Centralizer is Normal Subgroup of Normalizer

Theorem
Let $G$ be a group.

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

Let $\map {C_G} H$ be the centralizer of $H$ in $G$.

Let $\map {N_G} H$ be the normalizer of $H$ in $G$.

Let $\Aut G$ be the automorphism group of $G$.

Then:


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

where:
 * $\map {N_G} H / \map {C_G} H$ is the quotient group of $\map {N_G} H$ by $\map {C_G} H$
 * $K$ is a subgroup of $\Aut G$.

Proof
For each $x \in \map {N_G} H$, we invoke the inner automorphism $\kappa_x: H \to G$:


 * $\map {\kappa_x} h = x h x^{-1}$

From Inner Automorphism is Automorphism, $\kappa_x$ is an automorphism of $H$.

By Kernel of Inner Automorphism Group is Center, the kernel of $\kappa_x$ is $\map {C_G} H$.

The result follows from the First Isomorphism Theorem for Groups, Kernel of Group Homomorphism is Subgroup and Centralizer in Subgroup is Intersection.