Why is the intersection of two normal subgroups a normal subgroup?

Large circle with two overlapping smaller circles, the overlap highlighted with an arrow staying inside it — The intersection of two normal subgroups is again normal.

Let $(G, \cdot)$ be a group and $X \lhd G$ and $Y \lhd G$ be two normal subgroups.

I will show this in two steps:

Show that $X \cap Y$ is a group
Show that $X \cap Y$ is a normal group of $(G, \cdot)$

Intersection of two subgroups is a subgroup

Theorem: $(X \cap Y) \leq G$

Proof:

$X \cap Y$ is not empty: $e_G \in X \land e_G \in Y \Rightarrow e_G \in (X \cap Y)$

$X \cap Y$ has inverse elements. Let $a \in (X \cap Y)$. As a is in $X$ and $X$ is a group, $a^{-1} \in X$. The same is true for $Y$. So: $\forall a \in (X \cap Y) \exists a^{-1} \in (X \cap Y): a \cdot a^{-1} = a^{-1} \cdot a = e_G$

$\forall a,b \in (X \cap Y): a \cdot b^{-1} \in (X \cap Y)$, because both, $a$ and $b^{-1}$ are in $X$. As $X$ is a group, the result has to be in $X$. Same argumentation for $Y$. Then the result is in $X$ and $Y \blacksquare$

Intersection of two normal subgroups is normal

First the definition of a normal subgroup:

Let $N \leq G$ be a subgroup of $G$. $N \lhd G :\Leftrightarrow \forall n \in N \forall g \in G: g \cdot n \cdot g^{-1} \in N$

Theorem: $(X \cap Y) \lhd G$

Proof:

$X \cap Y$ is a subgroup of $G$ as I have proved above.

$\forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in X$ and $\forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in Y$

$\Rightarrow \forall n \in (X \cap Y) \forall g \in G: g \cdot n \cdot g^{-1} \in (X \cap Y)$

$\Rightarrow (X \cap Y) \lhd G \blacksquare$