Showing it is a normal subgroup

Question

Let $G$ - group, $H, N$ - its subgroups, $N$ is normal($H$ can be arbitrary). Prove that $H \cap N$ is a normal subgroup in $H$.

Here are my thoughts:

Bascically, we should prove that $\forall h \in H$ $ h (H \cap N) h^{-1} = H \cap N$. Since, we are dealing with intersection, we can represent any element in $H \cap N $ as $gNg^{-1}$, where $g \in G$. So, $h (H \cap N) h^{-1} = hgNg^{-1}h^{-1} = hgN(hg)^{-1}$. But what is $gh$? $gh$ is some element $\in G$. Unfortunately, i don't know how to show $gh \in H$.

Answer 1

Let $g\in H\cap N$, $h\in H$, $hgh^{-1}\in N$ since $N$ is normal, since $g\in H\cap N\subset H, hgh^{-1}\subset H$ since it is the product of three elements of $H$. This implies that $hgh^{-1}\in H\cap N$.

Answer 2

If $h\in H$, then $(H\cap N)^h=H^h\cap N^h=H\cap N$ is valid because $N$ is normal in $G$ and $H$ is normal in $H$. The intersection of two subgroups is a subgroup.

Answer 3

An alternative proof: the kernel of a homomorphism is a normal subgroup and since $N$ is normal, the canonical map $G\rightarrow G/N$ is a homomorphism. Thus the map $H\rightarrow G\rightarrow G/N$ is a homomorphism, where the first arrow is the inclusion and the kernel is $N\cap H$.