Is my proof correct?: Let $K$ be a normal subgroup of a finite group $G$ and assume $|K|$ and $|G/K|$ are coprime. Prove $K$ is characteristic in $G$.

Question

Let $f \in \mathrm{Aut}(G)$. Let $f(K)=H$.

Consider $\pi: G \to G/K$.

Since $\pi(H)=HK/K \le G/K$, then $|\pi(H)| \mid |G/K|$.

Also, since $HK/K \cong H/(H\cap K)$, then $|\pi(H)|=|H/(H\cap K)|$.

So, $|\pi(H)||H\cap K|=|H|=|f(K)|=|K|$. So, $|\pi(H)|\mid |K|$.

Therefore, $|\pi(H)|\mid (|K|, |G/K|)=1$. Thus, $|\pi(H)|=1$, and thus $\pi(H)=\{K\}$.

So, $H \subset \ker \pi =K$. Since $|H|=|K|$, then $H=K$.

Hence, $f(K)=K$. So, $K$ is characteristic.

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Is this a correct proof?

Is there an easier way to do this?

Answer 1

Your argument is fine, but you can condense it considerably:

Let $f|_K : K \to G$ be the restriction of $f$ to $K$. The composition $$h := \pi_K \circ f|_K : K \to G/K$$ is a homomorphism, so by the first isomorphism theorem, $|\operatorname{\ker}(h)| |\operatorname{im}(h)| = |K|$.

Thus $|\operatorname{im}(h)|$ must divide both $|K|$ and $|G/K|$, which are coprime, forcing $|\operatorname{im}(h)| = 1$, so $\operatorname{im}(h) = \{K\}$. This forces $\operatorname{im}(f|_K) \leq K$, hence $f(K) \leq K$. Since $f$ is an isomorphism, $|f(K)| = |K|$, so we conclude that $f(K) = K$.

Answer 2

There is a way that shows something even stronger: If $H$ is any subgroup of $G$ such that $|H|\mid |K|$ then $H\subseteq K$.

To see this, consider the order of $HK$, which is $\frac{|H||K|}{|H\cap K|}$, but it is also a divisor in $|G|$, so the only option is that it equals $|K|$ since there are no primes that can divide both $|K|$ and $|G|/|K|$. Thus, $|H\cap K| = |H|$ so $H\subseteq K$.