Proving that a normal, abelian subgroup of G is in the center of G if |G/N| and |Aut(N)| are relatively prime.
Solution 1:
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.