Subgroup is Normal Subgroup of Normalizer
From ProofWiki
Theorem
Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
- $H \le G \implies H \triangleleft N_G \left({H}\right)$
Proof
From Subgroup is Subgroup of Normalizer we have that $H \le N_G \left({H}\right)$.
It remains to show that $H$ is normal in $N_G \left({H}\right)$.
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Sources
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $8.12 \ \text{(i)}$
