TITLE: Group Theory Problem in Lang's Algebra
QUESTION [5 upvotes]: Refer to Lang's Algebra problem 14b, Chapter 1, p.76.
Let $G$ be a finite group and $N$ normal in $G$ such that the order
of $N$ and its index in $G$ are relatively prime. Let $g$ be an
automorphism of $G$. Then $g(N)=N$.
Any hints?

REPLY [3 votes]: Even more generally, such an $N$ is fully invariant: if $\phi$ is any endomorphism of $G$, then $\phi(N)\subseteq N$.
To prove this, consider $\pi\circ\phi\colon G\to G/N$, the composition of $\phi$ with the canonical projection $\pi\colon G\to G/N$. Since for any group homomorphism $f$ we know that $|f(x)|$ divides $|x|$, it follows that for any $x\in N$, $|\pi\circ\phi(x)|$ divides $|x|$ which divides $|N|$. On the other hand, $\pi\circ\phi(x)\in G/N$, so $|\pi\circ\phi(x)|$ divides $|G/N| = [G:N]$. Hence, $|\pi\circ\phi(x)|$ divides $\gcd(|N|,|G/N|) = \gcd(N,[G:N]) = 1$. 
Therefore, $\pi\circ\phi(x)=1$, so $\phi(x)\in\mathrm{ker}(\pi) = N$.