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 $\Aut H$ be the automorphism group of $H$.

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 $\Aut H$.

Proof
In order to invoke the First Isomorphism Theorem for Groups, we must construct a group homomorphism $\phi: \map {N_G} H \to \Aut H$.

Consider the mapping $\phi: x \mapsto \paren {g \mapsto x g x^{-1}}$.

From Inner Automorphism is Automorphism, $g \mapsto x g x^{-1}$ is an automorphism of $G$, so $\phi$ is well-defined.

To see that $\phi$ is a homomorphism, notice that for any $x, y \in \map {N_G} H$:

Hence $\phi$ is a homomorphism.

Now we prove that $\ker \phi = \map {C_G} H$.

Note that for $x \in \map {N_G} H$:

Hence $\ker \phi = \map {C_G} H$.

By Kernel is Normal Subgroup of Domain:
 * $\map {C_G} H \lhd \map {N_G} H$

By First Isomorphism Theorem for Groups:
 * $\map {N_G} H / \map {C_G} H \cong \Img \phi$

By Image of Group Homomorphism is Subgroup:
 * $\Img \phi \le \Aut H$

Hence the result.