Intersection of Normal Subgroups is Normal

Theorem
Let $$I$$ be an indexing set, and $$\left\{{N_i: i \in I}\right\}$$ be a 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: Generalized Result, $$N \le G$$.

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.