Centralizer is Normal Subgroup of Normalizer

Theorem
Let $G$ be a group.

Let $H \le G$ be a subgroup of $G$.

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 $\operatorname{Aut} \left({G}\right)$ be the group of automorphisms of $G$.

Then:


 * $(1): \quad C_G \left({H}\right) \lhd N_G \left({H}\right)$
 * $(2): \quad N_G \left({H}\right) / C_G \left({H}\right) \cong K$

where:
 * $N_G \left({H}\right) / C_G \left({H}\right)$ is the quotient group of $N_G \left({H}\right)$ by $C_G \left({H}\right)$
 * $K$ is a subgroup of $\operatorname{Aut} \left({G}\right)$.

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 for Groups, Kernel of Group Homomorphism is Subgroup and Centralizer in Subgroup is Intersection.