Show that a group is equal to the product of two subgroups

Question

I'm taking a course in abstract algebra, and I'm working through some old exams. I'm stuck at this problem and have no idea how to approach it. Any help is appriciated.

Let $G$ be a group and let $N$ be normal subgroup with index $p$ in $G$ where $p$ is prime.

Let $H$ be a subgroup in $G$, that is not contained in $N$.

Show that $G=HN$ and that $[H : H \cap N] = p$

Answer 1

Hints:

Let $h\in H$ with $h\notin N$.

Observe that $G/N$ has order $p$ so that it is cyclic and $hN$ will serve as generator.

Then $\{h^kN\mid k=0,1,\dots,p-1\}$ is a partition of $G$.

Also have a look at the second isomorphism theorem for groups.