I'm studying for a qualifying exam in algebra and I've come across the following problem:
Let $G$ be a finite group with a subgroup $N$. Let $Aut(G)$ be the group of automorphisms of $G$. Prove that if $|Aut(G)|$ and $|N|$ are relatively prime, then $N$ is contained in the center of $G$.
I'm struggling to find a good way to approach this. I'm able to prove a similar result that if $|G|$ and $|Aut(G)|$ are relatively prime, then $G$ is abelian:
Let $\theta$ be a homomorphism from $G$ to the inner automorphism group of $G$, denoted $\theta:G\rightarrow Inn(G)$. Then because inner automorphisms are conjugations by fixed elements, the kernel of $\theta$ is the center of $G$, denoted $ker(\theta)=Z(G)$. By the first isomorphism theorem, it follows that $G/Z(G)\cong Inn(G)\subseteq Aut(G)$. Hence, by Lagrange's theorem, $|G/Z(G)|$ divides $|Aut(G)|$. Similarly, by consequence of Lagrange's theorem, $|G/Z(G)||Z|=|G|$, as $Z(G)$ is a normal subgroup of $G$. Since $|G/Z(G)|$ divides $|G|$ and $|Aut(G)|$, and $|G|$ and $|Aut(G)|$ are relatively prime, it follows that $|G/Z(G)|=1$, implying that $G=Z(G)$. The desired result follows. $\blacksquare$
It's not obvious to me how to transform this proof into one of the desired problem (or if there even is a good way to simply modify what I already have). If I knew that $N$ were a normal subgroup, I could potentially apply the same kind of argument using Lagrange's theorem, but I don't have that assumption. Without more information on the structure of $G$, it doesn't seem likely that I'll be able to use an element argument to show that $N\subseteq Z(G)$.
Hint:
First, note that $|N|$ is coprime to $|\!\operatorname{Aut}(G)|$ if and only if the order of every $n\in N$ is coprime to $|\!\operatorname{Aut}(G)|$.
Second, let $\varphi_n$ be the inner automorphism corresponding to any $n\in N$. What happens if $\varphi_n$ has nontrivial order? What does it mean if $\varphi_n$ has trivial order?