Let $H$ and $K$ be subgroups of a finite group $G$, at least one of which is normal. Show that if $|H|$ and $[G:K]$ are relatively prime, then $H \leq K$.
In the case that $K$ is normal, let $\pi : G \rightarrow G/K$ be the quotient map. Then $\pi(H)$ is a subgroup of $G/K$ and so $|\pi(H)|$ must divide both $|G/K|=[G:K]$ and $|H|$. These are relatively prime, so that $|\pi(H)|=1$, or $H \leq \ker \pi = K$.
If $H$ is normal, I'm not sure what to try. A hint would be perfect. This question is similar, but omits the normality hypothesis.
Best Answer
Hint : $HK$ is a subgroup of $G$ and $[HK:K]$ divides $[G:K]$. Here's a reasonably complete answer, so look away if you don't want that yet:
Also $|HK|/|K| = [H:H \cap K]$ divides $|H|$. Thus $[HK:K]$ divides two relatively prime integers, so divides $1$. Hence $[H:H \cap K] = 1$ and $H \leq K$. You don't really need normality, just that $HK=KH$ is a subgroup ( by the way, whether or not $HK$ is a subgroup, its cardinality is $|H| |K|/|H \cap K|$).