Intersection of Normal Subgroups is Normal
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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 $7.9$