# 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 $x, y \in \map {N_G} H$:

\(\displaystyle \map \phi x \map \phi y\) | \(=\) | \(\displaystyle \paren {g \mapsto x g x^{-1} } \circ \paren {g \mapsto y g y^{-1} }\) | where $\circ$ denote composition of maps | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g \mapsto x \paren {y g y^{-1} } x^{-1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g \mapsto \paren {x y} g \paren {x y}^{-1}\) | Inverse of Group Product | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x y}\) |

Hence $\phi$ is a homomorphism.

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

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

\(\displaystyle x\) | \(\in\) | \(\displaystyle \ker \phi\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle g\) | \(=\) | \(\displaystyle x g x^{-1}\) | \(\displaystyle \forall g \in H\) | $g \mapsto g$ is the identity of $\Aut H$ | ||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle g x\) | \(=\) | \(\displaystyle x g\) | \(\displaystyle \forall g \in H\) | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle \map {C_G} H\) | Definition of Centralizer of Subgroup |

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.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.26$