Let G be a finite group with $\displaystyle K \lhd G$. If $\displaystyle (|K|,[G : K])=1$, prove that K is the unique subgroup of G having order $\displaystyle |K|$

I was thinking about approaching this problem by..

If $\displaystyle H \leq G$ and $\displaystyle |H|=|K|$, and I figure out what happens to elements of $\displaystyle H$ in $\displaystyle G/K$ then I might be able to see something happen?.... Or maybe I'm just approaching it all wrong