Centralizer is Normal Subgroup of Normalizer

Theorem
Let $H \le G$.

Let $C_G \left({H}\right)$ be the centralizer of $H$ in $G$.

Let $N_G \left({H}\right)$ be the normalizer of $H$ in $G$.

Let $K = \operatorname{Aut} \left({G}\right)$ be the Group of Automorphisms of $G$.

Then:


 * $C_G \left({H}\right) \triangleleft N_G \left({H}\right)$
 * $N_G \left({H}\right) / C_G \left({H}\right) \cong K$

where $N_G \left({H}\right) / C_G \left({H}\right)$ is the quotient group of $N_G \left({H}\right)$ by $C_G \left({H}\right)$.

Proof
For each $x \in N_G \left({H}\right)$, we invoke the Inner Automorphism $\theta_x: H \to G$:

$\theta_x \left({h}\right) = x h x^{-1}$

From the definition of Inner Automorphism, $\theta_x$ is an automorphism of $H$.

The kernel of $\theta_x$ can be shown to be $C_G \left({H}\right)$ (probably by using Kernel of Inner Automorphisms is Center).

The result follows from the First Isomorphism Theorem, Kernel is Subgroup and Centralizer in Subgroup is Intersection.