Subgroup is Normal Subgroup of Normalizer

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 $12 \ \text{(i)}$