I was trying to prove that a normal, abelian subgroup of $G$, $N$ is in the center of $G$ given that $|\operatorname{Aut}(N)|$ and $|G/N|$ are relatively prime.
The official question:
Let $N$ be an abelian normal subgroup of a finite group $G$. Assume that the orders $|G/N|$ and $|\operatorname{Aut}(N)|$ are relatively prime. Prove that $N$ is contained in the center of $G$.
My approach:
I considered the homomorphism: $f:G \to \operatorname{Aut}(N)$ such that $f(g)(n) = gng^{-1}$. This was given as a hint on the problem. So naturally, the first this I did was compute the kernel, and found the kernel to be $N$ since any element in $N$ maps to the identity automorphism since $N$ is abelian.
But I don't know how to use the fact that $\gcd(|G/N|, |\operatorname{Aut}(N)|) = 1$ since it is not necessarily true that $G/N$ and $\operatorname{Aut}(N)$ each have prime order.
Am I missing some trivial property about the kernel of this homomorphism that would enable me to say something about $N$ and or its relationship with $Z(G)$?
Any help would be thoroughly appreciated.
By the first isomorphism theorem, $Im(f)\cong G/Ker(f)$. Now $N<Ker(f)$ since $N$ is abelian, so $|G/Ker(f)|=|Im(f)|$ is relatively prime to $|Aut(N)|$. But $Im(f)<Aut(N)$, so $|Im(f)|$ divides $|Aut(N)|$. Therefore the only possibility is that $Im(f)=\{e\}$, so that $Ker(f)=G$, i.e. $N$ is in the center.