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.