Group $N \triangleleft G$, $(|N|,|G/N|) = 1$, then $N$ is a characteristic subgroup of $G$.

Question

$N$ is a normal subgroup of a group $G$. The orders of $N$ and $G/N$ are relatively prime. Prove that $N$ is a characteristic subgroup of $G$.

I don't know what I can get from the second condition, can anyone give me some hints?

Answer 1

If $\psi$ is an automorphism of $G$:

Let $x \in N$. We're assuming the groups are finite so let $x^n = e$, so that $n$ divides $|N|$.

Moreover, $\psi(x)^m \in N$ for some $m$, or $x^m = e$ as well. $m$ divides $|G/N|$ but also $m$ divides $n$ - it must be that $m = 1$.

Answer 2

Outline:

Take arbitrarily $\phi \in \text{Aut}(G)$. Call $K=\phi(N)$. Enough to show $K \subseteq N$ . Here $K$ is a subgroup and $N$ is normal, so $KN$ is also a subgroup and $KN/N \leq G/N$, so order of $KN/N$ divides the order of $G/N$ . Use second isomorphism theorem to see order of $KN/N$ divides the order of $N$ also. Now use hypothesis to conclude $KN=N$ and hence $K \leq N$