Intersection of Normal Subgroups is Normal

Theorem
Let $I$ be an indexing set, and $\left\{{N_i: i \in I}\right\}$ be a non-empty set of normal subgroups of the group $G$.

Then $\bigcap N_i$ is a normal subgroup of $G$.

Proof
Let $N = \bigcap N_i$.

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

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

Since $N \subseteq H$, we have $a N a^{-1} \subseteq a H a^{-1} \subseteq H$ from Normal Subgroup Equivalent Definitions: 3.

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 Normal Subgroup Equivalent Definitions: 3.