquad g^{ab} = (g^a)^b \quad\mbox{and}\quad (gh)^a = g^a h^a $$ for $g,h \in G$, $a,b \in A$ and $1 \in A$ denoting the identity element in $A$. An action is called coprime if the orders of $A$ and $G$ are coprime. Let $C_A(G) = \{ a \in A : g^a = g \mbox{ for all } g \in G \}$, the kernel of the action, we say that $A$ acts faithful if $C_A(G) = 1$.
In Kurzweil/Stellmacher The Theory of finite groups in chapter 8 about coprime action, they state as theorem 8.2.10
Let $G$ be a $p$-group and $\mathcal K$ the set of all $A$-composition factors of $G$. Suppose that the action of $A$ on $G$ is coprime. Then $$ \bigcap_{K\in \mathcal K} C_A(K) / C_A(G) = O_p(A / C_A(G)). $$
But by coprime action I guess we have $O_p(A) = 1$, hence the statement is the same as $$ \bigcap_{K\in \mathcal K} C_A(K) = C_G(A) $$ or for a faithful action $\bigcap_{K\in \mathcal K} C_A(K) = 1$. Am I right, or am I missing something? Just wondering why it is stated in that ways that seems to complicated...