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

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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:

\(\displaystyle H^a\) \(=\) \(\displaystyle \set {x \in \map {N_G} H: a x a^{-1} \in H}\) Definition of Conjugate of Group Subset
\(\displaystyle \) \(=\) \(\displaystyle H^a \cap \map {N_G} H\) Definition of Set Intersection
\(\displaystyle \) \(=\) \(\displaystyle H \cap \map {N_G} H\) Definition of Normalizer
\(\displaystyle \) \(=\) \(\displaystyle H\) 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.

$\blacksquare$


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