Prove that if $\displaystyle H$ is a normal subgroup of $\displaystyle G$ of prime index $\displaystyle p$ then for all subgroups $\displaystyle K$ of $\displaystyle G$ either

(i) $\displaystyle K$ is a subgroup of $\displaystyle H$, or

(ii) $\displaystyle G = HK$ and $\displaystyle |K : K$ intersect $\displaystyle H| = p$

I tried to do this in 2 cases. The first case is to let $\displaystyle K$ be a subest of $\displaystyle H$, then since $\displaystyle K$ is a subgroup of $\displaystyle G$, it is a subgroup of $\displaystyle H$. For case 2, I let $\displaystyle K$ not contained in $\displaystyle H$, then try to show $\displaystyle G = HK$ and $\displaystyle |K : K$ intersect $\displaystyle H| = p$, but don't know how... please help, thank you.