Let the subgroup $N$ be normal in the finite group $G$ with index a prime $p$. Let $H$ be a subgroup of $G$ which is not contained in $N$. I would like to show that $H \cap N$ is normal in $H$ with index $p$.
To begin, I consider a homomorphism of which $H \cap N$ is the kernel. $N$ is the kernel of some homomorphism $g : G \rightarrow K$. So I construct the homomorphism $f : H \rightarrow g(H)$ with $h \mapsto g(h)$. If $h \in N$ then $g(h) = e$, so $h \in \ker(f)$ in this case. So $\ker(f) = N \cap H$.
I do not know how to show that $H \cap N$ has index $p$ in $H$. By the second isomorphism theorem we have the index of $H \cap N$ in $H$ to be $|HN|/|N|$. How can we proceed from here?