From Rotman "Introduction to the Theory of Groups", ex. 2:54:
Let $ G $ be a finite group, and let $H$ be a normal subgroup with $(H,[G:H])=1$.
Prove that $H$ is the unique such subgroup in G.
That exercise was introduced here before:
link
It's easy that $HK$ is a subgroup of $G$, thus $|K| / |H\cap K|$ divides $|G|/|H|$.
But I don't see what to do in next step anyway, in spite of reading link above.
Best Answer
As stated, the question is a bit vague, and I am having trouble interpreting it in a way that makes it correct.
Here is something that does hold and which might be what is being asked:
Let $G$ be a finite group and $H$ a normal subgroup of $G$ with $\rm{gcd}(|H|,|G/H|) =1$. Let $K$ be any subgroup of $G$ such that $|K|$ divides $|H|$. Then $K\subseteq H$ (as a special case of this, if $|K| = |H|$ then $K =H$ and I think this might be what is asked).
Proof: Since $H$ is normal, $KH$ is a subgroup of $G$ of order $\frac{|K||H|}{|K\cap H|}$. If $K$ is not contained in $H$ then $\frac{|K|}{|K\cap H|} > 1$. Let $p$ be a prime dividing $\frac{|K|}{|K\cap H|}$. Since $|K|$ divides $|H|$ we know that $p$ divides $|H|$, so since $\rm{gcd}(|H|,|G/H|) =1$ we know that $p$ does not divide $|G/H|$. But then $p|H|$ does not divide $|G|$ which contradicts the fact that it should divide $|KH|$.