Suppose $G$ is finite, $K$ is a normal subgroup in $G$, $H$ is a subgroup of $G$, and $|K|$ is relatively prime to $[G:H]$. Show that $K$ is a subgroup of $H$.
I don't know where to begin…
Abstract Algebra – Understanding Normal Subgroups and Relatively Prime Index
abstract-algebrafinite-groups
Best Answer
You don't need $K$ to be normal, you just need $HK$ to be a subgroup (which it is when $K$ is normal). Note that $K\subseteq H$ if and only if $H\cap K=K$.
In general, we know that $|HK||H\cap K| = |H||K|$. Therefore, if all quantities are finite, we have $$\frac{|HK|}{|H|} = \frac{|K|}{|H\cap K|}.$$ If $HK$ is a subgroup and $G$ is finite, then $|HK|$ divides $|G|$, so $$\frac{|K|}{|H\cap K|} = \frac{|HK|}{|H|} \text{ divides }\frac{|G|}{|H|} = [G:H].$$
From this you should be able to finish.