Intersection of Normal Subgroups is Normal

Theorem

Let $G$ be a group.

Let $I$ be an indexing set.

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

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

Proof

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

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

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

We have that $N \subseteq H$.

Thus from Subgroup is 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 $\set {N_i: i \in I}$, and hence in their intersection $N$.

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

The result follows by Subgroup is Superset of Conjugate iff Normal.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $7$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.9$