Action on G via Automorphism Here is an exercise from Isaacs, Finite Group Theory, $4D.1$: 
Let $A$ act on $G$ via automorphism, and assume that $N \trianglelefteq G$ admits $A$ and that $N \geq C_G(N)$. Assume that $(|A|,|N|)=1$. If $A$ acts trivially on $N$, show that its action on $G$ is trivial.
Hint: Show that $[G,A]\leq N$ and consider $C_\Gamma(N)$ where $\Gamma=G\rtimes A$
I couldn't get the way how i can use the hint

I can only prove the conclusion is true if the assumption (|A|,|N|)=1 become to (|A|,|G|)=1.
Clearly, [A, N, G]=1, and [N,G,A]=1. By three subgroups lemma, [G,A,N]=1, which implies that $[G,A]\le C_G(N) \le N$. By the condition (|A|,|G|)=1, we get $G=C_G(A)[G,A]\le C_G(A)N=C_G(A)$, which implies $[G,A]=1$.
