# Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup

## Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.

## Proof

From Subgroup is Subgroup of Normalizer, we have that $H \le \map {N_G} H$.

Now we need to show that $H \lhd \map {N_G} H$.

For $a \in \map {N_G} H$, the conjugate of $H$ by $a$ in $\map {N_G} H$ is:

\(\ds H^a\) | \(=\) | \(\ds \set {x \in \map {N_G} H: a x a^{-1} \in H}\) | Definition of Conjugate of Group Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds H^a \cap \map {N_G} H\) | Definition of Set Intersection | |||||||||||

\(\ds \) | \(=\) | \(\ds H \cap \map {N_G} H\) | Definition of Normalizer | |||||||||||

\(\ds \) | \(=\) | \(\ds H\) | Intersection with Subset is Subsetâ€Ž |

so:

- $\forall a \in \map {N_G} H: H^a = H$

and so by definition of normal subgroup:

- $H \lhd \map {N_G} H$

Now we need to show that $\map {N_G} H$ is the largest subgroup of $G$ containing $H$ such that $H \lhd \map {N_G} H$.

That is, to show that any subgroup of $G$ in which $H$ is normal is also a subset of $\map {N_G} H$.

Take any $N$ such that $H \lhd N \le G$.

In $N$, the conjugate of $H$ by $a \in N$ is $N \cap H^a = H$.

Therefore:

- $H \subseteq H^a$

Similarly, $H \subseteq H^{a^{-1} }$, so:

- $H^a \subseteq \paren {H^a}^{a^{-1} } = H$

Thus:

- $\forall a \in N: H^a = H, a \in \map {N_G} H$

That is:

- $N \subseteq \map {N_G} H$

So what we have shown is that any subgroup of $G$ in which $H$ is normal is a subset of $\map {N_G} H$, which is another way of saying that $\map {N_G} H$ is the largest such subgroup.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$