# 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.