Intersection of Normal Subgroups is Normal

Theorem
Let $G$ be a group.

Let $I$ be an indexing set.

Let $\left\langle{N_i}\right\rangle_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$.

Then $\displaystyle \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$.

Proof
Let $\displaystyle N = \bigcap_{i \mathop \in I} N_i$.

From Intersection of Subgroups, $N \le G$.

Suppose $H \in \left\{{N_i: i \in I}\right\}$.

We have that $N \subseteq H$.

Thus from Subgroup Superset of Conjugate iff Normal:
 * $a N a^{-1} \subseteq a H a^{-1} \subseteq H$

Thus $a N a^{-1}$ is a subset of each one of the subgroups in $\left\{{N_i: i \in I}\right\}$, and hence in their intersection $N$.

That is, $a N a^{-1} \subseteq N$.

The result follows by Subgroup Superset of Conjugate iff Normal.