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 Subset is Intersection.