Let $G$ be a finite group and $H$ and $K$ subgroups such that $H \lhd G$ or $K \lhd G$. If $gcd(|K|;|G:H|)=1$, show that $K \subset H$.
I think I could prove it for the case $H \lhd G$
Let $k \in K$, if I take the quotient $G/H$, then $\overline{1}=\overline{k^{|K|}}={\overline{k}}^{|K|}$. Then, $ord(\overline{k})$ divides $|K|$, but it also divides $|G:H|$, so $ord(\overline{k})$ divides $gcd(|K|;|G:H|)=1$, which means $\overline{k}=\overline{1} \implies k \in H$.
I couldn't show the proposition for the case $K \lhd G$. I would like to show $K= K \cap H$, since $K \lhd G$, then $K \cap H \lhd H$. I don't know what to do from here. I would appreciate suggestions.
We shall prove that $|K:H\cap K|$ divides $|G:H|$. If it holds then $\gcd(|K|, |K:H\cap K|)=1$ and it is possible only when $K\subset H$.
Note that $KH\le G$ if $K$ is a normal subgroup of $G$. By third inequality in the post in Wikipedia $|K:H\cap K| = |KH:H|$. (Check the equality condition!) Since $KH\le G$, $|KH: H|$ divides $|G:H|$ (because $|KH|$ divides $|G|$.)