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

$\map {C_G} H \lhd \map {N_G} H$
$\map {N_G} H / \map {C_G} H \cong \Img \phi$
$\Img \phi \le \Aut H$

Hence the result.

$\blacksquare$