For a finite group $G$, if $N, G/N$ have relatively prime orders with $N$ normal, then for any automorphism on $g$, $g(N) =N$.

Question

For a finite group $G$, if $N, G/N$ have relatively prime orders with $N$ normal, then for any automorphism $g$, $g(N) =N$.

I can prove this for the case when there is a subgroup $H$ with the same order as $G/N$ because in this case I can set an isomorphism from $G$ to $H \times N$. But otherwise, don't know how; stuck. Any suggestion for break through?

Answer 1

Say $g(N)=H$, $H$ has same order than $N$, and let $\bar{H}$ be the image of $H$ in $G/N$. Clearly $\bar{H} \simeq H/(N \cap H)$, and is a sub group of $G/H$; so Lagrange Theorem gives you $\bar{H}=\lbrace 0 \rbrace$ and $H=N$.

Answer 2

Because automorphism are bijection, $|g(N)| = |N|=n$. If $|G| = nm$, then $gcd(n,m)=1$.

$|Ng(N)| = \frac{|N||g(N)|}{|N\cap g(N)|}$, so $|Ng(N)|$ divides $n^2$, and since $|Ng(N)|$ is a subgroup, $|Ng(N)|$ also divides $nm$. Since |Ng(N)| = rn^2, gcd(|Ng(N)|,m)=1 and so $|Ng(N)|$ divides $n$. But $|Ng(N)| = n [g(N):N\cap g(N)]$ and so $|Ng(N)|=n$

That leave us with $[g(N): N\cap g(N)] = 1$, so that $g(N) \subset N$. But $[g(N): N\cap g(N)] = [N: N\cap g(N)] = 1$ and so $N = g(N)$.