Intersection with Normal Subgroup is Normal

Theorem
Let $G$ be a group.

Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$.

Then $H \cap N$ is a normal subgroup of $H$.

Proof
Because $N \triangleleft G$, $\forall n \in N: \forall g \in G: g n g^{-1} \in N$.

Let $x \in H \cap N$.

Because $H \le G$ and therefore closed, $\forall x \in H \cap N: \forall g \in H: g x g^{-1} \in H$.

But since $x \in N$ and $N \triangleleft G$, $g x g^{-1} \in N$.

The result follows.