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 Subset‎ |
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.20$
- 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 $12 \ \text{(i)}$