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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?

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Let $n\in N$. The order of $g(n)$ is the same as that of $n$, so it divides the order of $N$. It is therefore coprime with the order of $G/N$, and the class of $g(n)$ in the latter quotient must be trivial. This means that $g(n)$ is in $N$.

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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 $\phi(x)\in\mathrm{ker}(\pi) = N$.

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