Centralizer 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 for Groups, Kernel is Subgroup and Centralizer in Subgroup is Intersection.
The kernel of $\theta_x$ can be shown to be $C_G \left({H}\right)$.
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The result follows from the First Isomorphism Theorem for Groups, Kernel is Subgroup and Centralizer in Subgroup is Intersection.
You can help Proof Wiki by completing it.
To discuss it in more detail, feel free to use the talk page.
When this has been done, {{WIP}} should be removed from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
Sources
- John F. Humphreys: A Course in Group Theory (1996): $\S 10$: Proposition $10.26$
