Subgroup is Normal Subgroup of Normalizer
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Theorem
Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
- $H \le G \implies H \lhd \map {N_G} H$
Proof
From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$.
It remains to show that $H$ is normal in $\map {N_G} H$.
Let $a \in H$ and $b \in \map {N_G} H$.
By the definition of normalizer:
- $b a b^{-1} \in H$
Thus $H$ is normal in $\map {N_G} H$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$